\(\int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx\) [69]

Optimal result

Integrand size = 24, antiderivative size = 407 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {15 d^2 \sqrt {c+d x} \cos (2 a+2 b x)}{128 b^3}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{8 b}-\frac {15 d^2 \sqrt {c+d x} \cos (4 a+4 b x)}{2048 b^3}+\frac {(c+d x)^{5/2} \cos (4 a+4 b x)}{32 b}+\frac {15 d^{5/2} \sqrt {\frac {\pi }{2}} \cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4096 b^{7/2}}-\frac {15 d^{5/2} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{256 b^{7/2}}-\frac {15 d^{5/2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (4 a-\frac {4 b c}{d}\right )}{4096 b^{7/2}}+\frac {15 d^{5/2} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{256 b^{7/2}}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{32 b^2}-\frac {5 d (c+d x)^{3/2} \sin (4 a+4 b x)}{256 b^2} \]

[Out]

-1/8*(d*x+c)^(5/2)*cos(2*b*x+2*a)/b+1/32*(d*x+c)^(5/2)*cos(4*b*x+4*a)/b+5/32*d*(d*x+c)^(3/2)*sin(2*b*x+2*a)/b^
2-5/256*d*(d*x+c)^(3/2)*sin(4*b*x+4*a)/b^2+15/8192*d^(5/2)*cos(4*a-4*b*c/d)*FresnelC(2*b^(1/2)*2^(1/2)/Pi^(1/2
)*(d*x+c)^(1/2)/d^(1/2))*2^(1/2)*Pi^(1/2)/b^(7/2)-15/8192*d^(5/2)*FresnelS(2*b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^
(1/2)/d^(1/2))*sin(4*a-4*b*c/d)*2^(1/2)*Pi^(1/2)/b^(7/2)-15/256*d^(5/2)*cos(2*a-2*b*c/d)*FresnelC(2*b^(1/2)*(d
*x+c)^(1/2)/d^(1/2)/Pi^(1/2))*Pi^(1/2)/b^(7/2)+15/256*d^(5/2)*FresnelS(2*b^(1/2)*(d*x+c)^(1/2)/d^(1/2)/Pi^(1/2
))*sin(2*a-2*b*c/d)*Pi^(1/2)/b^(7/2)+15/128*d^2*cos(2*b*x+2*a)*(d*x+c)^(1/2)/b^3-15/2048*d^2*cos(4*b*x+4*a)*(d
*x+c)^(1/2)/b^3

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {4491, 3377, 3387, 3386, 3432, 3385, 3433} \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4096 b^{7/2}}-\frac {15 \sqrt {\pi } d^{5/2} \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{256 b^{7/2}}-\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \sin \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4096 b^{7/2}}+\frac {15 \sqrt {\pi } d^{5/2} \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{256 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \cos (2 a+2 b x)}{128 b^3}-\frac {15 d^2 \sqrt {c+d x} \cos (4 a+4 b x)}{2048 b^3}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{32 b^2}-\frac {5 d (c+d x)^{3/2} \sin (4 a+4 b x)}{256 b^2}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{8 b}+\frac {(c+d x)^{5/2} \cos (4 a+4 b x)}{32 b} \]

[In]

Int[(c + d*x)^(5/2)*Cos[a + b*x]*Sin[a + b*x]^3,x]

[Out]

(15*d^2*Sqrt[c + d*x]*Cos[2*a + 2*b*x])/(128*b^3) - ((c + d*x)^(5/2)*Cos[2*a + 2*b*x])/(8*b) - (15*d^2*Sqrt[c
+ d*x]*Cos[4*a + 4*b*x])/(2048*b^3) + ((c + d*x)^(5/2)*Cos[4*a + 4*b*x])/(32*b) + (15*d^(5/2)*Sqrt[Pi/2]*Cos[4
*a - (4*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(4096*b^(7/2)) - (15*d^(5/2)*Sqrt[Pi]*
Cos[2*a - (2*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(256*b^(7/2)) - (15*d^(5/2)*Sqrt[
Pi/2]*FresnelS[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[4*a - (4*b*c)/d])/(4096*b^(7/2)) + (15*d^(5/2
)*Sqrt[Pi]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(256*b^(7/2)) + (5*d*(
c + d*x)^(3/2)*Sin[2*a + 2*b*x])/(32*b^2) - (5*d*(c + d*x)^(3/2)*Sin[4*a + 4*b*x])/(256*b^2)

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3385

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Cos[f*(x^2/d)],
x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3386

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[f*(x^2/d)], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3387

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) +
f*x]/Sqrt[c + d*x], x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c
, d, e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 4491

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{4} (c+d x)^{5/2} \sin (2 a+2 b x)-\frac {1}{8} (c+d x)^{5/2} \sin (4 a+4 b x)\right ) \, dx \\ & = -\left (\frac {1}{8} \int (c+d x)^{5/2} \sin (4 a+4 b x) \, dx\right )+\frac {1}{4} \int (c+d x)^{5/2} \sin (2 a+2 b x) \, dx \\ & = -\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{8 b}+\frac {(c+d x)^{5/2} \cos (4 a+4 b x)}{32 b}-\frac {(5 d) \int (c+d x)^{3/2} \cos (4 a+4 b x) \, dx}{64 b}+\frac {(5 d) \int (c+d x)^{3/2} \cos (2 a+2 b x) \, dx}{16 b} \\ & = -\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{8 b}+\frac {(c+d x)^{5/2} \cos (4 a+4 b x)}{32 b}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{32 b^2}-\frac {5 d (c+d x)^{3/2} \sin (4 a+4 b x)}{256 b^2}+\frac {\left (15 d^2\right ) \int \sqrt {c+d x} \sin (4 a+4 b x) \, dx}{512 b^2}-\frac {\left (15 d^2\right ) \int \sqrt {c+d x} \sin (2 a+2 b x) \, dx}{64 b^2} \\ & = \frac {15 d^2 \sqrt {c+d x} \cos (2 a+2 b x)}{128 b^3}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{8 b}-\frac {15 d^2 \sqrt {c+d x} \cos (4 a+4 b x)}{2048 b^3}+\frac {(c+d x)^{5/2} \cos (4 a+4 b x)}{32 b}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{32 b^2}-\frac {5 d (c+d x)^{3/2} \sin (4 a+4 b x)}{256 b^2}+\frac {\left (15 d^3\right ) \int \frac {\cos (4 a+4 b x)}{\sqrt {c+d x}} \, dx}{4096 b^3}-\frac {\left (15 d^3\right ) \int \frac {\cos (2 a+2 b x)}{\sqrt {c+d x}} \, dx}{256 b^3} \\ & = \frac {15 d^2 \sqrt {c+d x} \cos (2 a+2 b x)}{128 b^3}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{8 b}-\frac {15 d^2 \sqrt {c+d x} \cos (4 a+4 b x)}{2048 b^3}+\frac {(c+d x)^{5/2} \cos (4 a+4 b x)}{32 b}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{32 b^2}-\frac {5 d (c+d x)^{3/2} \sin (4 a+4 b x)}{256 b^2}+\frac {\left (15 d^3 \cos \left (4 a-\frac {4 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {4 b c}{d}+4 b x\right )}{\sqrt {c+d x}} \, dx}{4096 b^3}-\frac {\left (15 d^3 \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}} \, dx}{256 b^3}-\frac {\left (15 d^3 \sin \left (4 a-\frac {4 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {4 b c}{d}+4 b x\right )}{\sqrt {c+d x}} \, dx}{4096 b^3}+\frac {\left (15 d^3 \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{\sqrt {c+d x}} \, dx}{256 b^3} \\ & = \frac {15 d^2 \sqrt {c+d x} \cos (2 a+2 b x)}{128 b^3}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{8 b}-\frac {15 d^2 \sqrt {c+d x} \cos (4 a+4 b x)}{2048 b^3}+\frac {(c+d x)^{5/2} \cos (4 a+4 b x)}{32 b}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{32 b^2}-\frac {5 d (c+d x)^{3/2} \sin (4 a+4 b x)}{256 b^2}+\frac {\left (15 d^2 \cos \left (4 a-\frac {4 b c}{d}\right )\right ) \text {Subst}\left (\int \cos \left (\frac {4 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{2048 b^3}-\frac {\left (15 d^2 \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \text {Subst}\left (\int \cos \left (\frac {2 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{128 b^3}-\frac {\left (15 d^2 \sin \left (4 a-\frac {4 b c}{d}\right )\right ) \text {Subst}\left (\int \sin \left (\frac {4 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{2048 b^3}+\frac {\left (15 d^2 \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \text {Subst}\left (\int \sin \left (\frac {2 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{128 b^3} \\ & = \frac {15 d^2 \sqrt {c+d x} \cos (2 a+2 b x)}{128 b^3}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{8 b}-\frac {15 d^2 \sqrt {c+d x} \cos (4 a+4 b x)}{2048 b^3}+\frac {(c+d x)^{5/2} \cos (4 a+4 b x)}{32 b}+\frac {15 d^{5/2} \sqrt {\frac {\pi }{2}} \cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4096 b^{7/2}}-\frac {15 d^{5/2} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{256 b^{7/2}}-\frac {15 d^{5/2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (4 a-\frac {4 b c}{d}\right )}{4096 b^{7/2}}+\frac {15 d^{5/2} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{256 b^{7/2}}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{32 b^2}-\frac {5 d (c+d x)^{3/2} \sin (4 a+4 b x)}{256 b^2} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.03 (sec) , antiderivative size = 1332, normalized size of antiderivative = 3.27 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {\left (\frac {1}{128}+\frac {i}{128}\right ) c \sqrt {d} e^{-\frac {2 i (a d+b (c+d x))}{d}} \left ((2+2 i) \sqrt {b} \sqrt {d} e^{\frac {2 i b c}{d}} \sqrt {c+d x} \left (3+4 i b x+e^{4 i (a+b x)} (-3+4 i b x)\right )+i (4 b c+3 i d) e^{\frac {2 i b (2 c+d x)}{d}} \sqrt {\pi } \text {erf}\left (\frac {(1+i) \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )+(4 i b c+3 d) e^{2 i (2 a+b x)} \sqrt {\pi } \text {erfi}\left (\frac {(1+i) \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )\right )}{b^{5/2}}-\frac {c \sqrt {d} e^{-\frac {4 i (a d+b (c+d x))}{d}} \left (-4 \sqrt {b} \sqrt {d} e^{\frac {4 i b c}{d}} \sqrt {c+d x} \left (-3 i+8 b x+e^{8 i (a+b x)} (3 i+8 b x)\right )+(-1)^{3/4} (8 b c+3 i d) e^{\frac {4 i b (2 c+d x)}{d}} \sqrt {\pi } \text {erf}\left (\frac {2 \sqrt [4]{-1} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )+\sqrt [4]{-1} (8 i b c+3 d) e^{4 i (2 a+b x)} \sqrt {\pi } \text {erfi}\left (\frac {2 \sqrt [4]{-1} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )\right )}{1024 b^{5/2}}+\frac {1}{4} c^2 \left (-\frac {e^{2 i \left (a-\frac {b c}{d}\right )} \sqrt {c+d x} \Gamma \left (\frac {3}{2},-\frac {2 i b (c+d x)}{d}\right )}{4 \sqrt {2} b \sqrt {-\frac {i b (c+d x)}{d}}}-\frac {e^{-2 i \left (a-\frac {b c}{d}\right )} \sqrt {c+d x} \Gamma \left (\frac {3}{2},\frac {2 i b (c+d x)}{d}\right )}{4 \sqrt {2} b \sqrt {\frac {i b (c+d x)}{d}}}\right )-\frac {c^2 e^{-\frac {4 i (b c+a d)}{d}} \sqrt {c+d x} \left (-\frac {e^{8 i a} \Gamma \left (\frac {3}{2},-\frac {4 i b (c+d x)}{d}\right )}{\sqrt {-\frac {i b (c+d x)}{d}}}-\frac {e^{\frac {8 i b c}{d}} \Gamma \left (\frac {3}{2},\frac {4 i b (c+d x)}{d}\right )}{\sqrt {\frac {i b (c+d x)}{d}}}\right )}{128 b}+\frac {\sqrt {d} \left ((1-i) e^{2 i \left (a-\frac {b c}{d}\right )} \left ((2+2 i) \sqrt {b} \sqrt {d} e^{\frac {2 i b (c+d x)}{d}} \sqrt {c+d x} \left (15 d-16 b^2 d x^2+4 i b (c-5 d x)\right )+\left (16 b^2 c^2-24 i b c d-15 d^2\right ) \sqrt {\pi } \text {erfi}\left (\frac {(1+i) \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )\right )+i (\cos (2 (a+b x))-i \sin (2 (a+b x))) \left (4 \sqrt {b} \sqrt {d} \sqrt {c+d x} \left (-15 i d+16 i b^2 d x^2-4 b (c-5 d x)\right )-(1+i) \left (16 b^2 c^2+24 i b c d-15 d^2\right ) \sqrt {\pi } \text {erf}\left (\frac {(1+i) \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right ) \left (\cos \left (\frac {2 b (c+d x)}{d}\right )+i \sin \left (\frac {2 b (c+d x)}{d}\right )\right )\right )\right )}{1024 b^{7/2}}-\frac {\sqrt {d} \left (e^{4 i \left (a-\frac {b c}{d}\right )} \left (4 \sqrt {b} \sqrt {d} e^{\frac {4 i b (c+d x)}{d}} \sqrt {c+d x} \left (15 d-64 b^2 d x^2+8 i b (c-5 d x)\right )+(-1)^{3/4} \left (-64 b^2 c^2+48 i b c d+15 d^2\right ) \sqrt {\pi } \text {erfi}\left (\frac {2 \sqrt [4]{-1} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )\right )+i (\cos (4 (a+b x))-i \sin (4 (a+b x))) \left (4 \sqrt {b} \sqrt {d} \sqrt {c+d x} \left (-15 i d+64 i b^2 d x^2-8 b (c-5 d x)\right )-(1+i) \left (64 b^2 c^2+48 i b c d-15 d^2\right ) \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {(1+i) \sqrt {2} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right ) \left (\cos \left (\frac {4 b (c+d x)}{d}\right )+i \sin \left (\frac {4 b (c+d x)}{d}\right )\right )\right )\right )}{16384 b^{7/2}} \]

[In]

Integrate[(c + d*x)^(5/2)*Cos[a + b*x]*Sin[a + b*x]^3,x]

[Out]

((1/128 + I/128)*c*Sqrt[d]*((2 + 2*I)*Sqrt[b]*Sqrt[d]*E^(((2*I)*b*c)/d)*Sqrt[c + d*x]*(3 + (4*I)*b*x + E^((4*I
)*(a + b*x))*(-3 + (4*I)*b*x)) + I*(4*b*c + (3*I)*d)*E^(((2*I)*b*(2*c + d*x))/d)*Sqrt[Pi]*Erf[((1 + I)*Sqrt[b]
*Sqrt[c + d*x])/Sqrt[d]] + ((4*I)*b*c + 3*d)*E^((2*I)*(2*a + b*x))*Sqrt[Pi]*Erfi[((1 + I)*Sqrt[b]*Sqrt[c + d*x
])/Sqrt[d]]))/(b^(5/2)*E^(((2*I)*(a*d + b*(c + d*x)))/d)) - (c*Sqrt[d]*(-4*Sqrt[b]*Sqrt[d]*E^(((4*I)*b*c)/d)*S
qrt[c + d*x]*(-3*I + 8*b*x + E^((8*I)*(a + b*x))*(3*I + 8*b*x)) + (-1)^(3/4)*(8*b*c + (3*I)*d)*E^(((4*I)*b*(2*
c + d*x))/d)*Sqrt[Pi]*Erf[(2*(-1)^(1/4)*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]] + (-1)^(1/4)*((8*I)*b*c + 3*d)*E^((4*I
)*(2*a + b*x))*Sqrt[Pi]*Erfi[(2*(-1)^(1/4)*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]]))/(1024*b^(5/2)*E^(((4*I)*(a*d + b*
(c + d*x)))/d)) + (c^2*(-1/4*(E^((2*I)*(a - (b*c)/d))*Sqrt[c + d*x]*Gamma[3/2, ((-2*I)*b*(c + d*x))/d])/(Sqrt[
2]*b*Sqrt[((-I)*b*(c + d*x))/d]) - (Sqrt[c + d*x]*Gamma[3/2, ((2*I)*b*(c + d*x))/d])/(4*Sqrt[2]*b*E^((2*I)*(a
- (b*c)/d))*Sqrt[(I*b*(c + d*x))/d])))/4 - (c^2*Sqrt[c + d*x]*(-((E^((8*I)*a)*Gamma[3/2, ((-4*I)*b*(c + d*x))/
d])/Sqrt[((-I)*b*(c + d*x))/d]) - (E^(((8*I)*b*c)/d)*Gamma[3/2, ((4*I)*b*(c + d*x))/d])/Sqrt[(I*b*(c + d*x))/d
]))/(128*b*E^(((4*I)*(b*c + a*d))/d)) + (Sqrt[d]*((1 - I)*E^((2*I)*(a - (b*c)/d))*((2 + 2*I)*Sqrt[b]*Sqrt[d]*E
^(((2*I)*b*(c + d*x))/d)*Sqrt[c + d*x]*(15*d - 16*b^2*d*x^2 + (4*I)*b*(c - 5*d*x)) + (16*b^2*c^2 - (24*I)*b*c*
d - 15*d^2)*Sqrt[Pi]*Erfi[((1 + I)*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]]) + I*(Cos[2*(a + b*x)] - I*Sin[2*(a + b*x)]
)*(4*Sqrt[b]*Sqrt[d]*Sqrt[c + d*x]*((-15*I)*d + (16*I)*b^2*d*x^2 - 4*b*(c - 5*d*x)) - (1 + I)*(16*b^2*c^2 + (2
4*I)*b*c*d - 15*d^2)*Sqrt[Pi]*Erf[((1 + I)*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]]*(Cos[(2*b*(c + d*x))/d] + I*Sin[(2*
b*(c + d*x))/d]))))/(1024*b^(7/2)) - (Sqrt[d]*(E^((4*I)*(a - (b*c)/d))*(4*Sqrt[b]*Sqrt[d]*E^(((4*I)*b*(c + d*x
))/d)*Sqrt[c + d*x]*(15*d - 64*b^2*d*x^2 + (8*I)*b*(c - 5*d*x)) + (-1)^(3/4)*(-64*b^2*c^2 + (48*I)*b*c*d + 15*
d^2)*Sqrt[Pi]*Erfi[(2*(-1)^(1/4)*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]]) + I*(Cos[4*(a + b*x)] - I*Sin[4*(a + b*x)])*
(4*Sqrt[b]*Sqrt[d]*Sqrt[c + d*x]*((-15*I)*d + (64*I)*b^2*d*x^2 - 8*b*(c - 5*d*x)) - (1 + I)*(64*b^2*c^2 + (48*
I)*b*c*d - 15*d^2)*Sqrt[Pi/2]*Erf[((1 + I)*Sqrt[2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]]*(Cos[(4*b*(c + d*x))/d] + I
*Sin[(4*b*(c + d*x))/d]))))/(16384*b^(7/2))

Maple [A] (verified)

Time = 3.56 (sec) , antiderivative size = 470, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {-\frac {d \left (d x +c \right )^{\frac {5}{2}} \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{8 b}+\frac {5 d \left (\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{4 b}-\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{4 b}+\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}\right )}{4 b}\right )}{8 b}+\frac {d \left (d x +c \right )^{\frac {5}{2}} \cos \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{32 b}-\frac {5 d \left (\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{8 b}-\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{8 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{32 b \sqrt {\frac {b}{d}}}\right )}{8 b}\right )}{32 b}}{d}\)	\(470\)
default	\(\frac {-\frac {d \left (d x +c \right )^{\frac {5}{2}} \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{8 b}+\frac {5 d \left (\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{4 b}-\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{4 b}+\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}\right )}{4 b}\right )}{8 b}+\frac {d \left (d x +c \right )^{\frac {5}{2}} \cos \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{32 b}-\frac {5 d \left (\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{8 b}-\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{8 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{32 b \sqrt {\frac {b}{d}}}\right )}{8 b}\right )}{32 b}}{d}\)	\(470\)

[In]

int((d*x+c)^(5/2)*cos(b*x+a)*sin(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

2/d*(-1/16/b*d*(d*x+c)^(5/2)*cos(2*b/d*(d*x+c)+2*(a*d-b*c)/d)+5/16/b*d*(1/4/b*d*(d*x+c)^(3/2)*sin(2*b/d*(d*x+c
)+2*(a*d-b*c)/d)-3/4/b*d*(-1/4/b*d*(d*x+c)^(1/2)*cos(2*b/d*(d*x+c)+2*(a*d-b*c)/d)+1/8/b*d*Pi^(1/2)/(b/d)^(1/2)
*(cos(2*(a*d-b*c)/d)*FresnelC(2/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d)-sin(2*(a*d-b*c)/d)*FresnelS(2/Pi^(1/2)
/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d))))+1/64/b*d*(d*x+c)^(5/2)*cos(4*b/d*(d*x+c)+4*(a*d-b*c)/d)-5/64/b*d*(1/8/b*d*(
d*x+c)^(3/2)*sin(4*b/d*(d*x+c)+4*(a*d-b*c)/d)-3/8/b*d*(-1/8/b*d*(d*x+c)^(1/2)*cos(4*b/d*(d*x+c)+4*(a*d-b*c)/d)
+1/32/b*d*2^(1/2)*Pi^(1/2)/(b/d)^(1/2)*(cos(4*(a*d-b*c)/d)*FresnelC(2*2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(
1/2)/d)-sin(4*(a*d-b*c)/d)*FresnelS(2*2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d)))))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.00 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {15 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 15 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - 480 \, \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 480 \, \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 4 \, {\left (320 \, b^{3} d^{2} x^{2} + 640 \, b^{3} c d x + 320 \, b^{3} c^{2} + 8 \, {\left (64 \, b^{3} d^{2} x^{2} + 128 \, b^{3} c d x + 64 \, b^{3} c^{2} - 15 \, b d^{2}\right )} \cos \left (b x + a\right )^{4} - 255 \, b d^{2} - 8 \, {\left (128 \, b^{3} d^{2} x^{2} + 256 \, b^{3} c d x + 128 \, b^{3} c^{2} - 75 \, b d^{2}\right )} \cos \left (b x + a\right )^{2} - 160 \, {\left (2 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right )^{3} - 5 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )\right )} \sqrt {d x + c}}{8192 \, b^{4}} \]

[In]

integrate((d*x+c)^(5/2)*cos(b*x+a)*sin(b*x+a)^3,x, algorithm="fricas")

[Out]

1/8192*(15*sqrt(2)*pi*d^3*sqrt(b/(pi*d))*cos(-4*(b*c - a*d)/d)*fresnel_cos(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*
d))) - 15*sqrt(2)*pi*d^3*sqrt(b/(pi*d))*fresnel_sin(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-4*(b*c - a*d)
/d) - 480*pi*d^3*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d))) + 480*pi*d^3
*sqrt(b/(pi*d))*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) + 4*(320*b^3*d^2*x^2 + 640*b
^3*c*d*x + 320*b^3*c^2 + 8*(64*b^3*d^2*x^2 + 128*b^3*c*d*x + 64*b^3*c^2 - 15*b*d^2)*cos(b*x + a)^4 - 255*b*d^2
- 8*(128*b^3*d^2*x^2 + 256*b^3*c*d*x + 128*b^3*c^2 - 75*b*d^2)*cos(b*x + a)^2 - 160*(2*(b^2*d^2*x + b^2*c*d)*
cos(b*x + a)^3 - 5*(b^2*d^2*x + b^2*c*d)*cos(b*x + a))*sin(b*x + a))*sqrt(d*x + c))/b^4

Sympy [F(-1)]

Timed out. \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx=\text {Timed out} \]

[In]

integrate((d*x+c)**(5/2)*cos(b*x+a)*sin(b*x+a)**3,x)

[Out]

Timed out

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.40 (sec) , antiderivative size = 551, normalized size of antiderivative = 1.35 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx=-\frac {{\left (640 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} \sin \left (\frac {4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - 5120 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} \sin \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - 16 \, {\left (\frac {64 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4}}{d} - 15 \, \sqrt {d x + c} b^{2} d\right )} \cos \left (\frac {4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) + 256 \, {\left (\frac {16 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{4}}{d} - 15 \, \sqrt {d x + c} b^{2} d\right )} \cos \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - 240 \, {\left (\left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {2 i \, b}{d}}\right ) - 15 \, {\left (-\left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } b d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - \left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } b d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (2 \, \sqrt {d x + c} \sqrt {\frac {i \, b}{d}}\right ) - 15 \, {\left (\left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } b d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + \left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } b d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (2 \, \sqrt {d x + c} \sqrt {-\frac {i \, b}{d}}\right ) - 240 \, {\left (-\left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - \left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {2 i \, b}{d}}\right )\right )} d}{32768 \, b^{5}} \]

[In]

integrate((d*x+c)^(5/2)*cos(b*x+a)*sin(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/32768*(640*(d*x + c)^(3/2)*b^3*sin(4*((d*x + c)*b - b*c + a*d)/d) - 5120*(d*x + c)^(3/2)*b^3*sin(2*((d*x +
c)*b - b*c + a*d)/d) - 16*(64*(d*x + c)^(5/2)*b^4/d - 15*sqrt(d*x + c)*b^2*d)*cos(4*((d*x + c)*b - b*c + a*d)/
d) + 256*(16*(d*x + c)^(5/2)*b^4/d - 15*sqrt(d*x + c)*b^2*d)*cos(2*((d*x + c)*b - b*c + a*d)/d) - 240*((I - 1)
*4^(1/4)*sqrt(2)*sqrt(pi)*b*d^2*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) + (I + 1)*4^(1/4)*sqrt(2)*sqrt(pi)*b*d^2
*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(2*I*b/d)) - 15*(-(I - 1)*sqrt(2)*sqrt(pi)*b*d^2
*(b^2/d^2)^(1/4)*cos(-4*(b*c - a*d)/d) - (I + 1)*sqrt(2)*sqrt(pi)*b*d^2*(b^2/d^2)^(1/4)*sin(-4*(b*c - a*d)/d))
*erf(2*sqrt(d*x + c)*sqrt(I*b/d)) - 15*((I + 1)*sqrt(2)*sqrt(pi)*b*d^2*(b^2/d^2)^(1/4)*cos(-4*(b*c - a*d)/d) +
(I - 1)*sqrt(2)*sqrt(pi)*b*d^2*(b^2/d^2)^(1/4)*sin(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(-I*b/d)) - 240
*(-(I + 1)*4^(1/4)*sqrt(2)*sqrt(pi)*b*d^2*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) - (I - 1)*4^(1/4)*sqrt(2)*sqrt
(pi)*b*d^2*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-2*I*b/d)))*d/b^5

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.14 (sec) , antiderivative size = 2435, normalized size of antiderivative = 5.98 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx=\text {Too large to display} \]

[In]

integrate((d*x+c)^(5/2)*cos(b*x+a)*sin(b*x+a)^3,x, algorithm="giac")

[Out]

-1/16384*(512*(sqrt(2)*sqrt(pi)*d*erf(-I*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-4*(I
*b*c - I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)) + sqrt(2)*sqrt(pi)*d*erf(I*sqrt(2)*sqrt(b*d)*sqrt(d*x +
c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-4*(-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)) - 4*sqrt(pi
)*d*erf(-I*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(I*b*c - I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqr
t(b^2*d^2) + 1)) - 4*sqrt(pi)*d*erf(I*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(-I*b*c + I*
a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)))*c^3 + 24*c*d^2*((sqrt(2)*sqrt(pi)*(64*b^2*c^2 - 16*I*b*c*d - 3
*d^2)*d*erf(-I*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-4*(I*b*c - I*a*d)/d)/(sqrt(b*d
)*(I*b*d/sqrt(b^2*d^2) + 1)*b^2) - 4*I*(-8*I*(d*x + c)^(3/2)*b*d + 16*I*sqrt(d*x + c)*b*c*d + 3*sqrt(d*x + c)*
d^2)*e^(-4*(-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b^2)/d^2 + (sqrt(2)*sqrt(pi)*(64*b^2*c^2 + 16*I*b*c*d - 3*d^2)*
d*erf(I*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-4*(-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I
*b*d/sqrt(b^2*d^2) + 1)*b^2) - 4*I*(-8*I*(d*x + c)^(3/2)*b*d + 16*I*sqrt(d*x + c)*b*c*d - 3*sqrt(d*x + c)*d^2)
*e^(-4*(I*(d*x + c)*b - I*b*c + I*a*d)/d)/b^2)/d^2 - 16*(sqrt(pi)*(16*b^2*c^2 - 8*I*b*c*d - 3*d^2)*d*erf(-I*sq
rt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(I*b*c - I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) +
1)*b^2) + 2*I*(4*I*(d*x + c)^(3/2)*b*d - 8*I*sqrt(d*x + c)*b*c*d - 3*sqrt(d*x + c)*d^2)*e^(-2*(-I*(d*x + c)*b
+ I*b*c - I*a*d)/d)/b^2)/d^2 - 16*(sqrt(pi)*(16*b^2*c^2 + 8*I*b*c*d - 3*d^2)*d*erf(I*sqrt(b*d)*sqrt(d*x + c)*(
-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b^2) + 2*I*(4*I*(
d*x + c)^(3/2)*b*d - 8*I*sqrt(d*x + c)*b*c*d + 3*sqrt(d*x + c)*d^2)*e^(-2*(I*(d*x + c)*b - I*b*c + I*a*d)/d)/b
^2)/d^2) - d^3*((sqrt(2)*sqrt(pi)*(512*b^3*c^3 - 192*I*b^2*c^2*d - 72*b*c*d^2 + 15*I*d^3)*d*erf(-I*sqrt(2)*sqr
t(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-4*(I*b*c - I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1
)*b^3) + 4*I*(-64*I*(d*x + c)^(5/2)*b^2*d + 192*I*(d*x + c)^(3/2)*b^2*c*d - 192*I*sqrt(d*x + c)*b^2*c^2*d + 40
*(d*x + c)^(3/2)*b*d^2 - 72*sqrt(d*x + c)*b*c*d^2 + 15*I*sqrt(d*x + c)*d^3)*e^(-4*(-I*(d*x + c)*b + I*b*c - I*
a*d)/d)/b^3)/d^3 + (sqrt(2)*sqrt(pi)*(512*b^3*c^3 + 192*I*b^2*c^2*d - 72*b*c*d^2 - 15*I*d^3)*d*erf(I*sqrt(2)*s
qrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-4*(-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2
) + 1)*b^3) + 4*I*(-64*I*(d*x + c)^(5/2)*b^2*d + 192*I*(d*x + c)^(3/2)*b^2*c*d - 192*I*sqrt(d*x + c)*b^2*c^2*d
- 40*(d*x + c)^(3/2)*b*d^2 + 72*sqrt(d*x + c)*b*c*d^2 + 15*I*sqrt(d*x + c)*d^3)*e^(-4*(I*(d*x + c)*b - I*b*c
+ I*a*d)/d)/b^3)/d^3 - 32*(sqrt(pi)*(64*b^3*c^3 - 48*I*b^2*c^2*d - 36*b*c*d^2 + 15*I*d^3)*d*erf(-I*sqrt(b*d)*s
qrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(I*b*c - I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b^3) -
2*I*(16*I*(d*x + c)^(5/2)*b^2*d - 48*I*(d*x + c)^(3/2)*b^2*c*d + 48*I*sqrt(d*x + c)*b^2*c^2*d - 20*(d*x + c)^
(3/2)*b*d^2 + 36*sqrt(d*x + c)*b*c*d^2 - 15*I*sqrt(d*x + c)*d^3)*e^(-2*(-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b^3
)/d^3 - 32*(sqrt(pi)*(64*b^3*c^3 + 48*I*b^2*c^2*d - 36*b*c*d^2 - 15*I*d^3)*d*erf(I*sqrt(b*d)*sqrt(d*x + c)*(-I
*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b^3) - 2*I*(16*I*(d
*x + c)^(5/2)*b^2*d - 48*I*(d*x + c)^(3/2)*b^2*c*d + 48*I*sqrt(d*x + c)*b^2*c^2*d + 20*(d*x + c)^(3/2)*b*d^2 -
36*sqrt(d*x + c)*b*c*d^2 - 15*I*sqrt(d*x + c)*d^3)*e^(-2*(I*(d*x + c)*b - I*b*c + I*a*d)/d)/b^3)/d^3) - 192*(
sqrt(2)*sqrt(pi)*(8*b*c - I*d)*d*erf(-I*sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-4*(I*
b*c - I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) + sqrt(2)*sqrt(pi)*(8*b*c + I*d)*d*erf(I*sqrt(2)*sqrt(
b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-4*(-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) +
1)*b) - 8*sqrt(pi)*(4*b*c - I*d)*d*erf(-I*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(I*b*c -
I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) - 8*sqrt(pi)*(4*b*c + I*d)*d*erf(I*sqrt(b*d)*sqrt(d*x + c)*(
-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-2*(-I*b*c + I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b) - 16*sqrt(d*x
+ c)*d*e^(-2*(I*(d*x + c)*b - I*b*c + I*a*d)/d)/b + 4*sqrt(d*x + c)*d*e^(-4*(I*(d*x + c)*b - I*b*c + I*a*d)/d)
/b - 16*sqrt(d*x + c)*d*e^(-2*(-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b + 4*sqrt(d*x + c)*d*e^(-4*(-I*(d*x + c)*b
+ I*b*c - I*a*d)/d)/b)*c^2)/d

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx=\int \cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^{5/2} \,d x \]

[In]

int(cos(a + b*x)*sin(a + b*x)^3*(c + d*x)^(5/2),x)

[Out]

int(cos(a + b*x)*sin(a + b*x)^3*(c + d*x)^(5/2), x)