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

Optimal. Leaf size=351 \[ \frac {3 \sqrt {\frac {\pi }{2}} d^{3/2} \sin \left (4 a-\frac {4 b c}{d}\right ) C\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{512 b^{5/2}}-\frac {3 \sqrt {\pi } d^{3/2} \sin \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{64 b^{5/2}}+\frac {3 \sqrt {\frac {\pi }{2}} d^{3/2} \cos \left (4 a-\frac {4 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{512 b^{5/2}}-\frac {3 \sqrt {\pi } d^{3/2} \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{64 b^{5/2}}+\frac {3 d \sqrt {c+d x} \sin (2 a+2 b x)}{32 b^2}-\frac {3 d \sqrt {c+d x} \sin (4 a+4 b x)}{256 b^2}-\frac {(c+d x)^{3/2} \cos (2 a+2 b x)}{8 b}+\frac {(c+d x)^{3/2} \cos (4 a+4 b x)}{32 b} \]

[Out]

-1/8*(d*x+c)^(3/2)*cos(2*b*x+2*a)/b+1/32*(d*x+c)^(3/2)*cos(4*b*x+4*a)/b+3/1024*d^(3/2)*cos(4*a-4*b*c/d)*Fresne
lS(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^(5/2)+3/1024*d^(3/2)*FresnelC(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^(5/2)-3/64*d^(3/2)*cos(2*a-2*b*c
/d)*FresnelS(2*b^(1/2)*(d*x+c)^(1/2)/d^(1/2)/Pi^(1/2))*Pi^(1/2)/b^(5/2)-3/64*d^(3/2)*FresnelC(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^(5/2)+3/32*d*sin(2*b*x+2*a)*(d*x+c)^(1/2)/b^2-3/256*d*si
n(4*b*x+4*a)*(d*x+c)^(1/2)/b^2

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Rubi [A]  time = 0.67, antiderivative size = 351, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {4406, 3296, 3306, 3305, 3351, 3304, 3352} \[ \frac {3 \sqrt {\frac {\pi }{2}} d^{3/2} \sin \left (4 a-\frac {4 b c}{d}\right ) \text {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {b} \sqrt {c+d x}}{\sqrt {d}}\right )}{512 b^{5/2}}-\frac {3 \sqrt {\pi } d^{3/2} \sin \left (2 a-\frac {2 b c}{d}\right ) \text {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {\pi } \sqrt {d}}\right )}{64 b^{5/2}}+\frac {3 \sqrt {\frac {\pi }{2}} d^{3/2} \cos \left (4 a-\frac {4 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{512 b^{5/2}}-\frac {3 \sqrt {\pi } d^{3/2} \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{64 b^{5/2}}+\frac {3 d \sqrt {c+d x} \sin (2 a+2 b x)}{32 b^2}-\frac {3 d \sqrt {c+d x} \sin (4 a+4 b x)}{256 b^2}-\frac {(c+d x)^{3/2} \cos (2 a+2 b x)}{8 b}+\frac {(c+d x)^{3/2} \cos (4 a+4 b x)}{32 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c + d*x)^(3/2)*Cos[2*a + 2*b*x])/(8*b) + ((c + d*x)^(3/2)*Cos[4*a + 4*b*x])/(32*b) + (3*d^(3/2)*Sqrt[Pi/2]*
Cos[4*a - (4*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(512*b^(5/2)) - (3*d^(3/2)*Sqrt[P
i]*Cos[2*a - (2*b*c)/d]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(64*b^(5/2)) + (3*d^(3/2)*Sqrt
[Pi/2]*FresnelC[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[4*a - (4*b*c)/d])/(512*b^(5/2)) - (3*d^(3/2)
*Sqrt[Pi]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(64*b^(5/2)) + (3*d*Sqr
t[c + d*x]*Sin[2*a + 2*b*x])/(32*b^2) - (3*d*Sqrt[c + d*x]*Sin[4*a + 4*b*x])/(256*b^2)

Rule 3296

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 3304

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 3305

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 3306

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 3351

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

Rule 3352

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

Rule 4406

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*} \int (c+d x)^{3/2} \cos (a+b x) \sin ^3(a+b x) \, dx &=\int \left (\frac {1}{4} (c+d x)^{3/2} \sin (2 a+2 b x)-\frac {1}{8} (c+d x)^{3/2} \sin (4 a+4 b x)\right ) \, dx\\ &=-\left (\frac {1}{8} \int (c+d x)^{3/2} \sin (4 a+4 b x) \, dx\right )+\frac {1}{4} \int (c+d x)^{3/2} \sin (2 a+2 b x) \, dx\\ &=-\frac {(c+d x)^{3/2} \cos (2 a+2 b x)}{8 b}+\frac {(c+d x)^{3/2} \cos (4 a+4 b x)}{32 b}-\frac {(3 d) \int \sqrt {c+d x} \cos (4 a+4 b x) \, dx}{64 b}+\frac {(3 d) \int \sqrt {c+d x} \cos (2 a+2 b x) \, dx}{16 b}\\ &=-\frac {(c+d x)^{3/2} \cos (2 a+2 b x)}{8 b}+\frac {(c+d x)^{3/2} \cos (4 a+4 b x)}{32 b}+\frac {3 d \sqrt {c+d x} \sin (2 a+2 b x)}{32 b^2}-\frac {3 d \sqrt {c+d x} \sin (4 a+4 b x)}{256 b^2}+\frac {\left (3 d^2\right ) \int \frac {\sin (4 a+4 b x)}{\sqrt {c+d x}} \, dx}{512 b^2}-\frac {\left (3 d^2\right ) \int \frac {\sin (2 a+2 b x)}{\sqrt {c+d x}} \, dx}{64 b^2}\\ &=-\frac {(c+d x)^{3/2} \cos (2 a+2 b x)}{8 b}+\frac {(c+d x)^{3/2} \cos (4 a+4 b x)}{32 b}+\frac {3 d \sqrt {c+d x} \sin (2 a+2 b x)}{32 b^2}-\frac {3 d \sqrt {c+d x} \sin (4 a+4 b x)}{256 b^2}+\frac {\left (3 d^2 \cos \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}{512 b^2}-\frac {\left (3 d^2 \cos \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}{64 b^2}+\frac {\left (3 d^2 \sin \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}{512 b^2}-\frac {\left (3 d^2 \sin \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}{64 b^2}\\ &=-\frac {(c+d x)^{3/2} \cos (2 a+2 b x)}{8 b}+\frac {(c+d x)^{3/2} \cos (4 a+4 b x)}{32 b}+\frac {3 d \sqrt {c+d x} \sin (2 a+2 b x)}{32 b^2}-\frac {3 d \sqrt {c+d x} \sin (4 a+4 b x)}{256 b^2}+\frac {\left (3 d \cos \left (4 a-\frac {4 b c}{d}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {4 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{256 b^2}-\frac {\left (3 d \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {2 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{32 b^2}+\frac {\left (3 d \sin \left (4 a-\frac {4 b c}{d}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {4 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{256 b^2}-\frac {\left (3 d \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {2 b x^2}{d}\right ) \, dx,x,\sqrt {c+d x}\right )}{32 b^2}\\ &=-\frac {(c+d x)^{3/2} \cos (2 a+2 b x)}{8 b}+\frac {(c+d x)^{3/2} \cos (4 a+4 b x)}{32 b}+\frac {3 d^{3/2} \sqrt {\frac {\pi }{2}} \cos \left (4 a-\frac {4 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{512 b^{5/2}}-\frac {3 d^{3/2} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{64 b^{5/2}}+\frac {3 d^{3/2} \sqrt {\frac {\pi }{2}} C\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 )}{512 b^{5/2}}-\frac {3 d^{3/2} \sqrt {\pi } C\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{64 b^{5/2}}+\frac {3 d \sqrt {c+d x} \sin (2 a+2 b x)}{32 b^2}-\frac {3 d \sqrt {c+d x} \sin (4 a+4 b x)}{256 b^2}\\ \end {align*}

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Mathematica [A]  time = 3.08, size = 393, normalized size = 1.12 \[ \frac {3 \sqrt {2 \pi } d \sin \left (4 a-\frac {4 b c}{d}\right ) C\left (2 \sqrt {\frac {b}{d}} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}\right )-48 \sqrt {\pi } d \sin \left (2 a-\frac {2 b c}{d}\right ) C\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )+3 \sqrt {2 \pi } d \cos \left (4 a-\frac {4 b c}{d}\right ) S\left (2 \sqrt {\frac {b}{d}} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}\right )-48 \sqrt {\pi } d \cos \left (2 a-\frac {2 b c}{d}\right ) S\left (\frac {2 \sqrt {\frac {b}{d}} \sqrt {c+d x}}{\sqrt {\pi }}\right )+96 d \sqrt {\frac {b}{d}} \sqrt {c+d x} \sin (2 (a+b x))-12 d \sqrt {\frac {b}{d}} \sqrt {c+d x} \sin (4 (a+b x))-128 b d x \sqrt {\frac {b}{d}} \sqrt {c+d x} \cos (2 (a+b x))-128 b c \sqrt {\frac {b}{d}} \sqrt {c+d x} \cos (2 (a+b x))+32 b d x \sqrt {\frac {b}{d}} \sqrt {c+d x} \cos (4 (a+b x))+32 b c \sqrt {\frac {b}{d}} \sqrt {c+d x} \cos (4 (a+b x))}{1024 b^2 \sqrt {\frac {b}{d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-128*b*c*Sqrt[b/d]*Sqrt[c + d*x]*Cos[2*(a + b*x)] - 128*b*Sqrt[b/d]*d*x*Sqrt[c + d*x]*Cos[2*(a + b*x)] + 32*b
*c*Sqrt[b/d]*Sqrt[c + d*x]*Cos[4*(a + b*x)] + 32*b*Sqrt[b/d]*d*x*Sqrt[c + d*x]*Cos[4*(a + b*x)] + 3*d*Sqrt[2*P
i]*Cos[4*a - (4*b*c)/d]*FresnelS[2*Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]] - 48*d*Sqrt[Pi]*Cos[2*a - (2*b*c)/d]*Fr
esnelS[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]] + 3*d*Sqrt[2*Pi]*FresnelC[2*Sqrt[b/d]*Sqrt[2/Pi]*Sqrt[c + d*x]]*S
in[4*a - (4*b*c)/d] - 48*d*Sqrt[Pi]*FresnelC[(2*Sqrt[b/d]*Sqrt[c + d*x])/Sqrt[Pi]]*Sin[2*a - (2*b*c)/d] + 96*S
qrt[b/d]*d*Sqrt[c + d*x]*Sin[2*(a + b*x)] - 12*Sqrt[b/d]*d*Sqrt[c + d*x]*Sin[4*(a + b*x)])/(1024*b^2*Sqrt[b/d]
)

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fricas [A]  time = 0.58, size = 316, normalized size = 0.90 \[ \frac {3 \, \sqrt {2} \pi d^{2} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 3 \, \sqrt {2} \pi d^{2} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - 48 \, \pi d^{2} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 48 \, \pi d^{2} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 16 \, {\left (16 \, {\left (b^{2} d x + b^{2} c\right )} \cos \left (b x + a\right )^{4} + 10 \, b^{2} d x + 10 \, b^{2} c - 32 \, {\left (b^{2} d x + b^{2} c\right )} \cos \left (b x + a\right )^{2} - 3 \, {\left (2 \, b d \cos \left (b x + a\right )^{3} - 5 \, b d \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )\right )} \sqrt {d x + c}}{1024 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1024*(3*sqrt(2)*pi*d^2*sqrt(b/(pi*d))*cos(-4*(b*c - a*d)/d)*fresnel_sin(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d
))) + 3*sqrt(2)*pi*d^2*sqrt(b/(pi*d))*fresnel_cos(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-4*(b*c - a*d)/d
) - 48*pi*d^2*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d))) - 48*pi*d^2*sqr
t(b/(pi*d))*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-2*(b*c - a*d)/d) + 16*(16*(b^2*d*x + b^2*c)*cos(b
*x + a)^4 + 10*b^2*d*x + 10*b^2*c - 32*(b^2*d*x + b^2*c)*cos(b*x + a)^2 - 3*(2*b*d*cos(b*x + a)^3 - 5*b*d*cos(
b*x + a))*sin(b*x + a))*sqrt(d*x + c))/b^3

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giac [C]  time = 4.75, size = 1503, normalized size = 4.28 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2048*(64*(-I*sqrt(2)*sqrt(pi)*d*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((4*I*b
*c - 4*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)) + I*sqrt(2)*sqrt(pi)*d*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x
+ c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-4*I*b*c + 4*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)) + 4*I*sqr
t(pi)*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((2*I*b*c - 2*I*a*d)/d)/(sqrt(b*d)*(I*b*d/
sqrt(b^2*d^2) + 1)) - 4*I*sqrt(pi)*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-2*I*b*c +
2*I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)))*c^2 + d^2*((-I*sqrt(2)*sqrt(pi)*(64*b^2*c^2 + 16*I*b*c*d
- 3*d^2)*d*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((4*I*b*c - 4*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 + 4*I*b*c - 4*I*a*d)/d)/b^2)/d^2 + (I*sqrt(2)*sqrt(pi)*(64*b^2*c^2 - 16*I*b*c*d -
3*d^2)*d*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-4*I*b*c + 4*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 - 4*I*b*c + 4*I*a*d)/d)/b^2)/d^2 + 16*(I*sqrt(pi)*(16*b^2*c^2 + 8*I*b*c*d - 3*d^2)
*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((2*I*b*c - 2*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 + 2*I*b*c - 2*I*a*d)/d)/b^2)/d^2 + 16*(-I*sqrt(pi)*(16*b^2*c^2 - 8*I*b*c*d - 3*d^2)*d*erf(-sqrt(b*d)
*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-2*I*b*c + 2*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 - 2*I
*b*c + 2*I*a*d)/d)/b^2)/d^2) + 16*(I*sqrt(2)*sqrt(pi)*(8*b*c + I*d)*d*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(I*
b*d/sqrt(b^2*d^2) + 1)/d)*e^((4*I*b*c - 4*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) - I*sqrt(2)*sqrt(p
i)*(8*b*c - I*d)*d*erf(-sqrt(2)*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-4*I*b*c + 4*I*a*d)/
d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b) - 8*I*sqrt(pi)*(4*b*c + I*d)*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(I*b*d
/sqrt(b^2*d^2) + 1)/d)*e^((2*I*b*c - 2*I*a*d)/d)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b) + 8*I*sqrt(pi)*(4*b*c
- I*d)*d*erf(-sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-2*I*b*c + 2*I*a*d)/d)/(sqrt(b*d)*(-I
*b*d/sqrt(b^2*d^2) + 1)*b) - 4*sqrt(d*x + c)*d*e^((4*I*(d*x + c)*b - 4*I*b*c + 4*I*a*d)/d)/b + 16*sqrt(d*x + c
)*d*e^((2*I*(d*x + c)*b - 2*I*b*c + 2*I*a*d)/d)/b + 16*sqrt(d*x + c)*d*e^((-2*I*(d*x + c)*b + 2*I*b*c - 2*I*a*
d)/d)/b - 4*sqrt(d*x + c)*d*e^((-4*I*(d*x + c)*b + 4*I*b*c - 4*I*a*d)/d)/b)*c)/d

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maple [A]  time = 0.04, size = 376, normalized size = 1.07 \[ \frac {-\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{8 b}+\frac {3 d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {2 \left (d x +c \right ) b}{d}+\frac {2 d a -2 c b}{d}\right )}{4 b}-\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 d a -2 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {2 d a -2 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}\right )}{8 b}+\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {4 \left (d x +c \right ) b}{d}+\frac {4 d a -4 c b}{d}\right )}{32 b}-\frac {3 d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {4 \left (d x +c \right ) b}{d}+\frac {4 d a -4 c b}{d}\right )}{8 b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {4 d a -4 c b}{d}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {4 d a -4 c b}{d}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {d x +c}\, b}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{32 b \sqrt {\frac {b}{d}}}\right )}{32 b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.51, size = 503, normalized size = 1.43 \[ \frac {{\left (\frac {256 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} \cos \left (\frac {4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right )}{d} - \frac {1024 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} \cos \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right )}{d} - 96 \, \sqrt {d x + c} b^{2} \sin \left (\frac {4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) + 768 \, \sqrt {d x + c} b^{2} \sin \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - {\left (\left (48 i + 48\right ) \cdot 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - \left (48 i - 48\right ) \cdot 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b d \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 ) - {\left (-\left (6 i + 6\right ) \, \sqrt {2} \sqrt {\pi } b d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + \left (6 i - 6\right ) \, \sqrt {2} \sqrt {\pi } b d \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 ) - {\left (\left (6 i - 6\right ) \, \sqrt {2} \sqrt {\pi } b d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - \left (6 i + 6\right ) \, \sqrt {2} \sqrt {\pi } b d \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 ) - {\left (-\left (48 i - 48\right ) \cdot 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \left (48 i + 48\right ) \cdot 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b d \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}{8192 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8192*(256*(d*x + c)^(3/2)*b^3*cos(4*((d*x + c)*b - b*c + a*d)/d)/d - 1024*(d*x + c)^(3/2)*b^3*cos(2*((d*x +
c)*b - b*c + a*d)/d)/d - 96*sqrt(d*x + c)*b^2*sin(4*((d*x + c)*b - b*c + a*d)/d) + 768*sqrt(d*x + c)*b^2*sin(2
*((d*x + c)*b - b*c + a*d)/d) - ((48*I + 48)*4^(1/4)*sqrt(2)*sqrt(pi)*b*d*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d
) - (48*I - 48)*4^(1/4)*sqrt(2)*sqrt(pi)*b*d*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(2*I
*b/d)) - (-(6*I + 6)*sqrt(2)*sqrt(pi)*b*d*(b^2/d^2)^(1/4)*cos(-4*(b*c - a*d)/d) + (6*I - 6)*sqrt(2)*sqrt(pi)*b
*d*(b^2/d^2)^(1/4)*sin(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(I*b/d)) - ((6*I - 6)*sqrt(2)*sqrt(pi)*b*d*(
b^2/d^2)^(1/4)*cos(-4*(b*c - a*d)/d) - (6*I + 6)*sqrt(2)*sqrt(pi)*b*d*(b^2/d^2)^(1/4)*sin(-4*(b*c - a*d)/d))*e
rf(2*sqrt(d*x + c)*sqrt(-I*b/d)) - (-(48*I - 48)*4^(1/4)*sqrt(2)*sqrt(pi)*b*d*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*
d)/d) + (48*I + 48)*4^(1/4)*sqrt(2)*sqrt(pi)*b*d*(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^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^3\,{\left (c+d\,x\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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