\(\int (c+d x)^{3/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx\) [128]

Optimal result

Integrand size = 26, antiderivative size = 200 \[ \int (c+d x)^{3/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\frac {(c+d x)^{5/2}}{20 d}-\frac {3 d \sqrt {c+d x} \cos (4 a+4 b x)}{256 b^2}+\frac {3 d^{3/2} \sqrt {\frac {\pi }{2}} \cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{512 b^{5/2}}-\frac {3 d^{3/2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (4 a-\frac {4 b c}{d}\right )}{512 b^{5/2}}-\frac {(c+d x)^{3/2} \sin (4 a+4 b x)}{32 b} \]

[Out]

1/20*(d*x+c)^(5/2)/d-1/32*(d*x+c)^(3/2)*sin(4*b*x+4*a)/b+3/1024*d^(3/2)*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^(5/2)-3/1024*d^(3/2)*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^(5/2)-3/256*d*cos(4*b*x+4*a)*(d*x+c)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {4491, 3377, 3387, 3386, 3432, 3385, 3433} \[ \int (c+d x)^{3/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\frac {3 \sqrt {\frac {\pi }{2}} d^{3/2} \cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{512 b^{5/2}}-\frac {3 \sqrt {\frac {\pi }{2}} d^{3/2} \sin \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{512 b^{5/2}}-\frac {3 d \sqrt {c+d x} \cos (4 a+4 b x)}{256 b^2}-\frac {(c+d x)^{3/2} \sin (4 a+4 b x)}{32 b}+\frac {(c+d x)^{5/2}}{20 d} \]

[In]

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

[Out]

(c + d*x)^(5/2)/(20*d) - (3*d*Sqrt[c + d*x]*Cos[4*a + 4*b*x])/(256*b^2) + (3*d^(3/2)*Sqrt[Pi/2]*Cos[4*a - (4*b
*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(512*b^(5/2)) - (3*d^(3/2)*Sqrt[Pi/2]*FresnelS[
(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[4*a - (4*b*c)/d])/(512*b^(5/2)) - ((c + d*x)^(3/2)*Sin[4*a +
4*b*x])/(32*b)

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}{8} (c+d x)^{3/2}-\frac {1}{8} (c+d x)^{3/2} \cos (4 a+4 b x)\right ) \, dx \\ & = \frac {(c+d x)^{5/2}}{20 d}-\frac {1}{8} \int (c+d x)^{3/2} \cos (4 a+4 b x) \, dx \\ & = \frac {(c+d x)^{5/2}}{20 d}-\frac {(c+d x)^{3/2} \sin (4 a+4 b x)}{32 b}+\frac {(3 d) \int \sqrt {c+d x} \sin (4 a+4 b x) \, dx}{64 b} \\ & = \frac {(c+d x)^{5/2}}{20 d}-\frac {3 d \sqrt {c+d x} \cos (4 a+4 b x)}{256 b^2}-\frac {(c+d x)^{3/2} \sin (4 a+4 b x)}{32 b}+\frac {\left (3 d^2\right ) \int \frac {\cos (4 a+4 b x)}{\sqrt {c+d x}} \, dx}{512 b^2} \\ & = \frac {(c+d x)^{5/2}}{20 d}-\frac {3 d \sqrt {c+d x} \cos (4 a+4 b x)}{256 b^2}-\frac {(c+d x)^{3/2} \sin (4 a+4 b x)}{32 b}+\frac {\left (3 d^2 \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}{512 b^2}-\frac {\left (3 d^2 \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}{512 b^2} \\ & = \frac {(c+d x)^{5/2}}{20 d}-\frac {3 d \sqrt {c+d x} \cos (4 a+4 b x)}{256 b^2}-\frac {(c+d x)^{3/2} \sin (4 a+4 b x)}{32 b}+\frac {\left (3 d \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 )}{256 b^2}-\frac {\left (3 d \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 )}{256 b^2} \\ & = \frac {(c+d x)^{5/2}}{20 d}-\frac {3 d \sqrt {c+d x} \cos (4 a+4 b x)}{256 b^2}+\frac {3 d^{3/2} \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 )}{512 b^{5/2}}-\frac {3 d^{3/2} \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 )}{512 b^{5/2}}-\frac {(c+d x)^{3/2} \sin (4 a+4 b x)}{32 b} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.05 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.70 \[ \int (c+d x)^{3/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\frac {\sqrt {c+d x} \left (128 (c+d x)^2-\frac {5 d^2 e^{4 i \left (a-\frac {b c}{d}\right )} \Gamma \left (\frac {5}{2},-\frac {4 i b (c+d x)}{d}\right )}{b^2 \sqrt {-\frac {i b (c+d x)}{d}}}-\frac {5 d^2 e^{-4 i \left (a-\frac {b c}{d}\right )} \Gamma \left (\frac {5}{2},\frac {4 i b (c+d x)}{d}\right )}{b^2 \sqrt {\frac {i b (c+d x)}{d}}}\right )}{2560 d} \]

[In]

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

[Out]

(Sqrt[c + d*x]*(128*(c + d*x)^2 - (5*d^2*E^((4*I)*(a - (b*c)/d))*Gamma[5/2, ((-4*I)*b*(c + d*x))/d])/(b^2*Sqrt
[((-I)*b*(c + d*x))/d]) - (5*d^2*Gamma[5/2, ((4*I)*b*(c + d*x))/d])/(b^2*E^((4*I)*(a - (b*c)/d))*Sqrt[(I*b*(c
+ d*x))/d])))/(2560*d)

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(\frac {\frac {\left (d x +c \right )^{\frac {5}{2}}}{20}-\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{32 b}+\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{8 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{32 b \sqrt {\frac {b}{d}}}\right )}{32 b}}{d}\)	\(206\)
default	\(\frac {\frac {\left (d x +c \right )^{\frac {5}{2}}}{20}-\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{32 b}+\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{8 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{32 b \sqrt {\frac {b}{d}}}\right )}{32 b}}{d}\)	\(206\)

[In]

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

[Out]

2/d*(1/40*(d*x+c)^(5/2)-1/64/b*d*(d*x+c)^(3/2)*sin(4*b/d*(d*x+c)+4*(a*d-b*c)/d)+3/64/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 = 0.27 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.24 \[ \int (c+d x)^{3/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\frac {15 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 15 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + 4 \, {\left (64 \, b^{3} d^{2} x^{2} - 120 \, b d^{2} \cos \left (b x + a\right )^{4} + 128 \, b^{3} c d x + 64 \, b^{3} c^{2} + 120 \, b d^{2} \cos \left (b x + a\right )^{2} - 15 \, b d^{2} - 160 \, {\left (2 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right )^{3} - {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )\right )} \sqrt {d x + c}}{5120 \, b^{3} d} \]

[In]

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

[Out]

1/5120*(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) + 4*(64*b^3*d^2*x^2 - 120*b*d^2*cos(b*x + a)^4 + 128*b^3*c*d*x + 64*b^3*c^2 + 120*b*d^2*cos(b*x + a)^2 - 1
5*b*d^2 - 160*(2*(b^2*d^2*x + b^2*c*d)*cos(b*x + a)^3 - (b^2*d^2*x + b^2*c*d)*cos(b*x + a))*sin(b*x + a))*sqrt
(d*x + c))/(b^3*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.32 \[ \int (c+d x)^{3/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\frac {\sqrt {2} {\left (\frac {512 \, \sqrt {2} {\left (d x + c\right )}^{\frac {5}{2}} b^{3}}{d} - 320 \, \sqrt {2} {\left (d x + c\right )}^{\frac {3}{2}} b^{2} \sin \left (\frac {4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) - 120 \, \sqrt {2} \sqrt {d x + c} b d \cos \left (\frac {4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) + 15 \, {\left (-\left (i - 1\right ) \, \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - \left (i + 1\right ) \, \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (2 \, \sqrt {d x + c} \sqrt {\frac {i \, b}{d}}\right ) + 15 \, {\left (\left (i + 1\right ) \, \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + \left (i - 1\right ) \, \sqrt {\pi } d^{2} \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (2 \, \sqrt {d x + c} \sqrt {-\frac {i \, b}{d}}\right )\right )}}{20480 \, b^{3}} \]

[In]

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

[Out]

1/20480*sqrt(2)*(512*sqrt(2)*(d*x + c)^(5/2)*b^3/d - 320*sqrt(2)*(d*x + c)^(3/2)*b^2*sin(4*((d*x + c)*b - b*c
+ a*d)/d) - 120*sqrt(2)*sqrt(d*x + c)*b*d*cos(4*((d*x + c)*b - b*c + a*d)/d) + 15*(-(I - 1)*sqrt(pi)*d^2*(b^2/
d^2)^(1/4)*cos(-4*(b*c - a*d)/d) - (I + 1)*sqrt(pi)*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(pi)*d^2*(b^2/d^2)^(1/4)*cos(-4*(b*c - a*d)/d) + (I - 1)*sqrt(pi)*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)))/b^3

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.79 (sec) , antiderivative size = 857, normalized size of antiderivative = 4.28 \[ \int (c+d x)^{3/2} \cos ^2(a+b x) \sin ^2(a+b x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/30720*(960*(I*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)) - I*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)) - 8*sqr
t(d*x + c))*c^2 - d^2*(512*(3*(d*x + c)^(5/2) - 10*(d*x + c)^(3/2)*c + 15*sqrt(d*x + c)*c^2)/d^2 - 15*(I*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*(-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 - 15*(-I*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*(8*I*(d*x + c)^(3/2)*b*d - 16*I*s
qrt(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) + 80*(-3*I*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) + 3*I*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) - 64*(d*x + c)^(3/2) + 192*sqrt(d*x + c)*c + 12*I*sqrt(d*x + c)*d*e^(-4*(I*(d*x + c)*b - I*b*c + I*a*d)/d)
/b - 12*I*sqrt(d*x + c)*d*e^(-4*(-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b)*c)/d

Mupad [F(-1)]

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

[In]

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

[Out]

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