\(\int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx\) [64]

Optimal result

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 11.25 (sec) , antiderivative size = 1332, normalized size of antiderivative = 3.27 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

((1/128 + I/128)*c*Sqrt[d]*((2 + 2*I)*Sqrt[b]*Sqrt[d]*E^(((2*I)*b*c)/d)*Sq 
rt[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]*S 
qrt[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)*Sqrt[ 
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)*Sq 
rt[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]]))/(10 
24*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*Sqr 
t[((-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...
 

Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {4906, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 4906

\(\displaystyle \int \left (\frac {1}{4} (c+d x)^{5/2} \sin (2 a+2 b x)-\frac {1}{8} (c+d x)^{5/2} \sin (4 a+4 b x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4096 b^{7/2}}-\frac {15 \sqrt {\pi } d^{5/2} \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{256 b^{7/2}}-\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \sin \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4096 b^{7/2}}+\frac {15 \sqrt {\pi } d^{5/2} \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{256 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \cos (2 a+2 b x)}{128 b^3}-\frac {15 d^2 \sqrt {c+d x} \cos (4 a+4 b x)}{2048 b^3}+\frac {5 d (c+d x)^{3/2} \sin (2 a+2 b x)}{32 b^2}-\frac {5 d (c+d x)^{3/2} \sin (4 a+4 b x)}{256 b^2}-\frac {(c+d x)^{5/2} \cos (2 a+2 b x)}{8 b}+\frac {(c+d x)^{5/2} \cos (4 a+4 b x)}{32 b}\)

Input:

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

Output:

(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]])/(409 
6*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)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4906

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b 
_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(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] && IG 
tQ[p, 0]
 

Maple [A] (verified)

Time = 3.06 (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	\(\frac {-\frac {d \left (d x +c \right )^{\frac {5}{2}} \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{8 b}+\frac {5 d \left (\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{4 b}-\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{4 b}+\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}\right )}{4 b}\right )}{8 b}+\frac {d \left (d x +c \right )^{\frac {5}{2}} \cos \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{32 b}-\frac {5 d \left (\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{8 b}-\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{8 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{32 b \sqrt {\frac {b}{d}}}\right )}{8 b}\right )}{32 b}}{d}\)	\(470\)

Input:

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

Output:

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)*(co 
s(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)

Time = 0.10 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.00 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx=\frac {15 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 15 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) - 480 \, \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 480 \, \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 4 \, {\left (320 \, b^{3} d^{2} x^{2} + 640 \, b^{3} c d x + 320 \, b^{3} c^{2} + 8 \, {\left (64 \, b^{3} d^{2} x^{2} + 128 \, b^{3} c d x + 64 \, b^{3} c^{2} - 15 \, b d^{2}\right )} \cos \left (b x + a\right )^{4} - 255 \, b d^{2} - 8 \, {\left (128 \, b^{3} d^{2} x^{2} + 256 \, b^{3} c d x + 128 \, b^{3} c^{2} - 75 \, b d^{2}\right )} \cos \left (b x + a\right )^{2} - 160 \, {\left (2 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right )^{3} - 5 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )\right )} \sqrt {d x + c}}{8192 \, b^{4}} \] Input:

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

Output:

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 + 12 
8*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. \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 551, normalized size of antiderivative = 1.35 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/32768*(640*(d*x + c)^(3/2)*b^3*sin(4*((d*x + c)*b - b*c + a*d)/d) - 512 
0*(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 = 0.99 (sec) , antiderivative size = 2412, normalized size of antiderivative = 5.93 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^3(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

-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/sqr 
t(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/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/sqrt(b^2*d^2) + 
 1)))*c^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)*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) + 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)*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) + 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*...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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