\(\int (c+d x)^{5/2} \cos ^2(a+b x) \sin (a+b x) \, dx\) [118]

Optimal result

Integrand size = 24, antiderivative size = 406 \[ \int (c+d x)^{5/2} \cos ^2(a+b x) \sin (a+b x) \, dx=\frac {15 d^2 \sqrt {c+d x} \cos (a+b x)}{16 b^3}-\frac {(c+d x)^{5/2} \cos (a+b x)}{4 b}+\frac {5 d^2 \sqrt {c+d x} \cos (3 a+3 b x)}{144 b^3}-\frac {(c+d x)^{5/2} \cos (3 a+3 b x)}{12 b}-\frac {15 d^{5/2} \sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{16 b^{7/2}}-\frac {5 d^{5/2} \sqrt {\frac {\pi }{6}} \cos \left (3 a-\frac {3 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {6}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{144 b^{7/2}}+\frac {5 d^{5/2} \sqrt {\frac {\pi }{6}} \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {6}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (3 a-\frac {3 b c}{d}\right )}{144 b^{7/2}}+\frac {15 d^{5/2} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (a-\frac {b c}{d}\right )}{16 b^{7/2}}+\frac {5 d (c+d x)^{3/2} \sin (a+b x)}{8 b^2}+\frac {5 d (c+d x)^{3/2} \sin (3 a+3 b x)}{72 b^2} \] Output:

15/16*d^2*(d*x+c)^(1/2)*cos(b*x+a)/b^3-1/4*(d*x+c)^(5/2)*cos(b*x+a)/b+5/14 
4*d^2*(d*x+c)^(1/2)*cos(3*b*x+3*a)/b^3-1/12*(d*x+c)^(5/2)*cos(3*b*x+3*a)/b 
-15/32*d^(5/2)*2^(1/2)*Pi^(1/2)*cos(a-b*c/d)*FresnelC(b^(1/2)*2^(1/2)/Pi^( 
1/2)*(d*x+c)^(1/2)/d^(1/2))/b^(7/2)-5/864*d^(5/2)*6^(1/2)*Pi^(1/2)*cos(3*a 
-3*b*c/d)*FresnelC(b^(1/2)*6^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))/b^(7/2) 
+5/864*d^(5/2)*6^(1/2)*Pi^(1/2)*FresnelS(b^(1/2)*6^(1/2)/Pi^(1/2)*(d*x+c)^ 
(1/2)/d^(1/2))*sin(3*a-3*b*c/d)/b^(7/2)+15/32*d^(5/2)*2^(1/2)*Pi^(1/2)*Fre 
snelS(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2))*sin(a-b*c/d)/b^(7/2) 
+5/8*d*(d*x+c)^(3/2)*sin(b*x+a)/b^2+5/72*d*(d*x+c)^(3/2)*sin(3*b*x+3*a)/b^ 
2
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.29 (sec) , antiderivative size = 1445, normalized size of antiderivative = 3.56 \[ \int (c+d x)^{5/2} \cos ^2(a+b x) \sin (a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

(c*Sqrt[d]*(-12*Sqrt[b]*Sqrt[d]*E^(((3*I)*b*c)/d)*Sqrt[c + d*x]*(-I + 2*b* 
x + E^((6*I)*(a + b*x))*(I + 2*b*x)) - (1 - I)*(2*b*c + I*d)*E^(((3*I)*b*( 
2*c + d*x))/d)*Sqrt[6*Pi]*Erf[((1 + I)*Sqrt[3/2]*Sqrt[b]*Sqrt[c + d*x])/Sq 
rt[d]] + (1 + I)*((2*I)*b*c + d)*E^((3*I)*(2*a + b*x))*Sqrt[6*Pi]*Erfi[((1 
 + I)*Sqrt[3/2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]]))/(288*b^(5/2)*E^(((3*I)*( 
a*d + b*(c + d*x)))/d)) + ((I/8)*c^2*d*(-(E^((2*I)*a)*Sqrt[((-I)*b*(c + d* 
x))/d]*Gamma[3/2, ((-I)*b*(c + d*x))/d]) + E^(((2*I)*b*c)/d)*Sqrt[(I*b*(c 
+ d*x))/d]*Gamma[3/2, (I*b*(c + d*x))/d]))/(b^2*E^((I*(b*c + a*d))/d)*Sqrt 
[c + d*x]) + (c^2*(-1/6*(E^((3*I)*(a - (b*c)/d))*Sqrt[c + d*x]*Gamma[3/2, 
((-3*I)*b*(c + d*x))/d])/(Sqrt[3]*b*Sqrt[((-I)*b*(c + d*x))/d]) - (Sqrt[c 
+ d*x]*Gamma[3/2, ((3*I)*b*(c + d*x))/d])/(6*Sqrt[3]*b*E^((3*I)*(a - (b*c) 
/d))*Sqrt[(I*b*(c + d*x))/d])))/4 + (c*Sqrt[d]*(E^(I*(a - (b*c)/d))*(-2*Sq 
rt[b]*Sqrt[d]*E^((I*b*(c + d*x))/d)*(3*I + 2*b*x)*Sqrt[c + d*x] + (-1)^(1/ 
4)*((2*I)*b*c + 3*d)*Sqrt[Pi]*Erfi[((-1)^(1/4)*Sqrt[b]*Sqrt[c + d*x])/Sqrt 
[d]]) + I*(2*Sqrt[b]*Sqrt[d]*(3 + (2*I)*b*x)*Sqrt[c + d*x] + (1 + I)*(2*b* 
c + (3*I)*d)*Sqrt[Pi/2]*Erf[((1 + I)*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[2]*Sqrt[ 
d])]*(Cos[b*(c/d + x)] + I*Sin[b*(c/d + x)]))*(Cos[a + b*x] - I*Sin[a + b* 
x])))/(16*b^(5/2)) - ((I/64)*Sqrt[d]*((Cos[a - (b*c)/d] + I*Sin[a - (b*c)/ 
d])*((1 + I)*(4*b^2*c^2 - (12*I)*b*c*d - 15*d^2)*Sqrt[Pi/2]*Erfi[((1 + I)* 
Sqrt[b]*Sqrt[c + d*x])/(Sqrt[2]*Sqrt[d])] + 2*Sqrt[b]*Sqrt[d]*Sqrt[c + ...
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 406, 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.

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

\(\Big \downarrow \) 4906

\(\displaystyle \int \left (\frac {1}{4} (c+d x)^{5/2} \sin (a+b x)+\frac {1}{4} (c+d x)^{5/2} \sin (3 a+3 b x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \cos \left (a-\frac {b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{16 b^{7/2}}-\frac {5 \sqrt {\frac {\pi }{6}} d^{5/2} \cos \left (3 a-\frac {3 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {6}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{144 b^{7/2}}+\frac {5 \sqrt {\frac {\pi }{6}} d^{5/2} \sin \left (3 a-\frac {3 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {6}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{144 b^{7/2}}+\frac {15 \sqrt {\frac {\pi }{2}} d^{5/2} \sin \left (a-\frac {b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{16 b^{7/2}}+\frac {15 d^2 \sqrt {c+d x} \cos (a+b x)}{16 b^3}+\frac {5 d^2 \sqrt {c+d x} \cos (3 a+3 b x)}{144 b^3}+\frac {5 d (c+d x)^{3/2} \sin (a+b x)}{8 b^2}+\frac {5 d (c+d x)^{3/2} \sin (3 a+3 b x)}{72 b^2}-\frac {(c+d x)^{5/2} \cos (a+b x)}{4 b}-\frac {(c+d x)^{5/2} \cos (3 a+3 b x)}{12 b}\)

Input:

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

Output:

(15*d^2*Sqrt[c + d*x]*Cos[a + b*x])/(16*b^3) - ((c + d*x)^(5/2)*Cos[a + b* 
x])/(4*b) + (5*d^2*Sqrt[c + d*x]*Cos[3*a + 3*b*x])/(144*b^3) - ((c + d*x)^ 
(5/2)*Cos[3*a + 3*b*x])/(12*b) - (15*d^(5/2)*Sqrt[Pi/2]*Cos[a - (b*c)/d]*F 
resnelC[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(16*b^(7/2)) - (5*d^( 
5/2)*Sqrt[Pi/6]*Cos[3*a - (3*b*c)/d]*FresnelC[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + 
 d*x])/Sqrt[d]])/(144*b^(7/2)) + (5*d^(5/2)*Sqrt[Pi/6]*FresnelS[(Sqrt[b]*S 
qrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[3*a - (3*b*c)/d])/(144*b^(7/2)) + (1 
5*d^(5/2)*Sqrt[Pi/2]*FresnelS[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]* 
Sin[a - (b*c)/d])/(16*b^(7/2)) + (5*d*(c + d*x)^(3/2)*Sin[a + b*x])/(8*b^2 
) + (5*d*(c + d*x)^(3/2)*Sin[3*a + 3*b*x])/(72*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 = 1.36 (sec) , antiderivative size = 476, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {-\frac {d \left (d x +c \right )^{\frac {5}{2}} \cos \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{4 b}+\frac {5 d \left (\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{2 b}-\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{2 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -c b}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {a d -c b}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{4 b \sqrt {\frac {b}{d}}}\right )}{2 b}\right )}{4 b}-\frac {d \left (d x +c \right )^{\frac {5}{2}} \cos \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 c b}{d}\right )}{12 b}+\frac {5 d \left (\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 c b}{d}\right )}{6 b}-\frac {d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 c b}{d}\right )}{6 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \left (\cos \left (\frac {3 a d -3 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {3 a d -3 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{36 b \sqrt {\frac {b}{d}}}\right )}{2 b}\right )}{12 b}}{d}\)	\(476\)
default	\(\frac {-\frac {d \left (d x +c \right )^{\frac {5}{2}} \cos \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{4 b}+\frac {5 d \left (\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{2 b}-\frac {3 d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{2 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -c b}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {a d -c b}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{4 b \sqrt {\frac {b}{d}}}\right )}{2 b}\right )}{4 b}-\frac {d \left (d x +c \right )^{\frac {5}{2}} \cos \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 c b}{d}\right )}{12 b}+\frac {5 d \left (\frac {d \left (d x +c \right )^{\frac {3}{2}} \sin \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 c b}{d}\right )}{6 b}-\frac {d \left (-\frac {d \sqrt {d x +c}\, \cos \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 c b}{d}\right )}{6 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \left (\cos \left (\frac {3 a d -3 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {3 a d -3 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{36 b \sqrt {\frac {b}{d}}}\right )}{2 b}\right )}{12 b}}{d}\)	\(476\)

Input:

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

Output:

2/d*(-1/8/b*d*(d*x+c)^(5/2)*cos(b/d*(d*x+c)+(a*d-b*c)/d)+5/8/b*d*(1/2/b*d* 
(d*x+c)^(3/2)*sin(b/d*(d*x+c)+(a*d-b*c)/d)-3/2/b*d*(-1/2/b*d*(d*x+c)^(1/2) 
*cos(b/d*(d*x+c)+(a*d-b*c)/d)+1/4/b*d*2^(1/2)*Pi^(1/2)/(b/d)^(1/2)*(cos((a 
*d-b*c)/d)*FresnelC(2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d)-sin((a 
*d-b*c)/d)*FresnelS(2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d))))-1/2 
4/b*d*(d*x+c)^(5/2)*cos(3*b/d*(d*x+c)+3*(a*d-b*c)/d)+5/24/b*d*(1/6/b*d*(d* 
x+c)^(3/2)*sin(3*b/d*(d*x+c)+3*(a*d-b*c)/d)-1/2/b*d*(-1/6/b*d*(d*x+c)^(1/2 
)*cos(3*b/d*(d*x+c)+3*(a*d-b*c)/d)+1/36/b*d*2^(1/2)*Pi^(1/2)*3^(1/2)/(b/d) 
^(1/2)*(cos(3*(a*d-b*c)/d)*FresnelC(2^(1/2)/Pi^(1/2)*3^(1/2)/(b/d)^(1/2)*b 
*(d*x+c)^(1/2)/d)-sin(3*(a*d-b*c)/d)*FresnelS(2^(1/2)/Pi^(1/2)*3^(1/2)/(b/ 
d)^(1/2)*b*(d*x+c)^(1/2)/d)))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 341, normalized size of antiderivative = 0.84 \[ \int (c+d x)^{5/2} \cos ^2(a+b x) \sin (a+b x) \, dx=-\frac {5 \, \sqrt {6} \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (\sqrt {6} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) + 405 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {C}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 405 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {b c - a d}{d}\right ) - 5 \, \sqrt {6} \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (\sqrt {6} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) - 24 \, {\left (30 \, b d^{2} \cos \left (b x + a\right ) - {\left (12 \, b^{3} d^{2} x^{2} + 24 \, b^{3} c d x + 12 \, b^{3} c^{2} - 5 \, b d^{2}\right )} \cos \left (b x + a\right )^{3} + 10 \, {\left (2 \, b^{2} d^{2} x + 2 \, b^{2} c d + {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )\right )} \sqrt {d x + c}}{864 \, b^{4}} \] Input:

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

Output:

-1/864*(5*sqrt(6)*pi*d^3*sqrt(b/(pi*d))*cos(-3*(b*c - a*d)/d)*fresnel_cos( 
sqrt(6)*sqrt(d*x + c)*sqrt(b/(pi*d))) + 405*sqrt(2)*pi*d^3*sqrt(b/(pi*d))* 
cos(-(b*c - a*d)/d)*fresnel_cos(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d))) - 40 
5*sqrt(2)*pi*d^3*sqrt(b/(pi*d))*fresnel_sin(sqrt(2)*sqrt(d*x + c)*sqrt(b/( 
pi*d)))*sin(-(b*c - a*d)/d) - 5*sqrt(6)*pi*d^3*sqrt(b/(pi*d))*fresnel_sin( 
sqrt(6)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-3*(b*c - a*d)/d) - 24*(30*b*d^2 
*cos(b*x + a) - (12*b^3*d^2*x^2 + 24*b^3*c*d*x + 12*b^3*c^2 - 5*b*d^2)*cos 
(b*x + a)^3 + 10*(2*b^2*d^2*x + 2*b^2*c*d + (b^2*d^2*x + b^2*c*d)*cos(b*x 
+ a)^2)*sin(b*x + a))*sqrt(d*x + c))/b^4
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 547, normalized size of antiderivative = 1.35 \[ \int (c+d x)^{5/2} \cos ^2(a+b x) \sin (a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

1/3456*(240*(d*x + c)^(3/2)*b^3*sin(3*((d*x + c)*b - b*c + a*d)/d) + 2160* 
(d*x + c)^(3/2)*b^3*sin(((d*x + c)*b - b*c + a*d)/d) - 24*(12*(d*x + c)^(5 
/2)*b^4/d - 5*sqrt(d*x + c)*b^2*d)*cos(3*((d*x + c)*b - b*c + a*d)/d) - 21 
6*(4*(d*x + c)^(5/2)*b^4/d - 15*sqrt(d*x + c)*b^2*d)*cos(((d*x + c)*b - b* 
c + a*d)/d) - 5*(-(I - 1)*9^(1/4)*sqrt(2)*sqrt(pi)*b*d^2*(b^2/d^2)^(1/4)*c 
os(-3*(b*c - a*d)/d) - (I + 1)*9^(1/4)*sqrt(2)*sqrt(pi)*b*d^2*(b^2/d^2)^(1 
/4)*sin(-3*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(3*I*b/d)) - 405*(-(I - 1 
)*sqrt(2)*sqrt(pi)*b*d^2*(b^2/d^2)^(1/4)*cos(-(b*c - a*d)/d) - (I + 1)*sqr 
t(2)*sqrt(pi)*b*d^2*(b^2/d^2)^(1/4)*sin(-(b*c - a*d)/d))*erf(sqrt(d*x + c) 
*sqrt(I*b/d)) - 405*((I + 1)*sqrt(2)*sqrt(pi)*b*d^2*(b^2/d^2)^(1/4)*cos(-( 
b*c - a*d)/d) + (I - 1)*sqrt(2)*sqrt(pi)*b*d^2*(b^2/d^2)^(1/4)*sin(-(b*c - 
 a*d)/d))*erf(sqrt(d*x + c)*sqrt(-I*b/d)) - 5*((I + 1)*9^(1/4)*sqrt(2)*sqr 
t(pi)*b*d^2*(b^2/d^2)^(1/4)*cos(-3*(b*c - a*d)/d) + (I - 1)*9^(1/4)*sqrt(2 
)*sqrt(pi)*b*d^2*(b^2/d^2)^(1/4)*sin(-3*(b*c - a*d)/d))*erf(sqrt(d*x + c)* 
sqrt(-3*I*b/d)))*d/b^5
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.83 (sec) , antiderivative size = 2445, normalized size of antiderivative = 6.02 \[ \int (c+d x)^{5/2} \cos ^2(a+b x) \sin (a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/1728*(72*(3*sqrt(2)*sqrt(pi)*d*erf(1/2*I*sqrt(2)*sqrt(b*d)*sqrt(d*x + c) 
*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((I*b*c - I*a*d)/d)/(sqrt(b*d)*(-I*b*d/sq 
rt(b^2*d^2) + 1)) + sqrt(6)*sqrt(pi)*d*erf(-1/2*I*sqrt(6)*sqrt(b*d)*sqrt(d 
*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-3*(I*b*c - I*a*d)/d)/(sqrt(b*d)*( 
I*b*d/sqrt(b^2*d^2) + 1)) + 3*sqrt(2)*sqrt(pi)*d*erf(-1/2*I*sqrt(2)*sqrt(b 
*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((-I*b*c + I*a*d)/d)/(sqr 
t(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)) + sqrt(6)*sqrt(pi)*d*erf(1/2*I*sqrt(6)*s 
qrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-3*(-I*b*c + I*a*d 
)/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)))*c^3 - d^3*(27*(sqrt(2)*sqrt(p 
i)*(8*b^3*c^3 + 12*I*b^2*c^2*d - 18*b*c*d^2 - 15*I*d^3)*d*erf(1/2*I*sqrt(2 
)*sqrt(b*d)*sqrt(d*x + c)*(-I*b*d/sqrt(b^2*d^2) + 1)/d)*e^((I*b*c - I*a*d) 
/d)/(sqrt(b*d)*(-I*b*d/sqrt(b^2*d^2) + 1)*b^3) - 2*I*(4*I*(d*x + c)^(5/2)* 
b^2*d - 12*I*(d*x + c)^(3/2)*b^2*c*d + 12*I*sqrt(d*x + c)*b^2*c^2*d + 10*( 
d*x + c)^(3/2)*b*d^2 - 18*sqrt(d*x + c)*b*c*d^2 - 15*I*sqrt(d*x + c)*d^3)* 
e^((-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b^3)/d^3 + (sqrt(6)*sqrt(pi)*(72*b^ 
3*c^3 - 36*I*b^2*c^2*d - 18*b*c*d^2 + 5*I*d^3)*d*erf(-1/2*I*sqrt(6)*sqrt(b 
*d)*sqrt(d*x + c)*(I*b*d/sqrt(b^2*d^2) + 1)/d)*e^(-3*(I*b*c - I*a*d)/d)/(s 
qrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b^3) - 6*I*(12*I*(d*x + c)^(5/2)*b^2*d 
- 36*I*(d*x + c)^(3/2)*b^2*c*d + 36*I*sqrt(d*x + c)*b^2*c^2*d - 10*(d*x + 
c)^(3/2)*b*d^2 + 18*sqrt(d*x + c)*b*c*d^2 - 5*I*sqrt(d*x + c)*d^3)*e^(-...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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