\(\int (c+d x)^{5/2} \cos (a+b x) \sin ^2(a+b x) \, dx\) [63]

Optimal result

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

5/8*d*(d*x+c)^(3/2)*cos(b*x+a)/b^2-5/72*d*(d*x+c)^(3/2)*cos(3*b*x+3*a)/b^2 
+15/32*d^(5/2)*2^(1/2)*Pi^(1/2)*cos(a-b*c/d)*FresnelS(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)*FresnelS(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)*FresnelC(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 
snelC(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) 
-15/16*d^2*(d*x+c)^(1/2)*sin(b*x+a)/b^3+1/4*(d*x+c)^(5/2)*sin(b*x+a)/b+5/1 
44*d^2*(d*x+c)^(1/2)*sin(3*b*x+3*a)/b^3-1/12*(d*x+c)^(5/2)*sin(3*b*x+3*a)/ 
b
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.83 (sec) , antiderivative size = 1418, normalized size of antiderivative = 3.49 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^2(a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/288*(c*Sqrt[d]*(12*Sqrt[b]*Sqrt[d]*E^(((3*I)*b*c)/d)*Sqrt[c + d*x]*(1 + 
 (2*I)*b*x + E^((6*I)*(a + b*x))*(1 - (2*I)*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])/Sqrt[d]] - (1 + I)*(2*b*c - I*d)*E^((3*I)*(2*a + b*x))*Sqrt[6*P 
i]*Erfi[((1 + I)*Sqrt[3/2]*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]]))/(b^(5/2)*E^(( 
(3*I)*(a*d + b*(c + d*x)))/d)) + (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]))/(8*b^2*E^((I*(b*c + a*d))/d)*Sqrt 
[c + d*x]) - (c^2*(c + d*x)^(3/2)*(-((E^((6*I)*a)*Gamma[3/2, ((-3*I)*b*(c 
+ d*x))/d])/(((-I)*b*(c + d*x))/d)^(3/2)) - (E^(((6*I)*b*c)/d)*Gamma[3/2, 
((3*I)*b*(c + d*x))/d])/((I*b*(c + d*x))/d)^(3/2)))/(24*Sqrt[3]*d*E^(((3*I 
)*(b*c + a*d))/d)) + (c*Sqrt[d]*(E^(I*(a - (b*c)/d))*(2*Sqrt[b]*Sqrt[d]*E^ 
((I*b*(c + d*x))/d)*(3 - (2*I)*b*x)*Sqrt[c + d*x] + (-1)^(1/4)*(-2*b*c + ( 
3*I)*d)*Sqrt[Pi]*Erfi[((-1)^(1/4)*Sqrt[b]*Sqrt[c + d*x])/Sqrt[d]]) + (2*Sq 
rt[b]*Sqrt[d]*(3 + (2*I)*b*x)*Sqrt[c + d*x] + (1 + I)*(2*b*c + (3*I)*d)*Sq 
rt[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)) + (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 + d*x]*((15*I)*d - (4*I)...
 

Rubi [A] (verified)

Time = 1.00 (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.

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

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(5*d*(c + d*x)^(3/2)*Cos[a + b*x])/(8*b^2) - (5*d*(c + d*x)^(3/2)*Cos[3*a 
+ 3*b*x])/(72*b^2) + (15*d^(5/2)*Sqrt[Pi/2]*Cos[a - (b*c)/d]*FresnelS[(Sqr 
t[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]*FresnelS[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[ 
d]])/(144*b^(7/2)) - (5*d^(5/2)*Sqrt[Pi/6]*FresnelC[(Sqrt[b]*Sqrt[6/Pi]*Sq 
rt[c + d*x])/Sqrt[d]]*Sin[3*a - (3*b*c)/d])/(144*b^(7/2)) + (15*d^(5/2)*Sq 
rt[Pi/2]*FresnelC[(Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[a - (b*c 
)/d])/(16*b^(7/2)) - (15*d^2*Sqrt[c + d*x]*Sin[a + b*x])/(16*b^3) + ((c + 
d*x)^(5/2)*Sin[a + b*x])/(4*b) + (5*d^2*Sqrt[c + d*x]*Sin[3*a + 3*b*x])/(1 
44*b^3) - ((c + d*x)^(5/2)*Sin[3*a + 3*b*x])/(12*b)
 

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 = 1.11 (sec) , antiderivative size = 474, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {\frac {d \left (d x +c \right )^{\frac {5}{2}} \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{4 b}-\frac {5 d \left (-\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{2 b}+\frac {3 d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{2 b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -c b}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {a d -c b}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{4 b \sqrt {\frac {b}{d}}}\right )}{2 b}\right )}{4 b}-\frac {d \left (d x +c \right )^{\frac {5}{2}} \sin \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 c b}{d}\right )}{12 b}+\frac {5 d \left (-\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 c b}{d}\right )}{6 b}+\frac {d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 c b}{d}\right )}{6 b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \left (\cos \left (\frac {3 a d -3 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {3 a d -3 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{36 b \sqrt {\frac {b}{d}}}\right )}{2 b}\right )}{12 b}}{d}\)	\(474\)
default	\(\frac {\frac {d \left (d x +c \right )^{\frac {5}{2}} \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{4 b}-\frac {5 d \left (-\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{2 b}+\frac {3 d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{2 b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -c b}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {a d -c b}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{4 b \sqrt {\frac {b}{d}}}\right )}{2 b}\right )}{4 b}-\frac {d \left (d x +c \right )^{\frac {5}{2}} \sin \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 c b}{d}\right )}{12 b}+\frac {5 d \left (-\frac {d \left (d x +c \right )^{\frac {3}{2}} \cos \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 c b}{d}\right )}{6 b}+\frac {d \left (\frac {d \sqrt {d x +c}\, \sin \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 c b}{d}\right )}{6 b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \left (\cos \left (\frac {3 a d -3 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {3 a d -3 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{36 b \sqrt {\frac {b}{d}}}\right )}{2 b}\right )}{12 b}}{d}\)	\(474\)

Input:

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

Output:

2/d*(1/8/b*d*(d*x+c)^(5/2)*sin(b/d*(d*x+c)+(a*d-b*c)/d)-5/8/b*d*(-1/2/b*d* 
(d*x+c)^(3/2)*cos(b/d*(d*x+c)+(a*d-b*c)/d)+3/2/b*d*(1/2/b*d*(d*x+c)^(1/2)* 
sin(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)*FresnelS(2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d)+sin((a* 
d-b*c)/d)*FresnelC(2^(1/2)/Pi^(1/2)/(b/d)^(1/2)*b*(d*x+c)^(1/2)/d))))-1/24 
/b*d*(d*x+c)^(5/2)*sin(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)*cos(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) 
*sin(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)*FresnelS(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)*FresnelC(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 = 370, normalized size of antiderivative = 0.91 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^2(a+b x) \, dx=-\frac {5 \, \sqrt {6} \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (\sqrt {6} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 405 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {S}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 405 \, \sqrt {2} \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {b c - a d}{d}\right ) + 5 \, \sqrt {6} \pi d^{3} \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (\sqrt {6} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + 24 \, {\left (10 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right )^{3} - 30 \, {\left (b^{2} d^{2} x + b^{2} c d\right )} \cos \left (b x + a\right ) - {\left (12 \, b^{3} d^{2} x^{2} + 24 \, b^{3} c d x + 12 \, b^{3} c^{2} - 35 \, b d^{2} - {\left (12 \, b^{3} d^{2} x^{2} + 24 \, b^{3} c d x + 12 \, b^{3} c^{2} - 5 \, b d^{2}\right )} \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )\right )} \sqrt {d x + c}}{864 \, b^{4}} \] Input:

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

Output:

-1/864*(5*sqrt(6)*pi*d^3*sqrt(b/(pi*d))*cos(-3*(b*c - a*d)/d)*fresnel_sin( 
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_sin(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d))) - 40 
5*sqrt(2)*pi*d^3*sqrt(b/(pi*d))*fresnel_cos(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_cos( 
sqrt(6)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-3*(b*c - a*d)/d) + 24*(10*(b^2* 
d^2*x + b^2*c*d)*cos(b*x + a)^3 - 30*(b^2*d^2*x + b^2*c*d)*cos(b*x + a) - 
(12*b^3*d^2*x^2 + 24*b^3*c*d*x + 12*b^3*c^2 - 35*b*d^2 - (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)^2)*sin(b*x + a))*sqrt(d 
*x + c))/b^4
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 547, normalized size of antiderivative = 1.35 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^2(a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/3456*(240*(d*x + c)^(3/2)*b^3*cos(3*((d*x + c)*b - b*c + a*d)/d) - 2160 
*(d*x + c)^(3/2)*b^3*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)*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(s 
qrt(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)*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)) - 405*(-(I - 1)*sqr 
t(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)*sqrt(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)) + 24*(12*(d*x + 
c)^(5/2)*b^4/d - 5*sqrt(d*x + c)*b^2*d)*sin(3*((d*x + c)*b - b*c + a*d)/d) 
 - 216*(4*(d*x + c)^(5/2)*b^4/d - 15*sqrt(d*x + c)*b^2*d)*sin(((d*x + c)*b 
 - b*c + a*d)/d))*d/b^5
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.81 (sec) , antiderivative size = 2453, normalized size of antiderivative = 6.04 \[ \int (c+d x)^{5/2} \cos (a+b x) \sin ^2(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/1728*(72*(3*I*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 
/sqrt(b^2*d^2) + 1)) + I*sqrt(6)*sqrt(pi)*d*erf(-1/2*I*sqrt(6)*sqrt(b*d)*s 
qrt(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*I*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/sqrt(b^2*d^2) + 1)) - I*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)))*c^3 + d^3*(27*(-I*sq 
rt(2)*sqrt(pi)*(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*(-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 + (-I*sqrt(6)*s 
qrt(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)/(sqrt(b*d)*(I*b*d/sqrt(b^2*d^2) + 1)*b^3) + 6*(-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 +...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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