3.1.61 \(\int \sqrt {c+d x} \cos (a+b x) \sin ^2(a+b x) \, dx\) [61]

3.1.61.1 Optimal result
3.1.61.2 Mathematica [C] (verified)
3.1.61.3 Rubi [A] (verified)
3.1.61.4 Maple [A] (verified)
3.1.61.5 Fricas [A] (verification not implemented)
3.1.61.6 Sympy [F]
3.1.61.7 Maxima [C] (verification not implemented)
3.1.61.8 Giac [C] (verification not implemented)
3.1.61.9 Mupad [F(-1)]

3.1.61.1 Optimal result

Integrand size = 24, antiderivative size = 304 \[ \int \sqrt {c+d x} \cos (a+b x) \sin ^2(a+b x) \, dx=-\frac {\sqrt {d} \sqrt {\frac {\pi }{2}} \cos \left (a-\frac {b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{4 b^{3/2}}+\frac {\sqrt {d} \sqrt {\frac {\pi }{6}} \cos \left (3 a-\frac {3 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {b} \sqrt {\frac {6}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{12 b^{3/2}}+\frac {\sqrt {d} \sqrt {\frac {\pi }{6}} \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {6}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (3 a-\frac {3 b c}{d}\right )}{12 b^{3/2}}-\frac {\sqrt {d} \sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\frac {\sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (a-\frac {b c}{d}\right )}{4 b^{3/2}}+\frac {\sqrt {c+d x} \sin (a+b x)}{4 b}-\frac {\sqrt {c+d x} \sin (3 a+3 b x)}{12 b} \]

output
1/72*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))*d^(1/2)*6^(1/2)*Pi^(1/2)/b^(3/2)+1/72*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)*d^(1/2)*6^(1/2)*Pi^(1/2)/b^(3/2 
)-1/8*cos(a-b*c/d)*FresnelS(b^(1/2)*2^(1/2)/Pi^(1/2)*(d*x+c)^(1/2)/d^(1/2) 
)*d^(1/2)*2^(1/2)*Pi^(1/2)/b^(3/2)-1/8*FresnelC(b^(1/2)*2^(1/2)/Pi^(1/2)*( 
d*x+c)^(1/2)/d^(1/2))*sin(a-b*c/d)*d^(1/2)*2^(1/2)*Pi^(1/2)/b^(3/2)+1/4*si 
n(b*x+a)*(d*x+c)^(1/2)/b-1/12*sin(3*b*x+3*a)*(d*x+c)^(1/2)/b
 
3.1.61.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.01 (sec) , antiderivative size = 251, normalized size of antiderivative = 0.83 \[ \int \sqrt {c+d x} \cos (a+b x) \sin ^2(a+b x) \, dx=\frac {d e^{-\frac {i (b c+a d)}{d}} \left (e^{2 i a} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {3}{2},-\frac {i b (c+d x)}{d}\right )+e^{\frac {2 i b c}{d}} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {3}{2},\frac {i b (c+d x)}{d}\right )\right )}{8 b^2 \sqrt {c+d x}}-\frac {e^{-\frac {3 i (b c+a d)}{d}} (c+d x)^{3/2} \left (-\frac {e^{6 i a} \Gamma \left (\frac {3}{2},-\frac {3 i b (c+d x)}{d}\right )}{\left (-\frac {i b (c+d x)}{d}\right )^{3/2}}-\frac {e^{\frac {6 i b c}{d}} \Gamma \left (\frac {3}{2},\frac {3 i b (c+d x)}{d}\right )}{\left (\frac {i b (c+d x)}{d}\right )^{3/2}}\right )}{24 \sqrt {3} d} \]

input
Integrate[Sqrt[c + d*x]*Cos[a + b*x]*Sin[a + b*x]^2,x]
 
output
(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 + 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))
 
3.1.61.3 Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 304, 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.

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

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 2009

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

input
Int[Sqrt[c + d*x]*Cos[a + b*x]*Sin[a + b*x]^2,x]
 
output
-1/4*(Sqrt[d]*Sqrt[Pi/2]*Cos[a - (b*c)/d]*FresnelS[(Sqrt[b]*Sqrt[2/Pi]*Sqr 
t[c + d*x])/Sqrt[d]])/b^(3/2) + (Sqrt[d]*Sqrt[Pi/6]*Cos[3*a - (3*b*c)/d]*F 
resnelS[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(12*b^(3/2)) + (Sqrt[ 
d]*Sqrt[Pi/6]*FresnelC[(Sqrt[b]*Sqrt[6/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[3*a 
 - (3*b*c)/d])/(12*b^(3/2)) - (Sqrt[d]*Sqrt[Pi/2]*FresnelC[(Sqrt[b]*Sqrt[2 
/Pi]*Sqrt[c + d*x])/Sqrt[d]]*Sin[a - (b*c)/d])/(4*b^(3/2)) + (Sqrt[c + d*x 
]*Sin[a + b*x])/(4*b) - (Sqrt[c + d*x]*Sin[3*a + 3*b*x])/(12*b)
 

3.1.61.3.1 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]
 
3.1.61.4 Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.97

method result size
derivativedivides \(\frac {\frac {d \sqrt {d x +c}\, \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{4 b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -c b}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {a d -c b}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}-\frac {d \sqrt {d x +c}\, \sin \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 c b}{d}\right )}{12 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \left (\cos \left (\frac {3 a d -3 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {3 a d -3 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{72 b \sqrt {\frac {b}{d}}}}{d}\) \(294\)
default \(\frac {\frac {d \sqrt {d x +c}\, \sin \left (\frac {b \left (d x +c \right )}{d}+\frac {a d -c b}{d}\right )}{4 b}-\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {a d -c b}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {a d -c b}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}-\frac {d \sqrt {d x +c}\, \sin \left (\frac {3 b \left (d x +c \right )}{d}+\frac {3 a d -3 c b}{d}\right )}{12 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \sqrt {3}\, \left (\cos \left (\frac {3 a d -3 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )+\sin \left (\frac {3 a d -3 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {\sqrt {2}\, \sqrt {3}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{72 b \sqrt {\frac {b}{d}}}}{d}\) \(294\)

input
int((d*x+c)^(1/2)*cos(b*x+a)*sin(b*x+a)^2,x,method=_RETURNVERBOSE)
 
output
2/d*(1/8/b*d*(d*x+c)^(1/2)*sin(b/d*(d*x+c)+(a*d-b*c)/d)-1/16/b*d*2^(1/2)*P 
i^(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)^(1/2)*sin(3*b/d*(d*x+c)+3*(a*d-b*c) 
/d)+1/144/b*d*2^(1/2)*Pi^(1/2)*3^(1/2)/(b/d)^(1/2)*(cos(3*(a*d-b*c)/d)*Fre 
snelS(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)))
 
3.1.61.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.81 \[ \int \sqrt {c+d x} \cos (a+b x) \sin ^2(a+b x) \, dx=\frac {\sqrt {6} \pi d \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {S}\left (\sqrt {6} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 9 \, \sqrt {2} \pi d \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {b c - a d}{d}\right ) \operatorname {S}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 9 \, \sqrt {2} \pi d \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (\sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {b c - a d}{d}\right ) + \sqrt {6} \pi d \sqrt {\frac {b}{\pi d}} \operatorname {C}\left (\sqrt {6} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) - 24 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sqrt {d x + c} \sin \left (b x + a\right )}{72 \, b^{2}} \]

input
integrate((d*x+c)^(1/2)*cos(b*x+a)*sin(b*x+a)^2,x, algorithm="fricas")
 
output
1/72*(sqrt(6)*pi*d*sqrt(b/(pi*d))*cos(-3*(b*c - a*d)/d)*fresnel_sin(sqrt(6 
)*sqrt(d*x + c)*sqrt(b/(pi*d))) - 9*sqrt(2)*pi*d*sqrt(b/(pi*d))*cos(-(b*c 
- a*d)/d)*fresnel_sin(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d))) - 9*sqrt(2)*pi 
*d*sqrt(b/(pi*d))*fresnel_cos(sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-( 
b*c - a*d)/d) + sqrt(6)*pi*d*sqrt(b/(pi*d))*fresnel_cos(sqrt(6)*sqrt(d*x + 
 c)*sqrt(b/(pi*d)))*sin(-3*(b*c - a*d)/d) - 24*(b*cos(b*x + a)^2 - b)*sqrt 
(d*x + c)*sin(b*x + a))/b^2
 
3.1.61.6 Sympy [F]

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

input
integrate((d*x+c)**(1/2)*cos(b*x+a)*sin(b*x+a)**2,x)
 
output
Integral(sqrt(c + d*x)*sin(a + b*x)**2*cos(a + b*x), x)
 
3.1.61.7 Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.38 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.39 \[ \int \sqrt {c+d x} \cos (a+b x) \sin ^2(a+b x) \, dx=-\frac {{\left (\frac {24 \, \sqrt {d x + c} b^{2} \sin \left (\frac {3 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right )}{d} - \frac {72 \, \sqrt {d x + c} b^{2} \sin \left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right )}{d} + {\left (-\left (i + 1\right ) \cdot 9^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) + \left (i - 1\right ) \cdot 9^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {3 i \, b}{d}}\right ) - 9 \, {\left (-\left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } b \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {b c - a d}{d}\right ) + \left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } b \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {b c - a d}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {i \, b}{d}}\right ) - 9 \, {\left (\left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } b \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {b c - a d}{d}\right ) - \left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } b \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {b c - a d}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {i \, b}{d}}\right ) + {\left (\left (i - 1\right ) \cdot 9^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right ) - \left (i + 1\right ) \cdot 9^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {3 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {3 i \, b}{d}}\right )\right )} d}{288 \, b^{3}} \]

input
integrate((d*x+c)^(1/2)*cos(b*x+a)*sin(b*x+a)^2,x, algorithm="maxima")
 
output
-1/288*(24*sqrt(d*x + c)*b^2*sin(3*((d*x + c)*b - b*c + a*d)/d)/d - 72*sqr 
t(d*x + c)*b^2*sin(((d*x + c)*b - b*c + a*d)/d)/d + (-(I + 1)*9^(1/4)*sqrt 
(2)*sqrt(pi)*b*(b^2/d^2)^(1/4)*cos(-3*(b*c - a*d)/d) + (I - 1)*9^(1/4)*sqr 
t(2)*sqrt(pi)*b*(b^2/d^2)^(1/4)*sin(-3*(b*c - a*d)/d))*erf(sqrt(d*x + c)*s 
qrt(3*I*b/d)) - 9*(-(I + 1)*sqrt(2)*sqrt(pi)*b*(b^2/d^2)^(1/4)*cos(-(b*c - 
 a*d)/d) + (I - 1)*sqrt(2)*sqrt(pi)*b*(b^2/d^2)^(1/4)*sin(-(b*c - a*d)/d)) 
*erf(sqrt(d*x + c)*sqrt(I*b/d)) - 9*((I - 1)*sqrt(2)*sqrt(pi)*b*(b^2/d^2)^ 
(1/4)*cos(-(b*c - a*d)/d) - (I + 1)*sqrt(2)*sqrt(pi)*b*(b^2/d^2)^(1/4)*sin 
(-(b*c - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-I*b/d)) + ((I - 1)*9^(1/4)*sqrt( 
2)*sqrt(pi)*b*(b^2/d^2)^(1/4)*cos(-3*(b*c - a*d)/d) - (I + 1)*9^(1/4)*sqrt 
(2)*sqrt(pi)*b*(b^2/d^2)^(1/4)*sin(-3*(b*c - a*d)/d))*erf(sqrt(d*x + c)*sq 
rt(-3*I*b/d)))*d/b^3
 
3.1.61.8 Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.55 (sec) , antiderivative size = 848, normalized size of antiderivative = 2.79 \[ \int \sqrt {c+d x} \cos (a+b x) \sin ^2(a+b x) \, dx=\text {Too large to display} \]

input
integrate((d*x+c)^(1/2)*cos(b*x+a)*sin(b*x+a)^2,x, algorithm="giac")
 
output
-1/144*(-9*I*sqrt(2)*sqrt(pi)*(2*b*c + I*d)*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) - I*sqrt(6)*sqrt(pi)*(6*b*c - I*d)*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) + 9*I*sqrt(2)*s 
qrt(pi)*(2*b*c - I*d)*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) + I*sqrt(6)*sqrt(pi)*(6*b*c + I*d)*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) + 6*(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)) - 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/sqr 
t(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 + 18*I*sqrt(d*x + c)*d*e^((I*(d*x + c)*b - I*b*c + I*a*d)/d)/ 
b + 6*I*sqrt(d*x + c)*d*e^(-3*(I*(d*x + c)*b - I*b*c + I*a*d)/d)/b - 18...
 
3.1.61.9 Mupad [F(-1)]

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

input
int(cos(a + b*x)*sin(a + b*x)^2*(c + d*x)^(1/2),x)
 
output
int(cos(a + b*x)*sin(a + b*x)^2*(c + d*x)^(1/2), x)