\(\int \sqrt {c+d x} \cos ^3(a+b x) \sin (a+b x) \, dx\) [187]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [C] (verification not implemented)
Giac [C] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 24, antiderivative size = 299 \[ \int \sqrt {c+d x} \cos ^3(a+b x) \sin (a+b x) \, dx=-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{8 b}-\frac {\sqrt {c+d x} \cos (4 a+4 b x)}{32 b}+\frac {\sqrt {d} \sqrt {\frac {\pi }{2}} \cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{64 b^{3/2}}+\frac {\sqrt {d} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{16 b^{3/2}}-\frac {\sqrt {d} \sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right ) \sin \left (4 a-\frac {4 b c}{d}\right )}{64 b^{3/2}}-\frac {\sqrt {d} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{16 b^{3/2}} \] Output:

-1/8*(d*x+c)^(1/2)*cos(2*b*x+2*a)/b-1/32*(d*x+c)^(1/2)*cos(4*b*x+4*a)/b+1/ 
128*d^(1/2)*2^(1/2)*Pi^(1/2)*cos(4*a-4*b*c/d)*FresnelC(2*b^(1/2)*2^(1/2)/P 
i^(1/2)*(d*x+c)^(1/2)/d^(1/2))/b^(3/2)+1/16*d^(1/2)*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^(3/2)-1/128*d^(1/ 
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^(3/2)-1/16*d^(1/2)*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^(3/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.84 \[ \int \sqrt {c+d x} \cos ^3(a+b x) \sin (a+b x) \, dx=-\frac {e^{-\frac {4 i (b c+a d)}{d}} \sqrt {c+d x} \left (4 \sqrt {2} e^{2 i \left (3 a+\frac {b c}{d}\right )} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {3}{2},-\frac {2 i b (c+d x)}{d}\right )+4 \sqrt {2} e^{2 i \left (a+\frac {3 b c}{d}\right )} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {3}{2},\frac {2 i b (c+d x)}{d}\right )+e^{8 i a} \sqrt {\frac {i b (c+d x)}{d}} \Gamma \left (\frac {3}{2},-\frac {4 i b (c+d x)}{d}\right )+e^{\frac {8 i b c}{d}} \sqrt {-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {3}{2},\frac {4 i b (c+d x)}{d}\right )\right )}{128 b \sqrt {\frac {b^2 (c+d x)^2}{d^2}}} \] Input:

Integrate[Sqrt[c + d*x]*Cos[a + b*x]^3*Sin[a + b*x],x]
 

Output:

-1/128*(Sqrt[c + d*x]*(4*Sqrt[2]*E^((2*I)*(3*a + (b*c)/d))*Sqrt[(I*b*(c + 
d*x))/d]*Gamma[3/2, ((-2*I)*b*(c + d*x))/d] + 4*Sqrt[2]*E^((2*I)*(a + (3*b 
*c)/d))*Sqrt[((-I)*b*(c + d*x))/d]*Gamma[3/2, ((2*I)*b*(c + d*x))/d] + E^( 
(8*I)*a)*Sqrt[(I*b*(c + d*x))/d]*Gamma[3/2, ((-4*I)*b*(c + d*x))/d] + E^(( 
(8*I)*b*c)/d)*Sqrt[((-I)*b*(c + d*x))/d]*Gamma[3/2, ((4*I)*b*(c + d*x))/d] 
))/(b*E^(((4*I)*(b*c + a*d))/d)*Sqrt[(b^2*(c + d*x)^2)/d^2])
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 299, 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.

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

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {\frac {\pi }{2}} \sqrt {d} \cos \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{64 b^{3/2}}+\frac {\sqrt {\pi } \sqrt {d} \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{16 b^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} \sqrt {d} \sin \left (4 a-\frac {4 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {\frac {2}{\pi }} \sqrt {c+d x}}{\sqrt {d}}\right )}{64 b^{3/2}}-\frac {\sqrt {\pi } \sqrt {d} \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{16 b^{3/2}}-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{8 b}-\frac {\sqrt {c+d x} \cos (4 a+4 b x)}{32 b}\)

Input:

Int[Sqrt[c + d*x]*Cos[a + b*x]^3*Sin[a + b*x],x]
 

Output:

-1/8*(Sqrt[c + d*x]*Cos[2*a + 2*b*x])/b - (Sqrt[c + d*x]*Cos[4*a + 4*b*x]) 
/(32*b) + (Sqrt[d]*Sqrt[Pi/2]*Cos[4*a - (4*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqr 
t[2/Pi]*Sqrt[c + d*x])/Sqrt[d]])/(64*b^(3/2)) + (Sqrt[d]*Sqrt[Pi]*Cos[2*a 
- (2*b*c)/d]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(16*b 
^(3/2)) - (Sqrt[d]*Sqrt[Pi/2]*FresnelS[(2*Sqrt[b]*Sqrt[2/Pi]*Sqrt[c + d*x] 
)/Sqrt[d]]*Sin[4*a - (4*b*c)/d])/(64*b^(3/2)) - (Sqrt[d]*Sqrt[Pi]*FresnelS 
[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin[2*a - (2*b*c)/d])/(16*b 
^(3/2))
 

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 = 0.57 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.96

method result size
derivativedivides \(\frac {-\frac {d \sqrt {d x +c}\, \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{8 b}+\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{16 b \sqrt {\frac {b}{d}}}-\frac {d \sqrt {d x +c}\, \cos \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{32 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{128 b \sqrt {\frac {b}{d}}}}{d}\) \(286\)
default \(\frac {-\frac {d \sqrt {d x +c}\, \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{8 b}+\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{16 b \sqrt {\frac {b}{d}}}-\frac {d \sqrt {d x +c}\, \cos \left (\frac {4 b \left (d x +c \right )}{d}+\frac {4 a d -4 c b}{d}\right )}{32 b}+\frac {d \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {4 a d -4 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {2}\, b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{128 b \sqrt {\frac {b}{d}}}}{d}\) \(286\)

Input:

int((d*x+c)^(1/2)*cos(b*x+a)^3*sin(b*x+a),x,method=_RETURNVERBOSE)
 

Output:

2/d*(-1/16/b*d*(d*x+c)^(1/2)*cos(2*b/d*(d*x+c)+2*(a*d-b*c)/d)+1/32/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)^(1/2)*cos(4*b/d*(d*x+c)+4*(a*d-b*c)/d)+1/256/ 
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)

Time = 0.09 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.78 \[ \int \sqrt {c+d x} \cos ^3(a+b x) \sin (a+b x) \, dx=\frac {\sqrt {2} \pi d \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - \sqrt {2} \pi d \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {2} \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {4 \, {\left (b c - a d\right )}}{d}\right ) + 8 \, \pi d \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - 8 \, \pi d \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 4 \, {\left (8 \, b \cos \left (b x + a\right )^{4} - 3 \, b\right )} \sqrt {d x + c}}{128 \, b^{2}} \] Input:

integrate((d*x+c)^(1/2)*cos(b*x+a)^3*sin(b*x+a),x, algorithm="fricas")
 

Output:

1/128*(sqrt(2)*pi*d*sqrt(b/(pi*d))*cos(-4*(b*c - a*d)/d)*fresnel_cos(2*sqr 
t(2)*sqrt(d*x + c)*sqrt(b/(pi*d))) - sqrt(2)*pi*d*sqrt(b/(pi*d))*fresnel_s 
in(2*sqrt(2)*sqrt(d*x + c)*sqrt(b/(pi*d)))*sin(-4*(b*c - a*d)/d) + 8*pi*d* 
sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_cos(2*sqrt(d*x + c)*sqrt(b/(p 
i*d))) - 8*pi*d*sqrt(b/(pi*d))*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/(pi*d))) 
*sin(-2*(b*c - a*d)/d) - 4*(8*b*cos(b*x + a)^4 - 3*b)*sqrt(d*x + c))/b^2
 

Sympy [F]

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

integrate((d*x+c)**(1/2)*cos(b*x+a)**3*sin(b*x+a),x)
 

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 429, normalized size of antiderivative = 1.43 \[ \int \sqrt {c+d x} \cos ^3(a+b x) \sin (a+b x) \, dx=-\frac {{\left (\frac {16 \, \sqrt {d x + c} b^{2} \cos \left (\frac {4 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right )}{d} + \frac {64 \, \sqrt {d x + c} b^{2} \cos \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right )}{d} - 4 \, {\left (-\left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - \left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {2 i \, b}{d}}\right ) - {\left (-\left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } b \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 {2} \sqrt {\pi } b \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 ) - {\left (\left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } b \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 {2} \sqrt {\pi } b \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 ) - 4 \, {\left (\left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } b \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {2 i \, b}{d}}\right )\right )} d}{512 \, b^{3}} \] Input:

integrate((d*x+c)^(1/2)*cos(b*x+a)^3*sin(b*x+a),x, algorithm="maxima")
 

Output:

-1/512*(16*sqrt(d*x + c)*b^2*cos(4*((d*x + c)*b - b*c + a*d)/d)/d + 64*sqr 
t(d*x + c)*b^2*cos(2*((d*x + c)*b - b*c + a*d)/d)/d - 4*(-(I - 1)*4^(1/4)* 
sqrt(2)*sqrt(pi)*b*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) - (I + 1)*4^(1/4) 
*sqrt(2)*sqrt(pi)*b*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sqrt(d*x + 
c)*sqrt(2*I*b/d)) - (-(I - 1)*sqrt(2)*sqrt(pi)*b*(b^2/d^2)^(1/4)*cos(-4*(b 
*c - a*d)/d) - (I + 1)*sqrt(2)*sqrt(pi)*b*(b^2/d^2)^(1/4)*sin(-4*(b*c - a* 
d)/d))*erf(2*sqrt(d*x + c)*sqrt(I*b/d)) - ((I + 1)*sqrt(2)*sqrt(pi)*b*(b^2 
/d^2)^(1/4)*cos(-4*(b*c - a*d)/d) + (I - 1)*sqrt(2)*sqrt(pi)*b*(b^2/d^2)^( 
1/4)*sin(-4*(b*c - a*d)/d))*erf(2*sqrt(d*x + c)*sqrt(-I*b/d)) - 4*((I + 1) 
*4^(1/4)*sqrt(2)*sqrt(pi)*b*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) + (I - 1 
)*4^(1/4)*sqrt(2)*sqrt(pi)*b*(b^2/d^2)^(1/4)*sin(-2*(b*c - a*d)/d))*erf(sq 
rt(d*x + c)*sqrt(-2*I*b/d)))*d/b^3
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.42 (sec) , antiderivative size = 826, normalized size of antiderivative = 2.76 \[ \int \sqrt {c+d x} \cos ^3(a+b x) \sin (a+b x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^(1/2)*cos(b*x+a)^3*sin(b*x+a),x, algorithm="giac")
 

Output:

-1/256*(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) + 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) - 8*(sqrt(2)*sqrt(pi)*d*er 
f(-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)) + 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*sq 
rt(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 + 8*sqrt(pi)*(4*b*c - I*d)*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)*b) + 8*sqrt(pi)*(4*b*c + 
I*d)*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)*b) + 16*sqrt(d*x 
+ c)*d*e^(-2*(I*(d*x + c)*b - I*b*c + I*a*d)/d)/b + 4*sqrt(d*x + c)*d*e^(- 
4*(I*(d*x + c)*b - I*b*c + I*a*d)/d)/b + 16*sqrt(d*x + c)*d*e^(-2*(-I*(d*x 
 + c)*b + I*b*c - I*a*d)/d)/b + 4*sqrt(d*x + c)*d*e^(-4*(-I*(d*x + c)*b...
 

Mupad [F(-1)]

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

int(cos(a + b*x)^3*sin(a + b*x)*(c + d*x)^(1/2),x)
 

Output:

int(cos(a + b*x)^3*sin(a + b*x)*(c + d*x)^(1/2), x)
 

Reduce [F]

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

int((d*x+c)^(1/2)*cos(b*x+a)^3*sin(b*x+a),x)
 

Output:

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