3.1.55 \(\int \sqrt {c+d x} \cos (a+b x) \sin (a+b x) \, dx\) [55]

3.1.55.1 Optimal result

Integrand size = 22, antiderivative size = 142 \[ \int \sqrt {c+d x} \cos (a+b x) \sin (a+b x) \, dx=-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{4 b}+\frac {\sqrt {d} \sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{8 b^{3/2}}-\frac {\sqrt {d} \sqrt {\pi } \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {2 b c}{d}\right )}{8 b^{3/2}} \]

output

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

3.1.55.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.01 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.02 \[ \int \sqrt {c+d x} \cos (a+b x) \sin (a+b x) \, dx=\frac {1}{2} \left (-\frac {e^{2 i \left (a-\frac {b c}{d}\right )} \sqrt {c+d x} \Gamma \left (\frac {3}{2},-\frac {2 i b (c+d x)}{d}\right )}{4 \sqrt {2} b \sqrt {-\frac {i b (c+d x)}{d}}}-\frac {e^{-2 i \left (a-\frac {b c}{d}\right )} \sqrt {c+d x} \Gamma \left (\frac {3}{2},\frac {2 i b (c+d x)}{d}\right )}{4 \sqrt {2} b \sqrt {\frac {i b (c+d x)}{d}}}\right ) \]

input

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

output

(-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*Sqrt[((-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]))/2
 

3.1.55.3 Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4906, 27, 3042, 3777, 3042, 3787, 3042, 3785, 3786, 3832, 3833}

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.

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

\(\Big \downarrow \) 4906

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3787

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3785

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

\(\Big \downarrow \) 3786

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

\(\Big \downarrow \) 3832

\(\displaystyle \frac {1}{2} \left (\frac {d \left (\frac {2 \cos \left (2 a-\frac {2 b c}{d}\right ) \int \cos \left (\frac {2 b (c+d x)}{d}\right )d\sqrt {c+d x}}{d}-\frac {\sqrt {\pi } \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{\sqrt {b} \sqrt {d}}\right )}{4 b}-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{2 b}\right )\)

\(\Big \downarrow \) 3833

\(\displaystyle \frac {1}{2} \left (\frac {d \left (\frac {\sqrt {\pi } \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelC}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{\sqrt {b} \sqrt {d}}-\frac {\sqrt {\pi } \sin \left (2 a-\frac {2 b c}{d}\right ) \operatorname {FresnelS}\left (\frac {2 \sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {\pi }}\right )}{\sqrt {b} \sqrt {d}}\right )}{4 b}-\frac {\sqrt {c+d x} \cos (2 a+2 b x)}{2 b}\right )\)

input

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

output

(-1/2*(Sqrt[c + d*x]*Cos[2*a + 2*b*x])/b + (d*((Sqrt[Pi]*Cos[2*a - (2*b*c) 
/d]*FresnelC[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])])/(Sqrt[b]*Sqrt[ 
d]) - (Sqrt[Pi]*FresnelS[(2*Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[Pi])]*Sin 
[2*a - (2*b*c)/d])/(Sqrt[b]*Sqrt[d])))/(4*b))/2
 

3.1.55.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3777

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( 
-(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1)*C 
os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 3785

Int[sin[Pi/2 + (e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> S 
imp[2/d   Subst[Int[Cos[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, 
d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
 

rule 3786

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[2/d 
   Subst[Int[Sin[f*(x^2/d)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f 
}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]
 

rule 3787

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Cos 
[(d*e - c*f)/d]   Int[Sin[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] + Simp[Sin[( 
d*e - c*f)/d]   Int[Cos[c*(f/d) + f*x]/Sqrt[c + d*x], x], x] /; FreeQ[{c, d 
, e, f}, x] && ComplexFreeQ[f] && NeQ[d*e - c*f, 0]
 

rule 3832

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

rule 3833

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ 
d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
 

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]
 

3.1.55.4 Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {-\frac {d \sqrt {d x +c}\, \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{4 b}+\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}}{d}\)	\(142\)
default	\(\frac {-\frac {d \sqrt {d x +c}\, \cos \left (\frac {2 b \left (d x +c \right )}{d}+\frac {2 a d -2 c b}{d}\right )}{4 b}+\frac {d \sqrt {\pi }\, \left (\cos \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelC}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )-\sin \left (\frac {2 a d -2 c b}{d}\right ) \operatorname {FresnelS}\left (\frac {2 b \sqrt {d x +c}}{\sqrt {\pi }\, \sqrt {\frac {b}{d}}\, d}\right )\right )}{8 b \sqrt {\frac {b}{d}}}}{d}\)	\(142\)

input

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

output

2/d*(-1/8/b*d*(d*x+c)^(1/2)*cos(2*b/d*(d*x+c)+2*(a*d-b*c)/d)+1/16/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)))
 

3.1.55.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.88 \[ \int \sqrt {c+d x} \cos (a+b x) \sin (a+b x) \, dx=\frac {\pi d \sqrt {\frac {b}{\pi d}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) \operatorname {C}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) - \pi d \sqrt {\frac {b}{\pi d}} \operatorname {S}\left (2 \, \sqrt {d x + c} \sqrt {\frac {b}{\pi d}}\right ) \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - 2 \, {\left (2 \, b \cos \left (b x + a\right )^{2} - b\right )} \sqrt {d x + c}}{8 \, b^{2}} \]

input

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

output

1/8*(pi*d*sqrt(b/(pi*d))*cos(-2*(b*c - a*d)/d)*fresnel_cos(2*sqrt(d*x + c) 
*sqrt(b/(pi*d))) - pi*d*sqrt(b/(pi*d))*fresnel_sin(2*sqrt(d*x + c)*sqrt(b/ 
(pi*d)))*sin(-2*(b*c - a*d)/d) - 2*(2*b*cos(b*x + a)^2 - b)*sqrt(d*x + c)) 
/b^2
 

3.1.55.6 Sympy [F]

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

input

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

output

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

3.1.55.7 Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.47 \[ \int \sqrt {c+d x} \cos (a+b x) \sin (a+b x) \, dx=-\frac {\sqrt {2} {\left (8 \, \sqrt {2} \sqrt {d x + c} b \cos \left (\frac {2 \, {\left ({\left (d x + c\right )} b - b c + a d\right )}}{d}\right ) + {\left (\left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + \left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {\frac {2 i \, b}{d}}\right ) + {\left (-\left (i + 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) - \left (i - 1\right ) \cdot 4^{\frac {1}{4}} \sqrt {\pi } d \left (\frac {b^{2}}{d^{2}}\right )^{\frac {1}{4}} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )\right )} \operatorname {erf}\left (\sqrt {d x + c} \sqrt {-\frac {2 i \, b}{d}}\right )\right )}}{64 \, b^{2}} \]

input

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

output

-1/64*sqrt(2)*(8*sqrt(2)*sqrt(d*x + c)*b*cos(2*((d*x + c)*b - b*c + a*d)/d 
) + ((I - 1)*4^(1/4)*sqrt(pi)*d*(b^2/d^2)^(1/4)*cos(-2*(b*c - a*d)/d) + (I 
 + 1)*4^(1/4)*sqrt(pi)*d*(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)*4^(1/4)*sqrt(pi)*d*(b^2/d^2)^(1/4)*cos( 
-2*(b*c - a*d)/d) - (I - 1)*4^(1/4)*sqrt(pi)*d*(b^2/d^2)^(1/4)*sin(-2*(b*c 
 - a*d)/d))*erf(sqrt(d*x + c)*sqrt(-2*I*b/d)))/b^2
 

3.1.55.8 Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.38 (sec) , antiderivative size = 406, normalized size of antiderivative = 2.86 \[ \int \sqrt {c+d x} \cos (a+b x) \sin (a+b x) \, dx=\frac {4 \, {\left (\frac {\sqrt {\pi } d \operatorname {erf}\left (-\frac {i \, \sqrt {b d} \sqrt {d x + c} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (-\frac {2 \, {\left (i \, b c - i \, a d\right )}}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}} + \frac {\sqrt {\pi } d \operatorname {erf}\left (\frac {i \, \sqrt {b d} \sqrt {d x + c} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (-\frac {2 \, {\left (-i \, b c + i \, a d\right )}}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}\right )} c - \frac {\sqrt {\pi } {\left (4 \, b c - i \, d\right )} d \operatorname {erf}\left (-\frac {i \, \sqrt {b d} \sqrt {d x + c} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (-\frac {2 \, {\left (i \, b c - i \, a d\right )}}{d}\right )}}{\sqrt {b d} {\left (\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b} - \frac {\sqrt {\pi } {\left (4 \, b c + i \, d\right )} d \operatorname {erf}\left (\frac {i \, \sqrt {b d} \sqrt {d x + c} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )}}{d}\right ) e^{\left (-\frac {2 \, {\left (-i \, b c + i \, a d\right )}}{d}\right )}}{\sqrt {b d} {\left (-\frac {i \, b d}{\sqrt {b^{2} d^{2}}} + 1\right )} b} - \frac {2 \, \sqrt {d x + c} d e^{\left (-\frac {2 \, {\left (i \, {\left (d x + c\right )} b - i \, b c + i \, a d\right )}}{d}\right )}}{b} - \frac {2 \, \sqrt {d x + c} d e^{\left (-\frac {2 \, {\left (-i \, {\left (d x + c\right )} b + i \, b c - i \, a d\right )}}{d}\right )}}{b}}{16 \, d} \]

input

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

output

1/16*(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)) + sqr 
t(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 - 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) - sqrt(p 
i)*(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) - 
2*sqrt(d*x + c)*d*e^(-2*(I*(d*x + c)*b - I*b*c + I*a*d)/d)/b - 2*sqrt(d*x 
+ c)*d*e^(-2*(-I*(d*x + c)*b + I*b*c - I*a*d)/d)/b)/d
 

3.1.55.9 Mupad [F(-1)]

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

input

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

output

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