\(\int \sin ^3(a+b x) \sqrt {\sin (c+d x)} \, dx\) [136]

Optimal result

Integrand size = 19, antiderivative size = 477 \[ \int \sin ^3(a+b x) \sqrt {\sin (c+d x)} \, dx=-\frac {3 e^{\frac {1}{2} i (2 a-c)+\frac {1}{2} i (2 b-d) x+\frac {1}{2} i (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} \left (-1+\frac {2 b}{d}\right ),\frac {1}{4} \left (3+\frac {2 b}{d}\right ),e^{2 i (c+d x)}\right ) \sqrt {\sin (c+d x)}}{4 (2 b-d) \sqrt {1-e^{2 i (c+d x)}}}+\frac {e^{\frac {1}{2} i (6 a-c)+\frac {1}{2} i (6 b-d) x+\frac {1}{2} i (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} \left (-1+\frac {6 b}{d}\right ),\frac {3 (2 b+d)}{4 d},e^{2 i (c+d x)}\right ) \sqrt {\sin (c+d x)}}{4 (6 b-d) \sqrt {1-e^{2 i (c+d x)}}}-\frac {3 e^{-\frac {1}{2} i (2 a+c)-\frac {1}{2} i (2 b+d) x+\frac {1}{2} i (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {2 b+d}{4 d},\frac {1}{4} \left (3-\frac {2 b}{d}\right ),e^{2 i (c+d x)}\right ) \sqrt {\sin (c+d x)}}{4 (2 b+d) \sqrt {1-e^{2 i (c+d x)}}}+\frac {e^{-\frac {1}{2} i (6 a+c)-\frac {1}{2} i (6 b+d) x+\frac {1}{2} i (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {6 b+d}{4 d},\frac {3}{4} \left (1-\frac {2 b}{d}\right ),e^{2 i (c+d x)}\right ) \sqrt {\sin (c+d x)}}{4 (6 b+d) \sqrt {1-e^{2 i (c+d x)}}} \] Output:

-3/4*exp(1/2*I*(2*a-c)+1/2*I*(2*b-d)*x+1/2*I*(d*x+c))*hypergeom([-1/2, -1/ 
4+1/2*b/d],[3/4+1/2*b/d],exp(2*I*(d*x+c)))*sin(d*x+c)^(1/2)/(2*b-d)/(1-exp 
(2*I*(d*x+c)))^(1/2)+1/4*exp(1/2*I*(6*a-c)+1/2*I*(6*b-d)*x+1/2*I*(d*x+c))* 
hypergeom([-1/2, -1/4+3/2*b/d],[3/4*(2*b+d)/d],exp(2*I*(d*x+c)))*sin(d*x+c 
)^(1/2)/(6*b-d)/(1-exp(2*I*(d*x+c)))^(1/2)-3/4*exp(-1/2*I*(2*a+c)-1/2*I*(2 
*b+d)*x+1/2*I*(d*x+c))*hypergeom([-1/2, -1/4*(2*b+d)/d],[3/4-1/2*b/d],exp( 
2*I*(d*x+c)))*sin(d*x+c)^(1/2)/(2*b+d)/(1-exp(2*I*(d*x+c)))^(1/2)+1/4*exp( 
-1/2*I*(6*a+c)-1/2*I*(6*b+d)*x+1/2*I*(d*x+c))*hypergeom([-1/2, -1/4*(6*b+d 
)/d],[3/4-3/2*b/d],exp(2*I*(d*x+c)))*sin(d*x+c)^(1/2)/(6*b+d)/(1-exp(2*I*( 
d*x+c)))^(1/2)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 1.75 (sec) , antiderivative size = 590, normalized size of antiderivative = 1.24, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {5064, 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.

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

\(\Big \downarrow \) 5064

\(\displaystyle \frac {\int \left (3 i e^{-i a-i b x} \sqrt {i e^{-i (c+d x)}-i e^{i (c+d x)}}-3 i e^{i a+i b x} \sqrt {i e^{-i (c+d x)}-i e^{i (c+d x)}}-i e^{-3 i a-3 i b x} \sqrt {i e^{-i (c+d x)}-i e^{i (c+d x)}}+i e^{3 i a+3 i b x} \sqrt {i e^{-i (c+d x)}-i e^{i (c+d x)}}\right )dx}{8 \sqrt {2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {6 \sqrt {i e^{-i (c+d x)}-i e^{i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} \left (\frac {2 b}{d}-1\right ),\frac {1}{4} \left (\frac {2 b}{d}+3\right ),e^{2 i (c+d x)}\right ) \exp \left (\frac {1}{2} i (2 a-c)+\frac {1}{2} i x (2 b-d)+\frac {1}{2} i (c+d x)\right )}{(2 b-d) \sqrt {1-e^{2 i c+2 i d x}}}+\frac {2 \sqrt {i e^{-i (c+d x)}-i e^{i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4} \left (\frac {6 b}{d}-1\right ),\frac {3 (2 b+d)}{4 d},e^{2 i (c+d x)}\right ) \exp \left (\frac {1}{2} i (6 a-c)+\frac {1}{2} i x (6 b-d)+\frac {1}{2} i (c+d x)\right )}{(6 b-d) \sqrt {1-e^{2 i c+2 i d x}}}-\frac {6 \sqrt {i e^{-i (c+d x)}-i e^{i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {2 b+d}{4 d},\frac {1}{4} \left (3-\frac {2 b}{d}\right ),e^{2 i (c+d x)}\right ) \exp \left (-\frac {1}{2} i (2 a+c)-\frac {1}{2} i x (2 b+d)+\frac {1}{2} i (c+d x)\right )}{(2 b+d) \sqrt {1-e^{2 i c+2 i d x}}}+\frac {2 \sqrt {i e^{-i (c+d x)}-i e^{i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {6 b+d}{4 d},\frac {3}{4} \left (1-\frac {2 b}{d}\right ),e^{2 i (c+d x)}\right ) \exp \left (-\frac {1}{2} i (6 a+c)-\frac {1}{2} i x (6 b+d)+\frac {1}{2} i (c+d x)\right )}{(6 b+d) \sqrt {1-e^{2 i c+2 i d x}}}}{8 \sqrt {2}}\)

Input:

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

Output:

((-6*E^((I/2)*(2*a - c) + (I/2)*(2*b - d)*x + (I/2)*(c + d*x))*Sqrt[I/E^(I 
*(c + d*x)) - I*E^(I*(c + d*x))]*Hypergeometric2F1[-1/2, (-1 + (2*b)/d)/4, 
 (3 + (2*b)/d)/4, E^((2*I)*(c + d*x))])/((2*b - d)*Sqrt[1 - E^((2*I)*c + ( 
2*I)*d*x)]) + (2*E^((I/2)*(6*a - c) + (I/2)*(6*b - d)*x + (I/2)*(c + d*x)) 
*Sqrt[I/E^(I*(c + d*x)) - I*E^(I*(c + d*x))]*Hypergeometric2F1[-1/2, (-1 + 
 (6*b)/d)/4, (3*(2*b + d))/(4*d), E^((2*I)*(c + d*x))])/((6*b - d)*Sqrt[1 
- E^((2*I)*c + (2*I)*d*x)]) - (6*E^((-1/2*I)*(2*a + c) - (I/2)*(2*b + d)*x 
 + (I/2)*(c + d*x))*Sqrt[I/E^(I*(c + d*x)) - I*E^(I*(c + d*x))]*Hypergeome 
tric2F1[-1/2, -1/4*(2*b + d)/d, (3 - (2*b)/d)/4, E^((2*I)*(c + d*x))])/((2 
*b + d)*Sqrt[1 - E^((2*I)*c + (2*I)*d*x)]) + (2*E^((-1/2*I)*(6*a + c) - (I 
/2)*(6*b + d)*x + (I/2)*(c + d*x))*Sqrt[I/E^(I*(c + d*x)) - I*E^(I*(c + d* 
x))]*Hypergeometric2F1[-1/2, -1/4*(6*b + d)/d, (3*(1 - (2*b)/d))/4, E^((2* 
I)*(c + d*x))])/((6*b + d)*Sqrt[1 - E^((2*I)*c + (2*I)*d*x)]))/(8*Sqrt[2])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5064

Int[Sin[(a_.) + (b_.)*(x_)]^(p_.)*Sin[(c_.) + (d_.)*(x_)]^(q_.), x_Symbol] 
:> Simp[1/2^(p + q)   Int[ExpandIntegrand[(I/E^(I*(c + d*x)) - I*E^(I*(c + 
d*x)))^q, (I/E^(I*(a + b*x)) - I*E^(I*(a + b*x)))^p, x], x], x] /; FreeQ[{a 
, b, c, d, q}, x] && IGtQ[p, 0] &&  !IntegerQ[q]
 

Maple [F]

\[\int \sin \left (b x +a \right )^{3} \sqrt {\sin \left (d x +c \right )}d x\]

Input:

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

Output:

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

Fricas [F(-2)]

Exception generated. \[ \int \sin ^3(a+b x) \sqrt {\sin (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (has polynomial part)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate(sin(b*x + a)^3*sqrt(sin(d*x + c)), x)
 

Giac [F]

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

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

Output:

integrate(sin(b*x + a)^3*sqrt(sin(d*x + c)), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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