\(\int \sin (c+d x) (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x))^2 \, dx\) [882]

Optimal result

Integrand size = 31, antiderivative size = 148 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right )^2 \, dx=b^2 x+a c x-\frac {a^2 \cos (c+d x)}{d}-\frac {c^2 \cos (c+d x)}{d}+\frac {c^2 \cos ^3(c+d x)}{3 d}+\frac {4 a b E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d}+\frac {4 b c \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right )}{3 d}-\frac {4 b c \cos (c+d x) \sqrt {\sin (c+d x)}}{3 d}-\frac {a c \cos (c+d x) \sin (c+d x)}{d} \] Output:

b^2*x+a*c*x-a^2*cos(d*x+c)/d-c^2*cos(d*x+c)/d+1/3*c^2*cos(d*x+c)^3/d-4*a*b 
*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))/d+4/3*b*c*InverseJacobiAM(1/ 
2*c-1/4*Pi+1/2*d*x,2^(1/2))/d-4/3*b*c*cos(d*x+c)*sin(d*x+c)^(1/2)/d-a*c*co 
s(d*x+c)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.07 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.93 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right )^2 \, dx=\frac {12 b^2 c+12 a c^2+12 b^2 d x+12 a c d x-12 a^2 \cos (c+d x)-9 c^2 \cos (c+d x)+c^2 \cos (3 (c+d x))-48 a b E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )-16 b c \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right )-16 b c \cos (c+d x) \sqrt {\sin (c+d x)}-6 a c \sin (2 (c+d x))}{12 d} \] Input:

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

Output:

(12*b^2*c + 12*a*c^2 + 12*b^2*d*x + 12*a*c*d*x - 12*a^2*Cos[c + d*x] - 9*c 
^2*Cos[c + d*x] + c^2*Cos[3*(c + d*x)] - 48*a*b*EllipticE[(-2*c + Pi - 2*d 
*x)/4, 2] - 16*b*c*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] - 16*b*c*Cos[c + d* 
x]*Sqrt[Sin[c + d*x]] - 6*a*c*Sin[2*(c + d*x)])/(12*d)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 4895, 3042, 4901, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4895

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a \sqrt {\sin (c+d x)}+b+c \sin (c+d x)^{3/2}\right )^2dx\)

\(\Big \downarrow \) 4901

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a^2 \cos (c+d x)}{d}+\frac {4 a b E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d}-\frac {a c \sin (c+d x) \cos (c+d x)}{d}+a c x+b^2 x+\frac {4 b c \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d}-\frac {4 b c \sqrt {\sin (c+d x)} \cos (c+d x)}{3 d}+\frac {c^2 \cos ^3(c+d x)}{3 d}-\frac {c^2 \cos (c+d x)}{d}\)

Input:

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

Output:

b^2*x + a*c*x - (a^2*Cos[c + d*x])/d - (c^2*Cos[c + d*x])/d + (c^2*Cos[c + 
 d*x]^3)/(3*d) + (4*a*b*EllipticE[(c - Pi/2 + d*x)/2, 2])/d + (4*b*c*Ellip 
ticF[(c - Pi/2 + d*x)/2, 2])/(3*d) - (4*b*c*Cos[c + d*x]*Sqrt[Sin[c + d*x] 
])/(3*d) - (a*c*Cos[c + d*x]*Sin[c + d*x])/d
 

Defintions of rubi rules used

rule 2009

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

rule 3042

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

rule 4895

Int[(u_)*((a_) + (b_.)*(F_)[(d_.) + (e_.)*(x_)]^(p_.) + (c_.)*(F_)[(d_.) + 
(e_.)*(x_)]^(q_.))^(n_.), x_Symbol] :> Int[ActivateTrig[u*F[d + e*x]^(n*p)* 
(b + a/F[d + e*x]^p + c*F[d + e*x]^(q - p))^n], x] /; FreeQ[{a, b, c, d, e, 
 p, q}, x] && InertTrigQ[F] && IntegerQ[n] && NegQ[p]
 

rule 4901

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /; 
 !InertTrigFreeQ[u]
 

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.73

method	result	size
parts	\(b^{2} x -\frac {c^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3 d}-\frac {a^{2} \cos \left (d x +c \right )}{d}+\frac {2 a c \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a b \sqrt {\sin \left (d x +c \right )+1}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \left (2 \operatorname {EllipticE}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right )}{\cos \left (d x +c \right ) \sqrt {\sin \left (d x +c \right )}\, d}+\frac {2 b c \left (\frac {\sqrt {\sin \left (d x +c \right )+1}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{3}-\frac {2 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{2}}{3}\right )}{\cos \left (d x +c \right ) \sqrt {\sin \left (d x +c \right )}\, d}\)	\(256\)
default	\(b^{2} x -\frac {c^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3 d}-\frac {a^{2} \cos \left (d x +c \right )}{d}+\frac {2 a c \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 b \left (6 a \sqrt {\sin \left (d x +c \right )+1}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 a \sqrt {\sin \left (d x +c \right )+1}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {\sin \left (d x +c \right )+1}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) c +2 \cos \left (d x +c \right )^{2} c \sin \left (d x +c \right )\right )}{3 \cos \left (d x +c \right ) \sqrt {\sin \left (d x +c \right )}\, d}\)	\(267\)

Input:

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

Output:

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.34 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right )^2 \, dx=\frac {c^{2} \cos \left (d x + c\right )^{3} - 3 \, a c \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 4 \, b c \cos \left (d x + c\right ) \sqrt {\sin \left (d x + c\right )} + 4 \, \sqrt {-\frac {1}{2} i} b c {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 4 \, \sqrt {\frac {1}{2} i} b c {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 12 i \, \sqrt {-\frac {1}{2} i} a b {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 i \, \sqrt {\frac {1}{2} i} a b {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (b^{2} + a c\right )} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right ) - 3 \, {\left (a^{2} + c^{2}\right )} \cos \left (d x + c\right )}{3 \, d} \] Input:

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

Output:

1/3*(c^2*cos(d*x + c)^3 - 3*a*c*cos(d*x + c)*sin(d*x + c) - 4*b*c*cos(d*x 
+ c)*sqrt(sin(d*x + c)) + 4*sqrt(-1/2*I)*b*c*weierstrassPInverse(4, 0, cos 
(d*x + c) + I*sin(d*x + c)) + 4*sqrt(1/2*I)*b*c*weierstrassPInverse(4, 0, 
cos(d*x + c) - I*sin(d*x + c)) + 12*I*sqrt(-1/2*I)*a*b*weierstrassZeta(4, 
0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) - 12*I*sqrt(1 
/2*I)*a*b*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I 
*sin(d*x + c))) + 3*(b^2 + a*c)*arctan(sin(d*x + c)/cos(d*x + c)) - 3*(a^2 
 + c^2)*cos(d*x + c))/d
 

Sympy [F]

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

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

Output:

Integral((a + b/sqrt(sin(c + d*x)) + c*sin(c + d*x))**2*sin(c + d*x), x)
 

Maxima [F]

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

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

Output:

1/12*(12*b^2*c + 12*(b^2 + a*c)*d*x + c^2*cos(3*d*x + 3*c) - 6*a*c*sin(2*d 
*x + 2*c) - 3*sqrt(2)*d*integrate((((sqrt(2)*b*c*cos(5/2*d*x + 5/2*c) + sq 
rt(2)*b*c*cos(3/2*d*x + 3/2*c) - 2*sqrt(2)*b*c*cos(1/2*d*x + 1/2*c) - 2*sq 
rt(2)*a*b*sin(3/2*d*x + 3/2*c) - 2*sqrt(2)*a*b*sin(1/2*d*x + 1/2*c))*cos(1 
/2*arctan2(sin(d*x + c), -cos(d*x + c) + 1)) - (2*sqrt(2)*a*b*cos(3/2*d*x 
+ 3/2*c) - 2*sqrt(2)*a*b*cos(1/2*d*x + 1/2*c) + sqrt(2)*b*c*sin(5/2*d*x + 
5/2*c) - sqrt(2)*b*c*sin(3/2*d*x + 3/2*c) - 2*sqrt(2)*b*c*sin(1/2*d*x + 1/ 
2*c))*sin(1/2*arctan2(sin(d*x + c), -cos(d*x + c) + 1)))*cos(1/2*arctan2(s 
in(d*x + c), cos(d*x + c) + 1)) + ((2*sqrt(2)*a*b*cos(3/2*d*x + 3/2*c) - 2 
*sqrt(2)*a*b*cos(1/2*d*x + 1/2*c) + sqrt(2)*b*c*sin(5/2*d*x + 5/2*c) - sqr 
t(2)*b*c*sin(3/2*d*x + 3/2*c) - 2*sqrt(2)*b*c*sin(1/2*d*x + 1/2*c))*cos(1/ 
2*arctan2(sin(d*x + c), -cos(d*x + c) + 1)) + (sqrt(2)*b*c*cos(5/2*d*x + 5 
/2*c) + sqrt(2)*b*c*cos(3/2*d*x + 3/2*c) - 2*sqrt(2)*b*c*cos(1/2*d*x + 1/2 
*c) - 2*sqrt(2)*a*b*sin(3/2*d*x + 3/2*c) - 2*sqrt(2)*a*b*sin(1/2*d*x + 1/2 
*c))*sin(1/2*arctan2(sin(d*x + c), -cos(d*x + c) + 1)))*sin(1/2*arctan2(si 
n(d*x + c), cos(d*x + c) + 1)))/((cos(d*x + c)^2 + sin(d*x + c)^2 + 2*cos( 
d*x + c) + 1)^(1/4)*(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*cos(d*x + c) + 1) 
^(1/4)), x) - 3*sqrt(2)*d*integrate((((2*sqrt(2)*a*b*cos(3/2*d*x + 3/2*c) 
- 2*sqrt(2)*a*b*cos(1/2*d*x + 1/2*c) + sqrt(2)*b*c*sin(5/2*d*x + 5/2*c) - 
sqrt(2)*b*c*sin(3/2*d*x + 3/2*c) - 2*sqrt(2)*b*c*sin(1/2*d*x + 1/2*c))*...
 

Giac [F]

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

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

Output:

integrate((c*sin(d*x + c) + a + b/sqrt(sin(d*x + c)))^2*sin(d*x + c), x)
 

Mupad [B] (verification not implemented)

Time = 20.24 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.87 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right )^2 \, dx=b^2\,x-\frac {a^2\,\cos \left (c+d\,x\right )}{d}-\frac {a\,c\,\left (\sin \left (2\,c+2\,d\,x\right )-2\,d\,x\right )}{2\,d}+\frac {4\,a\,b\,\mathrm {E}\left (\frac {c}{2}-\frac {\pi }{4}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {c^2\,\cos \left (c+d\,x\right )\,\left ({\cos \left (c+d\,x\right )}^2-3\right )}{3\,d}-\frac {2\,b\,c\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\left ({\sin \left (c+d\,x\right )}^2\right )}^{5/4}} \] Input:

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

Output:

b^2*x - (a^2*cos(c + d*x))/d - (a*c*(sin(2*c + 2*d*x) - 2*d*x))/(2*d) + (4 
*a*b*ellipticE(c/2 - pi/4 + (d*x)/2, 2))/d + (c^2*cos(c + d*x)*(cos(c + d* 
x)^2 - 3))/(3*d) - (2*b*c*cos(c + d*x)*sin(c + d*x)^(5/2)*hypergeom([-1/4, 
 1/2], 3/2, cos(c + d*x)^2))/(d*(sin(c + d*x)^2)^(5/4))
 

Reduce [F]

\[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right )^2 \, dx=\frac {-\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c^{2}-3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a c -3 \cos \left (d x +c \right ) a^{2}-2 \cos \left (d x +c \right ) c^{2}+6 \left (\int \sqrt {\sin \left (d x +c \right )}d x \right ) a b d +6 \left (\int \sqrt {\sin \left (d x +c \right )}\, \sin \left (d x +c \right )d x \right ) b c d +3 a c d x +3 b^{2} d x +2 c^{2}}{3 d} \] Input:

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

Output:

( - cos(c + d*x)*sin(c + d*x)**2*c**2 - 3*cos(c + d*x)*sin(c + d*x)*a*c - 
3*cos(c + d*x)*a**2 - 2*cos(c + d*x)*c**2 + 6*int(sqrt(sin(c + d*x)),x)*a* 
b*d + 6*int(sqrt(sin(c + d*x))*sin(c + d*x),x)*b*c*d + 3*a*c*d*x + 3*b**2* 
d*x + 2*c**2)/(3*d)