Integrand size = 31, antiderivative size = 148 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right )^2 \, dx=b^2 x+a c x-\frac {a^2 \cos (c+d x)}{d}-\frac {c^2 \cos (c+d x)}{d}+\frac {c^2 \cos ^3(c+d x)}{3 d}+\frac {4 a b E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d}+\frac {4 b c \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right )}{3 d}-\frac {4 b c \cos (c+d x) \sqrt {\sin (c+d x)}}{3 d}-\frac {a c \cos (c+d x) \sin (c+d x)}{d} \]
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 *(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(c os(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))/d-4/3*b*c*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^ (1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2 ))/d-a*c*cos(d*x+c)*sin(d*x+c)/d-4/3*b*c*cos(d*x+c)*sin(d*x+c)^(1/2)/d
Time = 1.13 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.93 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right )^2 \, dx=\frac {12 b^2 c+12 a c^2+12 b^2 d x+12 a c d x-12 a^2 \cos (c+d x)-9 c^2 \cos (c+d x)+c^2 \cos (3 (c+d x))-48 a b E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )-16 b c \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right )-16 b c \cos (c+d x) \sqrt {\sin (c+d x)}-6 a c \sin (2 (c+d x))}{12 d} \]
(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)
Time = 0.49 (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 |
| \(\displaystyle -\frac {a^2 \cos (c+d x)}{d}+\frac {4 a b E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d}-\frac {a c \sin (c+d x) \cos (c+d x)}{d}+a c x+b^2 x+\frac {4 b c \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{3 d}-\frac {4 b c \sqrt {\sin (c+d x)} \cos (c+d x)}{3 d}+\frac {c^2 \cos ^3(c+d x)}{3 d}-\frac {c^2 \cos (c+d x)}{d}\) |
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
3.10.39.3.1 Defintions of rubi rules used
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]
Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /; !InertTrigFreeQ[u]
Time = 2.60 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.73
| method | result | size |
| parts | \(x \,b^{2}-\frac {c^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3 d}-\frac {a^{2} \cos \left (d x +c \right )}{d}+\frac {2 a c \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a b \sqrt {\sin \left (d x +c \right )+1}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \left (2 \operatorname {EllipticE}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right )}{\cos \left (d x +c \right ) \sqrt {\sin \left (d x +c \right )}\, d}+\frac {2 c b \left (\frac {\sqrt {\sin \left (d x +c \right )+1}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{3}-\frac {2 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )}{3}\right )}{\cos \left (d x +c \right ) \sqrt {\sin \left (d x +c \right )}\, d}\) | \(256\) |
| default | \(x \,b^{2}-\frac {c^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3 d}-\frac {a^{2} \cos \left (d x +c \right )}{d}+\frac {2 a c \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 b \left (6 a \sqrt {\sin \left (d x +c \right )+1}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-3 a \sqrt {\sin \left (d x +c \right )+1}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {\sin \left (d x +c \right )+1}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) c +2 \cos \left (d x +c \right )^{2} c \sin \left (d x +c \right )\right )}{3 \cos \left (d x +c \right ) \sqrt {\sin \left (d x +c \right )}\, d}\) | \(267\) |
x*b^2-1/3*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 *sin(d*x+c)*cos(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*c*b*(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*cos(d*x+c)^2*si n(d*x+c))/cos(d*x+c)/sin(d*x+c)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.42 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right )^2 \, dx=\frac {c^{2} \cos \left (d x + c\right )^{3} - 3 \, a c \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {-i} b c {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 2 \, \sqrt {2} \sqrt {i} b c {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 6 i \, \sqrt {2} \sqrt {-i} a b {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 6 i \, \sqrt {2} \sqrt {i} a b {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 4 \, b c \cos \left (d x + c\right ) \sqrt {\sin \left (d x + c\right )} + 3 \, {\left (b^{2} + a c\right )} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right ) - 3 \, {\left (a^{2} + c^{2}\right )} \cos \left (d x + c\right )}{3 \, d} \]
1/3*(c^2*cos(d*x + c)^3 - 3*a*c*cos(d*x + c)*sin(d*x + c) + 2*sqrt(2)*sqrt (-I)*b*c*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c)) + 2*sqrt (2)*sqrt(I)*b*c*weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c)) + 6*I*sqrt(2)*sqrt(-I)*a*b*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) - 6*I*sqrt(2)*sqrt(I)*a*b*weierstrassZeta( 4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) - 4*b*c*co s(d*x + c)*sqrt(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
\[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right )^2 \, dx=\int \left (a + \frac {b}{\sqrt {\sin {\left (c + d x \right )}}} + c \sin {\left (c + d x \right )}\right )^{2} \sin {\left (c + d x \right )}\, dx \]
\[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right )^2 \, dx=\int { {\left (c \sin \left (d x + c\right ) + a + \frac {b}{\sqrt {\sin \left (d x + c\right )}}\right )}^{2} \sin \left (d x + c\right ) \,d x } \]
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))*...
\[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right )^2 \, dx=\int { {\left (c \sin \left (d x + c\right ) + a + \frac {b}{\sqrt {\sin \left (d x + c\right )}}\right )}^{2} \sin \left (d x + c\right ) \,d x } \]
Time = 31.51 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.87 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right )^2 \, dx=b^2\,x-\frac {a^2\,\cos \left (c+d\,x\right )}{d}-\frac {a\,c\,\left (\sin \left (2\,c+2\,d\,x\right )-2\,d\,x\right )}{2\,d}+\frac {4\,a\,b\,\mathrm {E}\left (\frac {c}{2}-\frac {\pi }{4}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {c^2\,\cos \left (c+d\,x\right )\,\left ({\cos \left (c+d\,x\right )}^2-3\right )}{3\,d}-\frac {2\,b\,c\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\left ({\sin \left (c+d\,x\right )}^2\right )}^{5/4}} \]