Integrand size = 29, antiderivative size = 61 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=\frac {c x}{2}-\frac {a \cos (c+d x)}{d}+\frac {2 b E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right )}{d}-\frac {c \cos (c+d x) \sin (c+d x)}{2 d} \] Output:
1/2*c*x-a*cos(d*x+c)/d-2*b*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))/d- 1/2*c*cos(d*x+c)*sin(d*x+c)/d
Time = 0.58 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.90 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=\frac {-4 a \cos (c+d x)-8 b E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )+c (2 c+2 d x-\sin (2 (c+d x)))}{4 d} \] Input:
Integrate[Sin[c + d*x]*(a + b/Sqrt[Sin[c + d*x]] + c*Sin[c + d*x]),x]
Output:
(-4*a*Cos[c + d*x] - 8*b*EllipticE[(-2*c + Pi - 2*d*x)/4, 2] + c*(2*c + 2* d*x - Sin[2*(c + d*x)]))/(4*d)
Time = 0.50 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, 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.
\(\displaystyle \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right )dx\) |
\(\Big \downarrow \) 4895 |
\(\displaystyle \int \sqrt {\sin (c+d x)} \left (a \sqrt {\sin (c+d x)}+b+c \sin ^{\frac {3}{2}}(c+d x)\right )dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {\sin (c+d x)} \left (a \sqrt {\sin (c+d x)}+b+c \sin (c+d x)^{3/2}\right )dx\) |
\(\Big \downarrow \) 4901 |
\(\displaystyle \int \left (a \sin (c+d x)+b \sqrt {\sin (c+d x)}+c \sin ^2(c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a \cos (c+d x)}{d}+\frac {2 b E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right )}{d}-\frac {c \sin (c+d x) \cos (c+d x)}{2 d}+\frac {c x}{2}\) |
Input:
Int[Sin[c + d*x]*(a + b/Sqrt[Sin[c + d*x]] + c*Sin[c + d*x]),x]
Output:
(c*x)/2 - (a*Cos[c + d*x])/d + (2*b*EllipticE[(c - Pi/2 + d*x)/2, 2])/d - (c*Cos[c + d*x]*Sin[c + d*x])/(2*d)
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]
Leaf count of result is larger than twice the leaf count of optimal. \(131\) vs. \(2(57)=114\).
Time = 0.40 (sec) , antiderivative size = 132, normalized size of antiderivative = 2.16
method | result | size |
default | \(-\frac {a \cos \left (d x +c \right )}{d}+\frac {c \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {b \sqrt {\sin \left (d x +c \right )+1}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \left (2 \operatorname {EllipticE}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right )}{\cos \left (d x +c \right ) \sqrt {\sin \left (d x +c \right )}\, d}\) | \(132\) |
parts | \(-\frac {a \cos \left (d x +c \right )}{d}+\frac {c \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {b \sqrt {\sin \left (d x +c \right )+1}\, \sqrt {-2 \sin \left (d x +c \right )+2}\, \sqrt {-\sin \left (d x +c \right )}\, \left (2 \operatorname {EllipticE}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right )}{\cos \left (d x +c \right ) \sqrt {\sin \left (d x +c \right )}\, d}\) | \(132\) |
Input:
int(sin(d*x+c)*(a+b/sin(d*x+c)^(1/2)+c*sin(d*x+c)),x,method=_RETURNVERBOSE )
Output:
-a*cos(d*x+c)/d+c/d*(-1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)-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
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.67 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=-\frac {c \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 4 i \, \sqrt {-\frac {1}{2} i} b {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 4 i \, \sqrt {\frac {1}{2} i} b {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - c \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right )}\right ) + 2 \, a \cos \left (d x + c\right )}{2 \, d} \] Input:
integrate(sin(d*x+c)*(a+b/sin(d*x+c)^(1/2)+c*sin(d*x+c)),x, algorithm="fri cas")
Output:
-1/2*(c*cos(d*x + c)*sin(d*x + c) - 4*I*sqrt(-1/2*I)*b*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 4*I*sqrt(1/ 2*I)*b*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*si n(d*x + c))) - c*arctan(sin(d*x + c)/cos(d*x + c)) + 2*a*cos(d*x + c))/d
\[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=\int \left (a \sqrt {\sin {\left (c + d x \right )}} + b + c \sin ^{\frac {3}{2}}{\left (c + d x \right )}\right ) \sqrt {\sin {\left (c + d x \right )}}\, dx \] Input:
integrate(sin(d*x+c)*(a+b/sin(d*x+c)**(1/2)+c*sin(d*x+c)),x)
Output:
Integral((a*sqrt(sin(c + d*x)) + b + c*sin(c + d*x)**(3/2))*sqrt(sin(c + d *x)), x)
\[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=\int { {\left (c \sin \left (d x + c\right ) + a + \frac {b}{\sqrt {\sin \left (d x + c\right )}}\right )} \sin \left (d x + c\right ) \,d x } \] Input:
integrate(sin(d*x+c)*(a+b/sin(d*x+c)^(1/2)+c*sin(d*x+c)),x, algorithm="max ima")
Output:
1/4*(2*c*d*x - 4*a*cos(d*x + c) + 2*d*integrate(-(((b*cos(3/2*d*x + 3/2*c) - b*cos(1/2*d*x + 1/2*c) - b*sin(3/2*d*x + 3/2*c) - b*sin(1/2*d*x + 1/2*c ))*cos(1/2*arctan2(sin(d*x + c), -cos(d*x + c) + 1)) - (b*cos(3/2*d*x + 3/ 2*c) - b*cos(1/2*d*x + 1/2*c) + b*sin(3/2*d*x + 3/2*c) + b*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( sin(d*x + c), cos(d*x + c) + 1)) + ((b*cos(3/2*d*x + 3/2*c) - b*cos(1/2*d* x + 1/2*c) + b*sin(3/2*d*x + 3/2*c) + b*sin(1/2*d*x + 1/2*c))*cos(1/2*arct an2(sin(d*x + c), -cos(d*x + c) + 1)) + (b*cos(3/2*d*x + 3/2*c) - b*cos(1/ 2*d*x + 1/2*c) - b*sin(3/2*d*x + 3/2*c) - 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(sin(d*x + c), c os(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) - c*sin(2*d*x + 2*c))/d
\[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=\int { {\left (c \sin \left (d x + c\right ) + a + \frac {b}{\sqrt {\sin \left (d x + c\right )}}\right )} \sin \left (d x + c\right ) \,d x } \] Input:
integrate(sin(d*x+c)*(a+b/sin(d*x+c)^(1/2)+c*sin(d*x+c)),x, algorithm="gia c")
Output:
integrate((c*sin(d*x + c) + a + b/sqrt(sin(d*x + c)))*sin(d*x + c), x)
Time = 16.44 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.84 \[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=\frac {c\,x}{2}-\frac {c\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}-\frac {a\,\cos \left (c+d\,x\right )}{d}+\frac {2\,b\,\mathrm {E}\left (\frac {c}{2}-\frac {\pi }{4}+\frac {d\,x}{2}\middle |2\right )}{d} \] Input:
int(sin(c + d*x)*(a + c*sin(c + d*x) + b/sin(c + d*x)^(1/2)),x)
Output:
(c*x)/2 - (c*sin(2*c + 2*d*x))/(4*d) - (a*cos(c + d*x))/d + (2*b*ellipticE (c/2 - pi/4 + (d*x)/2, 2))/d
\[ \int \sin (c+d x) \left (a+\frac {b}{\sqrt {\sin (c+d x)}}+c \sin (c+d x)\right ) \, dx=\frac {-\cos \left (d x +c \right ) \sin \left (d x +c \right ) c -2 \cos \left (d x +c \right ) a +2 \left (\int \sqrt {\sin \left (d x +c \right )}d x \right ) b d +c d x}{2 d} \] Input:
int(sin(d*x+c)*(a+b/sin(d*x+c)^(1/2)+c*sin(d*x+c)),x)
Output:
( - cos(c + d*x)*sin(c + d*x)*c - 2*cos(c + d*x)*a + 2*int(sqrt(sin(c + d* x)),x)*b*d + c*d*x)/(2*d)