Integrand size = 27, antiderivative size = 220 \[ \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=-\frac {2 d^2 \cos (e+f x)}{(b c-a d) \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}-\frac {2 d E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{(b c-a d) \left (c^2-d^2\right ) f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {2 b \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{(a+b) (b c-a d) f \sqrt {c+d \sin (e+f x)}} \] Output:
-2*d^2*cos(f*x+e)/(-a*d+b*c)/(c^2-d^2)/f/(c+d*sin(f*x+e))^(1/2)+2*d*Ellipt icE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1 /2)/(-a*d+b*c)/(c^2-d^2)/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-2*b*EllipticPi(c os(1/2*e+1/4*Pi+1/2*f*x),2*b/(a+b),2^(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x+ e))/(c+d))^(1/2)/(a+b)/(-a*d+b*c)/f/(c+d*sin(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 5.07 (sec) , antiderivative size = 617, normalized size of antiderivative = 2.80 \[ \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=-\frac {\frac {4 d^2 \cos (e+f x)}{\left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}}+\frac {\frac {4 i (b c+a d) \left ((-b c+a d) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{c+d}} \sqrt {c+d \sin (e+f x)}\right ),\frac {c+d}{c-d}\right )-a d \operatorname {EllipticPi}\left (\frac {b (c+d)}{b c-a d},i \text {arcsinh}\left (\sqrt {-\frac {1}{c+d}} \sqrt {c+d \sin (e+f x)}\right ),\frac {c+d}{c-d}\right )\right ) \sec (e+f x) \sqrt {-\frac {d (-1+\sin (e+f x))}{c+d}} \sqrt {\frac {d (1+\sin (e+f x))}{-c+d}}}{b \sqrt {-\frac {1}{c+d}} (b c-a d)}-\frac {2 i \left (-2 b (c-d) (b c-a d) E\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{c+d}} \sqrt {c+d \sin (e+f x)}\right )|\frac {c+d}{c-d}\right )+d \left (2 (a+b) (-b c+a d) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {1}{c+d}} \sqrt {c+d \sin (e+f x)}\right ),\frac {c+d}{c-d}\right )+\left (-2 a^2+b^2\right ) d \operatorname {EllipticPi}\left (\frac {b (c+d)}{b c-a d},i \text {arcsinh}\left (\sqrt {-\frac {1}{c+d}} \sqrt {c+d \sin (e+f x)}\right ),\frac {c+d}{c-d}\right )\right )\right ) \sec (e+f x) \sqrt {-\frac {d (-1+\sin (e+f x))}{c+d}} \sqrt {\frac {d (1+\sin (e+f x))}{-c+d}}}{b \sqrt {-\frac {1}{c+d}} (b c-a d)}+\frac {2 \left (2 b c^2-2 a c d-3 b d^2\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{(a+b) \sqrt {c+d \sin (e+f x)}}}{(c-d) (c+d)}}{2 (b c-a d) f} \] Input:
Integrate[1/((a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^(3/2)),x]
Output:
-1/2*((4*d^2*Cos[e + f*x])/((c^2 - d^2)*Sqrt[c + d*Sin[e + f*x]]) + (((4*I )*(b*c + a*d)*((-(b*c) + a*d)*EllipticF[I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt [c + d*Sin[e + f*x]]], (c + d)/(c - d)] - a*d*EllipticPi[(b*(c + d))/(b*c - a*d), I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/( c - d)])*Sec[e + f*x]*Sqrt[-((d*(-1 + Sin[e + f*x]))/(c + d))]*Sqrt[(d*(1 + Sin[e + f*x]))/(-c + d)])/(b*Sqrt[-(c + d)^(-1)]*(b*c - a*d)) - ((2*I)*( -2*b*(c - d)*(b*c - a*d)*EllipticE[I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)] + d*(2*(a + b)*(-(b*c) + a*d)*EllipticF [I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)] + (-2*a^2 + b^2)*d*EllipticPi[(b*(c + d))/(b*c - a*d), I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)]))*Sec[e + f*x]*Sqr t[-((d*(-1 + Sin[e + f*x]))/(c + d))]*Sqrt[(d*(1 + Sin[e + f*x]))/(-c + d) ])/(b*Sqrt[-(c + d)^(-1)]*(b*c - a*d)) + (2*(2*b*c^2 - 2*a*c*d - 3*b*d^2)* EllipticPi[(2*b)/(a + b), (-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/((a + b)*Sqrt[c + d*Sin[e + f*x]]))/((c - d)*(c + d)))/((b*c - a*d)*f)
Time = 1.42 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.481, Rules used = {3042, 3281, 27, 3042, 3538, 27, 3042, 3134, 3042, 3132, 3286, 3042, 3284}
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.
\(\displaystyle \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 3281 |
\(\displaystyle \frac {2 \int -\frac {b d^2 \sin ^2(e+f x)+d (b c+a d) \sin (e+f x)+a c d-b \left (c^2-d^2\right )}{2 (a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}dx}{\left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt {c+d \sin (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {b d^2 \sin ^2(e+f x)+d (b c+a d) \sin (e+f x)+a c d-b \left (c^2-d^2\right )}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}dx}{\left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt {c+d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {b d^2 \sin (e+f x)^2+d (b c+a d) \sin (e+f x)+a c d-b \left (c^2-d^2\right )}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}dx}{\left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt {c+d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle -\frac {d \int \sqrt {c+d \sin (e+f x)}dx-\frac {\int \frac {b^2 d \left (c^2-d^2\right )}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}dx}{b d}}{\left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt {c+d \sin (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {d \int \sqrt {c+d \sin (e+f x)}dx-b \left (c^2-d^2\right ) \int \frac {1}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}dx}{\left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt {c+d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {d \int \sqrt {c+d \sin (e+f x)}dx-b \left (c^2-d^2\right ) \int \frac {1}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}dx}{\left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt {c+d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3134 |
\(\displaystyle -\frac {\frac {d \sqrt {c+d \sin (e+f x)} \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}dx}{\sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-b \left (c^2-d^2\right ) \int \frac {1}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}dx}{\left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt {c+d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {d \sqrt {c+d \sin (e+f x)} \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}dx}{\sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-b \left (c^2-d^2\right ) \int \frac {1}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}dx}{\left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt {c+d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3132 |
\(\displaystyle -\frac {\frac {2 d \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-b \left (c^2-d^2\right ) \int \frac {1}{(a+b \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}dx}{\left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt {c+d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3286 |
\(\displaystyle -\frac {\frac {2 d \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {b \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{(a+b \sin (e+f x)) \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{\sqrt {c+d \sin (e+f x)}}}{\left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt {c+d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {2 d \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {b \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{(a+b \sin (e+f x)) \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{\sqrt {c+d \sin (e+f x)}}}{\left (c^2-d^2\right ) (b c-a d)}-\frac {2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt {c+d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle -\frac {2 d^2 \cos (e+f x)}{f \left (c^2-d^2\right ) (b c-a d) \sqrt {c+d \sin (e+f x)}}-\frac {\frac {2 d \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 b \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{f (a+b) \sqrt {c+d \sin (e+f x)}}}{\left (c^2-d^2\right ) (b c-a d)}\) |
Input:
Int[1/((a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^(3/2)),x]
Output:
(-2*d^2*Cos[e + f*x])/((b*c - a*d)*(c^2 - d^2)*f*Sqrt[c + d*Sin[e + f*x]]) - ((2*d*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f *x]])/(f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) - (2*b*(c^2 - d^2)*EllipticPi [(2*b)/(a + b), (e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x ])/(c + d)])/((a + b)*f*Sqrt[c + d*Sin[e + f*x]]))/((b*c - a*d)*(c^2 - d^2 ))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[Sqrt[a/(a + b) + ( b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 , 0] && !GtQ[a + b, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 ))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2* n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt [c + d*Sin[e + f*x]] Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !GtQ[c + d, 0]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(606\) vs. \(2(216)=432\).
Time = 1.66 (sec) , antiderivative size = 607, normalized size of antiderivative = 2.76
method | result | size |
default | \(\frac {\sqrt {-\left (-c -d \sin \left (f x +e \right )\right ) \cos \left (f x +e \right )^{2}}\, \left (\frac {d \left (\frac {2 d \cos \left (f x +e \right )^{2}}{\left (c^{2}-d^{2}\right ) \sqrt {-\left (-c -d \sin \left (f x +e \right )\right ) \cos \left (f x +e \right )^{2}}}+\frac {2 c \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )}{\left (c^{2}-d^{2}\right ) \sqrt {-\left (-c -d \sin \left (f x +e \right )\right ) \cos \left (f x +e \right )^{2}}}+\frac {2 d \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \left (\left (-\frac {c}{d}-1\right ) \operatorname {EllipticE}\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )+\operatorname {EllipticF}\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )\right )}{\left (c^{2}-d^{2}\right ) \sqrt {-\left (-c -d \sin \left (f x +e \right )\right ) \cos \left (f x +e \right )^{2}}}\right )}{a d -b c}-\frac {2 \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {-\frac {d \left (1+\sin \left (f x +e \right )\right )}{c -d}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \frac {-\frac {c}{d}+1}{-\frac {c}{d}+\frac {a}{b}}, \sqrt {\frac {c -d}{c +d}}\right )}{\left (a d -b c \right ) \sqrt {-\left (-c -d \sin \left (f x +e \right )\right ) \cos \left (f x +e \right )^{2}}\, \left (-\frac {c}{d}+\frac {a}{b}\right )}\right )}{\cos \left (f x +e \right ) \sqrt {c +d \sin \left (f x +e \right )}\, f}\) | \(607\) |
Input:
int(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
Output:
(-(-c-d*sin(f*x+e))*cos(f*x+e)^2)^(1/2)*(d/(a*d-b*c)*(2*d*cos(f*x+e)^2/(c^ 2-d^2)/(-(-c-d*sin(f*x+e))*cos(f*x+e)^2)^(1/2)+2/(c^2-d^2)*c*(c/d-1)*((c+d *sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*(-d*(1+sin(f*x+e) )/(c-d))^(1/2)/(-(-c-d*sin(f*x+e))*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin (f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+2/(c^2-d^2)*d*(c/d-1)*((c+d*sin (f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c -d))^(1/2)/(-(-c-d*sin(f*x+e))*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c +d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e ))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))))-2/(a*d-b*c)*(c/d-1)*((c+d*sin(f*x+e ))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*(-d*(1+sin(f*x+e))/(c-d))^( 1/2)/(-(-c-d*sin(f*x+e))*cos(f*x+e)^2)^(1/2)/(-c/d+a/b)*EllipticPi(((c+d*s in(f*x+e))/(c-d))^(1/2),(-c/d+1)/(-c/d+a/b),((c-d)/(c+d))^(1/2)))/cos(f*x+ e)/(c+d*sin(f*x+e))^(1/2)/f
Timed out. \[ \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(3/2),x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))**(3/2),x)
Output:
Timed out
\[ \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(3/2),x, algorithm="maxima")
Output:
integrate(1/((b*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(3/2)), x)
\[ \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=\int { \frac {1}{{\left (b \sin \left (f x + e\right ) + a\right )} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(3/2),x, algorithm="giac")
Output:
integrate(1/((b*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(3/2)), x)
Timed out. \[ \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {1}{\left (a+b\,\sin \left (e+f\,x\right )\right )\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:
int(1/((a + b*sin(e + f*x))*(c + d*sin(e + f*x))^(3/2)),x)
Output:
int(1/((a + b*sin(e + f*x))*(c + d*sin(e + f*x))^(3/2)), x)
\[ \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}}{\sin \left (f x +e \right )^{3} b \,d^{2}+\sin \left (f x +e \right )^{2} a \,d^{2}+2 \sin \left (f x +e \right )^{2} b c d +2 \sin \left (f x +e \right ) a c d +\sin \left (f x +e \right ) b \,c^{2}+a \,c^{2}}d x \] Input:
int(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(3/2),x)
Output:
int(sqrt(sin(e + f*x)*d + c)/(sin(e + f*x)**3*b*d**2 + sin(e + f*x)**2*a*d **2 + 2*sin(e + f*x)**2*b*c*d + 2*sin(e + f*x)*a*c*d + sin(e + f*x)*b*c**2 + a*c**2),x)