Integrand size = 27, antiderivative size = 247 \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx=\frac {2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 (c+3 d) \cos (e+f x)}{3 d (c+d)^2 f \sqrt {c+d \sin (e+f x)}}-\frac {4 a^2 (c+3 d) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{3 d^2 (c+d)^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^2 (c+2 d) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{3 d^2 (c+d) f \sqrt {c+d \sin (e+f x)}} \] Output:
2/3*a^2*(c-d)*cos(f*x+e)/d/(c+d)/f/(c+d*sin(f*x+e))^(3/2)-4/3*a^2*(c+3*d)* cos(f*x+e)/d/(c+d)^2/f/(c+d*sin(f*x+e))^(1/2)+4/3*a^2*(c+3*d)*EllipticE(co s(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)/d^ 2/(c+d)^2/f/((c+d*sin(f*x+e))/(c+d))^(1/2)+4/3*a^2*(c+2*d)*InverseJacobiAM (1/2*e-1/4*Pi+1/2*f*x,2^(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1 /2)/d^2/(c+d)/f/(c+d*sin(f*x+e))^(1/2)
Time = 4.05 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.84 \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx=-\frac {2 a^2 (1+\sin (e+f x))^2 \left (-2 (c+d)^2 (c+3 d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right ) \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{3/2}+2 (c+d)^2 (c+2 d) \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right ) \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{3/2}+d \cos (e+f x) \left (c^2+6 c d+d^2+2 d (c+3 d) \sin (e+f x)\right )\right )}{3 d^2 (c+d)^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (c+d \sin (e+f x))^{3/2}} \] Input:
Integrate[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^(5/2),x]
Output:
(-2*a^2*(1 + Sin[e + f*x])^2*(-2*(c + d)^2*(c + 3*d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*((c + d*Sin[e + f*x])/(c + d))^(3/2) + 2*(c + d )^2*(c + 2*d)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*((c + d*Sin[ e + f*x])/(c + d))^(3/2) + d*Cos[e + f*x]*(c^2 + 6*c*d + d^2 + 2*d*(c + 3* d)*Sin[e + f*x])))/(3*d^2*(c + d)^2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2] )^4*(c + d*Sin[e + f*x])^(3/2))
Time = 1.25 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.06, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 3241, 25, 3042, 3233, 27, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}
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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{(c+d \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{(c+d \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3241 |
| \(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}-\frac {2 a \int -\frac {3 a d+a (c+2 d) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}}dx}{3 d (c+d)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 a \int \frac {3 a d+a (c+2 d) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}}dx}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \int \frac {3 a d+a (c+2 d) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}}dx}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {2 a \left (-\frac {2 \int -\frac {2 a (c-d) d-a (c-d) (c+3 d) \sin (e+f x)}{2 \sqrt {c+d \sin (e+f x)}}dx}{c^2-d^2}-\frac {2 a (c+3 d) \cos (e+f x)}{f (c+d) \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 a \left (\frac {\int \frac {2 a (c-d) d-a (c-d) (c+3 d) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}}dx}{c^2-d^2}-\frac {2 a (c+3 d) \cos (e+f x)}{f (c+d) \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \left (\frac {\int \frac {2 a (c-d) d-a (c-d) (c+3 d) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}}dx}{c^2-d^2}-\frac {2 a (c+3 d) \cos (e+f x)}{f (c+d) \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3231 |
| \(\displaystyle \frac {2 a \left (\frac {\frac {a (c+2 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}-\frac {a (c-d) (c+3 d) \int \sqrt {c+d \sin (e+f x)}dx}{d}}{c^2-d^2}-\frac {2 a (c+3 d) \cos (e+f x)}{f (c+d) \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \left (\frac {\frac {a (c+2 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}-\frac {a (c-d) (c+3 d) \int \sqrt {c+d \sin (e+f x)}dx}{d}}{c^2-d^2}-\frac {2 a (c+3 d) \cos (e+f x)}{f (c+d) \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {2 a \left (\frac {\frac {a (c+2 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}-\frac {a (c-d) (c+3 d) \sqrt {c+d \sin (e+f x)} \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}dx}{d \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{c^2-d^2}-\frac {2 a (c+3 d) \cos (e+f x)}{f (c+d) \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \left (\frac {\frac {a (c+2 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}-\frac {a (c-d) (c+3 d) \sqrt {c+d \sin (e+f x)} \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}dx}{d \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{c^2-d^2}-\frac {2 a (c+3 d) \cos (e+f x)}{f (c+d) \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {2 a \left (\frac {\frac {a (c+2 d) \left (c^2-d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}}dx}{d}-\frac {2 a (c-d) (c+3 d) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{c^2-d^2}-\frac {2 a (c+3 d) \cos (e+f x)}{f (c+d) \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {2 a \left (\frac {\frac {a (c+2 d) \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{d \sqrt {c+d \sin (e+f x)}}-\frac {2 a (c-d) (c+3 d) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{c^2-d^2}-\frac {2 a (c+3 d) \cos (e+f x)}{f (c+d) \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \left (\frac {\frac {a (c+2 d) \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \int \frac {1}{\sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}}}dx}{d \sqrt {c+d \sin (e+f x)}}-\frac {2 a (c-d) (c+3 d) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{c^2-d^2}-\frac {2 a (c+3 d) \cos (e+f x)}{f (c+d) \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}}+\frac {2 a \left (\frac {\frac {2 a (c+2 d) \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{d f \sqrt {c+d \sin (e+f x)}}-\frac {2 a (c-d) (c+3 d) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{d f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}}{c^2-d^2}-\frac {2 a (c+3 d) \cos (e+f x)}{f (c+d) \sqrt {c+d \sin (e+f x)}}\right )}{3 d (c+d)}\) |
Input:
Int[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^(5/2),x]
Output:
(2*a^2*(c - d)*Cos[e + f*x])/(3*d*(c + d)*f*(c + d*Sin[e + f*x])^(3/2)) + (2*a*((-2*a*(c + 3*d)*Cos[e + f*x])/((c + d)*f*Sqrt[c + d*Sin[e + f*x]]) + ((-2*a*(c - d)*(c + 3*d)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqr t[c + d*Sin[e + f*x]])/(d*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (2*a*(c + 2*d)*(c^2 - d^2)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(d*f*Sqrt[c + d*Sin[e + f*x]]))/(c^2 - d^2)))/(3 *d*(c + d))
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[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(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[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]] Int[1/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[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b Int[1/Sqrt[a + b*Sin[e + f*x ]], x], x] + Simp[d/b Int[Sqrt[a + b*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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b *Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))
Leaf count of result is larger than twice the leaf count of optimal. \(1220\) vs. \(2(232)=464\).
Time = 3.11 (sec) , antiderivative size = 1221, normalized size of antiderivative = 4.94
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1221\) |
| parts | \(\text {Expression too large to display}\) | \(2389\) |
Input:
int((a+sin(f*x+e)*a)^2/(c+d*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3*a^2*((2*c*d^3+6*d^4)*cos(f*x+e)^2*sin(f*x+e)+(c^2*d^2+6*c*d^3+d^4)*co s(f*x+e)^2+2*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/( c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*d*(EllipticF((d/(c-d)*sin( f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^2*d-EllipticF((d/(c-d)*sin( f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*d^3-EllipticE((d/(c-d)*sin(f* x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^3-3*EllipticE((d/(c-d)*sin(f* x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^2*d+EllipticE((d/(c-d)*sin(f* x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c*d^2+3*EllipticE((d/(c-d)*sin( f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*d^3)*sin(f*x+e)+2*(d/(c-d)*si n(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*si n(f*x+e)-d/(c-d))^(1/2)*EllipticF((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c -d)/(c+d))^(1/2))*c^3*d-2*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*s in(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticF((d/ (c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c*d^3-2*(d/(c-d)*si n(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*si n(f*x+e)-d/(c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c -d)/(c+d))^(1/2))*c^4-6*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin (f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticE((d/(c -d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^3*d+2*(d/(c-d)*sin( f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*s...
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 947, normalized size of antiderivative = 3.83 \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(5/2),x, algorithm="fricas")
Output:
-2/9*(4*(a^2*c^4 + 3*a^2*c^3*d + 4*a^2*c^2*d^2 + 3*a^2*c*d^3 + 3*a^2*d^4 - (a^2*c^2*d^2 + 3*a^2*c*d^3 + 3*a^2*d^4)*cos(f*x + e)^2 + 2*(a^2*c^3*d + 3 *a^2*c^2*d^2 + 3*a^2*c*d^3)*sin(f*x + e))*sqrt(1/2*I*d)*weierstrassPInvers e(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos( f*x + e) - 3*I*d*sin(f*x + e) - 2*I*c)/d) + 4*(a^2*c^4 + 3*a^2*c^3*d + 4*a ^2*c^2*d^2 + 3*a^2*c*d^3 + 3*a^2*d^4 - (a^2*c^2*d^2 + 3*a^2*c*d^3 + 3*a^2* d^4)*cos(f*x + e)^2 + 2*(a^2*c^3*d + 3*a^2*c^2*d^2 + 3*a^2*c*d^3)*sin(f*x + e))*sqrt(-1/2*I*d)*weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*( -8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e) + 2* I*c)/d) - 6*(-I*a^2*c^3*d - 3*I*a^2*c^2*d^2 - I*a^2*c*d^3 - 3*I*a^2*d^4 + (I*a^2*c*d^3 + 3*I*a^2*d^4)*cos(f*x + e)^2 + 2*(-I*a^2*c^2*d^2 - 3*I*a^2*c *d^3)*sin(f*x + e))*sqrt(1/2*I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2)/d^2 , -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, weierstrassPInverse(-4/3*(4*c^2 - 3*d^2 )/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d*sin( f*x + e) - 2*I*c)/d)) - 6*(I*a^2*c^3*d + 3*I*a^2*c^2*d^2 + I*a^2*c*d^3 + 3 *I*a^2*d^4 + (-I*a^2*c*d^3 - 3*I*a^2*d^4)*cos(f*x + e)^2 + 2*(I*a^2*c^2*d^ 2 + 3*I*a^2*c*d^3)*sin(f*x + e))*sqrt(-1/2*I*d)*weierstrassZeta(-4/3*(4*c^ 2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, weierstrassPInverse(-4/3 *(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e) + 2*I*c)/d)) - 3*(2*(a^2*c*d^3 + 3*a^2*d^4)*co...
Timed out. \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**2/(c+d*sin(f*x+e))**(5/2),x)
Output:
Timed out
\[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(5/2),x, algorithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)^2/(d*sin(f*x + e) + c)^(5/2), x)
\[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")
Output:
integrate((a*sin(f*x + e) + a)^2/(d*sin(f*x + e) + c)^(5/2), x)
Timed out. \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int((a + a*sin(e + f*x))^2/(c + d*sin(e + f*x))^(5/2),x)
Output:
int((a + a*sin(e + f*x))^2/(c + d*sin(e + f*x))^(5/2), x)
\[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx=a^{2} \left (\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}}{\sin \left (f x +e \right )^{3} d^{3}+3 \sin \left (f x +e \right )^{2} c \,d^{2}+3 \sin \left (f x +e \right ) c^{2} d +c^{3}}d x +\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3} d^{3}+3 \sin \left (f x +e \right )^{2} c \,d^{2}+3 \sin \left (f x +e \right ) c^{2} d +c^{3}}d x +2 \left (\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3} d^{3}+3 \sin \left (f x +e \right )^{2} c \,d^{2}+3 \sin \left (f x +e \right ) c^{2} d +c^{3}}d x \right )\right ) \] Input:
int((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(5/2),x)
Output:
a**2*(int(sqrt(sin(e + f*x)*d + c)/(sin(e + f*x)**3*d**3 + 3*sin(e + f*x)* *2*c*d**2 + 3*sin(e + f*x)*c**2*d + c**3),x) + int((sqrt(sin(e + f*x)*d + c)*sin(e + f*x)**2)/(sin(e + f*x)**3*d**3 + 3*sin(e + f*x)**2*c*d**2 + 3*s in(e + f*x)*c**2*d + c**3),x) + 2*int((sqrt(sin(e + f*x)*d + c)*sin(e + f* x))/(sin(e + f*x)**3*d**3 + 3*sin(e + f*x)**2*c*d**2 + 3*sin(e + f*x)*c**2 *d + c**3),x))