3.514 \(\int \frac{1}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=326 \[ -\frac{d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 f (c-d)^3 (c+d) \sqrt{c+d \sin (e+f x)}}-\frac{\left (c^2-5 c d-12 d^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 a^2 f (c-d)^3 (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{(c-5 d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1) \sqrt{c+d \sin (e+f x)}}+\frac{(c-5 d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 a^2 f (c-d)^2 \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 \sqrt{c+d \sin (e+f x)}} \]

[Out]

-(d*(c^2 - 5*c*d - 12*d^2)*Cos[e + f*x])/(3*a^2*(c - d)^3*(c + d)*f*Sqrt[c + d*Sin[e + f*x]]) - ((c - 5*d)*Cos
[e + f*x])/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x]]) - Cos[e + f*x]/(3*(c - d)*f*(a + a*
Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]]) - ((c^2 - 5*c*d - 12*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d
)]*Sqrt[c + d*Sin[e + f*x]])/(3*a^2*(c - d)^3*(c + d)*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + ((c - 5*d)*Ellip
ticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*a^2*(c - d)^2*f*Sqrt[c + d*Sin[
e + f*x]])

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Rubi [A]  time = 0.641699, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2766, 2978, 2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac{d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 f (c-d)^3 (c+d) \sqrt{c+d \sin (e+f x)}}-\frac{\left (c^2-5 c d-12 d^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 a^2 f (c-d)^3 (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{(c-5 d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1) \sqrt{c+d \sin (e+f x)}}+\frac{(c-5 d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 a^2 f (c-d)^2 \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^(3/2)),x]

[Out]

-(d*(c^2 - 5*c*d - 12*d^2)*Cos[e + f*x])/(3*a^2*(c - d)^3*(c + d)*f*Sqrt[c + d*Sin[e + f*x]]) - ((c - 5*d)*Cos
[e + f*x])/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x]]) - Cos[e + f*x]/(3*(c - d)*f*(a + a*
Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]]) - ((c^2 - 5*c*d - 12*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d
)]*Sqrt[c + d*Sin[e + f*x]])/(3*a^2*(c - d)^3*(c + d)*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + ((c - 5*d)*Ellip
ticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*a^2*(c - d)^2*f*Sqrt[c + d*Sin[
e + f*x]])

Rule 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 2754

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] + Dist[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]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[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]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2}} \, dx &=-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)}}-\frac{\int \frac{-\frac{1}{2} a (2 c-7 d)-\frac{3}{2} a d \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx}{3 a^2 (c-d)}\\ &=-\frac{(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)}}+\frac{\int \frac{6 a^2 d^2+\frac{1}{2} a^2 (c-5 d) d \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 a^4 (c-d)^2}\\ &=-\frac{d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)}}-\frac{2 \int \frac{-\frac{1}{4} a^2 d^2 (11 c+5 d)+\frac{1}{4} a^2 d \left (c^2-5 c d-12 d^2\right ) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{3 a^4 (c-d)^3 (c+d)}\\ &=-\frac{d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)}}+\frac{(c-5 d) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{6 a^2 (c-d)^2}-\frac{\left (c^2-5 c d-12 d^2\right ) \int \sqrt{c+d \sin (e+f x)} \, dx}{6 a^2 (c-d)^3 (c+d)}\\ &=-\frac{d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)}}-\frac{\left (\left (c^2-5 c d-12 d^2\right ) \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{6 a^2 (c-d)^3 (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left ((c-5 d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}\right ) \int \frac{1}{\sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}}} \, dx}{6 a^2 (c-d)^2 \sqrt{c+d \sin (e+f x)}}\\ &=-\frac{d \left (c^2-5 c d-12 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{(c-5 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)}}-\frac{\left (c^2-5 c d-12 d^2\right ) 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 a^2 (c-d)^3 (c+d) f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{(c-5 d) F\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 a^2 (c-d)^2 f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 4.77666, size = 405, normalized size = 1.24 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4 \left ((c+d \sin (e+f x)) \left (-\frac{2 \left (c^2-5 c d-9 d^2\right )}{c+d}+\frac{6 d^3 \cos (e+f x)}{(c+d) (c+d \sin (e+f x))}+\frac{2 (c-6 d) \sin \left (\frac{1}{2} (e+f x)\right )}{\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )}+\frac{d-c}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}+\frac{2 (c-d) \sin \left (\frac{1}{2} (e+f x)\right )}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}\right )+\frac{\left (c^2-5 c d-12 d^2\right ) (c+d \sin (e+f x))+\left (c^2-5 c d-12 d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \left ((c+d) E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-c F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )+d^2 (-(11 c+5 d)) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )}{c+d}\right )}{3 a^2 f (c-d)^3 (\sin (e+f x)+1)^2 \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^(3/2)),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*((c + d*Sin[e + f*x])*((-2*(c^2 - 5*c*d - 9*d^2))/(c + d) + (2*(c - d
)*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + (-c + d)/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2
+ (2*(c - 6*d)*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + (6*d^3*Cos[e + f*x])/((c + d)*(c + d
*Sin[e + f*x]))) + ((c^2 - 5*c*d - 12*d^2)*(c + d*Sin[e + f*x]) - d^2*(11*c + 5*d)*EllipticF[(-2*e + Pi - 2*f*
x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] + (c^2 - 5*c*d - 12*d^2)*((c + d)*EllipticE[(-2*e + Pi
- 2*f*x)/4, (2*d)/(c + d)] - c*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)])*Sqrt[(c + d*Sin[e + f*x])/(c
+ d)])/(c + d)))/(3*a^2*(c - d)^3*f*(1 + Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]])

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Maple [B]  time = 5.844, size = 1299, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(3/2),x)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/a^2*(1/(c-d)*(-1/3/(c-d)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(1+si
n(f*x+e))^2-1/3*(-sin(f*x+e)^2*d-c*sin(f*x+e)+d*sin(f*x+e)+c)/(c-d)^2*(c-3*d)/((-d*sin(f*x+e)-c)*(-1+sin(f*x+e
))*(1+sin(f*x+e)))^(1/2)+2*d^2/(3*c^2-6*c*d+3*d^2)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c
+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))
/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-1/3*d*(c-3*d)/(c-d)^2*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+
e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*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
))))+d^2/(c-d)^2*(2*d*cos(f*x+e)^2/(c^2-d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2*c/(c^2-d^2)*(c/d-1)*((c
+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)
*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)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*c
os(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))))-d/(c-d)^2*(-(-sin(f*x+e)^2*d-c*sin(f*x+e)+d*sin(f*x+e)+c)/(c-d)/(
(-d*sin(f*x+e)-c)*(-1+sin(f*x+e))*(1+sin(f*x+e)))^(1/2)-2*d/(2*c-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)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*Elliptic
F(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-d/(c-d)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin
(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*Ellipt
icE(((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)))))/cos(f*x+e)/(c+d*sin(f*x+e))^(1/2)/f

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \sin \left (f x + e\right ) + c}}{a^{2} d^{2} \cos \left (f x + e\right )^{4} + 2 \, a^{2} c^{2} + 4 \, a^{2} c d + 2 \, a^{2} d^{2} -{\left (a^{2} c^{2} + 4 \, a^{2} c d + 3 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \,{\left (a^{2} c^{2} + 2 \, a^{2} c d + a^{2} d^{2} -{\left (a^{2} c d + a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*sin(f*x + e) + c)/(a^2*d^2*cos(f*x + e)^4 + 2*a^2*c^2 + 4*a^2*c*d + 2*a^2*d^2 - (a^2*c^2 + 4*a
^2*c*d + 3*a^2*d^2)*cos(f*x + e)^2 + 2*(a^2*c^2 + 2*a^2*c*d + a^2*d^2 - (a^2*c*d + a^2*d^2)*cos(f*x + e)^2)*si
n(f*x + e)), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{1}{c \sqrt{c + d \sin{\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )} + 2 c \sqrt{c + d \sin{\left (e + f x \right )}} \sin{\left (e + f x \right )} + c \sqrt{c + d \sin{\left (e + f x \right )}} + d \sqrt{c + d \sin{\left (e + f x \right )}} \sin ^{3}{\left (e + f x \right )} + 2 d \sqrt{c + d \sin{\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )} + d \sqrt{c + d \sin{\left (e + f x \right )}} \sin{\left (e + f x \right )}}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))**2/(c+d*sin(f*x+e))**(3/2),x)

[Out]

Integral(1/(c*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**2 + 2*c*sqrt(c + d*sin(e + f*x))*sin(e + f*x) + c*sqrt(c
+ d*sin(e + f*x)) + d*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**3 + 2*d*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**2
+ d*sqrt(c + d*sin(e + f*x))*sin(e + f*x)), x)/a**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((a*sin(f*x + e) + a)^2*(d*sin(f*x + e) + c)^(3/2)), x)