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

Optimal. Leaf size=351 \[ \frac{\left (a^2 d^2+2 a b c d+b^2 \left (-\left (c^2+2 d^2\right )\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{b^2 f \left (a^2-b^2\right ) \sqrt{c+d \sin (e+f x)}}+\frac{(b c-a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}+\frac{(b c-a d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{b f \left (a^2-b^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{(b c-a d) \left (a^2 d+2 a b c-3 b^2 d\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{b^2 f (a-b) (a+b)^2 \sqrt{c+d \sin (e+f x)}} \]

[Out]

((b*c - a*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/((a^2 - b^2)*f*(a + b*Sin[e + f*x])) + ((b*c - a*d)*Ellipt
icE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(b*(a^2 - b^2)*f*Sqrt[(c + d*Sin[e + f*x])/(c
+ d)]) + ((2*a*b*c*d + a^2*d^2 - b^2*(c^2 + 2*d^2))*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*
Sin[e + f*x])/(c + d)])/(b^2*(a^2 - b^2)*f*Sqrt[c + d*Sin[e + f*x]]) + ((b*c - a*d)*(2*a*b*c + a^2*d - 3*b^2*d
)*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)*b^
2*(a + b)^2*f*Sqrt[c + d*Sin[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.03696, antiderivative size = 351, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2799, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ \frac{\left (a^2 d^2+2 a b c d+b^2 \left (-\left (c^2+2 d^2\right )\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{b^2 f \left (a^2-b^2\right ) \sqrt{c+d \sin (e+f x)}}+\frac{(b c-a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}+\frac{(b c-a d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{b f \left (a^2-b^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{(b c-a d) \left (a^2 d+2 a b c-3 b^2 d\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{b^2 f (a-b) (a+b)^2 \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/((a^2 - b^2)*f*(a + b*Sin[e + f*x])) + ((b*c - a*d)*Ellipt
icE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(b*(a^2 - b^2)*f*Sqrt[(c + d*Sin[e + f*x])/(c
+ d)]) + ((2*a*b*c*d + a^2*d^2 - b^2*(c^2 + 2*d^2))*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*
Sin[e + f*x])/(c + d)])/(b^2*(a^2 - b^2)*f*Sqrt[c + d*Sin[e + f*x]]) + ((b*c - a*d)*(2*a*b*c + a^2*d - 3*b^2*d
)*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)*b^
2*(a + b)^2*f*Sqrt[c + d*Sin[e + f*x]])

Rule 2799

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 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)*(c + d*Sin[e + f*x])^(n - 2)*Simp[c*(a
*c - b*d)*(m + 1) + d*(b*c - a*d)*(n - 1) + (d*(a*c - b*d)*(m + 1) - c*(b*c - a*d)*(m + 2))*Sin[e + f*x] - d*(
b*c - a*d)*(m + n + 1)*Sin[e + f*x]^2, 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] && LtQ[m, -1] && LtQ[1, n, 2] && IntegersQ[2*m, 2*n]

Rule 3059

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] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[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] /; FreeQ[{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]

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]

Rule 3002

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^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 2807

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

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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]

Rubi steps

\begin{align*} \int \frac{(c+d \sin (e+f x))^{3/2}}{(a+b \sin (e+f x))^2} \, dx &=\frac{(b c-a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{\int \frac{\frac{1}{2} \left (3 b c d-a \left (2 c^2+d^2\right )\right )-d (a c-b d) \sin (e+f x)-\frac{1}{2} d (b c-a d) \sin ^2(e+f x)}{(a+b \sin (e+f x)) \sqrt{c+d \sin (e+f x)}} \, dx}{-a^2+b^2}\\ &=\frac{(b c-a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{\int \frac{\frac{1}{2} d \left (a^2 c d-3 b^2 c d+a b \left (c^2+d^2\right )\right )+\frac{1}{2} d \left (2 a b c d+a^2 d^2-b^2 \left (c^2+2 d^2\right )\right ) \sin (e+f x)}{(a+b \sin (e+f x)) \sqrt{c+d \sin (e+f x)}} \, dx}{b \left (a^2-b^2\right ) d}+\frac{(b c-a d) \int \sqrt{c+d \sin (e+f x)} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{(b c-a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{\left ((b c-a d) \left (2 a b c+a^2 d-3 b^2 d\right )\right ) \int \frac{1}{(a+b \sin (e+f x)) \sqrt{c+d \sin (e+f x)}} \, dx}{2 b^2 \left (a^2-b^2\right )}+\frac{\left (2 a b c d+a^2 d^2-b^2 \left (c^2+2 d^2\right )\right ) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{2 b^2 \left (a^2-b^2\right )}+\frac{\left ((b c-a d) \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{2 b \left (a^2-b^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}\\ &=\frac{(b c-a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{(b c-a 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 \left (a^2-b^2\right ) f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left ((b c-a d) \left (2 a b c+a^2 d-3 b^2 d\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}\right ) \int \frac{1}{(a+b \sin (e+f x)) \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}}} \, dx}{2 b^2 \left (a^2-b^2\right ) \sqrt{c+d \sin (e+f x)}}+\frac{\left (\left (2 a b c d+a^2 d^2-b^2 \left (c^2+2 d^2\right )\right ) \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}{2 b^2 \left (a^2-b^2\right ) \sqrt{c+d \sin (e+f x)}}\\ &=\frac{(b c-a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{(b c-a 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 \left (a^2-b^2\right ) f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (2 a b c d+a^2 d^2-b^2 \left (c^2+2 d^2\right )\right ) 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}}}{b^2 \left (a^2-b^2\right ) f \sqrt{c+d \sin (e+f x)}}+\frac{(b c-a d) \left (2 a b c+a^2 d-3 b^2 d\right ) \Pi \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^2 (a+b)^2 f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 7.2076, size = 891, normalized size = 2.54 \[ \frac{\sqrt{c+d \sin (e+f x)} (b c \cos (e+f x)-a d \cos (e+f x))}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{-\frac{2 \left (4 a c^2-5 b d c+a d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{(a+b) \sqrt{c+d \sin (e+f x)}}-\frac{2 i \left (4 a c d-4 b d^2\right ) \cos (e+f x) \left ((b c-a d) F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )+a d \Pi \left (\frac{b (c+d)}{b c-a d};i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )\right ) \sqrt{\frac{d-d \sin (e+f x)}{c+d}} \sqrt{-\frac{\sin (e+f x) d+d}{c-d}} (-b c+a d+b (c+d \sin (e+f x)))}{b d^2 \sqrt{-\frac{1}{c+d}} (b c-a d) (a+b \sin (e+f x)) \sqrt{1-\sin ^2(e+f x)} \sqrt{-\frac{c^2-2 (c+d \sin (e+f x)) c-d^2+(c+d \sin (e+f x))^2}{d^2}}}-\frac{2 i \left (a d^2-b c d\right ) \cos (e+f x) \cos (2 (e+f x)) \left (2 b (c-d) (b c-a d) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )+d \left (\left (2 a^2-b^2\right ) d \Pi \left (\frac{b (c+d)}{b c-a d};i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )-2 (a+b) (a d-b c) F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )\right )\right ) \sqrt{\frac{d-d \sin (e+f x)}{c+d}} \sqrt{-\frac{\sin (e+f x) d+d}{c-d}} (-b c+a d+b (c+d \sin (e+f x)))}{b^2 d \sqrt{-\frac{1}{c+d}} (b c-a d) (a+b \sin (e+f x)) \sqrt{1-\sin ^2(e+f x)} \left (-2 c^2+4 (c+d \sin (e+f x)) c+d^2-2 (c+d \sin (e+f x))^2\right ) \sqrt{-\frac{c^2-2 (c+d \sin (e+f x)) c-d^2+(c+d \sin (e+f x))^2}{d^2}}}}{4 (a-b) (a+b) f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b*c*Cos[e + f*x] - a*d*Cos[e + f*x])*Sqrt[c + d*Sin[e + f*x]])/((a^2 - b^2)*f*(a + b*Sin[e + f*x])) + ((-2*(
4*a*c^2 - 5*b*c*d + a*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)*Sqrt[c + d*Sin[e + f*x]]) - ((2*I)*(4*a*c*d - 4*b*d^2)*Cos[e + f*x]*((b*c - a*d)*Ellip
ticF[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)])*Sqrt[(d - d*Sin[e + f*x]
)/(c + d)]*Sqrt[-((d + d*Sin[e + f*x])/(c - d))]*(-(b*c) + a*d + b*(c + d*Sin[e + f*x])))/(b*d^2*Sqrt[-(c + d)
^(-1)]*(b*c - a*d)*(a + b*Sin[e + f*x])*Sqrt[1 - Sin[e + f*x]^2]*Sqrt[-((c^2 - d^2 - 2*c*(c + d*Sin[e + f*x])
+ (c + d*Sin[e + f*x])^2)/d^2)]) - ((2*I)*(-(b*c*d) + a*d^2)*Cos[e + f*x]*Cos[2*(e + f*x)]*(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
)]))*Sqrt[(d - d*Sin[e + f*x])/(c + d)]*Sqrt[-((d + d*Sin[e + f*x])/(c - d))]*(-(b*c) + a*d + b*(c + d*Sin[e +
f*x])))/(b^2*d*Sqrt[-(c + d)^(-1)]*(b*c - a*d)*(a + b*Sin[e + f*x])*Sqrt[1 - Sin[e + f*x]^2]*(-2*c^2 + d^2 +
4*c*(c + d*Sin[e + f*x]) - 2*(c + d*Sin[e + f*x])^2)*Sqrt[-((c^2 - d^2 - 2*c*(c + d*Sin[e + f*x]) + (c + d*Sin
[e + f*x])^2)/d^2)]))/(4*(a - b)*(a + b)*f)

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Maple [B]  time = 4.618, size = 1027, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*(2*d^2/b^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))-4/b^3*d*(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)*((-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+a/b)*EllipticPi(((c
+d*sin(f*x+e))/(c-d))^(1/2),(-c/d+1)/(-c/d+a/b),((c-d)/(c+d))^(1/2))+1/b^2*(a^2*d^2-2*a*b*c*d+b^2*c^2)*(-b^2/(
a^3*d-a^2*b*c-a*b^2*d+b^3*c)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(a+b*sin(f*x+e))-a*d/(a^3*d-a^2*b*c-a*b^2
*d+b^3*c)*(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))-b*d/(a
^3*d-a^2*b*c-a*b^2*d+b^3*c)*(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)))+(3*a^2*d-2*a*b*c-b^2*d)/(
a^3*d-a^2*b*c-a*b^2*d+b^3*c)/b*(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+a/b)*EllipticPi(((c+d*sin(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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Sympy [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((c+d*sin(f*x+e))**(3/2)/(a+b*sin(f*x+e))**2,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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