3.750 \(\int \frac{1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=399 \[ \frac{2 b^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{f (a+b) (b c-a d)^2 \sqrt{c+d \sin (e+f x)}}-\frac{2 d^2 \left (-4 a c d+7 b c^2-3 b d^2\right ) \cos (e+f x)}{3 f \left (c^2-d^2\right )^2 (b c-a d)^2 \sqrt{c+d \sin (e+f x)}}-\frac{2 d^2 \cos (e+f x)}{3 f \left (c^2-d^2\right ) (b c-a d) (c+d \sin (e+f x))^{3/2}}+\frac{2 d \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 f \left (c^2-d^2\right ) (b c-a d) \sqrt{c+d \sin (e+f x)}}-\frac{2 d \left (-4 a c d+7 b c^2-3 b d^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 f \left (c^2-d^2\right )^2 (b c-a d)^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]

[Out]

(-2*d^2*Cos[e + f*x])/(3*(b*c - a*d)*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^(3/2)) - (2*d^2*(7*b*c^2 - 4*a*c*d - 3
*b*d^2)*Cos[e + f*x])/(3*(b*c - a*d)^2*(c^2 - d^2)^2*f*Sqrt[c + d*Sin[e + f*x]]) - (2*d*(7*b*c^2 - 4*a*c*d - 3
*b*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(3*(b*c - a*d)^2*(c^2 - d^2)^2*
f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (2*d*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e +
f*x])/(c + d)])/(3*(b*c - a*d)*(c^2 - d^2)*f*Sqrt[c + d*Sin[e + f*x]]) + (2*b^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)*(b*c - a*d)^2*f*Sqrt[c + d*Sin[e +
f*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.65294, antiderivative size = 399, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.37, Rules used = {2802, 3055, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ \frac{2 b^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{f (a+b) (b c-a d)^2 \sqrt{c+d \sin (e+f x)}}-\frac{2 d^2 \left (-4 a c d+7 b c^2-3 b d^2\right ) \cos (e+f x)}{3 f \left (c^2-d^2\right )^2 (b c-a d)^2 \sqrt{c+d \sin (e+f x)}}-\frac{2 d^2 \cos (e+f x)}{3 f \left (c^2-d^2\right ) (b c-a d) (c+d \sin (e+f x))^{3/2}}+\frac{2 d \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 f \left (c^2-d^2\right ) (b c-a d) \sqrt{c+d \sin (e+f x)}}-\frac{2 d \left (-4 a c d+7 b c^2-3 b d^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 f \left (c^2-d^2\right )^2 (b c-a d)^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^(5/2)),x]

[Out]

(-2*d^2*Cos[e + f*x])/(3*(b*c - a*d)*(c^2 - d^2)*f*(c + d*Sin[e + f*x])^(3/2)) - (2*d^2*(7*b*c^2 - 4*a*c*d - 3
*b*d^2)*Cos[e + f*x])/(3*(b*c - a*d)^2*(c^2 - d^2)^2*f*Sqrt[c + d*Sin[e + f*x]]) - (2*d*(7*b*c^2 - 4*a*c*d - 3
*b*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(3*(b*c - a*d)^2*(c^2 - d^2)^2*
f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (2*d*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e +
f*x])/(c + d)])/(3*(b*c - a*d)*(c^2 - d^2)*f*Sqrt[c + d*Sin[e + f*x]]) + (2*b^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)*(b*c - a*d)^2*f*Sqrt[c + d*Sin[e +
f*x]])

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(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] + Dist[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)*Sin[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])))

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*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] + Dis
t[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[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

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

Mathematica [C]  time = 7.17685, size = 1079, normalized size = 2.7 \[ \frac{\sqrt{c+d \sin (e+f x)} \left (\frac{2 \left (3 b \cos (e+f x) d^4+4 a c \cos (e+f x) d^3-7 b c^2 \cos (e+f x) d^2\right )}{3 (b c-a d)^2 \left (c^2-d^2\right )^2 (c+d \sin (e+f x))}-\frac{2 d^2 \cos (e+f x)}{3 (b c-a d) \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}\right )}{f}+\frac{-\frac{2 \left (6 b^2 c^4-12 a b d c^3+6 a^2 d^2 c^2-19 b^2 d^2 c^2+8 a b d^3 c+2 a^2 d^4+9 b^2 d^4\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 (8 a b d^4+8 a^2 c d^3+4 b^2 c d^3-8 a b c^2 d^2-12 b^2 c^3 d\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 (-3 b^2 d^4-4 a b c d^3+7 b^2 c^2 d^2\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}}}}{6 (c-d)^2 (c+d)^2 (b c-a d)^2 f} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])^(5/2)),x]

[Out]

(Sqrt[c + d*Sin[e + f*x]]*((-2*d^2*Cos[e + f*x])/(3*(b*c - a*d)*(c^2 - d^2)*(c + d*Sin[e + f*x])^2) + (2*(-7*b
*c^2*d^2*Cos[e + f*x] + 4*a*c*d^3*Cos[e + f*x] + 3*b*d^4*Cos[e + f*x]))/(3*(b*c - a*d)^2*(c^2 - d^2)^2*(c + d*
Sin[e + f*x]))))/f + ((-2*(6*b^2*c^4 - 12*a*b*c^3*d + 6*a^2*c^2*d^2 - 19*b^2*c^2*d^2 + 8*a*b*c*d^3 + 2*a^2*d^4
+ 9*b^2*d^4)*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)*(-12*b^2*c^3*d - 8*a*b*c^2*d^2 + 8*a^2*c*d^3 + 4*b^2*c*d^3 + 8*a
*b*d^4)*Cos[e + f*x]*((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)])*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)*(7*b^2*c^2*d^2 - 4*a*b*c*d^3
- 3*b^2*d^4)*Cos[e + f*x]*Cos[2*(e + f*x)]*(2*b*(c - d)*(b*c - a*d)*EllipticE[I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sq
rt[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)]))/(6*(c - d)^2*(c + d)^2*(b*
c - a*d)^2*f)

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Maple [B]  time = 5.758, size = 1072, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(5/2),x)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*(2*b/(a*d-b*c)^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+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))+d/(a*d-b*c)*(2/3/(c^2-d^2)/d*(-(-d*si
n(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^2+8/3*d*cos(f*x+e)^2/(c^2-d^2)^2*c/(-(-d*sin(f*x+e)-c)*cos(f*
x+e)^2)^(1/2)+2*(3*c^2+d^2)/(3*c^4-6*c^2*d^2+3*d^4)*(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))+8/3*c*d/(c^2-d^2)^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*b/(a*d-b*c)^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
)*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)))))/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+b*sin(f*x+e))/(c+d*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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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(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(5/2),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(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sin \left (f x + e\right ) + a\right )}{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((b*sin(f*x + e) + a)*(d*sin(f*x + e) + c)^(5/2)), x)