3.6.15 \(\int \frac {1}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{5/2}} \, dx\) [515]

Optimal. Leaf size=405 \[ -\frac {d \left (c^2-7 c d-10 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {(c-7 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))^{3/2}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2}}-\frac {d (c+3 d) \left (c^2-10 c d-7 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^4 (c+d)^2 f \sqrt {c+d \sin (e+f x)}}-\frac {(c+3 d) \left (c^2-10 c d-7 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{3 a^2 (c-d)^4 (c+d)^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (c^2-7 c d-10 d^2\right ) F\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{3 a^2 (c-d)^3 (c+d) f \sqrt {c+d \sin (e+f x)}} \]

[Out]

-1/3*d*(c^2-7*c*d-10*d^2)*cos(f*x+e)/a^2/(c-d)^3/(c+d)/f/(c+d*sin(f*x+e))^(3/2)-1/3*(c-7*d)*cos(f*x+e)/a^2/(c-
d)^2/f/(1+sin(f*x+e))/(c+d*sin(f*x+e))^(3/2)-1/3*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(3/2)-
1/3*d*(c+3*d)*(c^2-10*c*d-7*d^2)*cos(f*x+e)/a^2/(c-d)^4/(c+d)^2/f/(c+d*sin(f*x+e))^(1/2)+1/3*(c+3*d)*(c^2-10*c
*d-7*d^2)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticE(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^2/(c-d)^4/(c+d)^2/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-1/3*(c^2-7*
c*d-10*d^2)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticF(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))/(c+d))^(1/2)/a^2/(c-d)^3/(c+d)/f/(c+d*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.56, antiderivative size = 405, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2845, 3057, 2833, 2831, 2742, 2740, 2734, 2732} \begin {gather*} -\frac {d (c+3 d) \left (c^2-10 c d-7 d^2\right ) \cos (e+f x)}{3 a^2 f (c-d)^4 (c+d)^2 \sqrt {c+d \sin (e+f x)}}-\frac {d \left (c^2-7 c d-10 d^2\right ) \cos (e+f x)}{3 a^2 f (c-d)^3 (c+d) (c+d \sin (e+f x))^{3/2}}+\frac {\left (c^2-7 c d-10 d^2\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 )}{3 a^2 f (c-d)^3 (c+d) \sqrt {c+d \sin (e+f x)}}-\frac {(c+3 d) \left (c^2-10 c d-7 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)^4 (c+d)^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {(c-7 d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1) (c+d \sin (e+f x))^{3/2}}-\frac {\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(d*(c^2 - 7*c*d - 10*d^2)*Cos[e + f*x])/(a^2*(c - d)^3*(c + d)*f*(c + d*Sin[e + f*x])^(3/2)) - ((c - 7*d)
*Cos[e + f*x])/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])*(c + d*Sin[e + f*x])^(3/2)) - Cos[e + f*x]/(3*(c - d)*f*(
a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^(3/2)) - (d*(c + 3*d)*(c^2 - 10*c*d - 7*d^2)*Cos[e + f*x])/(3*a^2*(
c - d)^4*(c + d)^2*f*Sqrt[c + d*Sin[e + f*x]]) - ((c + 3*d)*(c^2 - 10*c*d - 7*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)^4*(c + d)^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) +
((c^2 - 7*c*d - 10*d^2)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*a
^2*(c - d)^3*(c + d)*f*Sqrt[c + d*Sin[e + f*x]])

Rule 2732

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]

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

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

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2831

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 2833

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 2845

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 3057

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])

Rubi steps

\begin {align*} \int \frac {1}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{5/2}} \, dx &=-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2}}-\frac {\int \frac {-\frac {1}{2} a (2 c-9 d)-\frac {5}{2} a d \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^{5/2}} \, dx}{3 a^2 (c-d)}\\ &=-\frac {(c-7 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))^{3/2}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2}}+\frac {\int \frac {15 a^2 d^2+\frac {3}{2} a^2 (c-7 d) d \sin (e+f x)}{(c+d \sin (e+f x))^{5/2}} \, dx}{3 a^4 (c-d)^2}\\ &=-\frac {d \left (c^2-7 c d-10 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {(c-7 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))^{3/2}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2}}-\frac {2 \int \frac {-\frac {9}{4} a^2 d^2 (9 c+7 d)-\frac {3}{4} a^2 d \left (c^2-7 c d-10 d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{9 a^4 (c-d)^3 (c+d)}\\ &=-\frac {d \left (c^2-7 c d-10 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {(c-7 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))^{3/2}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2}}-\frac {d (c+3 d) \left (c^2-10 c d-7 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^4 (c+d)^2 f \sqrt {c+d \sin (e+f x)}}+\frac {4 \int \frac {\frac {3}{4} a^2 d^2 \left (13 c^2+14 c d+5 d^2\right )-\frac {3}{8} a^2 d (c+3 d) \left (c^2-10 c d-7 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{9 a^4 (c-d)^4 (c+d)^2}\\ &=-\frac {d \left (c^2-7 c d-10 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {(c-7 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))^{3/2}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2}}-\frac {d (c+3 d) \left (c^2-10 c d-7 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^4 (c+d)^2 f \sqrt {c+d \sin (e+f x)}}+\frac {\left (c^2-7 c d-10 d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{6 a^2 (c-d)^3 (c+d)}-\frac {\left ((c+3 d) \left (c^2-10 c d-7 d^2\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{6 a^2 (c-d)^4 (c+d)^2}\\ &=-\frac {d \left (c^2-7 c d-10 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {(c-7 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))^{3/2}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2}}-\frac {d (c+3 d) \left (c^2-10 c d-7 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^4 (c+d)^2 f \sqrt {c+d \sin (e+f x)}}-\frac {\left ((c+3 d) \left (c^2-10 c d-7 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)^4 (c+d)^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (\left (c^2-7 c d-10 d^2\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}{6 a^2 (c-d)^3 (c+d) \sqrt {c+d \sin (e+f x)}}\\ &=-\frac {d \left (c^2-7 c d-10 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {(c-7 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))^{3/2}}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2}}-\frac {d (c+3 d) \left (c^2-10 c d-7 d^2\right ) \cos (e+f x)}{3 a^2 (c-d)^4 (c+d)^2 f \sqrt {c+d \sin (e+f x)}}-\frac {(c+3 d) \left (c^2-10 c d-7 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)^4 (c+d)^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (c^2-7 c d-10 d^2\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}}}{3 a^2 (c-d)^3 (c+d) f \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 6.54, size = 674, normalized size = 1.66 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sqrt {c+d \sin (e+f x)} \left (-\frac {2 \left (c^3-7 c^2 d-27 c d^2-15 d^3\right )}{3 (c-d)^4 (c+d)^2}+\frac {2 \sin \left (\frac {1}{2} (e+f x)\right )}{3 (c-d)^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}-\frac {1}{3 (c-d)^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {2 \left (c \sin \left (\frac {1}{2} (e+f x)\right )-9 d \sin \left (\frac {1}{2} (e+f x)\right )\right )}{3 (c-d)^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}+\frac {2 d^3 \cos (e+f x)}{3 (c-d)^3 (c+d) (c+d \sin (e+f x))^2}+\frac {4 \left (5 c d^3 \cos (e+f x)+3 d^4 \cos (e+f x)\right )}{3 (c-d)^4 (c+d)^2 (c+d \sin (e+f x))}\right )}{f (a+a \sin (e+f x))^2}+\frac {d \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (-\frac {2 \left (26 c^2 d+28 c d^2+10 d^3\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}}}{\sqrt {c+d \sin (e+f x)}}+\frac {2 \left (c^3-7 c^2 d-37 c d^2-21 d^3\right ) \cos ^2(e+f x) \sqrt {c+d \sin (e+f x)}}{d \left (1-\sin ^2(e+f x)\right )}-\frac {\left (-c^3+7 c^2 d+37 c d^2+21 d^3\right ) \left (\frac {2 (c+d) E\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}}}{\sqrt {c+d \sin (e+f x)}}-\frac {2 c 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}}}{\sqrt {c+d \sin (e+f x)}}\right )}{d}\right )}{6 (c-d)^4 (c+d)^2 f (a+a \sin (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*Sqrt[c + d*Sin[e + f*x]]*((-2*(c^3 - 7*c^2*d - 27*c*d^2 - 15*d^3))/(3
*(c - d)^4*(c + d)^2) + (2*Sin[(e + f*x)/2])/(3*(c - d)^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3) - 1/(3*(c -
d)^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2) + (2*(c*Sin[(e + f*x)/2] - 9*d*Sin[(e + f*x)/2]))/(3*(c - d)^4*
(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])) + (2*d^3*Cos[e + f*x])/(3*(c - d)^3*(c + d)*(c + d*Sin[e + f*x])^2) + (
4*(5*c*d^3*Cos[e + f*x] + 3*d^4*Cos[e + f*x]))/(3*(c - d)^4*(c + d)^2*(c + d*Sin[e + f*x]))))/(f*(a + a*Sin[e
+ f*x])^2) + (d*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*((-2*(26*c^2*d + 28*c*d^2 + 10*d^3)*EllipticF[(-e + Pi
/2 - f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/Sqrt[c + d*Sin[e + f*x]] + (2*(c^3 - 7*c^2*d -
37*c*d^2 - 21*d^3)*Cos[e + f*x]^2*Sqrt[c + d*Sin[e + f*x]])/(d*(1 - Sin[e + f*x]^2)) - ((-c^3 + 7*c^2*d + 37*
c*d^2 + 21*d^3)*((2*(c + d)*EllipticE[(-e + Pi/2 - f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/
Sqrt[c + d*Sin[e + f*x]] - (2*c*EllipticF[(-e + Pi/2 - f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d
)])/Sqrt[c + d*Sin[e + f*x]]))/d))/(6*(c - d)^4*(c + d)^2*f*(a + a*Sin[e + f*x])^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1757\) vs. \(2(443)=886\).
time = 33.35, size = 1758, normalized size = 4.34

method	result	size
default	\(\text {Expression too large to display}\)	\(1758\)

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))^(5/2),x,method=_RETURNVERBOSE)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/a^2*(1/(c-d)^2*(-1/3/(c-d)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(1+
sin(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)*(sin(f*x+e)
-1)*(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)*((-1-sin(f*x+e))*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)*((-1-sin(f*x+e))*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*(2/3/(c^2-d^2)/d*(-(-d*sin(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*s
in(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-1-sin(f*x+e))*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)*((-1-sin(f*x+e))*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*s
in(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))))-2/(c-d)^3*d*(-(-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)*(sin(f*x+e)-1)*(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)*((-1-sin(f*x+e))*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*Ellip
ticF(((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)*((-1-sin(f*x+e))*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*Ell
ipticE(((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*d^2/(c-d)^3*(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)*((-1-sin(f*x+e))*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)*((-1-sin(f*x+e))*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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.34, size = 3328, normalized size = 8.22 \begin {gather*} \text {Too large to display} \end {gather*}

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))^(5/2),x, algorithm="fricas")

[Out]

1/18*(2*(sqrt(2)*(c^4*d^2 - 7*c^3*d^3 + 2*c^2*d^4 + 21*c*d^5 + 15*d^6)*cos(f*x + e)^4 - sqrt(2)*(2*c^5*d - 13*
c^4*d^2 - 3*c^3*d^3 + 44*c^2*d^4 + 51*c*d^5 + 15*d^6)*cos(f*x + e)^3 - sqrt(2)*(c^6 - 3*c^5*d - 23*c^4*d^2 + 8
*c^3*d^3 + 105*c^2*d^4 + 123*c*d^5 + 45*d^6)*cos(f*x + e)^2 + sqrt(2)*(c^6 - 5*c^5*d - 11*c^4*d^2 + 18*c^3*d^3
+ 59*c^2*d^4 + 51*c*d^5 + 15*d^6)*cos(f*x + e) - (sqrt(2)*(c^4*d^2 - 7*c^3*d^3 + 2*c^2*d^4 + 21*c*d^5 + 15*d^
6)*cos(f*x + e)^3 + 2*sqrt(2)*(c^5*d - 6*c^4*d^2 - 5*c^3*d^3 + 23*c^2*d^4 + 36*c*d^5 + 15*d^6)*cos(f*x + e)^2
- sqrt(2)*(c^6 - 5*c^5*d - 11*c^4*d^2 + 18*c^3*d^3 + 59*c^2*d^4 + 51*c*d^5 + 15*d^6)*cos(f*x + e) - 2*sqrt(2)*
(c^6 - 5*c^5*d - 11*c^4*d^2 + 18*c^3*d^3 + 59*c^2*d^4 + 51*c*d^5 + 15*d^6))*sin(f*x + e) + 2*sqrt(2)*(c^6 - 5*
c^5*d - 11*c^4*d^2 + 18*c^3*d^3 + 59*c^2*d^4 + 51*c*d^5 + 15*d^6))*sqrt(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) + 2*(sqrt
(2)*(c^4*d^2 - 7*c^3*d^3 + 2*c^2*d^4 + 21*c*d^5 + 15*d^6)*cos(f*x + e)^4 - sqrt(2)*(2*c^5*d - 13*c^4*d^2 - 3*c
^3*d^3 + 44*c^2*d^4 + 51*c*d^5 + 15*d^6)*cos(f*x + e)^3 - sqrt(2)*(c^6 - 3*c^5*d - 23*c^4*d^2 + 8*c^3*d^3 + 10
5*c^2*d^4 + 123*c*d^5 + 45*d^6)*cos(f*x + e)^2 + sqrt(2)*(c^6 - 5*c^5*d - 11*c^4*d^2 + 18*c^3*d^3 + 59*c^2*d^4
+ 51*c*d^5 + 15*d^6)*cos(f*x + e) - (sqrt(2)*(c^4*d^2 - 7*c^3*d^3 + 2*c^2*d^4 + 21*c*d^5 + 15*d^6)*cos(f*x +
e)^3 + 2*sqrt(2)*(c^5*d - 6*c^4*d^2 - 5*c^3*d^3 + 23*c^2*d^4 + 36*c*d^5 + 15*d^6)*cos(f*x + e)^2 - sqrt(2)*(c^
6 - 5*c^5*d - 11*c^4*d^2 + 18*c^3*d^3 + 59*c^2*d^4 + 51*c*d^5 + 15*d^6)*cos(f*x + e) - 2*sqrt(2)*(c^6 - 5*c^5*
d - 11*c^4*d^2 + 18*c^3*d^3 + 59*c^2*d^4 + 51*c*d^5 + 15*d^6))*sin(f*x + e) + 2*sqrt(2)*(c^6 - 5*c^5*d - 11*c^
4*d^2 + 18*c^3*d^3 + 59*c^2*d^4 + 51*c*d^5 + 15*d^6))*sqrt(-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) + 3*(sqrt(2)*(I*c^3*
d^3 - 7*I*c^2*d^4 - 37*I*c*d^5 - 21*I*d^6)*cos(f*x + e)^4 + sqrt(2)*(-2*I*c^4*d^2 + 13*I*c^3*d^3 + 81*I*c^2*d^
4 + 79*I*c*d^5 + 21*I*d^6)*cos(f*x + e)^3 + sqrt(2)*(-I*c^5*d + 3*I*c^4*d^2 + 62*I*c^3*d^3 + 190*I*c^2*d^4 + 1
95*I*c*d^5 + 63*I*d^6)*cos(f*x + e)^2 + sqrt(2)*(I*c^5*d - 5*I*c^4*d^2 - 50*I*c^3*d^3 - 102*I*c^2*d^4 - 79*I*c
*d^5 - 21*I*d^6)*cos(f*x + e) + (sqrt(2)*(-I*c^3*d^3 + 7*I*c^2*d^4 + 37*I*c*d^5 + 21*I*d^6)*cos(f*x + e)^3 + 2
*sqrt(2)*(-I*c^4*d^2 + 6*I*c^3*d^3 + 44*I*c^2*d^4 + 58*I*c*d^5 + 21*I*d^6)*cos(f*x + e)^2 + sqrt(2)*(I*c^5*d -
5*I*c^4*d^2 - 50*I*c^3*d^3 - 102*I*c^2*d^4 - 79*I*c*d^5 - 21*I*d^6)*cos(f*x + e) + 2*sqrt(2)*(I*c^5*d - 5*I*c
^4*d^2 - 50*I*c^3*d^3 - 102*I*c^2*d^4 - 79*I*c*d^5 - 21*I*d^6))*sin(f*x + e) + 2*sqrt(2)*(I*c^5*d - 5*I*c^4*d^
2 - 50*I*c^3*d^3 - 102*I*c^2*d^4 - 79*I*c*d^5 - 21*I*d^6))*sqrt(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*(sqrt(2)*(-I*c^3*d^3 + 7*I*c^2*d^4 + 37*I*c*d^5
+ 21*I*d^6)*cos(f*x + e)^4 + sqrt(2)*(2*I*c^4*d^2 - 13*I*c^3*d^3 - 81*I*c^2*d^4 - 79*I*c*d^5 - 21*I*d^6)*cos(
f*x + e)^3 + sqrt(2)*(I*c^5*d - 3*I*c^4*d^2 - 62*I*c^3*d^3 - 190*I*c^2*d^4 - 195*I*c*d^5 - 63*I*d^6)*cos(f*x +
e)^2 + sqrt(2)*(-I*c^5*d + 5*I*c^4*d^2 + 50*I*c^3*d^3 + 102*I*c^2*d^4 + 79*I*c*d^5 + 21*I*d^6)*cos(f*x + e) +
(sqrt(2)*(I*c^3*d^3 - 7*I*c^2*d^4 - 37*I*c*d^5 - 21*I*d^6)*cos(f*x + e)^3 + 2*sqrt(2)*(I*c^4*d^2 - 6*I*c^3*d^
3 - 44*I*c^2*d^4 - 58*I*c*d^5 - 21*I*d^6)*cos(f*x + e)^2 + sqrt(2)*(-I*c^5*d + 5*I*c^4*d^2 + 50*I*c^3*d^3 + 10
2*I*c^2*d^4 + 79*I*c*d^5 + 21*I*d^6)*cos(f*x + e) + 2*sqrt(2)*(-I*c^5*d + 5*I*c^4*d^2 + 50*I*c^3*d^3 + 102*I*c
^2*d^4 + 79*I*c*d^5 + 21*I*d^6))*sin(f*x + e) + 2*sqrt(2)*(-I*c^5*d + 5*I*c^4*d^2 + 50*I*c^3*d^3 + 102*I*c^2*d
^4 + 79*I*c*d^5 + 21*I*d^6))*sqrt(-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*(c^5*d - c^4*d^2 - 2*c^3*d^3 + 2*c^2*d^4 + c*d^5 - d^6 - (c^3*d^3 - 7*c^2*
d^4 - 37*c*d^5 - 21*d^6)*cos(f*x + e)^4 - 2*(c^4*d^2 - 6*c^3*d^3 - 31*c^2*d^4 - 44*c*d^5 - 16*d^6)*cos(f*x + e
)^3 + (c^5*d - 5*c^4*d^2 - 15*c^3*d^3 - 41*c^2*d^4 - 50*c*d^5 - 18*d^6)*cos(f*x + e)^2 + 2*(c^5*d - 2*c^4*d^2
- 15*c^3*d^3 - 47*c^2*d^4 - 50*c*d^5 - 15*d^6)*cos(f*x + e) - (c^5*d - c^4*d^2 - 2*c^3*d^3 + 2*c^2*d^4 + c*d^5
- d^6 + (c^3*d^3 - 7*c^2*d^4 - 37*c*d^5 - 21*d^6)*cos(f*x + e)^3 - (2*c^4*d^2 - 13*c^3*d^3 - 55*c^2*d^4 - 51*
c*d^5 - 11*d^6)*cos(f*x + e)^2 - (c^5*d - 3*c^4*d^2 - 28*c^3*d^3 - 96*c^2*d^4 - 101*c*d^5 - 29*d^6)*cos(f*x +
e))*sin(f*x + e))*sqrt(d*sin(f*x + e) + c))/((a^2*c^6*d^3 - 2*a^2*c^5*d^4 - a^2*c^4*d^5 + 4*a^2*c^3*d^6 - a^2*
c^2*d^7 - 2*a^2*c*d^8 + a^2*d^9)*f*cos(f*x + e)^4 - (2*a^2*c^7*d^2 - 3*a^2*c^6*d^3 - 4*a^2*c^5*d^4 + 7*a^2*c^4
*d^5 + 2*a^2*c^3*d^6 - 5*a^2*c^2*d^7 + a^2*d^9)...

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {1}{c^{2} \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )} + 2 c^{2} \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )} + c^{2} \sqrt {c + d \sin {\left (e + f x \right )}} + 2 c d \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{3}{\left (e + f x \right )} + 4 c d \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )} + 2 c d \sqrt {c + d \sin {\left (e + f x \right )}} \sin {\left (e + f x \right )} + d^{2} \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{4}{\left (e + f x \right )} + 2 d^{2} \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{3}{\left (e + f x \right )} + d^{2} \sqrt {c + d \sin {\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )}}\, dx}{a^{2}} \end {gather*}

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))**(5/2),x)

[Out]

Integral(1/(c**2*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**2 + 2*c**2*sqrt(c + d*sin(e + f*x))*sin(e + f*x) + c**
2*sqrt(c + d*sin(e + f*x)) + 2*c*d*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**3 + 4*c*d*sqrt(c + d*sin(e + f*x))*s
in(e + f*x)**2 + 2*c*d*sqrt(c + d*sin(e + f*x))*sin(e + f*x) + d**2*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**4 +
2*d**2*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**3 + d**2*sqrt(c + d*sin(e + f*x))*sin(e + f*x)**2), x)/a**2

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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