3.599 \(\int \frac{1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=271 \[ -\frac{(c-11 d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f (c-d)^{7/2}}-\frac{d \left (3 c^2+38 c d+19 d^2\right ) \cos (e+f x)}{6 a f (c-d)^3 (c+d)^2 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{d (3 c+7 d) \cos (e+f x)}{6 a f (c-d)^2 (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}-\frac{\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}} \]

[Out]

-((c - 11*d)*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f
*x]])])/(2*Sqrt[2]*a^(3/2)*(c - d)^(7/2)*f) - Cos[e + f*x]/(2*(c - d)*f*(a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[
e + f*x])^(3/2)) - (d*(3*c + 7*d)*Cos[e + f*x])/(6*a*(c - d)^2*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e
+ f*x])^(3/2)) - (d*(3*c^2 + 38*c*d + 19*d^2)*Cos[e + f*x])/(6*a*(c - d)^3*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x
]]*Sqrt[c + d*Sin[e + f*x]])

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Rubi [A]  time = 0.854785, antiderivative size = 271, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2766, 2984, 12, 2782, 208} \[ -\frac{(c-11 d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} f (c-d)^{7/2}}-\frac{d \left (3 c^2+38 c d+19 d^2\right ) \cos (e+f x)}{6 a f (c-d)^3 (c+d)^2 \sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}-\frac{d (3 c+7 d) \cos (e+f x)}{6 a f (c-d)^2 (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}-\frac{\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x])^(5/2)),x]

[Out]

-((c - 11*d)*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f
*x]])])/(2*Sqrt[2]*a^(3/2)*(c - d)^(7/2)*f) - Cos[e + f*x]/(2*(c - d)*f*(a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[
e + f*x])^(3/2)) - (d*(3*c + 7*d)*Cos[e + f*x])/(6*a*(c - d)^2*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e
+ f*x])^(3/2)) - (d*(3*c^2 + 38*c*d + 19*d^2)*Cos[e + f*x])/(6*a*(c - d)^3*(c + d)^2*f*Sqrt[a + a*Sin[e + f*x
]]*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 2984

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*Sin[e + f*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]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2}} \, dx &=-\frac{\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}}-\frac{\int \frac{-\frac{1}{2} a (c-7 d)-2 a d \sin (e+f x)}{\sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}} \, dx}{2 a^2 (c-d)}\\ &=-\frac{\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}}-\frac{d (3 c+7 d) \cos (e+f x)}{6 a (c-d)^2 (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac{\int \frac{\frac{1}{4} a^2 \left (3 c^2-24 c d-19 d^2\right )+\frac{1}{2} a^2 d (3 c+7 d) \sin (e+f x)}{\sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx}{3 a^3 (c-d)^2 (c+d)}\\ &=-\frac{\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}}-\frac{d (3 c+7 d) \cos (e+f x)}{6 a (c-d)^2 (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac{d \left (3 c^2+38 c d+19 d^2\right ) \cos (e+f x)}{6 a (c-d)^3 (c+d)^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}-\frac{2 \int -\frac{3 a^3 (c-11 d) (c+d)^2}{8 \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx}{3 a^4 (c-d)^3 (c+d)^2}\\ &=-\frac{\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}}-\frac{d (3 c+7 d) \cos (e+f x)}{6 a (c-d)^2 (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac{d \left (3 c^2+38 c d+19 d^2\right ) \cos (e+f x)}{6 a (c-d)^3 (c+d)^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}+\frac{(c-11 d) \int \frac{1}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}} \, dx}{4 a (c-d)^3}\\ &=-\frac{\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}}-\frac{d (3 c+7 d) \cos (e+f x)}{6 a (c-d)^2 (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac{d \left (3 c^2+38 c d+19 d^2\right ) \cos (e+f x)}{6 a (c-d)^3 (c+d)^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}-\frac{(c-11 d) \operatorname{Subst}\left (\int \frac{1}{2 a^2-(a c-a d) x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\right )}{2 (c-d)^3 f}\\ &=-\frac{(c-11 d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{c-d} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} (c-d)^{7/2} f}-\frac{\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{3/2}}-\frac{d (3 c+7 d) \cos (e+f x)}{6 a (c-d)^2 (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac{d \left (3 c^2+38 c d+19 d^2\right ) \cos (e+f x)}{6 a (c-d)^3 (c+d)^2 f \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 8.99197, size = 478, normalized size = 1.76 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (\frac{3 (c-11 d) \left (\log \left (\tan \left (\frac{1}{2} (e+f x)\right )+1\right )-\log \left ((d-c) \tan \left (\frac{1}{2} (e+f x)\right )+2 \sqrt{c-d} \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{c+d \sin (e+f x)}+c-d\right )\right )}{\frac{\sec ^2\left (\frac{1}{2} (e+f x)\right )}{2 \tan \left (\frac{1}{2} (e+f x)\right )+2}-\frac{\frac{\sqrt{c-d} \left (\frac{1}{\cos (e+f x)+1}\right )^{3/2} (c \sin (e+f x)+d \cos (e+f x)+d)}{\sqrt{c+d \sin (e+f x)}}-\frac{1}{2} (c-d) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{(d-c) \tan \left (\frac{1}{2} (e+f x)\right )+2 \sqrt{c-d} \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{c+d \sin (e+f x)}+c-d}}-\frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (12 d \left (8 c^2 d+c^3+9 c d^2+2 d^3\right ) \sin (e+f x)-d^2 \left (3 c^2+38 c d+19 d^2\right ) \cos (2 (e+f x))+81 c^2 d^2+12 c^3 d+6 c^4+70 c d^3+11 d^4\right )}{(c+d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))}\right )}{12 f (c-d)^3 (a (\sin (e+f x)+1))^{3/2} \sqrt{c+d \sin (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x])^(5/2)),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*(-(((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(6*c^4 + 12*c^3*d + 81*c^2*
d^2 + 70*c*d^3 + 11*d^4 - d^2*(3*c^2 + 38*c*d + 19*d^2)*Cos[2*(e + f*x)] + 12*d*(c^3 + 8*c^2*d + 9*c*d^2 + 2*d
^3)*Sin[e + f*x]))/((c + d)^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c + d*Sin[e + f*x]))) + (3*(c - 11*d)*(Lo
g[1 + Tan[(e + f*x)/2]] - Log[c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (
-c + d)*Tan[(e + f*x)/2]]))/(Sec[(e + f*x)/2]^2/(2 + 2*Tan[(e + f*x)/2]) - (-((c - d)*Sec[(e + f*x)/2]^2)/2 +
(Sqrt[c - d]*((1 + Cos[e + f*x])^(-1))^(3/2)*(d + d*Cos[e + f*x] + c*Sin[e + f*x]))/Sqrt[c + d*Sin[e + f*x]])/
(c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2]))))/
(12*(c - d)^3*f*(a*(1 + Sin[e + f*x]))^(3/2)*Sqrt[c + d*Sin[e + f*x]])

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Maple [B]  time = 0.286, size = 5040, normalized size = 18.6 \begin{align*} \text{output 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))^(3/2)/(c+d*sin(f*x+e))^(5/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}{\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+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e) + c)^(5/2)), x)

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Fricas [B]  time = 5.95996, size = 6974, normalized size = 25.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[1/48*(3*(2*c^5 - 14*c^4*d - 76*c^3*d^2 - 124*c^2*d^3 - 86*c*d^4 - 22*d^5 + (c^3*d^2 - 9*c^2*d^3 - 21*c*d^4 -
11*d^5)*cos(f*x + e)^4 - (2*c^4*d - 17*c^3*d^2 - 51*c^2*d^3 - 43*c*d^4 - 11*d^5)*cos(f*x + e)^3 - (c^5 - 5*c^4
*d - 54*c^3*d^2 - 122*c^2*d^3 - 107*c*d^4 - 33*d^5)*cos(f*x + e)^2 + (c^5 - 7*c^4*d - 38*c^3*d^2 - 62*c^2*d^3
- 43*c*d^4 - 11*d^5)*cos(f*x + e) + (2*c^5 - 14*c^4*d - 76*c^3*d^2 - 124*c^2*d^3 - 86*c*d^4 - 22*d^5 - (c^3*d^
2 - 9*c^2*d^3 - 21*c*d^4 - 11*d^5)*cos(f*x + e)^3 - 2*(c^4*d - 8*c^3*d^2 - 30*c^2*d^3 - 32*c*d^4 - 11*d^5)*cos
(f*x + e)^2 + (c^5 - 7*c^4*d - 38*c^3*d^2 - 62*c^2*d^3 - 43*c*d^4 - 11*d^5)*cos(f*x + e))*sin(f*x + e))*sqrt(2
*a*c - 2*a*d)*log(((a*c^2 - 14*a*c*d + 17*a*d^2)*cos(f*x + e)^3 - 4*a*c^2 - 8*a*c*d - 4*a*d^2 - (13*a*c^2 - 22
*a*c*d - 3*a*d^2)*cos(f*x + e)^2 - 4*((c - 3*d)*cos(f*x + e)^2 - (3*c - d)*cos(f*x + e) + ((c - 3*d)*cos(f*x +
e) + 4*c - 4*d)*sin(f*x + e) - 4*c + 4*d)*sqrt(2*a*c - 2*a*d)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) +
c) - 2*(9*a*c^2 - 14*a*c*d + 9*a*d^2)*cos(f*x + e) - (4*a*c^2 + 8*a*c*d + 4*a*d^2 - (a*c^2 - 14*a*c*d + 17*a*d
^2)*cos(f*x + e)^2 - 2*(7*a*c^2 - 18*a*c*d + 7*a*d^2)*cos(f*x + e))*sin(f*x + e))/(cos(f*x + e)^3 + 3*cos(f*x
+ e)^2 + (cos(f*x + e)^2 - 2*cos(f*x + e) - 4)*sin(f*x + e) - 2*cos(f*x + e) - 4)) - 8*(3*c^5 - 3*c^4*d - 6*c^
3*d^2 + 6*c^2*d^3 + 3*c*d^4 - 3*d^5 - (3*c^3*d^2 + 35*c^2*d^3 - 19*c*d^4 - 19*d^5)*cos(f*x + e)^3 + (6*c^4*d +
39*c^3*d^2 - 29*c^2*d^3 - 23*c*d^4 + 7*d^5)*cos(f*x + e)^2 + 3*(c^5 + c^4*d + 12*c^3*d^2 + 4*c^2*d^3 - 13*c*d
^4 - 5*d^5)*cos(f*x + e) - (3*c^5 - 3*c^4*d - 6*c^3*d^2 + 6*c^2*d^3 + 3*c*d^4 - 3*d^5 - (3*c^3*d^2 + 35*c^2*d^
3 - 19*c*d^4 - 19*d^5)*cos(f*x + e)^2 - 6*(c^4*d + 7*c^3*d^2 + c^2*d^3 - 7*c*d^4 - 2*d^5)*cos(f*x + e))*sin(f*
x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/((a^2*c^6*d^2 - 2*a^2*c^5*d^3 - a^2*c^4*d^4 + 4*a^2
*c^3*d^5 - a^2*c^2*d^6 - 2*a^2*c*d^7 + a^2*d^8)*f*cos(f*x + e)^4 - (2*a^2*c^7*d - 3*a^2*c^6*d^2 - 4*a^2*c^5*d^
3 + 7*a^2*c^4*d^4 + 2*a^2*c^3*d^5 - 5*a^2*c^2*d^6 + a^2*d^8)*f*cos(f*x + e)^3 - (a^2*c^8 + 2*a^2*c^7*d - 6*a^2
*c^6*d^2 - 6*a^2*c^5*d^3 + 12*a^2*c^4*d^4 + 6*a^2*c^3*d^5 - 10*a^2*c^2*d^6 - 2*a^2*c*d^7 + 3*a^2*d^8)*f*cos(f*
x + e)^2 + (a^2*c^8 - 4*a^2*c^6*d^2 + 6*a^2*c^4*d^4 - 4*a^2*c^2*d^6 + a^2*d^8)*f*cos(f*x + e) + 2*(a^2*c^8 - 4
*a^2*c^6*d^2 + 6*a^2*c^4*d^4 - 4*a^2*c^2*d^6 + a^2*d^8)*f - ((a^2*c^6*d^2 - 2*a^2*c^5*d^3 - a^2*c^4*d^4 + 4*a^
2*c^3*d^5 - a^2*c^2*d^6 - 2*a^2*c*d^7 + a^2*d^8)*f*cos(f*x + e)^3 + 2*(a^2*c^7*d - a^2*c^6*d^2 - 3*a^2*c^5*d^3
+ 3*a^2*c^4*d^4 + 3*a^2*c^3*d^5 - 3*a^2*c^2*d^6 - a^2*c*d^7 + a^2*d^8)*f*cos(f*x + e)^2 - (a^2*c^8 - 4*a^2*c^
6*d^2 + 6*a^2*c^4*d^4 - 4*a^2*c^2*d^6 + a^2*d^8)*f*cos(f*x + e) - 2*(a^2*c^8 - 4*a^2*c^6*d^2 + 6*a^2*c^4*d^4 -
4*a^2*c^2*d^6 + a^2*d^8)*f)*sin(f*x + e)), -1/24*(3*(2*c^5 - 14*c^4*d - 76*c^3*d^2 - 124*c^2*d^3 - 86*c*d^4 -
22*d^5 + (c^3*d^2 - 9*c^2*d^3 - 21*c*d^4 - 11*d^5)*cos(f*x + e)^4 - (2*c^4*d - 17*c^3*d^2 - 51*c^2*d^3 - 43*c
*d^4 - 11*d^5)*cos(f*x + e)^3 - (c^5 - 5*c^4*d - 54*c^3*d^2 - 122*c^2*d^3 - 107*c*d^4 - 33*d^5)*cos(f*x + e)^2
+ (c^5 - 7*c^4*d - 38*c^3*d^2 - 62*c^2*d^3 - 43*c*d^4 - 11*d^5)*cos(f*x + e) + (2*c^5 - 14*c^4*d - 76*c^3*d^2
- 124*c^2*d^3 - 86*c*d^4 - 22*d^5 - (c^3*d^2 - 9*c^2*d^3 - 21*c*d^4 - 11*d^5)*cos(f*x + e)^3 - 2*(c^4*d - 8*c
^3*d^2 - 30*c^2*d^3 - 32*c*d^4 - 11*d^5)*cos(f*x + e)^2 + (c^5 - 7*c^4*d - 38*c^3*d^2 - 62*c^2*d^3 - 43*c*d^4
- 11*d^5)*cos(f*x + e))*sin(f*x + e))*sqrt(-2*a*c + 2*a*d)*arctan(1/4*sqrt(-2*a*c + 2*a*d)*sqrt(a*sin(f*x + e)
+ a)*((c - 3*d)*sin(f*x + e) - 3*c + d)*sqrt(d*sin(f*x + e) + c)/((a*c*d - a*d^2)*cos(f*x + e)*sin(f*x + e) +
(a*c^2 - a*c*d)*cos(f*x + e))) + 4*(3*c^5 - 3*c^4*d - 6*c^3*d^2 + 6*c^2*d^3 + 3*c*d^4 - 3*d^5 - (3*c^3*d^2 +
35*c^2*d^3 - 19*c*d^4 - 19*d^5)*cos(f*x + e)^3 + (6*c^4*d + 39*c^3*d^2 - 29*c^2*d^3 - 23*c*d^4 + 7*d^5)*cos(f*
x + e)^2 + 3*(c^5 + c^4*d + 12*c^3*d^2 + 4*c^2*d^3 - 13*c*d^4 - 5*d^5)*cos(f*x + e) - (3*c^5 - 3*c^4*d - 6*c^3
*d^2 + 6*c^2*d^3 + 3*c*d^4 - 3*d^5 - (3*c^3*d^2 + 35*c^2*d^3 - 19*c*d^4 - 19*d^5)*cos(f*x + e)^2 - 6*(c^4*d +
7*c^3*d^2 + c^2*d^3 - 7*c*d^4 - 2*d^5)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e
) + c))/((a^2*c^6*d^2 - 2*a^2*c^5*d^3 - a^2*c^4*d^4 + 4*a^2*c^3*d^5 - a^2*c^2*d^6 - 2*a^2*c*d^7 + a^2*d^8)*f*c
os(f*x + e)^4 - (2*a^2*c^7*d - 3*a^2*c^6*d^2 - 4*a^2*c^5*d^3 + 7*a^2*c^4*d^4 + 2*a^2*c^3*d^5 - 5*a^2*c^2*d^6 +
a^2*d^8)*f*cos(f*x + e)^3 - (a^2*c^8 + 2*a^2*c^7*d - 6*a^2*c^6*d^2 - 6*a^2*c^5*d^3 + 12*a^2*c^4*d^4 + 6*a^2*c
^3*d^5 - 10*a^2*c^2*d^6 - 2*a^2*c*d^7 + 3*a^2*d^8)*f*cos(f*x + e)^2 + (a^2*c^8 - 4*a^2*c^6*d^2 + 6*a^2*c^4*d^4
- 4*a^2*c^2*d^6 + a^2*d^8)*f*cos(f*x + e) + 2*(a^2*c^8 - 4*a^2*c^6*d^2 + 6*a^2*c^4*d^4 - 4*a^2*c^2*d^6 + a^2*
d^8)*f - ((a^2*c^6*d^2 - 2*a^2*c^5*d^3 - a^2*c^4*d^4 + 4*a^2*c^3*d^5 - a^2*c^2*d^6 - 2*a^2*c*d^7 + a^2*d^8)*f*
cos(f*x + e)^3 + 2*(a^2*c^7*d - a^2*c^6*d^2 - 3*a^2*c^5*d^3 + 3*a^2*c^4*d^4 + 3*a^2*c^3*d^5 - 3*a^2*c^2*d^6 -
a^2*c*d^7 + a^2*d^8)*f*cos(f*x + e)^2 - (a^2*c^8 - 4*a^2*c^6*d^2 + 6*a^2*c^4*d^4 - 4*a^2*c^2*d^6 + a^2*d^8)*f*
cos(f*x + e) - 2*(a^2*c^8 - 4*a^2*c^6*d^2 + 6*a^2*c^4*d^4 - 4*a^2*c^2*d^6 + a^2*d^8)*f)*sin(f*x + e))]

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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+a*sin(f*x+e))**(3/2)/(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 (a \sin \left (f x + e\right ) + a\right )}^{\frac{3}{2}}{\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+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e) + c)^(5/2)), x)