3.542 \(\int \frac{\cot ^4(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=348 \[ \frac{8 (a+2 b) \cot (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 a^4 f}-\frac{(3 a+8 b) \cot (e+f x) \csc ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 a^3 b f}+\frac{2 (a+3 b) \cot (e+f x) \csc ^2(e+f x)}{3 a^2 b f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(5 a+8 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 a^3 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{8 (a+2 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 a^4 f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}+\frac{(a+b) \cot (e+f x) \csc ^2(e+f x)}{3 a b f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[Out]

((a + b)*Cot[e + f*x]*Csc[e + f*x]^2)/(3*a*b*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (2*(a + 3*b)*Cot[e + f*x]*Csc[e
+ f*x]^2)/(3*a^2*b*f*Sqrt[a + b*Sin[e + f*x]^2]) + (8*(a + 2*b)*Cot[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*a
^4*f) - ((3*a + 8*b)*Cot[e + f*x]*Csc[e + f*x]^2*Sqrt[a + b*Sin[e + f*x]^2])/(3*a^3*b*f) + (8*(a + 2*b)*Sqrt[C
os[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*a^4*f*Sqrt[
1 + (b*Sin[e + f*x]^2)/a]) - ((5*a + 8*b)*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e +
f*x]*Sqrt[1 + (b*Sin[e + f*x]^2)/a])/(3*a^3*f*Sqrt[a + b*Sin[e + f*x]^2])

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Rubi [A]  time = 0.522608, antiderivative size = 348, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {3196, 468, 579, 583, 524, 426, 424, 421, 419} \[ \frac{8 (a+2 b) \cot (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 a^4 f}-\frac{(3 a+8 b) \cot (e+f x) \csc ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 a^3 b f}+\frac{2 (a+3 b) \cot (e+f x) \csc ^2(e+f x)}{3 a^2 b f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(5 a+8 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 a^3 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{8 (a+2 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 a^4 f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}+\frac{(a+b) \cot (e+f x) \csc ^2(e+f x)}{3 a b f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^4/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

((a + b)*Cot[e + f*x]*Csc[e + f*x]^2)/(3*a*b*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (2*(a + 3*b)*Cot[e + f*x]*Csc[e
+ f*x]^2)/(3*a^2*b*f*Sqrt[a + b*Sin[e + f*x]^2]) + (8*(a + 2*b)*Cot[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*a
^4*f) - ((3*a + 8*b)*Cot[e + f*x]*Csc[e + f*x]^2*Sqrt[a + b*Sin[e + f*x]^2])/(3*a^3*b*f) + (8*(a + 2*b)*Sqrt[C
os[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*a^4*f*Sqrt[
1 + (b*Sin[e + f*x]^2)/a]) - ((5*a + 8*b)*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e +
f*x]*Sqrt[1 + (b*Sin[e + f*x]^2)/a])/(3*a^3*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 3196

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[(ff^(m + 1)*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]), Subst[Int[(x^m*(a + b*ff^2*
x^2)^p)/(1 - ff^2*x^2)^((m + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2]
&&  !IntegerQ[p]

Rule 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[((c*b -
a*d)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(a*b*e*n*(p + 1)), x] + Dist[1/(a*b*n*(p + 1)), I
nt[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c*b - a*d)*(m + 1)) + d*(c*b*n*(p
+ 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 579

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*g*n*(b*c - a*d)*(p +
1)), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)*(
m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 426

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

Rule 424

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

Rule 421

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

Rule 419

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

Rubi steps

\begin{align*} \int \frac{\cot ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{3/2}}{x^4 \left (a+b x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{(a+b) \cot (e+f x) \csc ^2(e+f x)}{3 a b f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{-3 (a+2 b)+(2 a+5 b) x^2}{x^4 \sqrt{1-x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 a b f}\\ &=\frac{(a+b) \cot (e+f x) \csc ^2(e+f x)}{3 a b f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 (a+3 b) \cot (e+f x) \csc ^2(e+f x)}{3 a^2 b f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{3 (a+b) (3 a+8 b)-6 (a+b) (a+3 b) x^2}{x^4 \sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 a^2 b (a+b) f}\\ &=\frac{(a+b) \cot (e+f x) \csc ^2(e+f x)}{3 a b f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 (a+3 b) \cot (e+f x) \csc ^2(e+f x)}{3 a^2 b f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(3 a+8 b) \cot (e+f x) \csc ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 a^3 b f}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{24 b (a+b) (a+2 b)-3 b (a+b) (3 a+8 b) x^2}{x^2 \sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{9 a^3 b (a+b) f}\\ &=\frac{(a+b) \cot (e+f x) \csc ^2(e+f x)}{3 a b f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 (a+3 b) \cot (e+f x) \csc ^2(e+f x)}{3 a^2 b f \sqrt{a+b \sin ^2(e+f x)}}+\frac{8 (a+2 b) \cot (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 a^4 f}-\frac{(3 a+8 b) \cot (e+f x) \csc ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 a^3 b f}+\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{3 a b (a+b) (3 a+8 b)+24 b^2 (a+b) (a+2 b) x^2}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{9 a^4 b (a+b) f}\\ &=\frac{(a+b) \cot (e+f x) \csc ^2(e+f x)}{3 a b f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 (a+3 b) \cot (e+f x) \csc ^2(e+f x)}{3 a^2 b f \sqrt{a+b \sin ^2(e+f x)}}+\frac{8 (a+2 b) \cot (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 a^4 f}-\frac{(3 a+8 b) \cot (e+f x) \csc ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 a^3 b f}+\frac{\left (8 (a+2 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{\sqrt{1-x^2}} \, dx,x,\sin (e+f x)\right )}{3 a^4 f}-\frac{\left ((5 a+8 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 a^3 f}\\ &=\frac{(a+b) \cot (e+f x) \csc ^2(e+f x)}{3 a b f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 (a+3 b) \cot (e+f x) \csc ^2(e+f x)}{3 a^2 b f \sqrt{a+b \sin ^2(e+f x)}}+\frac{8 (a+2 b) \cot (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 a^4 f}-\frac{(3 a+8 b) \cot (e+f x) \csc ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 a^3 b f}+\frac{\left (8 (a+2 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{b x^2}{a}}}{\sqrt{1-x^2}} \, dx,x,\sin (e+f x)\right )}{3 a^4 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}-\frac{\left ((5 a+8 b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+\frac{b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{3 a^3 f \sqrt{a+b \sin ^2(e+f x)}}\\ &=\frac{(a+b) \cot (e+f x) \csc ^2(e+f x)}{3 a b f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{2 (a+3 b) \cot (e+f x) \csc ^2(e+f x)}{3 a^2 b f \sqrt{a+b \sin ^2(e+f x)}}+\frac{8 (a+2 b) \cot (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 a^4 f}-\frac{(3 a+8 b) \cot (e+f x) \csc ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 a^3 b f}+\frac{8 (a+2 b) \sqrt{\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right ) \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 a^4 f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}-\frac{(5 a+8 b) \sqrt{\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right ) \sec (e+f x) \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}{3 a^3 f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 3.11643, size = 226, normalized size = 0.65 \[ \frac{2 a^2 b \left (\frac{2 a-b \cos (2 (e+f x))+b}{a}\right )^{3/2} \left (8 (a+2 b) E\left (e+f x\left |-\frac{b}{a}\right .\right )-(5 a+8 b) F\left (e+f x\left |-\frac{b}{a}\right .\right )\right )+\sqrt{2} b \left (2 a b (a+b) \sin (2 (e+f x))+4 b (a+2 b) \sin (2 (e+f x)) (2 a-b \cos (2 (e+f x))+b)+4 (a+2 b) \cot (e+f x) (2 a-b \cos (2 (e+f x))+b)^2-a \cot (e+f x) \csc ^2(e+f x) (2 a-b \cos (2 (e+f x))+b)^2\right )}{6 a^4 b f (2 a-b \cos (2 (e+f x))+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^4/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

(2*a^2*b*((2*a + b - b*Cos[2*(e + f*x)])/a)^(3/2)*(8*(a + 2*b)*EllipticE[e + f*x, -(b/a)] - (5*a + 8*b)*Ellipt
icF[e + f*x, -(b/a)]) + Sqrt[2]*b*(4*(a + 2*b)*(2*a + b - b*Cos[2*(e + f*x)])^2*Cot[e + f*x] - a*(2*a + b - b*
Cos[2*(e + f*x)])^2*Cot[e + f*x]*Csc[e + f*x]^2 + 2*a*b*(a + b)*Sin[2*(e + f*x)] + 4*b*(a + 2*b)*(2*a + b - b*
Cos[2*(e + f*x)])*Sin[2*(e + f*x)]))/(6*a^4*b*f*(2*a + b - b*Cos[2*(e + f*x)])^(3/2))

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Maple [A]  time = 1.55, size = 633, normalized size = 1.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^4/(a+b*sin(f*x+e)^2)^(5/2),x)

[Out]

-1/3*(5*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^2*b*sin(f*x+e
)^5+8*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2*sin(f*x+e)^
5-8*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^2*b*sin(f*x+e)^5-
16*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2*sin(f*x+e)^5+8
*a*b^2*sin(f*x+e)^8+16*b^3*sin(f*x+e)^8+5*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+
e),(-1/a*b)^(1/2))*a^3*sin(f*x+e)^3+8*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(
-1/a*b)^(1/2))*a^2*b*sin(f*x+e)^3-8*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1
/a*b)^(1/2))*a^3*sin(f*x+e)^3-16*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*
b)^(1/2))*a^2*b*sin(f*x+e)^3+13*a^2*b*sin(f*x+e)^6+16*a*b^2*sin(f*x+e)^6-16*b^3*sin(f*x+e)^6+4*a^3*sin(f*x+e)^
4-7*a^2*b*sin(f*x+e)^4-24*a*b^2*sin(f*x+e)^4-5*a^3*sin(f*x+e)^2-6*a^2*b*sin(f*x+e)^2+a^3)/sin(f*x+e)^3/a^4/(a+
b*sin(f*x+e)^2)^(3/2)/cos(f*x+e)/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(cot(f*x+e)^4/(a+b*sin(f*x+e)^2)^(5/2),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{-b \cos \left (f x + e\right )^{2} + a + b} \cot \left (f x + e\right )^{4}}{b^{3} \cos \left (f x + e\right )^{6} - 3 \,{\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} - 3 \, a^{2} b - 3 \, a b^{2} - b^{3} + 3 \,{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-b*cos(f*x + e)^2 + a + b)*cot(f*x + e)^4/(b^3*cos(f*x + e)^6 - 3*(a*b^2 + b^3)*cos(f*x + e)^4
- a^3 - 3*a^2*b - 3*a*b^2 - b^3 + 3*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)^2), x)

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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(cot(f*x+e)**4/(a+b*sin(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )^{4}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

integrate(cot(f*x + e)^4/(b*sin(f*x + e)^2 + a)^(5/2), x)