3.238 \(\int (e \cot (c+d x))^{5/2} (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=357 \[ -\frac{2 a^2 \sqrt{\sin (2 c+2 d x)} \tan ^2(c+d x) \sec (c+d x) \text{EllipticF}\left (c+d x-\frac{\pi }{4},2\right ) (e \cot (c+d x))^{5/2}}{3 d}+\frac{a^2 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right ) \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}{\sqrt{2} d}-\frac{a^2 \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right ) \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}{\sqrt{2} d}-\frac{4 a^2 \tan (c+d x) (e \cot (c+d x))^{5/2}}{3 d}+\frac{a^2 \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2} \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2} \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{4 a^2 \tan (c+d x) \sec (c+d x) (e \cot (c+d x))^{5/2}}{3 d} \]

[Out]

(-4*a^2*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x])/(3*d) - (4*a^2*(e*Cot[c + d*x])^(5/2)*Sec[c + d*x]*Tan[c + d*x])/
(3*d) - (2*a^2*(e*Cot[c + d*x])^(5/2)*EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]]*Tan[c +
d*x]^2)/(3*d) + (a^2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2))/(Sqrt[
2]*d) - (a^2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2))/(Sqrt[2]*d) + (
a^2*(e*Cot[c + d*x])^(5/2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]*Tan[c + d*x]^(5/2))/(2*Sqrt[2]*d
) - (a^2*(e*Cot[c + d*x])^(5/2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]*Tan[c + d*x]^(5/2))/(2*Sqrt
[2]*d)

________________________________________________________________________________________

Rubi [A]  time = 0.335922, antiderivative size = 357, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 17, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.68, Rules used = {3900, 3886, 3474, 3476, 329, 211, 1165, 628, 1162, 617, 204, 2609, 2614, 2573, 2641, 2607, 30} \[ \frac{a^2 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right ) \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}{\sqrt{2} d}-\frac{a^2 \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right ) \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}{\sqrt{2} d}-\frac{4 a^2 \tan (c+d x) (e \cot (c+d x))^{5/2}}{3 d}+\frac{a^2 \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2} \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{a^2 \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2} \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{4 a^2 \tan (c+d x) \sec (c+d x) (e \cot (c+d x))^{5/2}}{3 d}-\frac{2 a^2 \sqrt{\sin (2 c+2 d x)} \tan ^2(c+d x) \sec (c+d x) F\left (\left .c+d x-\frac{\pi }{4}\right |2\right ) (e \cot (c+d x))^{5/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(e*Cot[c + d*x])^(5/2)*(a + a*Sec[c + d*x])^2,x]

[Out]

(-4*a^2*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x])/(3*d) - (4*a^2*(e*Cot[c + d*x])^(5/2)*Sec[c + d*x]*Tan[c + d*x])/
(3*d) - (2*a^2*(e*Cot[c + d*x])^(5/2)*EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]]*Tan[c +
d*x]^2)/(3*d) + (a^2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2))/(Sqrt[
2]*d) - (a^2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2))/(Sqrt[2]*d) + (
a^2*(e*Cot[c + d*x])^(5/2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]*Tan[c + d*x]^(5/2))/(2*Sqrt[2]*d
) - (a^2*(e*Cot[c + d*x])^(5/2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]*Tan[c + d*x]^(5/2))/(2*Sqrt
[2]*d)

Rule 3900

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*((a_) + (b_.)*sec[(c_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Dist[(e*Co
t[c + d*x])^m*Tan[c + d*x]^m, Int[(a + b*Sec[c + d*x])^n/Tan[c + d*x]^m, x], x] /; FreeQ[{a, b, c, d, e, m, n}
, x] &&  !IntegerQ[m]

Rule 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rule 3474

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 2609

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1))/(b*f*(n + 1)), x] - Dist[(m + n + 1)/(b^2*(n + 1)), Int[(a*Sec[e + f*x])^m*(
b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && IntegersQ[2*m, 2*n]

Rule 2614

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int (e \cot (c+d x))^{5/2} (a+a \sec (c+d x))^2 \, dx &=\left ((e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \int \frac{(a+a \sec (c+d x))^2}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\left ((e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \int \left (\frac{a^2}{\tan ^{\frac{5}{2}}(c+d x)}+\frac{2 a^2 \sec (c+d x)}{\tan ^{\frac{5}{2}}(c+d x)}+\frac{a^2 \sec ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x)}\right ) \, dx\\ &=\left (a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \int \frac{1}{\tan ^{\frac{5}{2}}(c+d x)} \, dx+\left (a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \int \frac{\sec ^2(c+d x)}{\tan ^{\frac{5}{2}}(c+d x)} \, dx+\left (2 a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \int \frac{\sec (c+d x)}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 a^2 (e \cot (c+d x))^{5/2} \tan (c+d x)}{3 d}-\frac{4 a^2 (e \cot (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{3 d}-\frac{1}{3} \left (2 a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \int \frac{\sec (c+d x)}{\sqrt{\tan (c+d x)}} \, dx-\left (a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \int \frac{1}{\sqrt{\tan (c+d x)}} \, dx+\frac{\left (a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{x^{5/2}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{4 a^2 (e \cot (c+d x))^{5/2} \tan (c+d x)}{3 d}-\frac{4 a^2 (e \cot (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{3 d}-\frac{\left (2 a^2 (e \cot (c+d x))^{5/2} \sin ^{\frac{5}{2}}(c+d x)\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{\sin (c+d x)}} \, dx}{3 \cos ^{\frac{5}{2}}(c+d x)}-\frac{\left (a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{4 a^2 (e \cot (c+d x))^{5/2} \tan (c+d x)}{3 d}-\frac{4 a^2 (e \cot (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{3 d}-\frac{1}{3} \left (2 a^2 (e \cot (c+d x))^{5/2} \sec (c+d x) \sqrt{\sin (2 c+2 d x)} \tan ^2(c+d x)\right ) \int \frac{1}{\sqrt{\sin (2 c+2 d x)}} \, dx-\frac{\left (2 a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{4 a^2 (e \cot (c+d x))^{5/2} \tan (c+d x)}{3 d}-\frac{4 a^2 (e \cot (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{3 d}-\frac{2 a^2 (e \cot (c+d x))^{5/2} F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)} \tan ^2(c+d x)}{3 d}-\frac{\left (a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}-\frac{\left (a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{4 a^2 (e \cot (c+d x))^{5/2} \tan (c+d x)}{3 d}-\frac{4 a^2 (e \cot (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{3 d}-\frac{2 a^2 (e \cot (c+d x))^{5/2} F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)} \tan ^2(c+d x)}{3 d}-\frac{\left (a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}-\frac{\left (a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}+\frac{\left (a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} d}+\frac{\left (a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} d}\\ &=-\frac{4 a^2 (e \cot (c+d x))^{5/2} \tan (c+d x)}{3 d}-\frac{4 a^2 (e \cot (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{3 d}-\frac{2 a^2 (e \cot (c+d x))^{5/2} F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)} \tan ^2(c+d x)}{3 d}+\frac{a^2 (e \cot (c+d x))^{5/2} \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right ) \tan ^{\frac{5}{2}}(c+d x)}{2 \sqrt{2} d}-\frac{a^2 (e \cot (c+d x))^{5/2} \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right ) \tan ^{\frac{5}{2}}(c+d x)}{2 \sqrt{2} d}-\frac{\left (a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2 (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}\\ &=-\frac{4 a^2 (e \cot (c+d x))^{5/2} \tan (c+d x)}{3 d}-\frac{4 a^2 (e \cot (c+d x))^{5/2} \sec (c+d x) \tan (c+d x)}{3 d}-\frac{2 a^2 (e \cot (c+d x))^{5/2} F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)} \tan ^2(c+d x)}{3 d}+\frac{a^2 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right ) (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)}{\sqrt{2} d}-\frac{a^2 \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right ) (e \cot (c+d x))^{5/2} \tan ^{\frac{5}{2}}(c+d x)}{\sqrt{2} d}+\frac{a^2 (e \cot (c+d x))^{5/2} \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right ) \tan ^{\frac{5}{2}}(c+d x)}{2 \sqrt{2} d}-\frac{a^2 (e \cot (c+d x))^{5/2} \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right ) \tan ^{\frac{5}{2}}(c+d x)}{2 \sqrt{2} d}\\ \end{align*}

Mathematica [C]  time = 2.21904, size = 93, normalized size = 0.26 \[ -\frac{2 a^2 e \cos ^4\left (\frac{1}{2} (c+d x)\right ) (e \cot (c+d x))^{3/2} \sec ^4\left (\frac{1}{2} \cot ^{-1}(\cot (c+d x))\right ) \left (2 \text{Hypergeometric2F1}\left (-\frac{3}{4},\frac{1}{2},\frac{1}{4},-\tan ^2(c+d x)\right )-\text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{7}{4},-\cot ^2(c+d x)\right )+2\right )}{3 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Cot[c + d*x])^(5/2)*(a + a*Sec[c + d*x])^2,x]

[Out]

(-2*a^2*e*Cos[(c + d*x)/2]^4*(e*Cot[c + d*x])^(3/2)*(2 + 2*Hypergeometric2F1[-3/4, 1/2, 1/4, -Tan[c + d*x]^2]
- Hypergeometric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2])*Sec[ArcCot[Cot[c + d*x]]/2]^4)/(3*d)

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Maple [C]  time = 0.27, size = 650, 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((e*cot(d*x+c))^(5/2)*(a+a*sec(d*x+c))^2,x)

[Out]

1/6*a^2/d*2^(1/2)*(-1+cos(d*x+c))*(3*I*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2
^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*
x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-3*I*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c
))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*
x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+3*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))
/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+
c))^(1/2),1/2-1/2*I,1/2*2^(1/2))+3*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/s
in(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c)
)^(1/2),1/2+1/2*I,1/2*2^(1/2))-2*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin
(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(
1/2),1/2*2^(1/2))+4*cos(d*x+c)*2^(1/2))*(e*cos(d*x+c)/sin(d*x+c))^(5/2)*(cos(d*x+c)+1)^2/sin(d*x+c)/cos(d*x+c)
^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))^(5/2)*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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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((e*cot(d*x+c))^(5/2)*(a+a*sec(d*x+c))^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((e*cot(d*x+c))**(5/2)*(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cot(d*x+c))^(5/2)*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((e*cot(d*x + c))^(5/2)*(a*sec(d*x + c) + a)^2, x)