3.247 \(\int \frac{1}{(e \cot (c+d x))^{5/2} (a+a \sec (c+d x))} \, dx\)

Optimal. Leaf size=325 \[ \frac{2 \cos (c+d x) \cot (c+d x)}{a d (e \cot (c+d x))^{5/2}}+\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a d \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}-\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a d \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}-\frac{\log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} a d \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}+\frac{\log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} a d \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}-\frac{2 \cos (c+d x) \cot ^2(c+d x) E\left (\left .c+d x-\frac{\pi }{4}\right |2\right )}{a d \sqrt{\sin (2 c+2 d x)} (e \cot (c+d x))^{5/2}} \]

[Out]

(2*Cos[c + d*x]*Cot[c + d*x])/(a*d*(e*Cot[c + d*x])^(5/2)) - (2*Cos[c + d*x]*Cot[c + d*x]^2*EllipticE[c - Pi/4
+ d*x, 2])/(a*d*(e*Cot[c + d*x])^(5/2)*Sqrt[Sin[2*c + 2*d*x]]) + ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/(Sqrt
[2]*a*d*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2)) - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/(Sqrt[2]*a*d*(e*Co
t[c + d*x])^(5/2)*Tan[c + d*x]^(5/2)) - Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2]*a*d*(e*C
ot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2)) + Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2]*a*d*(e*
Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.323062, antiderivative size = 325, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 15, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {3900, 3888, 3884, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2613, 2615, 2572, 2639} \[ \frac{2 \cos (c+d x) \cot (c+d x)}{a d (e \cot (c+d x))^{5/2}}+\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} a d \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}-\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} a d \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}-\frac{\log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} a d \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}+\frac{\log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} a d \tan ^{\frac{5}{2}}(c+d x) (e \cot (c+d x))^{5/2}}-\frac{2 \cos (c+d x) \cot ^2(c+d x) E\left (\left .c+d x-\frac{\pi }{4}\right |2\right )}{a d \sqrt{\sin (2 c+2 d x)} (e \cot (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[c + d*x]*Cot[c + d*x])/(a*d*(e*Cot[c + d*x])^(5/2)) - (2*Cos[c + d*x]*Cot[c + d*x]^2*EllipticE[c - Pi/4
+ d*x, 2])/(a*d*(e*Cot[c + d*x])^(5/2)*Sqrt[Sin[2*c + 2*d*x]]) + ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/(Sqrt
[2]*a*d*(e*Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2)) - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/(Sqrt[2]*a*d*(e*Co
t[c + d*x])^(5/2)*Tan[c + d*x]^(5/2)) - Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2]*a*d*(e*C
ot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2)) + Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2]*a*d*(e*
Cot[c + d*x])^(5/2)*Tan[c + d*x]^(5/2))

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 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rule 3884

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

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 297

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

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 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 2613

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(a^2*(m - 2))/(m + n - 1), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 2615

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

Rule 2572

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

Rule 2639

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

Rubi steps

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

Mathematica [C]  time = 56.8628, size = 194, normalized size = 0.6 \[ -\frac{\sin ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (\sqrt{\sec ^2(c+d x)}+1\right ) \sqrt{e \cot (c+d x)} \left (3 \sqrt{2} \cot ^{\frac{3}{2}}(c+d x) \left (\log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )-\log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )+2 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )-2 \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )\right )-8 \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{3}{4},\frac{7}{4},-\tan ^2(c+d x)\right )\right )}{6 a d e^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(Sqrt[e*Cot[c + d*x]]*(-8*Hypergeometric2F1[1/2, 3/4, 7/4, -Tan[c + d*x]^2] + 3*Sqrt[2]*Cot[c + d*x]^(3/2)*(2
*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]] + Log[1 - Sqrt[2]*Sqrt[Cot[
c + d*x]] + Cot[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]))*Sec[c + d*x]*(1 + Sqrt[Sec[c
+ d*x]^2])*Sin[(c + d*x)/2]^2)/(6*a*d*e^3)

________________________________________________________________________________________

Maple [C]  time = 0.287, size = 1419, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a/d*2^(1/2)*(-1+cos(d*x+c))^2*(I*((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)*((-1+cos(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))*cos(d*x+c)-I*cos(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+si
n(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(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))-cos(d*x+c)*((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)*((-1+cos(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))-cos(d*x+c)*((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)*((-1+cos(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))-4*cos(d*x+c)*((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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticE(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/
2*2^(1/2))+2*cos(d*x+c)*((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)*((-1+cos(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))+I*
((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)*((-1+cos(d*x+c))/si
n(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))-I*((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)*((-1+cos(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))-((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)*((-1+cos(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))-((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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+s
in(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-4*((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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticE(((1-cos(d*x+c)+sin(d*x+c))/sin
(d*x+c))^(1/2),1/2*2^(1/2))+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)*((-1+cos(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))+2*cos(d*x+c)*2^(1/2)-2*2^(1/2))*cos(d*x+c)^2*(cos(d*x+c)+1)^2/sin(d*x+c)^7/(e*cos(d*x+c)/sin(d*x+c))^(5/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cot(d*x+c))^(5/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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