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(d*x+c)*cot(d*x+c)/a/d/(e*cot(d*x+c))^(5/2)+2*cos(d*x+c)*cot(d*x+c)^2*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1
/4*Pi+d*x)*EllipticE(cos(c+1/4*Pi+d*x),2^(1/2))/a/d/(e*cot(d*x+c))^(5/2)/sin(2*d*x+2*c)^(1/2)-1/2*arctan(-1+2^
(1/2)*tan(d*x+c)^(1/2))/a/d/(e*cot(d*x+c))^(5/2)*2^(1/2)/tan(d*x+c)^(5/2)-1/2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2
))/a/d/(e*cot(d*x+c))^(5/2)*2^(1/2)/tan(d*x+c)^(5/2)-1/4*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a/d/(e*cot(
d*x+c))^(5/2)*2^(1/2)/tan(d*x+c)^(5/2)+1/4*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/a/d/(e*cot(d*x+c))^(5/2)*
2^(1/2)/tan(d*x+c)^(5/2)

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Rubi [A]  time = 0.32, antiderivative size = 325, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 15, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, 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 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 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 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 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 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 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 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 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 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]

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 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 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 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]

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*}

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Mathematica [C]  time = 62.67, size = 194, normalized size = 0.60 \[ -\frac {\sin ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\sqrt {\sec ^2(c+d x)}+1\right ) \sqrt {e \cot (c+d x)} \left (3 \sqrt {2} \cot ^{\frac {3}{2}}(c+d x) \left (\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+2 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-2 \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )\right )-8 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\tan ^2(c+d x)\right )\right )}{6 a d e^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/6*(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[Se
c[c + d*x]^2])*Sin[(c + d*x)/2]^2)/(a*d*e^3)

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \cot \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sec \left (d x + c\right ) + a\right )}}\,{d x} \]

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)

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maple [C]  time = 2.40, size = 1419, normalized size = 4.37 \[ \text {result too large to display} \]

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*(-1+cos(d*x+c))^2*(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))*c
os(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)-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))*cos
(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))*cos(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)+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)-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))^(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)-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)+2*Elliptic
F(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(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)-4*EllipticE(((1-cos(d*x
+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+s
in(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))*((-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))*((-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)+2*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),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)-4*EllipticE(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),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)+2*co
s(d*x+c)*2^(1/2)-2*2^(1/2))*cos(d*x+c)^2*(1+cos(d*x+c))^2/sin(d*x+c)^7/(e*cos(d*x+c)/sin(d*x+c))^(5/2)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (e \cot \left (d x + c\right )\right )^{\frac {5}{2}} {\left (a \sec \left (d x + c\right ) + a\right )}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\cos \left (c+d\,x\right )}{a\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{5/2}\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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