3.199 \(\int (d \cot (e+f x))^{3/2} \tan ^5(e+f x) \, dx\)

Optimal. Leaf size=234 \[ -\frac {d^{3/2} \log \left (\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}+\frac {d^{3/2} \log \left (\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}+\frac {d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}-\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} f}+\frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}} \]

[Out]

2/5*d^4/f/(d*cot(f*x+e))^(5/2)+1/2*d^(3/2)*arctan(1-2^(1/2)*(d*cot(f*x+e))^(1/2)/d^(1/2))/f*2^(1/2)-1/2*d^(3/2
)*arctan(1+2^(1/2)*(d*cot(f*x+e))^(1/2)/d^(1/2))/f*2^(1/2)-1/4*d^(3/2)*ln(d^(1/2)+cot(f*x+e)*d^(1/2)-2^(1/2)*(
d*cot(f*x+e))^(1/2))/f*2^(1/2)+1/4*d^(3/2)*ln(d^(1/2)+cot(f*x+e)*d^(1/2)+2^(1/2)*(d*cot(f*x+e))^(1/2))/f*2^(1/
2)-2*d^2/f/(d*cot(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {16, 3474, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}-\frac {d^{3/2} \log \left (\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}+\frac {d^{3/2} \log \left (\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}+\frac {d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}-\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cot[e + f*x])^(3/2)*Tan[e + f*x]^5,x]

[Out]

(d^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Cot[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) - (d^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[d
*Cot[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) + (2*d^4)/(5*f*(d*Cot[e + f*x])^(5/2)) - (2*d^2)/(f*Sqrt[d*Cot[e + f*x]]
) - (d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Cot[e + f*x] - Sqrt[2]*Sqrt[d*Cot[e + f*x]]])/(2*Sqrt[2]*f) + (d^(3/2)*Log[
Sqrt[d] + Sqrt[d]*Cot[e + f*x] + Sqrt[2]*Sqrt[d*Cot[e + f*x]]])/(2*Sqrt[2]*f)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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 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]

Rubi steps

\begin {align*} \int (d \cot (e+f x))^{3/2} \tan ^5(e+f x) \, dx &=d^5 \int \frac {1}{(d \cot (e+f x))^{7/2}} \, dx\\ &=\frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-d^3 \int \frac {1}{(d \cot (e+f x))^{3/2}} \, dx\\ &=\frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}+d \int \sqrt {d \cot (e+f x)} \, dx\\ &=\frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}-\frac {d^2 \operatorname {Subst}\left (\int \frac {\sqrt {x}}{d^2+x^2} \, dx,x,d \cot (e+f x)\right )}{f}\\ &=\frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}-\frac {\left (2 d^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{d^2+x^4} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{f}\\ &=\frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}+\frac {d^2 \operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{f}-\frac {d^2 \operatorname {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{f}\\ &=\frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}-\frac {d^{3/2} \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}-\frac {d^{3/2} \operatorname {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}-\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{2 f}-\frac {d^2 \operatorname {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{2 f}\\ &=\frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}-\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}+\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}-\frac {d^{3/2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}+\frac {d^{3/2} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}\\ &=\frac {d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}-\frac {d^{3/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}+\frac {2 d^4}{5 f (d \cot (e+f x))^{5/2}}-\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}-\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}+\frac {d^{3/2} \log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.07, size = 45, normalized size = 0.19 \[ \frac {2 \tan ^4(e+f x) (d \cot (e+f x))^{3/2} \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\cot ^2(e+f x)\right )}{5 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cot[e + f*x])^(3/2)*Tan[e + f*x]^5,x]

[Out]

(2*(d*Cot[e + f*x])^(3/2)*Hypergeometric2F1[-5/4, 1, -1/4, -Cot[e + f*x]^2]*Tan[e + f*x]^4)/(5*f)

________________________________________________________________________________________

fricas [B]  time = 0.62, size = 617, normalized size = 2.64 \[ \frac {20 \, \sqrt {2} \left (\frac {d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \arctan \left (-\frac {\sqrt {2} \left (\frac {d^{6}}{f^{4}}\right )^{\frac {1}{4}} d^{4} f \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} + d^{6} - \sqrt {2} \left (\frac {d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \sqrt {\frac {d^{9} \cos \left (f x + e\right ) + \sqrt {2} \left (\frac {d^{6}}{f^{4}}\right )^{\frac {3}{4}} d^{4} f^{3} \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \sin \left (f x + e\right ) + \sqrt {\frac {d^{6}}{f^{4}}} d^{6} f^{2} \sin \left (f x + e\right )}{\sin \left (f x + e\right )}}}{d^{6}}\right ) \cos \left (f x + e\right )^{3} + 20 \, \sqrt {2} \left (\frac {d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \arctan \left (-\frac {\sqrt {2} \left (\frac {d^{6}}{f^{4}}\right )^{\frac {1}{4}} d^{4} f \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} - d^{6} - \sqrt {2} \left (\frac {d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \sqrt {\frac {d^{9} \cos \left (f x + e\right ) - \sqrt {2} \left (\frac {d^{6}}{f^{4}}\right )^{\frac {3}{4}} d^{4} f^{3} \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \sin \left (f x + e\right ) + \sqrt {\frac {d^{6}}{f^{4}}} d^{6} f^{2} \sin \left (f x + e\right )}{\sin \left (f x + e\right )}}}{d^{6}}\right ) \cos \left (f x + e\right )^{3} + 5 \, \sqrt {2} \left (\frac {d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \cos \left (f x + e\right )^{3} \log \left (\frac {d^{9} \cos \left (f x + e\right ) + \sqrt {2} \left (\frac {d^{6}}{f^{4}}\right )^{\frac {3}{4}} d^{4} f^{3} \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \sin \left (f x + e\right ) + \sqrt {\frac {d^{6}}{f^{4}}} d^{6} f^{2} \sin \left (f x + e\right )}{\sin \left (f x + e\right )}\right ) - 5 \, \sqrt {2} \left (\frac {d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \cos \left (f x + e\right )^{3} \log \left (\frac {d^{9} \cos \left (f x + e\right ) - \sqrt {2} \left (\frac {d^{6}}{f^{4}}\right )^{\frac {3}{4}} d^{4} f^{3} \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \sin \left (f x + e\right ) + \sqrt {\frac {d^{6}}{f^{4}}} d^{6} f^{2} \sin \left (f x + e\right )}{\sin \left (f x + e\right )}\right ) - 8 \, {\left (6 \, d \cos \left (f x + e\right )^{2} - d\right )} \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \sin \left (f x + e\right )}{20 \, f \cos \left (f x + e\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cot(f*x+e))^(3/2)*tan(f*x+e)^5,x, algorithm="fricas")

[Out]

1/20*(20*sqrt(2)*(d^6/f^4)^(1/4)*f*arctan(-(sqrt(2)*(d^6/f^4)^(1/4)*d^4*f*sqrt(d*cos(f*x + e)/sin(f*x + e)) +
d^6 - sqrt(2)*(d^6/f^4)^(1/4)*f*sqrt((d^9*cos(f*x + e) + sqrt(2)*(d^6/f^4)^(3/4)*d^4*f^3*sqrt(d*cos(f*x + e)/s
in(f*x + e))*sin(f*x + e) + sqrt(d^6/f^4)*d^6*f^2*sin(f*x + e))/sin(f*x + e)))/d^6)*cos(f*x + e)^3 + 20*sqrt(2
)*(d^6/f^4)^(1/4)*f*arctan(-(sqrt(2)*(d^6/f^4)^(1/4)*d^4*f*sqrt(d*cos(f*x + e)/sin(f*x + e)) - d^6 - sqrt(2)*(
d^6/f^4)^(1/4)*f*sqrt((d^9*cos(f*x + e) - sqrt(2)*(d^6/f^4)^(3/4)*d^4*f^3*sqrt(d*cos(f*x + e)/sin(f*x + e))*si
n(f*x + e) + sqrt(d^6/f^4)*d^6*f^2*sin(f*x + e))/sin(f*x + e)))/d^6)*cos(f*x + e)^3 + 5*sqrt(2)*(d^6/f^4)^(1/4
)*f*cos(f*x + e)^3*log((d^9*cos(f*x + e) + sqrt(2)*(d^6/f^4)^(3/4)*d^4*f^3*sqrt(d*cos(f*x + e)/sin(f*x + e))*s
in(f*x + e) + sqrt(d^6/f^4)*d^6*f^2*sin(f*x + e))/sin(f*x + e)) - 5*sqrt(2)*(d^6/f^4)^(1/4)*f*cos(f*x + e)^3*l
og((d^9*cos(f*x + e) - sqrt(2)*(d^6/f^4)^(3/4)*d^4*f^3*sqrt(d*cos(f*x + e)/sin(f*x + e))*sin(f*x + e) + sqrt(d
^6/f^4)*d^6*f^2*sin(f*x + e))/sin(f*x + e)) - 8*(6*d*cos(f*x + e)^2 - d)*sqrt(d*cos(f*x + e)/sin(f*x + e))*sin
(f*x + e))/(f*cos(f*x + e)^3)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \cot \left (f x + e\right )\right )^{\frac {3}{2}} \tan \left (f x + e\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cot(f*x+e))^(3/2)*tan(f*x+e)^5,x, algorithm="giac")

[Out]

integrate((d*cot(f*x + e))^(3/2)*tan(f*x + e)^5, x)

________________________________________________________________________________________

maple [C]  time = 0.68, size = 728, normalized size = 3.11 \[ \frac {\left (-1+\cos \left (f x +e \right )\right ) \left (5 i \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \EllipticPi \left (\sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}-5 i \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \EllipticPi \left (\sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}-5 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \EllipticPi \left (\sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}+10 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \EllipticF \left (\sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}-5 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \EllipticPi \left (\sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {-\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}-12 \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {2}+12 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}+2 \cos \left (f x +e \right ) \sqrt {2}-2 \sqrt {2}\right ) \left (1+\cos \left (f x +e \right )\right )^{2} \left (\frac {d \cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {2}}{10 f \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cot(f*x+e))^(3/2)*tan(f*x+e)^5,x)

[Out]

1/10/f*(-1+cos(f*x+e))*(5*I*EllipticPi((-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))*s
in(f*x+e)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(-sin(f*x+e)-1+co
s(f*x+e))/sin(f*x+e))^(1/2)*cos(f*x+e)^2-5*I*EllipticPi((-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2
*I,1/2*2^(1/2))*sin(f*x+e)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(-
(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2)*cos(f*x+e)^2-5*cos(f*x+e)^2*sin(f*x+e)*EllipticPi((-(-sin(f*x+e)-
1+cos(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(
f*x+e))/sin(f*x+e))^(1/2)*(-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2)+10*cos(f*x+e)^2*sin(f*x+e)*EllipticF(
(-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)
+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2)-5*cos(f*x+e)^2*sin(f*x+e)*Ellipt
icPi((-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2),1/2-1/2*I,1/2*2^(1/2))*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*
((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*(-(-sin(f*x+e)-1+cos(f*x+e))/sin(f*x+e))^(1/2)-12*cos(f*x+e)^3*2
^(1/2)+12*cos(f*x+e)^2*2^(1/2)+2*cos(f*x+e)*2^(1/2)-2*2^(1/2))*(1+cos(f*x+e))^2*(d*cos(f*x+e)/sin(f*x+e))^(3/2
)/sin(f*x+e)^2/cos(f*x+e)^4*2^(1/2)

________________________________________________________________________________________

maxima [A]  time = 0.55, size = 207, normalized size = 0.88 \[ -\frac {d^{6} {\left (\frac {5 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {\frac {d}{\tan \left (f x + e\right )}}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {\frac {d}{\tan \left (f x + e\right )}}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {d} \sqrt {\frac {d}{\tan \left (f x + e\right )}} + d + \frac {d}{\tan \left (f x + e\right )}\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {d} \sqrt {\frac {d}{\tan \left (f x + e\right )}} + d + \frac {d}{\tan \left (f x + e\right )}\right )}{\sqrt {d}}\right )}}{d^{4}} - \frac {8 \, {\left (d^{2} - \frac {5 \, d^{2}}{\tan \left (f x + e\right )^{2}}\right )}}{d^{4} \left (\frac {d}{\tan \left (f x + e\right )}\right )^{\frac {5}{2}}}\right )}}{20 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cot(f*x+e))^(3/2)*tan(f*x+e)^5,x, algorithm="maxima")

[Out]

-1/20*d^6*(5*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d/tan(f*x + e)))/sqrt(d))/sqrt(d) + 2*sqr
t(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d/tan(f*x + e)))/sqrt(d))/sqrt(d) - sqrt(2)*log(sqrt(2)*sqr
t(d)*sqrt(d/tan(f*x + e)) + d + d/tan(f*x + e))/sqrt(d) + sqrt(2)*log(-sqrt(2)*sqrt(d)*sqrt(d/tan(f*x + e)) +
d + d/tan(f*x + e))/sqrt(d))/d^4 - 8*(d^2 - 5*d^2/tan(f*x + e)^2)/(d^4*(d/tan(f*x + e))^(5/2)))/f

________________________________________________________________________________________

mupad [B]  time = 2.58, size = 97, normalized size = 0.41 \[ \frac {\frac {2\,d^4}{5}-\frac {2\,d^4}{{\mathrm {tan}\left (e+f\,x\right )}^2}}{f\,{\left (\frac {d}{\mathrm {tan}\left (e+f\,x\right )}\right )}^{5/2}}-\frac {{\left (-1\right )}^{1/4}\,d^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {\frac {d}{\mathrm {tan}\left (e+f\,x\right )}}}{\sqrt {d}}\right )}{f}+\frac {{\left (-1\right )}^{1/4}\,d^{3/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {\frac {d}{\mathrm {tan}\left (e+f\,x\right )}}}{\sqrt {d}}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^5*(d*cot(e + f*x))^(3/2),x)

[Out]

((2*d^4)/5 - (2*d^4)/tan(e + f*x)^2)/(f*(d/tan(e + f*x))^(5/2)) - ((-1)^(1/4)*d^(3/2)*atan(((-1)^(1/4)*(d/tan(
e + f*x))^(1/2))/d^(1/2)))/f + ((-1)^(1/4)*d^(3/2)*atanh(((-1)^(1/4)*(d/tan(e + f*x))^(1/2))/d^(1/2)))/f

________________________________________________________________________________________

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((d*cot(f*x+e))**(3/2)*tan(f*x+e)**5,x)

[Out]

Timed out

________________________________________________________________________________________