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

Optimal. Leaf size=212 \[ \frac {d^{3/2} \log \left (\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}-\frac {d^{3/2} \log \left (\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}-\frac {d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}+\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} f}+\frac {2 d^2}{f \sqrt {d \cot (e+f x)}} \]

[Out]

-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
)

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Rubi [A]  time = 0.17, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 13, 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} \[ \frac {2 d^2}{f \sqrt {d \cot (e+f x)}}+\frac {d^{3/2} \log \left (\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}-\frac {d^{3/2} \log \left (\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} f}-\frac {d^{3/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} f}+\frac {d^{3/2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} f} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cot[e + f*x])^(3/2)*Tan[e + f*x]^3,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]*Sqr
t[d*Cot[e + f*x]])/Sqrt[d]])/(Sqrt[2]*f) + (2*d^2)/(f*Sqrt[d*Cot[e + f*x]]) + (d^(3/2)*Log[Sqrt[d] + Sqrt[d]*C
ot[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] + Sqr
t[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 ^3(e+f x) \, dx &=d^3 \int \frac {1}{(d \cot (e+f x))^{3/2}} \, dx\\ &=\frac {2 d^2}{f \sqrt {d \cot (e+f x)}}-d \int \sqrt {d \cot (e+f x)} \, dx\\ &=\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^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^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^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^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^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*}

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Mathematica [C]  time = 0.05, size = 38, normalized size = 0.18 \[ \frac {2 d^2 \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\cot ^2(e+f x)\right )}{f \sqrt {d \cot (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.54, size = 594, normalized size = 2.80 \[ -\frac {4 \, \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 ) + 4 \, \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 ) + \sqrt {2} \left (\frac {d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \cos \left (f x + e\right ) \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 ) - \sqrt {2} \left (\frac {d^{6}}{f^{4}}\right )^{\frac {1}{4}} f \cos \left (f x + e\right ) \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 \, d \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \sin \left (f x + e\right )}{4 \, f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*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)/si
n(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) + 4*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))*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) + sqrt(2)*(d^6/f^4)^(1/4)*f*cos(
f*x + e)*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))*sin(f*x + e
) + sqrt(d^6/f^4)*d^6*f^2*sin(f*x + e))/sin(f*x + e)) - sqrt(2)*(d^6/f^4)^(1/4)*f*cos(f*x + e)*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))*sin(f*x + e) + sqrt(d^6/f^4)*d^6*f^
2*sin(f*x + e))/sin(f*x + e)) - 8*d*sqrt(d*cos(f*x + e)/sin(f*x + e))*sin(f*x + e))/(f*cos(f*x + e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.63, size = 660, normalized size = 3.11 \[ \frac {\left (i \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 )}}-i \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 )}}+\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 )}}+\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 )}}-2 \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 )}}+2 \cos \left (f x +e \right ) \sqrt {2}-2 \sqrt {2}\right ) \left (1+\cos \left (f x +e \right )\right )^{2} \left (-1+\cos \left (f x +e \right )\right ) \left (\frac {d \cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {2}}{2 f \sin \left (f x +e \right )^{2} \cos \left (f x +e \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*(I*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)-I*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+co
s(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)+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+co
s(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)+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+co
s(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)-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
)+2*cos(f*x+e)*2^(1/2)-2*2^(1/2))*(1+cos(f*x+e))^2*(-1+cos(f*x+e))*(d*cos(f*x+e)/sin(f*x+e))^(3/2)/sin(f*x+e)^
2/cos(f*x+e)^2*2^(1/2)

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maxima [A]  time = 0.49, size = 189, normalized size = 0.89 \[ \frac {d^{4} {\left (\frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {\frac {d}{\tan \left (f x + e\right )}}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {\frac {d}{\tan \left (f x + e\right )}}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {d} \sqrt {\frac {d}{\tan \left (f x + e\right )}} + d + \frac {d}{\tan \left (f x + e\right )}\right )}{\sqrt {d}} + \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {d} \sqrt {\frac {d}{\tan \left (f x + e\right )}} + d + \frac {d}{\tan \left (f x + e\right )}\right )}{\sqrt {d}}}{d^{2}} + \frac {8}{d^{2} \sqrt {\frac {d}{\tan \left (f x + e\right )}}}\right )}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*d^4*((2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d/tan(f*x + e)))/sqrt(d))/sqrt(d) + 2*sqrt(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)*sqrt(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^2 + 8/(d^2*sqrt(d/tan(f*x + e))))/f

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mupad [B]  time = 2.49, size = 82, normalized size = 0.39 \[ \frac {2\,d^2}{f\,\sqrt {\frac {d}{\mathrm {tan}\left (e+f\,x\right )}}}+\frac {{\left (-1\right )}^{1/4}\,d^{3/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {\frac {d}{\mathrm {tan}\left (e+f\,x\right )}}}{\sqrt {d}}\right )}{f}-\frac {{\left (-1\right )}^{1/4}\,d^{3/2}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {\frac {d}{\mathrm {tan}\left (e+f\,x\right )}}}{\sqrt {d}}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*d^2)/(f*(d/tan(e + f*x))^(1/2)) + ((-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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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