3.208 \(\int \frac {\tan ^2(e+f x)}{\sqrt {d \cot (e+f x)}} \, dx\)

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

[Out]

2/3*d/f/(d*cot(f*x+e))^(3/2)-1/2*arctan(1-2^(1/2)*(d*cot(f*x+e))^(1/2)/d^(1/2))/f*2^(1/2)/d^(1/2)+1/2*arctan(1
+2^(1/2)*(d*cot(f*x+e))^(1/2)/d^(1/2))/f*2^(1/2)/d^(1/2)-1/4*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)/d^(1/2)+1/4*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)/d^(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, 211, 1165, 628, 1162, 617, 204} \[ \frac {2 d}{3 f (d \cot (e+f x))^{3/2}}-\frac {\log \left (\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} \sqrt {d} f}+\frac {\log \left (\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}+\sqrt {d}\right )}{2 \sqrt {2} \sqrt {d} f}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d} f}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d} f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^2/Sqrt[d*Cot[e + f*x]],x]

[Out]

-(ArcTan[1 - (Sqrt[2]*Sqrt[d*Cot[e + f*x]])/Sqrt[d]]/(Sqrt[2]*Sqrt[d]*f)) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Cot[e +
f*x]])/Sqrt[d]]/(Sqrt[2]*Sqrt[d]*f) + (2*d)/(3*f*(d*Cot[e + f*x])^(3/2)) - Log[Sqrt[d] + Sqrt[d]*Cot[e + f*x]
- Sqrt[2]*Sqrt[d*Cot[e + f*x]]]/(2*Sqrt[2]*Sqrt[d]*f) + Log[Sqrt[d] + Sqrt[d]*Cot[e + f*x] + Sqrt[2]*Sqrt[d*C
ot[e + f*x]]]/(2*Sqrt[2]*Sqrt[d]*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 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 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 \frac {\tan ^2(e+f x)}{\sqrt {d \cot (e+f x)}} \, dx &=d^2 \int \frac {1}{(d \cot (e+f x))^{5/2}} \, dx\\ &=\frac {2 d}{3 f (d \cot (e+f x))^{3/2}}-\int \frac {1}{\sqrt {d \cot (e+f x)}} \, dx\\ &=\frac {2 d}{3 f (d \cot (e+f x))^{3/2}}+\frac {d \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (d^2+x^2\right )} \, dx,x,d \cot (e+f x)\right )}{f}\\ &=\frac {2 d}{3 f (d \cot (e+f x))^{3/2}}+\frac {(2 d) \operatorname {Subst}\left (\int \frac {1}{d^2+x^4} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{f}\\ &=\frac {2 d}{3 f (d \cot (e+f x))^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \cot (e+f x)}\right )}{f}+\frac {\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}{3 f (d \cot (e+f x))^{3/2}}+\frac {\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 {\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 {\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} \sqrt {d} f}-\frac {\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} \sqrt {d} f}\\ &=\frac {2 d}{3 f (d \cot (e+f x))^{3/2}}-\frac {\log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} \sqrt {d} f}+\frac {\log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} \sqrt {d} f}+\frac {\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} \sqrt {d} f}-\frac {\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} \sqrt {d} f}\\ &=-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d} f}+\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \cot (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d} f}+\frac {2 d}{3 f (d \cot (e+f x))^{3/2}}-\frac {\log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)-\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} \sqrt {d} f}+\frac {\log \left (\sqrt {d}+\sqrt {d} \cot (e+f x)+\sqrt {2} \sqrt {d \cot (e+f x)}\right )}{2 \sqrt {2} \sqrt {d} f}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 38, normalized size = 0.18 \[ \frac {2 d \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\cot ^2(e+f x)\right )}{3 f (d \cot (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^2/Sqrt[d*Cot[e + f*x]],x]

[Out]

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

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fricas [B]  time = 0.53, size = 579, normalized size = 2.73 \[ -\frac {12 \, \sqrt {2} d f \left (\frac {1}{d^{2} f^{4}}\right )^{\frac {1}{4}} \arctan \left (-\sqrt {2} d f^{3} \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {1}{d^{2} f^{4}}\right )^{\frac {3}{4}} + \sqrt {2} d f^{3} \sqrt {\frac {d^{2} f^{2} \sqrt {\frac {1}{d^{2} f^{4}}} \sin \left (f x + e\right ) + \sqrt {2} d f \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {1}{d^{2} f^{4}}\right )^{\frac {1}{4}} \sin \left (f x + e\right ) + d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {1}{d^{2} f^{4}}\right )^{\frac {3}{4}} - 1\right ) \cos \left (f x + e\right )^{2} + 12 \, \sqrt {2} d f \left (\frac {1}{d^{2} f^{4}}\right )^{\frac {1}{4}} \arctan \left (-\sqrt {2} d f^{3} \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {1}{d^{2} f^{4}}\right )^{\frac {3}{4}} + \sqrt {2} d f^{3} \sqrt {\frac {d^{2} f^{2} \sqrt {\frac {1}{d^{2} f^{4}}} \sin \left (f x + e\right ) - \sqrt {2} d f \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {1}{d^{2} f^{4}}\right )^{\frac {1}{4}} \sin \left (f x + e\right ) + d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {1}{d^{2} f^{4}}\right )^{\frac {3}{4}} + 1\right ) \cos \left (f x + e\right )^{2} - 3 \, \sqrt {2} d f \left (\frac {1}{d^{2} f^{4}}\right )^{\frac {1}{4}} \cos \left (f x + e\right )^{2} \log \left (\frac {d^{2} f^{2} \sqrt {\frac {1}{d^{2} f^{4}}} \sin \left (f x + e\right ) + \sqrt {2} d f \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {1}{d^{2} f^{4}}\right )^{\frac {1}{4}} \sin \left (f x + e\right ) + d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}\right ) + 3 \, \sqrt {2} d f \left (\frac {1}{d^{2} f^{4}}\right )^{\frac {1}{4}} \cos \left (f x + e\right )^{2} \log \left (\frac {d^{2} f^{2} \sqrt {\frac {1}{d^{2} f^{4}}} \sin \left (f x + e\right ) - \sqrt {2} d f \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}} \left (\frac {1}{d^{2} f^{4}}\right )^{\frac {1}{4}} \sin \left (f x + e\right ) + d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}\right ) + 8 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {\frac {d \cos \left (f x + e\right )}{\sin \left (f x + e\right )}}}{12 \, d f \cos \left (f x + e\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(d*cot(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/12*(12*sqrt(2)*d*f*(1/(d^2*f^4))^(1/4)*arctan(-sqrt(2)*d*f^3*sqrt(d*cos(f*x + e)/sin(f*x + e))*(1/(d^2*f^4)
)^(3/4) + sqrt(2)*d*f^3*sqrt((d^2*f^2*sqrt(1/(d^2*f^4))*sin(f*x + e) + sqrt(2)*d*f*sqrt(d*cos(f*x + e)/sin(f*x
+ e))*(1/(d^2*f^4))^(1/4)*sin(f*x + e) + d*cos(f*x + e))/sin(f*x + e))*(1/(d^2*f^4))^(3/4) - 1)*cos(f*x + e)^
2 + 12*sqrt(2)*d*f*(1/(d^2*f^4))^(1/4)*arctan(-sqrt(2)*d*f^3*sqrt(d*cos(f*x + e)/sin(f*x + e))*(1/(d^2*f^4))^(
3/4) + sqrt(2)*d*f^3*sqrt((d^2*f^2*sqrt(1/(d^2*f^4))*sin(f*x + e) - sqrt(2)*d*f*sqrt(d*cos(f*x + e)/sin(f*x +
e))*(1/(d^2*f^4))^(1/4)*sin(f*x + e) + d*cos(f*x + e))/sin(f*x + e))*(1/(d^2*f^4))^(3/4) + 1)*cos(f*x + e)^2 -
3*sqrt(2)*d*f*(1/(d^2*f^4))^(1/4)*cos(f*x + e)^2*log((d^2*f^2*sqrt(1/(d^2*f^4))*sin(f*x + e) + sqrt(2)*d*f*sq
rt(d*cos(f*x + e)/sin(f*x + e))*(1/(d^2*f^4))^(1/4)*sin(f*x + e) + d*cos(f*x + e))/sin(f*x + e)) + 3*sqrt(2)*d
*f*(1/(d^2*f^4))^(1/4)*cos(f*x + e)^2*log((d^2*f^2*sqrt(1/(d^2*f^4))*sin(f*x + e) - sqrt(2)*d*f*sqrt(d*cos(f*x
+ e)/sin(f*x + e))*(1/(d^2*f^4))^(1/4)*sin(f*x + e) + d*cos(f*x + e))/sin(f*x + e)) + 8*(cos(f*x + e)^2 - 1)*
sqrt(d*cos(f*x + e)/sin(f*x + e)))/(d*f*cos(f*x + e)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(d*cot(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.64, size = 548, normalized size = 2.58 \[ \frac {\left (1+\cos \left (f x +e \right )\right )^{2} \left (-1+\cos \left (f x +e \right )\right ) \left (3 i \cos \left (f x +e \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 )}}\, \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 )-3 i \cos \left (f x +e \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 )}}\, \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 )-3 \cos \left (f x +e \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 )}}\, \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 )-3 \cos \left (f x +e \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 )}}\, \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 )+2 \cos \left (f x +e \right ) \sqrt {2}-2 \sqrt {2}\right ) \sqrt {2}}{6 f \sin \left (f x +e \right )^{3} \sqrt {\frac {d \cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \cos \left (f x +e \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^2/(d*cot(f*x+e))^(1/2),x)

[Out]

1/6/f*(1+cos(f*x+e))^2*(-1+cos(f*x+e))*(3*I*cos(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)*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))-3*I*cos(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)*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))-3*cos(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)*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))-3*cos(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)*EllipticPi((-(-sin(f*x+e)-1+c
os(f*x+e))/sin(f*x+e))^(1/2),1/2+1/2*I,1/2*2^(1/2))+2*cos(f*x+e)*2^(1/2)-2*2^(1/2))/sin(f*x+e)^3/(d*cos(f*x+e)
/sin(f*x+e))^(1/2)/cos(f*x+e)*2^(1/2)

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maxima [A]  time = 0.77, size = 190, normalized size = 0.90 \[ \frac {d^{3} {\left (\frac {3 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {\frac {d}{\tan \left (f x + e\right )}}\right )}}{2 \, \sqrt {d}}\right )}{d^{\frac {3}{2}}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {\frac {d}{\tan \left (f x + e\right )}}\right )}}{2 \, \sqrt {d}}\right )}{d^{\frac {3}{2}}} + \frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {d} \sqrt {\frac {d}{\tan \left (f x + e\right )}} + d + \frac {d}{\tan \left (f x + e\right )}\right )}{d^{\frac {3}{2}}} - \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {d} \sqrt {\frac {d}{\tan \left (f x + e\right )}} + d + \frac {d}{\tan \left (f x + e\right )}\right )}{d^{\frac {3}{2}}}\right )}}{d^{2}} + \frac {8}{d^{2} \left (\frac {d}{\tan \left (f x + e\right )}\right )^{\frac {3}{2}}}\right )}}{12 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(d*cot(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

1/12*d^3*(3*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d/tan(f*x + e)))/sqrt(d))/d^(3/2) + 2*sqrt
(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d/tan(f*x + e)))/sqrt(d))/d^(3/2) + sqrt(2)*log(sqrt(2)*sqrt
(d)*sqrt(d/tan(f*x + e)) + d + d/tan(f*x + e))/d^(3/2) - sqrt(2)*log(-sqrt(2)*sqrt(d)*sqrt(d/tan(f*x + e)) + d
+ d/tan(f*x + e))/d^(3/2))/d^2 + 8/(d^2*(d/tan(f*x + e))^(3/2)))/f

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mupad [B]  time = 2.51, size = 81, normalized size = 0.38 \[ \frac {2\,d}{3\,f\,{\left (\frac {d}{\mathrm {tan}\left (e+f\,x\right )}\right )}^{3/2}}-\frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {\frac {d}{\mathrm {tan}\left (e+f\,x\right )}}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{\sqrt {d}\,f}-\frac {{\left (-1\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {\frac {d}{\mathrm {tan}\left (e+f\,x\right )}}}{\sqrt {d}}\right )\,1{}\mathrm {i}}{\sqrt {d}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^2/(d*cot(e + f*x))^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**2/(d*cot(f*x+e))**(1/2),x)

[Out]

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

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