3.401 \(\int \frac {\tan (x)}{(-1+\sqrt {\tan (x)})^2} \, dx\)

Optimal. Leaf size=84 \[ -\frac {x}{2}+\frac {\tan ^{-1}\left (\frac {1-\tan (x)}{\sqrt {2} \sqrt {\tan (x)}}\right )}{\sqrt {2}}+\frac {1}{1-\sqrt {\tan (x)}}+\log \left (1-\sqrt {\tan (x)}\right )+\frac {1}{2} \log (\cos (x))+\frac {\tanh ^{-1}\left (\frac {\tan (x)+1}{\sqrt {2} \sqrt {\tan (x)}}\right )}{\sqrt {2}} \]

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Rubi [A]  time = 0.37, antiderivative size = 133, normalized size of antiderivative = 1.58, number of steps used = 19, number of rules used = 13, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {3670, 6725, 1831, 297, 1162, 617, 204, 1165, 628, 1248, 635, 203, 260} \[ -\frac {x}{2}+\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (x)}\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {\tan (x)}+1\right )}{\sqrt {2}}+\frac {1}{1-\sqrt {\tan (x)}}+\log \left (1-\sqrt {\tan (x)}\right )-\frac {\log \left (\tan (x)-\sqrt {2} \sqrt {\tan (x)}+1\right )}{2 \sqrt {2}}+\frac {\log \left (\tan (x)+\sqrt {2} \sqrt {\tan (x)}+1\right )}{2 \sqrt {2}}+\frac {1}{2} \log (\cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/(-1 + Sqrt[Tan[x]])^2,x]

[Out]

-x/2 + ArcTan[1 - Sqrt[2]*Sqrt[Tan[x]]]/Sqrt[2] - ArcTan[1 + Sqrt[2]*Sqrt[Tan[x]]]/Sqrt[2] + Log[Cos[x]]/2 + L
og[1 - Sqrt[Tan[x]]] - Log[1 - Sqrt[2]*Sqrt[Tan[x]] + Tan[x]]/(2*Sqrt[2]) + Log[1 + Sqrt[2]*Sqrt[Tan[x]] + Tan
[x]]/(2*Sqrt[2]) + (1 - Sqrt[Tan[x]])^(-1)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 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 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

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 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 1831

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[((c*x)^(m + ii)*(Coeff[Pq,
x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2)))/(c^ii*(a + b*x^n)), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; Fr
eeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && Expon[Pq, x] < n

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\tan (x)}{\left (-1+\sqrt {\tan (x)}\right )^2} \, dx &=\operatorname {Subst}\left (\int \frac {x}{\left (-1+\sqrt {x}\right )^2 \left (1+x^2\right )} \, dx,x,\tan (x)\right )\\ &=2 \operatorname {Subst}\left (\int \frac {x^3}{(-1+x)^2 \left (1+x^4\right )} \, dx,x,\sqrt {\tan (x)}\right )\\ &=2 \operatorname {Subst}\left (\int \left (\frac {1}{2 (-1+x)^2}+\frac {1}{2 (-1+x)}-\frac {x (1+x)^2}{2 \left (1+x^4\right )}\right ) \, dx,x,\sqrt {\tan (x)}\right )\\ &=\log \left (1-\sqrt {\tan (x)}\right )+\frac {1}{1-\sqrt {\tan (x)}}-\operatorname {Subst}\left (\int \frac {x (1+x)^2}{1+x^4} \, dx,x,\sqrt {\tan (x)}\right )\\ &=\log \left (1-\sqrt {\tan (x)}\right )+\frac {1}{1-\sqrt {\tan (x)}}-\operatorname {Subst}\left (\int \left (\frac {2 x^2}{1+x^4}+\frac {x \left (1+x^2\right )}{1+x^4}\right ) \, dx,x,\sqrt {\tan (x)}\right )\\ &=\log \left (1-\sqrt {\tan (x)}\right )+\frac {1}{1-\sqrt {\tan (x)}}-2 \operatorname {Subst}\left (\int \frac {x^2}{1+x^4} \, dx,x,\sqrt {\tan (x)}\right )-\operatorname {Subst}\left (\int \frac {x \left (1+x^2\right )}{1+x^4} \, dx,x,\sqrt {\tan (x)}\right )\\ &=\log \left (1-\sqrt {\tan (x)}\right )+\frac {1}{1-\sqrt {\tan (x)}}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1+x}{1+x^2} \, dx,x,\tan (x)\right )+\operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (x)}\right )-\operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (x)}\right )\\ &=\log \left (1-\sqrt {\tan (x)}\right )+\frac {1}{1-\sqrt {\tan (x)}}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (x)\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\tan (x)\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (x)}\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (x)}\right )-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (x)}\right )}{2 \sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (x)}\right )}{2 \sqrt {2}}\\ &=-\frac {x}{2}+\frac {1}{2} \log (\cos (x))+\log \left (1-\sqrt {\tan (x)}\right )-\frac {\log \left (1-\sqrt {2} \sqrt {\tan (x)}+\tan (x)\right )}{2 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt {\tan (x)}+\tan (x)\right )}{2 \sqrt {2}}+\frac {1}{1-\sqrt {\tan (x)}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (x)}\right )}{\sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (x)}\right )}{\sqrt {2}}\\ &=-\frac {x}{2}+\frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (x)}\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (x)}\right )}{\sqrt {2}}+\frac {1}{2} \log (\cos (x))+\log \left (1-\sqrt {\tan (x)}\right )-\frac {\log \left (1-\sqrt {2} \sqrt {\tan (x)}+\tan (x)\right )}{2 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt {\tan (x)}+\tan (x)\right )}{2 \sqrt {2}}+\frac {1}{1-\sqrt {\tan (x)}}\\ \end {align*}

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Mathematica [C]  time = 0.28, size = 62, normalized size = 0.74 \[ -\frac {2}{3} \tan ^{\frac {3}{2}}(x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(x)\right )-\frac {1}{2} \tan ^{-1}(\tan (x))+\frac {1}{1-\sqrt {\tan (x)}}+\log \left (1-\sqrt {\tan (x)}\right )+\frac {1}{2} \log (\cos (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/(-1 + Sqrt[Tan[x]])^2,x]

[Out]

-1/2*ArcTan[Tan[x]] + Log[Cos[x]]/2 + Log[1 - Sqrt[Tan[x]]] + (1 - Sqrt[Tan[x]])^(-1) - (2*Hypergeometric2F1[3
/4, 1, 7/4, -Tan[x]^2]*Tan[x]^(3/2))/3

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan (x)}{\left (-1+\sqrt {\tan (x)}\right )^2} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Tan[x]/(-1 + Sqrt[Tan[x]])^2,x]

[Out]

Could not integrate

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fricas [C]  time = 3.25, size = 603, normalized size = 7.18 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(-1+tan(x)^(1/2))^2,x, algorithm="fricas")

[Out]

-1/8*(2*(2*sqrt(-I) - I + 1)*(tan(x) - 1)*log(-1/2*(2*sqrt(-I) - I + 1)^2*(4*(-1)^(1/4) + 2*I + 1) - (2*(-1)^(
1/4) + I + 1)^3 - ((2*(-1)^(1/4) + I + 1)^2 - 8*(-1)^(1/4) - 4*I - 3)*(2*sqrt(-I) - I + 1) + 4*(2*(-1)^(1/4) +
I + 1)^2 + 6*sqrt(tan(x)) - 16*(-1)^(1/4) - 8*I - 9) + 2*(2*(-1)^(1/4) + I + 1)*(tan(x) - 1)*log((2*(-1)^(1/4
) + I + 1)^3 - 7/2*(2*(-1)^(1/4) + I + 1)^2 + 6*sqrt(tan(x)) + 14*(-1)^(1/4) + 7*I + 14) - ((2*sqrt(-I) - I +
1)*(tan(x) - 1) + (2*(-1)^(1/4) + I + 1)*(tan(x) - 1) - 4*sqrt(-3/16*(2*sqrt(-I) - I + 1)^2 - 3/16*(2*(-1)^(1/
4) + I + 1)^2 - 1/8*(2*sqrt(-I) - I + 1)*(2*(-1)^(1/4) + I - 3) + (-1)^(1/4) + 1/2*I - 1/2)*(tan(x) - 1) - 4*t
an(x) + 4)*log(1/4*(2*sqrt(-I) - I + 1)^2*(4*(-1)^(1/4) + 2*I + 1) + 1/2*((2*(-1)^(1/4) + I + 1)^2 - 8*(-1)^(1
/4) - 4*I - 3)*(2*sqrt(-I) - I + 1) - 1/4*(2*(-1)^(1/4) + I + 1)^2 + sqrt(-3/16*(2*sqrt(-I) - I + 1)^2 - 3/16*
(2*(-1)^(1/4) + I + 1)^2 - 1/8*(2*sqrt(-I) - I + 1)*(2*(-1)^(1/4) + I - 3) + (-1)^(1/4) + 1/2*I - 1/2)*((2*sqr
t(-I) - I + 1)*(4*(-1)^(1/4) + 2*I + 1) - 2*(-1)^(1/4) - I + 1) + 6*sqrt(tan(x)) + (-1)^(1/4) + 1/2*I - 5/2) -
((2*sqrt(-I) - I + 1)*(tan(x) - 1) + (2*(-1)^(1/4) + I + 1)*(tan(x) - 1) + 4*sqrt(-3/16*(2*sqrt(-I) - I + 1)^
2 - 3/16*(2*(-1)^(1/4) + I + 1)^2 - 1/8*(2*sqrt(-I) - I + 1)*(2*(-1)^(1/4) + I - 3) + (-1)^(1/4) + 1/2*I - 1/2
)*(tan(x) - 1) - 4*tan(x) + 4)*log(1/4*(2*sqrt(-I) - I + 1)^2*(4*(-1)^(1/4) + 2*I + 1) + 1/2*((2*(-1)^(1/4) +
I + 1)^2 - 8*(-1)^(1/4) - 4*I - 3)*(2*sqrt(-I) - I + 1) - 1/4*(2*(-1)^(1/4) + I + 1)^2 - sqrt(-3/16*(2*sqrt(-I
) - I + 1)^2 - 3/16*(2*(-1)^(1/4) + I + 1)^2 - 1/8*(2*sqrt(-I) - I + 1)*(2*(-1)^(1/4) + I - 3) + (-1)^(1/4) +
1/2*I - 1/2)*((2*sqrt(-I) - I + 1)*(4*(-1)^(1/4) + 2*I + 1) - 2*(-1)^(1/4) - I + 1) + 6*sqrt(tan(x)) + (-1)^(1
/4) + 1/2*I - 5/2) - 8*(tan(x) - 1)*log(sqrt(tan(x)) - 1) + 8*sqrt(tan(x)) + 8)/(tan(x) - 1)

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giac [A]  time = 0.67, size = 111, normalized size = 1.32 \[ -\frac {1}{2} \, {\left (\sqrt {2} - 1\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \relax (x)}\right )}\right ) - \frac {1}{2} \, {\left (\sqrt {2} + 1\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \relax (x)}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \relax (x)} + \tan \relax (x) + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \relax (x)} + \tan \relax (x) + 1\right ) - \frac {1}{\sqrt {\tan \relax (x)} - 1} - \frac {1}{4} \, \log \left (\tan \relax (x)^{2} + 1\right ) + \log \left ({\left | \sqrt {\tan \relax (x)} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(-1+tan(x)^(1/2))^2,x, algorithm="giac")

[Out]

-1/2*(sqrt(2) - 1)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(x)))) - 1/2*(sqrt(2) + 1)*arctan(-1/2*sqrt(2)*(sqr
t(2) - 2*sqrt(tan(x)))) + 1/4*sqrt(2)*log(sqrt(2)*sqrt(tan(x)) + tan(x) + 1) - 1/4*sqrt(2)*log(-sqrt(2)*sqrt(t
an(x)) + tan(x) + 1) - 1/(sqrt(tan(x)) - 1) - 1/4*log(tan(x)^2 + 1) + log(abs(sqrt(tan(x)) - 1))

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maple [A]  time = 0.08, size = 94, normalized size = 1.12

method	result	size
derivativedivides	\(-\frac {\arctan \left (\tan \relax (x )\right )}{2}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\relax (x )\right )+\tan \relax (x )}{1+\sqrt {2}\, \left (\sqrt {\tan }\relax (x )\right )+\tan \relax (x )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\relax (x )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\relax (x )\right )\right )\right )}{4}-\frac {\ln \left (1+\tan ^{2}\relax (x )\right )}{4}-\frac {1}{-1+\sqrt {\tan }\relax (x )}+\ln \left (-1+\sqrt {\tan }\relax (x )\right )\)	\(94\)
default	\(-\frac {\arctan \left (\tan \relax (x )\right )}{2}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\relax (x )\right )+\tan \relax (x )}{1+\sqrt {2}\, \left (\sqrt {\tan }\relax (x )\right )+\tan \relax (x )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\relax (x )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\relax (x )\right )\right )\right )}{4}-\frac {\ln \left (1+\tan ^{2}\relax (x )\right )}{4}-\frac {1}{-1+\sqrt {\tan }\relax (x )}+\ln \left (-1+\sqrt {\tan }\relax (x )\right )\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(-1+tan(x)^(1/2))^2,x,method=_RETURNVERBOSE)

[Out]

-1/2*arctan(tan(x))-1/4*2^(1/2)*(ln((1-2^(1/2)*tan(x)^(1/2)+tan(x))/(1+2^(1/2)*tan(x)^(1/2)+tan(x)))+2*arctan(
1+2^(1/2)*tan(x)^(1/2))+2*arctan(-1+2^(1/2)*tan(x)^(1/2)))-1/4*ln(1+tan(x)^2)-1/(-1+tan(x)^(1/2))+ln(-1+tan(x)
^(1/2))

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maxima [A]  time = 0.97, size = 117, normalized size = 1.39 \[ \frac {1}{4} \, \sqrt {2} {\left (\sqrt {2} - 2\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \relax (x)}\right )}\right ) - \frac {1}{4} \, \sqrt {2} {\left (\sqrt {2} + 2\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \relax (x)}\right )}\right ) - \frac {1}{8} \, \sqrt {2} {\left (\sqrt {2} - 2\right )} \log \left (\sqrt {2} \sqrt {\tan \relax (x)} + \tan \relax (x) + 1\right ) - \frac {1}{8} \, \sqrt {2} {\left (\sqrt {2} + 2\right )} \log \left (-\sqrt {2} \sqrt {\tan \relax (x)} + \tan \relax (x) + 1\right ) - \frac {1}{\sqrt {\tan \relax (x)} - 1} + \log \left (\sqrt {\tan \relax (x)} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(-1+tan(x)^(1/2))^2,x, algorithm="maxima")

[Out]

1/4*sqrt(2)*(sqrt(2) - 2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(x)))) - 1/4*sqrt(2)*(sqrt(2) + 2)*arctan(-1
/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(x)))) - 1/8*sqrt(2)*(sqrt(2) - 2)*log(sqrt(2)*sqrt(tan(x)) + tan(x) + 1) - 1/
8*sqrt(2)*(sqrt(2) + 2)*log(-sqrt(2)*sqrt(tan(x)) + tan(x) + 1) - 1/(sqrt(tan(x)) - 1) + log(sqrt(tan(x)) - 1)

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mupad [B]  time = 1.27, size = 228, normalized size = 2.71 \[ \ln \left (612\,\sqrt {\mathrm {tan}\relax (x)}-612\right )-\frac {1}{\sqrt {\mathrm {tan}\relax (x)}-1}+\left (\sum _{k=1}^4\ln \left (4\,\sqrt {\mathrm {tan}\relax (x)}+{\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}-\frac {z}{8}+\frac {1}{64},z,k\right )}^2\,\sqrt {\mathrm {tan}\relax (x)}\,80+{\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}-\frac {z}{8}+\frac {1}{64},z,k\right )}^3\,\sqrt {\mathrm {tan}\relax (x)}\,448+{\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}-\frac {z}{8}+\frac {1}{64},z,k\right )}^4\,\sqrt {\mathrm {tan}\relax (x)}\,128+32\,{\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}-\frac {z}{8}+\frac {1}{64},z,k\right )}^2-384\,{\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}-\frac {z}{8}+\frac {1}{64},z,k\right )}^3-256\,{\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}-\frac {z}{8}+\frac {1}{64},z,k\right )}^4-\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}-\frac {z}{8}+\frac {1}{64},z,k\right )\,\sqrt {\mathrm {tan}\relax (x)}\,48-4\right )\,\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}-\frac {z}{8}+\frac {1}{64},z,k\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(tan(x)^(1/2) - 1)^2,x)

[Out]

log(612*tan(x)^(1/2) - 612) - 1/(tan(x)^(1/2) - 1) + symsum(log(4*tan(x)^(1/2) + 80*root(z^4 + z^3 + z^2/2 - z
/8 + 1/64, z, k)^2*tan(x)^(1/2) + 448*root(z^4 + z^3 + z^2/2 - z/8 + 1/64, z, k)^3*tan(x)^(1/2) + 128*root(z^4
+ z^3 + z^2/2 - z/8 + 1/64, z, k)^4*tan(x)^(1/2) + 32*root(z^4 + z^3 + z^2/2 - z/8 + 1/64, z, k)^2 - 384*root
(z^4 + z^3 + z^2/2 - z/8 + 1/64, z, k)^3 - 256*root(z^4 + z^3 + z^2/2 - z/8 + 1/64, z, k)^4 - 48*root(z^4 + z^
3 + z^2/2 - z/8 + 1/64, z, k)*tan(x)^(1/2) - 4)*root(z^4 + z^3 + z^2/2 - z/8 + 1/64, z, k), k, 1, 4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan {\relax (x )}}{\left (\sqrt {\tan {\relax (x )}} - 1\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(-1+tan(x)**(1/2))**2,x)

[Out]

Integral(tan(x)/(sqrt(tan(x)) - 1)**2, x)

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