3.183 \(\int \frac{1}{x \sqrt{\tan (a+b \log (c x^n))}} \, dx\)

Optimal. Leaf size=176 \[ -\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt{2} b n}+\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{\sqrt{2} b n}-\frac{\log \left (\tan \left (a+b \log \left (c x^n\right )\right )-\sqrt{2} \sqrt{\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{2 \sqrt{2} b n}+\frac{\log \left (\tan \left (a+b \log \left (c x^n\right )\right )+\sqrt{2} \sqrt{\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{2 \sqrt{2} b n} \]

[Out]

-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]]]/(Sqrt[2]*b*n)) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n
]]]]/(Sqrt[2]*b*n) - Log[1 - Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]] + Tan[a + b*Log[c*x^n]]]/(2*Sqrt[2]*b*n) + Lo
g[1 + Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]] + Tan[a + b*Log[c*x^n]]]/(2*Sqrt[2]*b*n)

________________________________________________________________________________________

Rubi [A]  time = 0.120569, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {3476, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt{2} b n}+\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{\sqrt{2} b n}-\frac{\log \left (\tan \left (a+b \log \left (c x^n\right )\right )-\sqrt{2} \sqrt{\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{2 \sqrt{2} b n}+\frac{\log \left (\tan \left (a+b \log \left (c x^n\right )\right )+\sqrt{2} \sqrt{\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{2 \sqrt{2} b n} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[Tan[a + b*Log[c*x^n]]]),x]

[Out]

-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]]]/(Sqrt[2]*b*n)) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n
]]]]/(Sqrt[2]*b*n) - Log[1 - Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]] + Tan[a + b*Log[c*x^n]]]/(2*Sqrt[2]*b*n) + Lo
g[1 + Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]] + Tan[a + b*Log[c*x^n]]]/(2*Sqrt[2]*b*n)

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 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 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 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 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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.13451, size = 142, normalized size = 0.81 \[ \frac{-2 \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan \left (a+b \log \left (c x^n\right )\right )}\right )+2 \tan ^{-1}\left (\sqrt{2} \sqrt{\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )-\log \left (\tan \left (a+b \log \left (c x^n\right )\right )-\sqrt{2} \sqrt{\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )+\log \left (\tan \left (a+b \log \left (c x^n\right )\right )+\sqrt{2} \sqrt{\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{2 \sqrt{2} b n} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[Tan[a + b*Log[c*x^n]]]),x]

[Out]

(-2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]]] + 2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]]] - Log[
1 - Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]] + Tan[a + b*Log[c*x^n]]] + Log[1 + Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]]
+ Tan[a + b*Log[c*x^n]]])/(2*Sqrt[2]*b*n)

________________________________________________________________________________________

Maple [A]  time = 0.031, size = 140, normalized size = 0.8 \begin{align*}{\frac{\sqrt{2}}{4\,bn}\ln \left ({ \left ( 1+\sqrt{2}\sqrt{\tan \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) }+\tan \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) \left ( 1-\sqrt{2}\sqrt{\tan \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) }+\tan \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{-1}} \right ) }+{\frac{\sqrt{2}}{2\,bn}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) } \right ) }+{\frac{\sqrt{2}}{2\,bn}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/tan(a+b*ln(c*x^n))^(1/2),x)

[Out]

1/4/b/n*2^(1/2)*ln((1+2^(1/2)*tan(a+b*ln(c*x^n))^(1/2)+tan(a+b*ln(c*x^n)))/(1-2^(1/2)*tan(a+b*ln(c*x^n))^(1/2)
+tan(a+b*ln(c*x^n))))+1/2*arctan(1+2^(1/2)*tan(a+b*ln(c*x^n))^(1/2))/b/n*2^(1/2)+1/2*arctan(-1+2^(1/2)*tan(a+b
*ln(c*x^n))^(1/2))/b/n*2^(1/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \sqrt{\tan \left (b \log \left (c x^{n}\right ) + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/tan(a+b*log(c*x^n))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(x*sqrt(tan(b*log(c*x^n) + a))), x)

________________________________________________________________________________________

Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/tan(a+b*log(c*x^n))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \sqrt{\tan{\left (a + b \log{\left (c x^{n} \right )} \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/tan(a+b*ln(c*x**n))**(1/2),x)

[Out]

Integral(1/(x*sqrt(tan(a + b*log(c*x**n)))), x)

________________________________________________________________________________________

Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/tan(a+b*log(c*x^n))^(1/2),x, algorithm="giac")

[Out]

Timed out