3.184 \(\int \frac {1}{x \tan ^{\frac {3}{2}}(a+b \log (c x^n))} \, dx\)

Optimal. Leaf size=199 \[ \frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt {2} b n}-\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{\sqrt {2} b n}-\frac {\log \left (\tan \left (a+b \log \left (c x^n\right )\right )-\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{2 \sqrt {2} b n}+\frac {\log \left (\tan \left (a+b \log \left (c x^n\right )\right )+\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{2 \sqrt {2} b n}-\frac {2}{b n \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3474, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac {\tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}\right )}{\sqrt {2} b n}-\frac {\tan ^{-1}\left (\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{\sqrt {2} b n}-\frac {\log \left (\tan \left (a+b \log \left (c x^n\right )\right )-\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{2 \sqrt {2} b n}+\frac {\log \left (\tan \left (a+b \log \left (c x^n\right )\right )+\sqrt {2} \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}+1\right )}{2 \sqrt {2} b n}-\frac {2}{b n \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Tan[a + b*Log[c*x^n]]^(3/2)),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) + Log[1
+ Sqrt[2]*Sqrt[Tan[a + b*Log[c*x^n]]] + Tan[a + b*Log[c*x^n]]]/(2*Sqrt[2]*b*n) - 2/(b*n*Sqrt[Tan[a + b*Log[c*
x^n]]])

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

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Mathematica [C]  time = 0.11, size = 46, normalized size = 0.23 \[ -\frac {2 \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\tan ^2\left (a+b \log \left (c x^n\right )\right )\right )}{b n \sqrt {\tan \left (a+b \log \left (c x^n\right )\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Tan[a + b*Log[c*x^n]]^(3/2)),x]

[Out]

(-2*Hypergeometric2F1[-1/4, 1, 3/4, -Tan[a + b*Log[c*x^n]]^2])/(b*n*Sqrt[Tan[a + b*Log[c*x^n]]])

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/tan(a+b*log(c*x^n))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   catdef: division by zero

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giac [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(1/x/tan(a+b*log(c*x^n))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.03, size = 161, normalized size = 0.81 \[ -\frac {\arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (a +b \ln \left (c \,x^{n}\right )\right )\right )\right ) \sqrt {2}}{2 b n}-\frac {\arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (a +b \ln \left (c \,x^{n}\right )\right )\right )\right ) \sqrt {2}}{2 b n}-\frac {\sqrt {2}\, \ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (a +b \ln \left (c \,x^{n}\right )\right )\right )+\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (a +b \ln \left (c \,x^{n}\right )\right )\right )+\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}\right )}{4 b n}-\frac {2}{b n \sqrt {\tan \left (a +b \ln \left (c \,x^{n}\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/tan(a+b*ln(c*x^n))^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \tan \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/tan(a+b*log(c*x^n))^(3/2),x, algorithm="maxima")

[Out]

integrate(1/(x*tan(b*log(c*x^n) + a)^(3/2)), x)

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mupad [B]  time = 2.92, size = 79, normalized size = 0.40 \[ \frac {{\left (-1\right )}^{1/4}\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}{b\,n}-\frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}{b\,n}-\frac {2}{b\,n\,\sqrt {\mathrm {tan}\left (a+b\,\ln \left (c\,x^n\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*tan(a + b*log(c*x^n))^(3/2)),x)

[Out]

((-1)^(1/4)*atanh((-1)^(1/4)*tan(a + b*log(c*x^n))^(1/2)))/(b*n) - ((-1)^(1/4)*atan((-1)^(1/4)*tan(a + b*log(c
*x^n))^(1/2)))/(b*n) - 2/(b*n*tan(a + b*log(c*x^n))^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \tan ^{\frac {3}{2}}{\left (a + b \log {\left (c x^{n} \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/tan(a+b*ln(c*x**n))**(3/2),x)

[Out]

Integral(1/(x*tan(a + b*log(c*x**n))**(3/2)), x)

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