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

Optimal. Leaf size=71 \[ \frac {\tanh ^{-1}\left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}-\frac {2}{b n \sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}}-\frac {\tan ^{-1}\left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )}{b n} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3474, 3476, 329, 298, 203, 206} \[ \frac {\tanh ^{-1}\left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )}{b n}-\frac {2}{b n \sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}}-\frac {\tan ^{-1}\left (\sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}\right )}{b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 206

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

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

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

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Mathematica [C]  time = 0.16, size = 44, normalized size = 0.62 \[ -\frac {2 \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};\tanh ^2\left (a+b \log \left (c x^n\right )\right )\right )}{b n \sqrt {\tanh \left (a+b \log \left (c x^n\right )\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.48, size = 625, normalized size = 8.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*(cosh(b*n*log(x) + b*log(c) + a)^2 + 2*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a)
+ sinh(b*n*log(x) + b*log(c) + a)^2 - 1)*arctan(-cosh(b*n*log(x) + b*log(c) + a)^2 - 2*cosh(b*n*log(x) + b*lo
g(c) + a)*sinh(b*n*log(x) + b*log(c) + a) - sinh(b*n*log(x) + b*log(c) + a)^2 + (cosh(b*n*log(x) + b*log(c) +
a)^2 + 2*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a) + sinh(b*n*log(x) + b*log(c) + a)^2 +
1)*sqrt(sinh(b*n*log(x) + b*log(c) + a)/cosh(b*n*log(x) + b*log(c) + a))) + 4*cosh(b*n*log(x) + b*log(c) + a)
^2 + (cosh(b*n*log(x) + b*log(c) + a)^2 + 2*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a) +
sinh(b*n*log(x) + b*log(c) + a)^2 - 1)*log(-cosh(b*n*log(x) + b*log(c) + a)^2 - 2*cosh(b*n*log(x) + b*log(c) +
a)*sinh(b*n*log(x) + b*log(c) + a) - sinh(b*n*log(x) + b*log(c) + a)^2 + (cosh(b*n*log(x) + b*log(c) + a)^2 +
2*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a) + sinh(b*n*log(x) + b*log(c) + a)^2 + 1)*sq
rt(sinh(b*n*log(x) + b*log(c) + a)/cosh(b*n*log(x) + b*log(c) + a))) + 8*cosh(b*n*log(x) + b*log(c) + a)*sinh(
b*n*log(x) + b*log(c) + a) + 4*sinh(b*n*log(x) + b*log(c) + a)^2 + 4*(cosh(b*n*log(x) + b*log(c) + a)^2 + 2*co
sh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a) + sinh(b*n*log(x) + b*log(c) + a)^2 + 1)*sqrt(si
nh(b*n*log(x) + b*log(c) + a)/cosh(b*n*log(x) + b*log(c) + a)) - 4)/(b*n*cosh(b*n*log(x) + b*log(c) + a)^2 + 2
*b*n*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a) + b*n*sinh(b*n*log(x) + b*log(c) + a)^2 -
b*n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 93, normalized size = 1.31 \[ -\frac {\ln \left (\sqrt {\tanh }\left (a +b \ln \left (c \,x^{n}\right )\right )-1\right )}{2 b n}+\frac {\ln \left (\sqrt {\tanh }\left (a +b \ln \left (c \,x^{n}\right )\right )+1\right )}{2 b n}-\frac {2}{b n \sqrt {\tanh \left (a +b \ln \left (c \,x^{n}\right )\right )}}-\frac {\arctan \left (\sqrt {\tanh }\left (a +b \ln \left (c \,x^{n}\right )\right )\right )}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.72, size = 65, normalized size = 0.92 \[ \frac {\mathrm {atanh}\left (\sqrt {\mathrm {tanh}\left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}{b\,n}-\frac {\mathrm {atan}\left (\sqrt {\mathrm {tanh}\left (a+b\,\ln \left (c\,x^n\right )\right )}\right )}{b\,n}-\frac {2}{b\,n\,\sqrt {\mathrm {tanh}\left (a+b\,\ln \left (c\,x^n\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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