3.2.49 \(\int \tanh (a+2 \log (x)) \, dx\) [149]

Optimal. Leaf size=145 \[ x+\frac {e^{-a/2} \text {ArcTan}\left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-a/2} \text {ArcTan}\left (1+\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}+\frac {e^{-a/2} \log \left (1-\sqrt {2} e^{a/2} x+e^a x^2\right )}{2 \sqrt {2}}-\frac {e^{-a/2} \log \left (1+\sqrt {2} e^{a/2} x+e^a x^2\right )}{2 \sqrt {2}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.143, Rules used = {5652, 396, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {e^{-a/2} \text {ArcTan}\left (1-\sqrt {2} e^{a/2} x\right )}{\sqrt {2}}-\frac {e^{-a/2} \text {ArcTan}\left (\sqrt {2} e^{a/2} x+1\right )}{\sqrt {2}}+\frac {e^{-a/2} \log \left (e^a x^2-\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}-\frac {e^{-a/2} \log \left (e^a x^2+\sqrt {2} e^{a/2} x+1\right )}{2 \sqrt {2}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tanh[a + 2*Log[x]],x]

[Out]

x + ArcTan[1 - Sqrt[2]*E^(a/2)*x]/(Sqrt[2]*E^(a/2)) - ArcTan[1 + Sqrt[2]*E^(a/2)*x]/(Sqrt[2]*E^(a/2)) + Log[1
- Sqrt[2]*E^(a/2)*x + E^a*x^2]/(2*Sqrt[2]*E^(a/2)) - Log[1 + Sqrt[2]*E^(a/2)*x + E^a*x^2]/(2*Sqrt[2]*E^(a/2))

Rule 210

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

Rule 217

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 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 631

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 642

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 1176

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 1179

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 5652

Int[Tanh[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Int[(-1 + E^(2*a*d)*x^(2*b*d))^p/(1 + E^(2*a*d)*x^
(2*b*d))^p, x] /; FreeQ[{a, b, d, p}, x]

Rubi steps

\begin {align*} \int \tanh (a+2 \log (x)) \, dx &=\int \tanh (a+2 \log (x)) \, dx\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 0.13, size = 58, normalized size = 0.40 \begin {gather*} x+\frac {1}{2} \text {RootSum}\left [\cosh (a)-\sinh (a)+\cosh (a) \text {$\#$1}^4+\sinh (a) \text {$\#$1}^4\&,\frac {\log (x)-\log (x-\text {$\#$1})}{\text {$\#$1}^3}\&\right ] (\cosh (2 a)-\sinh (2 a)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tanh[a + 2*Log[x]],x]

[Out]

x + (RootSum[Cosh[a] - Sinh[a] + Cosh[a]*#1^4 + Sinh[a]*#1^4 & , (Log[x] - Log[x - #1])/#1^3 & ]*(Cosh[2*a] -
Sinh[2*a]))/2

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.62, size = 33, normalized size = 0.23

method	result	size
risch	\(x -\frac {{\mathrm e}^{-2 a} \left (\munderset {\textit {\_R} =\RootOf \left ({\mathrm e}^{2 a} \textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{2}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a+2*ln(x)),x,method=_RETURNVERBOSE)

[Out]

x-1/2*exp(-2*a)*sum(1/_R^3*ln(x-_R),_R=RootOf(exp(2*a)*_Z^4+1))

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Maxima [A]
time = 0.50, size = 124, normalized size = 0.86 \begin {gather*} -\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} + \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x e^{a} - \sqrt {2} e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (-\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} + \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (x^{2} e^{a} - \sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + 1\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+2*log(x)),x, algorithm="maxima")

[Out]

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(2*x*e^a + sqrt(2)*e^(1/2*a))*e^(-1/2*a))*e^(-1/2*a) - 1/2*sqrt(2)*arctan(1/2*
sqrt(2)*(2*x*e^a - sqrt(2)*e^(1/2*a))*e^(-1/2*a))*e^(-1/2*a) - 1/4*sqrt(2)*e^(-1/2*a)*log(x^2*e^a + sqrt(2)*x*
e^(1/2*a) + 1) + 1/4*sqrt(2)*e^(-1/2*a)*log(x^2*e^a - sqrt(2)*x*e^(1/2*a) + 1) + x

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Fricas [A]
time = 0.37, size = 154, normalized size = 1.06 \begin {gather*} \sqrt {2} \arctan \left (-\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + \sqrt {2} \sqrt {\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}} e^{\left (\frac {1}{2} \, a\right )} - 1\right ) e^{\left (-\frac {1}{2} \, a\right )} + \sqrt {2} \arctan \left (-\sqrt {2} x e^{\left (\frac {1}{2} \, a\right )} + \sqrt {2} \sqrt {-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}} e^{\left (\frac {1}{2} \, a\right )} + 1\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+2*log(x)),x, algorithm="fricas")

[Out]

sqrt(2)*arctan(-sqrt(2)*x*e^(1/2*a) + sqrt(2)*sqrt(sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a))*e^(1/2*a) - 1)*e^(-1/2
*a) + sqrt(2)*arctan(-sqrt(2)*x*e^(1/2*a) + sqrt(2)*sqrt(-sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a))*e^(1/2*a) + 1)*
e^(-1/2*a) - 1/4*sqrt(2)*e^(-1/2*a)*log(sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a)) + 1/4*sqrt(2)*e^(-1/2*a)*log(-sqr
t(2)*x*e^(-1/2*a) + x^2 + e^(-a)) + x

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \tanh {\left (a + 2 \log {\left (x \right )} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+2*ln(x)),x)

[Out]

Integral(tanh(a + 2*log(x)), x)

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Giac [A]
time = 0.41, size = 119, normalized size = 0.82 \begin {gather*} -\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} + 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} - 2 \, x\right )} e^{\left (\frac {1}{2} \, a\right )}\right ) e^{\left (-\frac {1}{2} \, a\right )} - \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + \frac {1}{4} \, \sqrt {2} e^{\left (-\frac {1}{2} \, a\right )} \log \left (-\sqrt {2} x e^{\left (-\frac {1}{2} \, a\right )} + x^{2} + e^{\left (-a\right )}\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(a+2*log(x)),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*e^(-1/2*a) + 2*x)*e^(1/2*a))*e^(-1/2*a) - 1/2*sqrt(2)*arctan(-1/2*sqr
t(2)*(sqrt(2)*e^(-1/2*a) - 2*x)*e^(1/2*a))*e^(-1/2*a) - 1/4*sqrt(2)*e^(-1/2*a)*log(sqrt(2)*x*e^(-1/2*a) + x^2
+ e^(-a)) + 1/4*sqrt(2)*e^(-1/2*a)*log(-sqrt(2)*x*e^(-1/2*a) + x^2 + e^(-a)) + x

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Mupad [B]
time = 1.07, size = 44, normalized size = 0.30 \begin {gather*} x-\frac {\mathrm {atan}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}}-\frac {\mathrm {atanh}\left (x\,{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}\right )}{{\left (-{\mathrm {e}}^{2\,a}\right )}^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a + 2*log(x)),x)

[Out]

x - atan(x*(-exp(2*a))^(1/4))/(-exp(2*a))^(1/4) - atanh(x*(-exp(2*a))^(1/4))/(-exp(2*a))^(1/4)

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