\(\int x^3 \tanh (a+2 \log (x)) \, dx\) [146]

Optimal result

Integrand size = 11, antiderivative size = 29 \[ \int x^3 \tanh (a+2 \log (x)) \, dx=\frac {x^4}{4}-\frac {1}{2} e^{-2 a} \log \left (1+e^{2 a} x^4\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {5656, 455, 45} \[ \int x^3 \tanh (a+2 \log (x)) \, dx=\frac {x^4}{4}-\frac {1}{2} e^{-2 a} \log \left (e^{2 a} x^4+1\right ) \]

[In]

Int[x^3*Tanh[a + 2*Log[x]],x]

[Out]

x^4/4 - Log[1 + E^(2*a)*x^4]/(2*E^(2*a))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 455

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

Rule 5656

Int[((e_.)*(x_))^(m_.)*Tanh[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Int[(e*x)^m*((-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, e, m, p}, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^3 \left (-1+e^{2 a} x^4\right )}{1+e^{2 a} x^4} \, dx \\ & = \frac {1}{4} \text {Subst}\left (\int \frac {-1+e^{2 a} x}{1+e^{2 a} x} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (1-\frac {2}{1+e^{2 a} x}\right ) \, dx,x,x^4\right ) \\ & = \frac {x^4}{4}-\frac {1}{2} e^{-2 a} \log \left (1+e^{2 a} x^4\right ) \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(64\) vs. \(2(29)=58\).

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.21 \[ \int x^3 \tanh (a+2 \log (x)) \, dx=\frac {x^4}{4}-\frac {1}{2} \cosh (2 a) \log \left (\cosh (a)+x^4 \cosh (a)-\sinh (a)+x^4 \sinh (a)\right )+\frac {1}{2} \log \left (\cosh (a)+x^4 \cosh (a)-\sinh (a)+x^4 \sinh (a)\right ) \sinh (2 a) \]

[In]

Integrate[x^3*Tanh[a + 2*Log[x]],x]

[Out]

x^4/4 - (Cosh[2*a]*Log[Cosh[a] + x^4*Cosh[a] - Sinh[a] + x^4*Sinh[a]])/2 + (Log[Cosh[a] + x^4*Cosh[a] - Sinh[a
] + x^4*Sinh[a]]*Sinh[2*a])/2

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {x^{4}}{4}-\frac {{\mathrm e}^{-2 a} \ln \left (1+{\mathrm e}^{2 a} x^{4}\right )}{2}\)	\(24\)

[In]

int(x^3*tanh(a+2*ln(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int x^3 \tanh (a+2 \log (x)) \, dx=\frac {1}{4} \, {\left (x^{4} e^{\left (2 \, a\right )} - 2 \, \log \left (x^{4} e^{\left (2 \, a\right )} + 1\right )\right )} e^{\left (-2 \, a\right )} \]

[In]

integrate(x^3*tanh(a+2*log(x)),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int x^3 \tanh (a+2 \log (x)) \, dx=\int x^{3} \tanh {\left (a + 2 \log {\left (x \right )} \right )}\, dx \]

[In]

integrate(x**3*tanh(a+2*ln(x)),x)

[Out]

Integral(x**3*tanh(a + 2*log(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int x^3 \tanh (a+2 \log (x)) \, dx=\frac {1}{4} \, x^{4} - \frac {1}{2} \, e^{\left (-2 \, a\right )} \log \left (x^{4} e^{\left (2 \, a\right )} + 1\right ) \]

[In]

integrate(x^3*tanh(a+2*log(x)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int x^3 \tanh (a+2 \log (x)) \, dx=\frac {1}{4} \, x^{4} - \frac {1}{2} \, e^{\left (-2 \, a\right )} \log \left (x^{4} e^{\left (2 \, a\right )} + 1\right ) \]

[In]

integrate(x^3*tanh(a+2*log(x)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.71 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int x^3 \tanh (a+2 \log (x)) \, dx=\frac {x^4}{4}-\frac {{\mathrm {e}}^{-2\,a}\,\ln \left (x^4+{\mathrm {e}}^{-2\,a}\right )}{2} \]

[In]

int(x^3*tanh(a + 2*log(x)),x)

[Out]

x^4/4 - (exp(-2*a)*log(exp(-2*a) + x^4))/2