\(\int (e x)^m \tanh ^2(a+2 \log (x)) \, dx\) [161]

Optimal result

Integrand size = 15, antiderivative size = 79 \[ \int (e x)^m \tanh ^2(a+2 \log (x)) \, dx=\frac {(e x)^{1+m}}{e (1+m)}+\frac {(e x)^{1+m}}{e \left (1+e^{2 a} x^4\right )}-\frac {(e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-e^{2 a} x^4\right )}{e} \]

[Out]

(e*x)^(1+m)/e/(1+m)+(e*x)^(1+m)/e/(1+exp(2*a)*x^4)-(e*x)^(1+m)*hypergeom([1, 1/4+1/4*m],[5/4+1/4*m],-exp(2*a)*
x^4)/e

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {5656, 474, 470, 371} \[ \int (e x)^m \tanh ^2(a+2 \log (x)) \, dx=-\frac {(e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{4},\frac {m+5}{4},-e^{2 a} x^4\right )}{e}+\frac {(e x)^{m+1}}{e \left (e^{2 a} x^4+1\right )}+\frac {(e x)^{m+1}}{e (m+1)} \]

[In]

Int[(e*x)^m*Tanh[a + 2*Log[x]]^2,x]

[Out]

(e*x)^(1 + m)/(e*(1 + m)) + (e*x)^(1 + m)/(e*(1 + E^(2*a)*x^4)) - ((e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/
4, (5 + m)/4, -(E^(2*a)*x^4)])/e

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 470

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

Rule 474

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(-(b*c - a*
d)^2)*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b^2*e*n*(p + 1))), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a +
b*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a
, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

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 {(e x)^m \left (-1+e^{2 a} x^4\right )^2}{\left (1+e^{2 a} x^4\right )^2} \, dx \\ & = \frac {(e x)^{1+m}}{e \left (1+e^{2 a} x^4\right )}-\frac {1}{4} e^{-4 a} \int \frac {(e x)^m \left (4 e^{4 a} m-4 e^{6 a} x^4\right )}{1+e^{2 a} x^4} \, dx \\ & = \frac {(e x)^{1+m}}{e (1+m)}+\frac {(e x)^{1+m}}{e \left (1+e^{2 a} x^4\right )}-(1+m) \int \frac {(e x)^m}{1+e^{2 a} x^4} \, dx \\ & = \frac {(e x)^{1+m}}{e (1+m)}+\frac {(e x)^{1+m}}{e \left (1+e^{2 a} x^4\right )}-\frac {(e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-e^{2 a} x^4\right )}{e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00 \[ \int (e x)^m \tanh ^2(a+2 \log (x)) \, dx=-\frac {x (e x)^m \left (-1+4 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{4},\frac {5+m}{4},-x^4 (\cosh (2 a)+\sinh (2 a))\right )-4 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{4},\frac {5+m}{4},-x^4 (\cosh (2 a)+\sinh (2 a))\right )\right )}{1+m} \]

[In]

Integrate[(e*x)^m*Tanh[a + 2*Log[x]]^2,x]

[Out]

-((x*(e*x)^m*(-1 + 4*Hypergeometric2F1[1, (1 + m)/4, (5 + m)/4, -(x^4*(Cosh[2*a] + Sinh[2*a]))] - 4*Hypergeome
tric2F1[2, (1 + m)/4, (5 + m)/4, -(x^4*(Cosh[2*a] + Sinh[2*a]))]))/(1 + m))

Maple [F]

\[\int \left (e x \right )^{m} \tanh \left (a +2 \ln \left (x \right )\right )^{2}d x\]

[In]

int((e*x)^m*tanh(a+2*ln(x))^2,x)

[Out]

int((e*x)^m*tanh(a+2*ln(x))^2,x)

Fricas [F]

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

[In]

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

[Out]

integral((e*x)^m*tanh(a + 2*log(x))^2, x)

Sympy [F]

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

[In]

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

[Out]

Integral((e*x)**m*tanh(a + 2*log(x))**2, x)

Maxima [F]

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

[In]

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

[Out]

integrate((e*x)^m*tanh(a + 2*log(x))^2, x)

Giac [F]

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

[In]

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

[Out]

integrate((e*x)^m*tanh(a + 2*log(x))^2, x)

Mupad [F(-1)]

Timed out. \[ \int (e x)^m \tanh ^2(a+2 \log (x)) \, dx=\int {\mathrm {tanh}\left (a+2\,\ln \left (x\right )\right )}^2\,{\left (e\,x\right )}^m \,d x \]

[In]

int(tanh(a + 2*log(x))^2*(e*x)^m,x)

[Out]

int(tanh(a + 2*log(x))^2*(e*x)^m, x)