\(\int (e x)^m \tanh ^p(d (a+b \log (c x^n))) \, dx\) [193]

Optimal result

Integrand size = 21, antiderivative size = 135 \[ \int (e x)^m \tanh ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^{1+m} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^{-p} \left (-1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )^p \operatorname {AppellF1}\left (\frac {1+m}{2 b d n},-p,p,1+\frac {1+m}{2 b d n},e^{2 a d} \left (c x^n\right )^{2 b d},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{e (1+m)} \]

[Out]

(e*x)^(1+m)*(-1+exp(2*a*d)*(c*x^n)^(2*b*d))^p*AppellF1(1/2*(1+m)/b/d/n,-p,p,1+1/2*(1+m)/b/d/n,exp(2*a*d)*(c*x^
n)^(2*b*d),-exp(2*a*d)*(c*x^n)^(2*b*d))/e/(1+m)/((1-exp(2*a*d)*(c*x^n)^(2*b*d))^p)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5658, 5656, 525, 524} \[ \int (e x)^m \tanh ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(e x)^{m+1} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^{-p} \left (e^{2 a d} \left (c x^n\right )^{2 b d}-1\right )^p \operatorname {AppellF1}\left (\frac {m+1}{2 b d n},-p,p,\frac {m+1}{2 b d n}+1,e^{2 a d} \left (c x^n\right )^{2 b d},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{e (m+1)} \]

[In]

Int[(e*x)^m*Tanh[d*(a + b*Log[c*x^n])]^p,x]

[Out]

((e*x)^(1 + m)*(-1 + E^(2*a*d)*(c*x^n)^(2*b*d))^p*AppellF1[(1 + m)/(2*b*d*n), -p, p, 1 + (1 + m)/(2*b*d*n), E^
(2*a*d)*(c*x^n)^(2*b*d), -(E^(2*a*d)*(c*x^n)^(2*b*d))])/(e*(1 + m)*(1 - E^(2*a*d)*(c*x^n)^(2*b*d))^p)

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

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

Rule 5658

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left ((e x)^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int x^{-1+\frac {1+m}{n}} \tanh ^p(d (a+b \log (x))) \, dx,x,c x^n\right )}{e n} \\ & = \frac {\left ((e x)^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}}\right ) \text {Subst}\left (\int x^{-1+\frac {1+m}{n}} \left (-1+e^{2 a d} x^{2 b d}\right )^p \left (1+e^{2 a d} x^{2 b d}\right )^{-p} \, dx,x,c x^n\right )}{e n} \\ & = \frac {\left ((e x)^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^{-p} \left (-1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )^p\right ) \text {Subst}\left (\int x^{-1+\frac {1+m}{n}} \left (1-e^{2 a d} x^{2 b d}\right )^p \left (1+e^{2 a d} x^{2 b d}\right )^{-p} \, dx,x,c x^n\right )}{e n} \\ & = \frac {(e x)^{1+m} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^{-p} \left (-1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )^p \operatorname {AppellF1}\left (\frac {1+m}{2 b d n},-p,p,1+\frac {1+m}{2 b d n},e^{2 a d} \left (c x^n\right )^{2 b d},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{e (1+m)} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 1.28 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.29 \[ \int (e x)^m \tanh ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x (e x)^m \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^{-p} \left (\frac {-1+e^{2 a d} \left (c x^n\right )^{2 b d}}{1+e^{2 a d} \left (c x^n\right )^{2 b d}}\right )^p \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )^p \operatorname {AppellF1}\left (\frac {1+m}{2 b d n},-p,p,1+\frac {1+m}{2 b d n},e^{2 a d} \left (c x^n\right )^{2 b d},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{1+m} \]

[In]

Integrate[(e*x)^m*Tanh[d*(a + b*Log[c*x^n])]^p,x]

[Out]

(x*(e*x)^m*((-1 + E^(2*a*d)*(c*x^n)^(2*b*d))/(1 + E^(2*a*d)*(c*x^n)^(2*b*d)))^p*(1 + E^(2*a*d)*(c*x^n)^(2*b*d)
)^p*AppellF1[(1 + m)/(2*b*d*n), -p, p, 1 + (1 + m)/(2*b*d*n), E^(2*a*d)*(c*x^n)^(2*b*d), -(E^(2*a*d)*(c*x^n)^(
2*b*d))])/((1 + m)*(1 - E^(2*a*d)*(c*x^n)^(2*b*d))^p)

Maple [F]

\[\int \left (e x \right )^{m} {\tanh \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{p}d x\]

[In]

int((e*x)^m*tanh(d*(a+b*ln(c*x^n)))^p,x)

[Out]

int((e*x)^m*tanh(d*(a+b*ln(c*x^n)))^p,x)

Fricas [F]

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

[In]

integrate((e*x)^m*tanh(d*(a+b*log(c*x^n)))^p,x, algorithm="fricas")

[Out]

integral((e*x)^m*tanh(b*d*log(c*x^n) + a*d)^p, x)

Sympy [F(-1)]

Timed out. \[ \int (e x)^m \tanh ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {Timed out} \]

[In]

integrate((e*x)**m*tanh(d*(a+b*ln(c*x**n)))**p,x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate((e*x)^m*tanh(d*(a+b*log(c*x^n)))^p,x, algorithm="maxima")

[Out]

integrate((e*x)^m*tanh((b*log(c*x^n) + a)*d)^p, x)

Giac [F]

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

[In]

integrate((e*x)^m*tanh(d*(a+b*log(c*x^n)))^p,x, algorithm="giac")

[Out]

integrate((e*x)^m*tanh((b*log(c*x^n) + a)*d)^p, x)

Mupad [F(-1)]

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

[In]

int(tanh(d*(a + b*log(c*x^n)))^p*(e*x)^m,x)

[Out]

int(tanh(d*(a + b*log(c*x^n)))^p*(e*x)^m, x)