\(\int (e x)^m \tanh ^p(d (a+b \log (c x^n))) \, dx\) [203]

Optimal result

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

(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)
 

Mathematica [A] (warning: unable to verify)

Time = 0.75 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.29 \[ \int (e x)^m \tanh ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x (e x)^m \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^{-p} \left (\frac {-1+e^{2 a d} \left (c x^n\right )^{2 b d}}{1+e^{2 a d} \left (c x^n\right )^{2 b d}}\right )^p \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )^p \operatorname {AppellF1}\left (\frac {1+m}{2 b d n},-p,p,1+\frac {1+m}{2 b d n},e^{2 a d} \left (c x^n\right )^{2 b d},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{1+m} \] Input:

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

Output:

(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)
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6073, 6071, 1013, 1012}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (e x)^m \tanh ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\)

\(\Big \downarrow \) 6073

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (c x^n\right )^{\frac {m+1}{n}-1} \tanh ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right )d\left (c x^n\right )}{e n}\)

\(\Big \downarrow \) 6071

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (c x^n\right )^{\frac {m+1}{n}-1} \left (e^{2 a d} \left (c x^n\right )^{2 b d}-1\right )^p \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )^{-p}d\left (c x^n\right )}{e n}\)

\(\Big \downarrow \) 1013

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^{-p} \left (e^{2 a d} \left (c x^n\right )^{2 b d}-1\right )^p \int \left (c x^n\right )^{\frac {m+1}{n}-1} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^p \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )^{-p}d\left (c x^n\right )}{e n}\)

\(\Big \downarrow \) 1012

\(\displaystyle \frac {(e x)^{m+1} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^{-p} \left (e^{2 a d} \left (c x^n\right )^{2 b d}-1\right )^p \operatorname {AppellF1}\left (\frac {m+1}{2 b d n},-p,p,\frac {m+1}{2 b d n}+1,e^{2 a d} \left (c x^n\right )^{2 b d},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{e (m+1)}\)

Input:

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

Output:

((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)
 

Defintions of rubi rules used

rule 1012

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] /; 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]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1013

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^IntPart[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 6071

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 6073

Int[((e_.)*(x_))^(m_.)*Tanh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p 
_.), x_Symbol] :> Simp[(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])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (e x)^m \tanh ^p\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {e^{m} \left (x^{m} {\tanh \left (\mathrm {log}\left (x^{n} c \right ) b d +a d \right )}^{p} x -\left (\int \frac {x^{m} {\tanh \left (\mathrm {log}\left (x^{n} c \right ) b d +a d \right )}^{p}}{\tanh \left (\mathrm {log}\left (x^{n} c \right ) b d +a d \right )}d x \right ) b d n p +\left (\int x^{m} {\tanh \left (\mathrm {log}\left (x^{n} c \right ) b d +a d \right )}^{p} \tanh \left (\mathrm {log}\left (x^{n} c \right ) b d +a d \right )d x \right ) b d n p \right )}{m +1} \] Input:

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

Output:

(e**m*(x**m*tanh(log(x**n*c)*b*d + a*d)**p*x - int((x**m*tanh(log(x**n*c)* 
b*d + a*d)**p)/tanh(log(x**n*c)*b*d + a*d),x)*b*d*n*p + int(x**m*tanh(log( 
x**n*c)*b*d + a*d)**p*tanh(log(x**n*c)*b*d + a*d),x)*b*d*n*p))/(m + 1)