\(\int (e x)^m \tanh ^3(d (a+b \log (c x^n))) \, dx\) [201]

Optimal result

Integrand size = 21, antiderivative size = 307 \[ \int (e x)^m \tanh ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {(1+m+b d n) (1+m+2 b d n) (e x)^{1+m}}{2 b^2 d^2 e (1+m) n^2}-\frac {(e x)^{1+m} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^2}{2 b d e n \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )^2}+\frac {e^{-2 a d} (e x)^{1+m} \left (\frac {e^{2 a d} (1+m-2 b d n)}{n}-\frac {e^{4 a d} (1+m+2 b d n) \left (c x^n\right )^{2 b d}}{n}\right )}{2 b^2 d^2 e n \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )}-\frac {\left (1+2 m+m^2+2 b^2 d^2 n^2\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2 b d n},1+\frac {1+m}{2 b d n},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b^2 d^2 e (1+m) n^2} \] Output:

1/2*(b*d*n+m+1)*(2*b*d*n+m+1)*(e*x)^(1+m)/b^2/d^2/e/(1+m)/n^2-1/2*(e*x)^(1 
+m)*(1-exp(2*a*d)*(c*x^n)^(2*b*d))^2/b/d/e/n/(1+exp(2*a*d)*(c*x^n)^(2*b*d) 
)^2+1/2*(e*x)^(1+m)*(exp(2*a*d)*(-2*b*d*n+m+1)/n-exp(4*a*d)*(2*b*d*n+m+1)* 
(c*x^n)^(2*b*d)/n)/b^2/d^2/e/exp(2*a*d)/n/(1+exp(2*a*d)*(c*x^n)^(2*b*d))-( 
2*b^2*d^2*n^2+m^2+2*m+1)*(e*x)^(1+m)*hypergeom([1, 1/2*(1+m)/b/d/n],[1+1/2 
*(1+m)/b/d/n],-exp(2*a*d)*(c*x^n)^(2*b*d))/b^2/d^2/e/(1+m)/n^2
 

Mathematica [A] (verified)

Time = 13.10 (sec) , antiderivative size = 606, normalized size of antiderivative = 1.97 \[ \int (e x)^m \tanh ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x (e x)^m \text {sech}^2\left (b d n \log (x)+d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )}{2 b d n}-\frac {(1+m) x (e x)^m \text {sech}\left (d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \text {sech}\left (b d n \log (x)+d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \sinh (b d n \log (x))}{2 b^2 d^2 n^2}+\frac {\left (1+2 m+m^2+2 b^2 d^2 n^2\right ) x^{-m} (e x)^m \text {sech}\left (d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \left (\frac {x^{1+m} \text {sech}\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \sinh (b d n \log (x))}{1+m}-\frac {e^{-\frac {(1+2 m) \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{b n}} \cosh \left (d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right ) \left (e^{\frac {a+2 a m+b (1+m) n \log (x)+b (1+2 m) \left (-n \log (x)+\log \left (c x^n\right )\right )}{b n}} (1+m+2 b d n) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2 b d n},1+\frac {1+m}{2 b d n},-e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )-e^{\frac {a (1+2 m+2 b d n)}{b n}+(1+m+2 b d n) \log (x)+\frac {(1+2 m+2 b d n) \left (-n \log (x)+\log \left (c x^n\right )\right )}{n}} (1+m) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m+2 b d n}{2 b d n},\frac {1+m+4 b d n}{2 b d n},-e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )+e^{\frac {a+2 a m+b (1+m) n \log (x)+b (1+2 m) \left (-n \log (x)+\log \left (c x^n\right )\right )}{b n}} (1+m+2 b d n) \tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right )\right )}{(1+m) (1+m+2 b d n)}\right )}{2 b^2 d^2 n^2}+\frac {x (e x)^m \tanh \left (d \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )}{1+m} \] Input:

Integrate[(e*x)^m*Tanh[d*(a + b*Log[c*x^n])]^3,x]
 

Output:

(x*(e*x)^m*Sech[b*d*n*Log[x] + d*(a + b*(-(n*Log[x]) + Log[c*x^n]))]^2)/(2 
*b*d*n) - ((1 + m)*x*(e*x)^m*Sech[d*(a + b*(-(n*Log[x]) + Log[c*x^n]))]*Se 
ch[b*d*n*Log[x] + d*(a + b*(-(n*Log[x]) + Log[c*x^n]))]*Sinh[b*d*n*Log[x]] 
)/(2*b^2*d^2*n^2) + ((1 + 2*m + m^2 + 2*b^2*d^2*n^2)*(e*x)^m*Sech[d*(a + b 
*(-(n*Log[x]) + Log[c*x^n]))]*((x^(1 + m)*Sech[d*(a + b*Log[c*x^n])]*Sinh[ 
b*d*n*Log[x]])/(1 + m) - (Cosh[d*(a + b*(-(n*Log[x]) + Log[c*x^n]))]*(E^(( 
a + 2*a*m + b*(1 + m)*n*Log[x] + b*(1 + 2*m)*(-(n*Log[x]) + Log[c*x^n]))/( 
b*n))*(1 + m + 2*b*d*n)*Hypergeometric2F1[1, (1 + m)/(2*b*d*n), 1 + (1 + m 
)/(2*b*d*n), -E^(2*d*(a + b*Log[c*x^n]))] - E^((a*(1 + 2*m + 2*b*d*n))/(b* 
n) + (1 + m + 2*b*d*n)*Log[x] + ((1 + 2*m + 2*b*d*n)*(-(n*Log[x]) + Log[c* 
x^n]))/n)*(1 + m)*Hypergeometric2F1[1, (1 + m + 2*b*d*n)/(2*b*d*n), (1 + m 
 + 4*b*d*n)/(2*b*d*n), -E^(2*d*(a + b*Log[c*x^n]))] + E^((a + 2*a*m + b*(1 
 + m)*n*Log[x] + b*(1 + 2*m)*(-(n*Log[x]) + Log[c*x^n]))/(b*n))*(1 + m + 2 
*b*d*n)*Tanh[d*(a + b*Log[c*x^n])]))/(E^(((1 + 2*m)*(a + b*(-(n*Log[x]) + 
Log[c*x^n])))/(b*n))*(1 + m)*(1 + m + 2*b*d*n))))/(2*b^2*d^2*n^2*x^m) + (x 
*(e*x)^m*Tanh[d*(a + b*(-(n*Log[x]) + Log[c*x^n]))])/(1 + m)
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.18, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {6073, 6071, 1004, 27, 1064, 27, 959, 888}

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.

\(\displaystyle \int (e x)^m \tanh ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\)

\(\Big \downarrow \) 6073

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (c x^n\right )^{\frac {m+1}{n}-1} \tanh ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right )d\left (c x^n\right )}{e n}\)

\(\Big \downarrow \) 6071

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \frac {\left (c x^n\right )^{\frac {m+1}{n}-1} \left (e^{2 a d} \left (c x^n\right )^{2 b d}-1\right )^3}{\left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )^3}d\left (c x^n\right )}{e n}\)

\(\Big \downarrow \) 1004

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (-\frac {e^{-2 a d} \int -\frac {2 \left (c x^n\right )^{\frac {m+1}{n}-1} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right ) \left (\frac {e^{2 a d} (m-2 b d n+1)}{n}-\frac {e^{4 a d} (m+2 b d n+1) \left (c x^n\right )^{2 b d}}{n}\right )}{\left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )^2}d\left (c x^n\right )}{4 b d}-\frac {\left (c x^n\right )^{\frac {m+1}{n}} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^2}{2 b d \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )^2}\right )}{e n}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (\frac {e^{-2 a d} \int \frac {\left (c x^n\right )^{\frac {m+1}{n}-1} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right ) \left (\frac {e^{2 a d} (m-2 b d n+1)}{n}-\frac {e^{4 a d} (m+2 b d n+1) \left (c x^n\right )^{2 b d}}{n}\right )}{\left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )^2}d\left (c x^n\right )}{2 b d}-\frac {\left (c x^n\right )^{\frac {m+1}{n}} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^2}{2 b d \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )^2}\right )}{e n}\)

\(\Big \downarrow \) 1064

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (\frac {e^{-2 a d} \left (\frac {\left (c x^n\right )^{\frac {m+1}{n}} \left (\frac {e^{2 a d} (-2 b d n+m+1)}{n}-\frac {e^{4 a d} (2 b d n+m+1) \left (c x^n\right )^{2 b d}}{n}\right )}{b d \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )}-\frac {e^{-2 a d} \int \frac {2 \left (c x^n\right )^{\frac {m+1}{n}-1} \left (\frac {e^{4 a d} (m-2 b d n+1) (m-b d n+1)}{n^2}-\frac {e^{6 a d} (m+b d n+1) (m+2 b d n+1) \left (c x^n\right )^{2 b d}}{n^2}\right )}{e^{2 a d} \left (c x^n\right )^{2 b d}+1}d\left (c x^n\right )}{2 b d}\right )}{2 b d}-\frac {\left (c x^n\right )^{\frac {m+1}{n}} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^2}{2 b d \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )^2}\right )}{e n}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (\frac {e^{-2 a d} \left (\frac {\left (c x^n\right )^{\frac {m+1}{n}} \left (\frac {e^{2 a d} (-2 b d n+m+1)}{n}-\frac {e^{4 a d} (2 b d n+m+1) \left (c x^n\right )^{2 b d}}{n}\right )}{b d \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )}-\frac {e^{-2 a d} \int \frac {\left (c x^n\right )^{\frac {m+1}{n}-1} \left (\frac {e^{4 a d} (m-2 b d n+1) (m-b d n+1)}{n^2}-\frac {e^{6 a d} (m+b d n+1) (m+2 b d n+1) \left (c x^n\right )^{2 b d}}{n^2}\right )}{e^{2 a d} \left (c x^n\right )^{2 b d}+1}d\left (c x^n\right )}{b d}\right )}{2 b d}-\frac {\left (c x^n\right )^{\frac {m+1}{n}} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^2}{2 b d \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )^2}\right )}{e n}\)

\(\Big \downarrow \) 959

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (\frac {e^{-2 a d} \left (\frac {\left (c x^n\right )^{\frac {m+1}{n}} \left (\frac {e^{2 a d} (-2 b d n+m+1)}{n}-\frac {e^{4 a d} (2 b d n+m+1) \left (c x^n\right )^{2 b d}}{n}\right )}{b d \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )}-\frac {e^{-2 a d} \left (\frac {2 e^{4 a d} \left (2 b^2 d^2 n^2+m^2+2 m+1\right ) \int \frac {\left (c x^n\right )^{\frac {m+1}{n}-1}}{e^{2 a d} \left (c x^n\right )^{2 b d}+1}d\left (c x^n\right )}{n^2}-\frac {e^{4 a d} (b d n+m+1) (2 b d n+m+1) \left (c x^n\right )^{\frac {m+1}{n}}}{(m+1) n}\right )}{b d}\right )}{2 b d}-\frac {\left (c x^n\right )^{\frac {m+1}{n}} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^2}{2 b d \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )^2}\right )}{e n}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {(e x)^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \left (\frac {e^{-2 a d} \left (\frac {\left (c x^n\right )^{\frac {m+1}{n}} \left (\frac {e^{2 a d} (-2 b d n+m+1)}{n}-\frac {e^{4 a d} (2 b d n+m+1) \left (c x^n\right )^{2 b d}}{n}\right )}{b d \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )}-\frac {e^{-2 a d} \left (\frac {2 e^{4 a d} \left (2 b^2 d^2 n^2+m^2+2 m+1\right ) \left (c x^n\right )^{\frac {m+1}{n}} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2 b d n},\frac {m+1}{2 b d n}+1,-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{(m+1) n}-\frac {e^{4 a d} (b d n+m+1) (2 b d n+m+1) \left (c x^n\right )^{\frac {m+1}{n}}}{(m+1) n}\right )}{b d}\right )}{2 b d}-\frac {\left (c x^n\right )^{\frac {m+1}{n}} \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )^2}{2 b d \left (e^{2 a d} \left (c x^n\right )^{2 b d}+1\right )^2}\right )}{e n}\)

Input:

Int[(e*x)^m*Tanh[d*(a + b*Log[c*x^n])]^3,x]
 

Output:

((e*x)^(1 + m)*(-1/2*((c*x^n)^((1 + m)/n)*(1 - E^(2*a*d)*(c*x^n)^(2*b*d))^ 
2)/(b*d*(1 + E^(2*a*d)*(c*x^n)^(2*b*d))^2) + (((c*x^n)^((1 + m)/n)*((E^(2* 
a*d)*(1 + m - 2*b*d*n))/n - (E^(4*a*d)*(1 + m + 2*b*d*n)*(c*x^n)^(2*b*d))/ 
n))/(b*d*(1 + E^(2*a*d)*(c*x^n)^(2*b*d))) - (-((E^(4*a*d)*(1 + m + b*d*n)* 
(1 + m + 2*b*d*n)*(c*x^n)^((1 + m)/n))/((1 + m)*n)) + (2*E^(4*a*d)*(1 + 2* 
m + m^2 + 2*b^2*d^2*n^2)*(c*x^n)^((1 + m)/n)*Hypergeometric2F1[1, (1 + m)/ 
(2*b*d*n), 1 + (1 + m)/(2*b*d*n), -(E^(2*a*d)*(c*x^n)^(2*b*d))])/((1 + m)* 
n))/(b*d*E^(2*a*d)))/(2*b*d*E^(2*a*d))))/(e*n*(c*x^n)^((1 + m)/n))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 959

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] - Simp[(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 1004

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

rule 1064

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^( 
m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*b*g*n*(p + 1))), x] + Simp[1/( 
a*b*n*(p + 1))   Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c 
*(b*e*n*(p + 1) + (b*e - a*f)*(m + 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m 
+ n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && LtQ 
[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])
 

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]

\[\int \left (e x \right )^{m} {\tanh \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{3}d x\]

Input:

int((e*x)^m*tanh(d*(a+b*ln(c*x^n)))^3,x)
 

Output:

int((e*x)^m*tanh(d*(a+b*ln(c*x^n)))^3,x)
 

Fricas [F]

\[ \int (e x)^m \tanh ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{3} \,d x } \] Input:

integrate((e*x)^m*tanh(d*(a+b*log(c*x^n)))^3,x, algorithm="fricas")
 

Output:

integral((e*x)^m*tanh(b*d*log(c*x^n) + a*d)^3, x)
 

Sympy [F]

\[ \int (e x)^m \tanh ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int \left (e x\right )^{m} \tanh ^{3}{\left (a d + b d \log {\left (c x^{n} \right )} \right )}\, dx \] Input:

integrate((e*x)**m*tanh(d*(a+b*ln(c*x**n)))**3,x)
 

Output:

Integral((e*x)**m*tanh(a*d + b*d*log(c*x**n))**3, x)
 

Maxima [F]

\[ \int (e x)^m \tanh ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{3} \,d x } \] Input:

integrate((e*x)^m*tanh(d*(a+b*log(c*x^n)))^3,x, algorithm="maxima")
 

Output:

-(2*b^2*d^2*e^m*n^2 + (m^2 + 2*m + 1)*e^m)*integrate(x^m/(b^2*c^(2*b*d)*d^ 
2*n^2*e^(2*b*d*log(x^n) + 2*a*d) + b^2*d^2*n^2), x) + (b^2*c^(4*b*d)*d^2*e 
^m*n^2*x*e^(4*b*d*log(x^n) + 4*a*d + m*log(x)) + (b^2*d^2*e^m*n^2 + (m^2 + 
 2*m + 1)*e^m)*x*x^m + (2*b^2*c^(2*b*d)*d^2*e^m*n^2*e^(2*a*d) + 2*(m*n + n 
)*b*c^(2*b*d)*d*e^m*e^(2*a*d) + (m^2 + 2*m + 1)*c^(2*b*d)*e^m*e^(2*a*d))*x 
*e^(2*b*d*log(x^n) + m*log(x)))/((m*n^2 + n^2)*b^2*c^(4*b*d)*d^2*e^(4*b*d* 
log(x^n) + 4*a*d) + 2*(m*n^2 + n^2)*b^2*c^(2*b*d)*d^2*e^(2*b*d*log(x^n) + 
2*a*d) + (m*n^2 + n^2)*b^2*d^2)
 

Giac [F]

\[ \int (e x)^m \tanh ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \left (e x\right )^{m} \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{3} \,d x } \] Input:

integrate((e*x)^m*tanh(d*(a+b*log(c*x^n)))^3,x, algorithm="giac")
 

Output:

integrate((e*x)^m*tanh((b*log(c*x^n) + a)*d)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int (e x)^m \tanh ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int {\mathrm {tanh}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^3\,{\left (e\,x\right )}^m \,d x \] Input:

int(tanh(d*(a + b*log(c*x^n)))^3*(e*x)^m,x)
 

Output:

int(tanh(d*(a + b*log(c*x^n)))^3*(e*x)^m, x)
 

Reduce [F]

\[ \int (e x)^m \tanh ^3\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\text {too large to display} \] Input:

int((e*x)^m*tanh(d*(a+b*log(c*x^n)))^3,x)
 

Output:

(e**m*(8*x**(4*b*d*n + m)*e**(4*a*d)*c**(4*b*d)*b**2*d**2*n**2*x - 6*x**(4 
*b*d*n + m)*e**(4*a*d)*c**(4*b*d)*b*d*m*n*x - 6*x**(4*b*d*n + m)*e**(4*a*d 
)*c**(4*b*d)*b*d*n*x + x**(4*b*d*n + m)*e**(4*a*d)*c**(4*b*d)*m**2*x + 2*x 
**(4*b*d*n + m)*e**(4*a*d)*c**(4*b*d)*m*x + x**(4*b*d*n + m)*e**(4*a*d)*c* 
*(4*b*d)*x - 128*x**(4*b*d*n)*e**(4*a*d)*c**(4*b*d)*int(x**m/(8*x**(6*b*d* 
n)*e**(6*a*d)*c**(6*b*d)*b**2*d**2*n**2 - 6*x**(6*b*d*n)*e**(6*a*d)*c**(6* 
b*d)*b*d*m*n - 6*x**(6*b*d*n)*e**(6*a*d)*c**(6*b*d)*b*d*n + x**(6*b*d*n)*e 
**(6*a*d)*c**(6*b*d)*m**2 + 2*x**(6*b*d*n)*e**(6*a*d)*c**(6*b*d)*m + x**(6 
*b*d*n)*e**(6*a*d)*c**(6*b*d) + 24*x**(4*b*d*n)*e**(4*a*d)*c**(4*b*d)*b**2 
*d**2*n**2 - 18*x**(4*b*d*n)*e**(4*a*d)*c**(4*b*d)*b*d*m*n - 18*x**(4*b*d* 
n)*e**(4*a*d)*c**(4*b*d)*b*d*n + 3*x**(4*b*d*n)*e**(4*a*d)*c**(4*b*d)*m**2 
 + 6*x**(4*b*d*n)*e**(4*a*d)*c**(4*b*d)*m + 3*x**(4*b*d*n)*e**(4*a*d)*c**( 
4*b*d) + 24*x**(2*b*d*n)*e**(2*a*d)*c**(2*b*d)*b**2*d**2*n**2 - 18*x**(2*b 
*d*n)*e**(2*a*d)*c**(2*b*d)*b*d*m*n - 18*x**(2*b*d*n)*e**(2*a*d)*c**(2*b*d 
)*b*d*n + 3*x**(2*b*d*n)*e**(2*a*d)*c**(2*b*d)*m**2 + 6*x**(2*b*d*n)*e**(2 
*a*d)*c**(2*b*d)*m + 3*x**(2*b*d*n)*e**(2*a*d)*c**(2*b*d) + 8*b**2*d**2*n* 
*2 - 6*b*d*m*n - 6*b*d*n + m**2 + 2*m + 1),x)*b**4*d**4*m*n**4 - 128*x**(4 
*b*d*n)*e**(4*a*d)*c**(4*b*d)*int(x**m/(8*x**(6*b*d*n)*e**(6*a*d)*c**(6*b* 
d)*b**2*d**2*n**2 - 6*x**(6*b*d*n)*e**(6*a*d)*c**(6*b*d)*b*d*m*n - 6*x**(6 
*b*d*n)*e**(6*a*d)*c**(6*b*d)*b*d*n + x**(6*b*d*n)*e**(6*a*d)*c**(6*b*d...