\(\int \tanh ^2(d (a+b \log (c x^n))) \, dx\) [192]

Optimal result

Integrand size = 15, antiderivative size = 127 \[ \int \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\left (1+\frac {1}{b d n}\right ) x+\frac {x \left (1-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d n \left (1+e^{2 a d} \left (c x^n\right )^{2 b d}\right )}-\frac {2 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 b d n},1+\frac {1}{2 b d n},-e^{2 a d} \left (c x^n\right )^{2 b d}\right )}{b d n} \] Output:

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

Mathematica [A] (verified)

Time = 5.89 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.28 \[ \int \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\frac {x \left (e^{2 d \left (a+b \log \left (c x^n\right )\right )} \operatorname {Hypergeometric2F1}\left (1,1+\frac {1}{2 b d n},2+\frac {1}{2 b d n},-e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )+(1+2 b d n) \left (b d n-\operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 b d n},1+\frac {1}{2 b d n},-e^{2 d \left (a+b \log \left (c x^n\right )\right )}\right )-\tanh \left (d \left (a+b \log \left (c x^n\right )\right )\right )\right )\right )}{b d n (1+2 b d n)} \] Input:

Integrate[Tanh[d*(a + b*Log[c*x^n])]^2,x]
 

Output:

(x*(E^(2*d*(a + b*Log[c*x^n]))*Hypergeometric2F1[1, 1 + 1/(2*b*d*n), 2 + 1 
/(2*b*d*n), -E^(2*d*(a + b*Log[c*x^n]))] + (1 + 2*b*d*n)*(b*d*n - Hypergeo 
metric2F1[1, 1/(2*b*d*n), 1 + 1/(2*b*d*n), -E^(2*d*(a + b*Log[c*x^n]))] - 
Tanh[d*(a + b*Log[c*x^n])])))/(b*d*n*(1 + 2*b*d*n))
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.39, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6069, 6071, 1004, 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.

\(\displaystyle \int \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx\)

\(\Big \downarrow \) 6069

\(\displaystyle \frac {x \left (c x^n\right )^{-1/n} \int \left (c x^n\right )^{\frac {1}{n}-1} \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right )d\left (c x^n\right )}{n}\)

\(\Big \downarrow \) 6071

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

\(\Big \downarrow \) 1004

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 959

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

\(\Big \downarrow \) 888

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

Input:

Int[Tanh[d*(a + b*Log[c*x^n])]^2,x]
 

Output:

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

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 6069

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

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]
 

Maple [F]

\[\int {\tanh \left (d \left (a +b \ln \left (c \,x^{n}\right )\right )\right )}^{2}d x\]

Input:

int(tanh(d*(a+b*ln(c*x^n)))^2,x)
 

Output:

int(tanh(d*(a+b*ln(c*x^n)))^2,x)
 

Fricas [F]

\[ \int \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \] Input:

integrate(tanh(d*(a+b*log(c*x^n)))^2,x, algorithm="fricas")
 

Output:

integral(tanh(b*d*log(c*x^n) + a*d)^2, x)
 

Sympy [F]

\[ \int \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int \tanh ^{2}{\left (d \left (a + b \log {\left (c x^{n} \right )}\right ) \right )}\, dx \] Input:

integrate(tanh(d*(a+b*ln(c*x**n)))**2,x)
 

Output:

Integral(tanh(d*(a + b*log(c*x**n)))**2, x)
 

Maxima [F]

\[ \int \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \] Input:

integrate(tanh(d*(a+b*log(c*x^n)))^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int { \tanh \left ({\left (b \log \left (c x^{n}\right ) + a\right )} d\right )^{2} \,d x } \] Input:

integrate(tanh(d*(a+b*log(c*x^n)))^2,x, algorithm="giac")
 

Output:

integrate(tanh((b*log(c*x^n) + a)*d)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=\int {\mathrm {tanh}\left (d\,\left (a+b\,\ln \left (c\,x^n\right )\right )\right )}^2 \,d x \] Input:

int(tanh(d*(a + b*log(c*x^n)))^2,x)
 

Output:

int(tanh(d*(a + b*log(c*x^n)))^2, x)
 

Reduce [F]

\[ \int \tanh ^2\left (d \left (a+b \log \left (c x^n\right )\right )\right ) \, dx=-4 e^{2 a d} c^{2 b d} \left (\int \frac {x^{2 b d n}}{x^{4 b d n} e^{4 a d} c^{4 b d}+2 x^{2 b d n} e^{2 a d} c^{2 b d}+1}d x \right )+x \] Input:

int(tanh(d*(a+b*log(c*x^n)))^2,x)
                                                                                    
                                                                                    
 

Output:

 - 4*e**(2*a*d)*c**(2*b*d)*int(x**(2*b*d*n)/(x**(4*b*d*n)*e**(4*a*d)*c**(4 
*b*d) + 2*x**(2*b*d*n)*e**(2*a*d)*c**(2*b*d) + 1),x) + x