\(\int (c+d x)^2 \tanh ^2(e+f x) \, dx\) [7]

Optimal result

Integrand size = 16, antiderivative size = 88 \[ \int (c+d x)^2 \tanh ^2(e+f x) \, dx=-\frac {(c+d x)^2}{f}+\frac {(c+d x)^3}{3 d}+\frac {2 d (c+d x) \log \left (1+e^{2 (e+f x)}\right )}{f^2}+\frac {d^2 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3}-\frac {(c+d x)^2 \tanh (e+f x)}{f} \] Output:

-(d*x+c)^2/f+1/3*(d*x+c)^3/d+2*d*(d*x+c)*ln(1+exp(2*f*x+2*e))/f^2+d^2*poly 
log(2,-exp(2*f*x+2*e))/f^3-(d*x+c)^2*tanh(f*x+e)/f
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.31 \[ \int (c+d x)^2 \tanh ^2(e+f x) \, dx=c^2 x+c d x^2+\frac {d^2 x^3}{3}+\frac {\frac {2 f (c+d x) \left (f (c+d x)+d \left (1+e^{2 e}\right ) \log \left (1+e^{-2 (e+f x)}\right )\right )}{1+e^{2 e}}-d^2 \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )}{f^3}-\frac {(c+d x)^2 \text {sech}(e) \text {sech}(e+f x) \sinh (f x)}{f} \] Input:

Integrate[(c + d*x)^2*Tanh[e + f*x]^2,x]
 

Output:

c^2*x + c*d*x^2 + (d^2*x^3)/3 + ((2*f*(c + d*x)*(f*(c + d*x) + d*(1 + E^(2 
*e))*Log[1 + E^(-2*(e + f*x))]))/(1 + E^(2*e)) - d^2*PolyLog[2, -E^(-2*(e 
+ f*x))])/f^3 - ((c + d*x)^2*Sech[e]*Sech[e + f*x]*Sinh[f*x])/f
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.52 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.23, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {3042, 25, 4203, 17, 26, 3042, 26, 4201, 2620, 2715, 2838}

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.

\(\displaystyle \int (c+d x)^2 \tanh ^2(e+f x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -(c+d x)^2 \tan (i e+i f x)^2dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int (c+d x)^2 \tan (i e+i f x)^2dx\)

\(\Big \downarrow \) 4203

\(\displaystyle -\frac {2 i d \int i (c+d x) \tanh (e+f x)dx}{f}+\int (c+d x)^2dx-\frac {(c+d x)^2 \tanh (e+f x)}{f}\)

\(\Big \downarrow \) 17

\(\displaystyle -\frac {2 i d \int i (c+d x) \tanh (e+f x)dx}{f}-\frac {(c+d x)^2 \tanh (e+f x)}{f}+\frac {(c+d x)^3}{3 d}\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {2 d \int (c+d x) \tanh (e+f x)dx}{f}-\frac {(c+d x)^2 \tanh (e+f x)}{f}+\frac {(c+d x)^3}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 d \int -i (c+d x) \tan (i e+i f x)dx}{f}-\frac {(c+d x)^2 \tanh (e+f x)}{f}+\frac {(c+d x)^3}{3 d}\)

\(\Big \downarrow \) 26

\(\displaystyle -\frac {2 i d \int (c+d x) \tan (i e+i f x)dx}{f}-\frac {(c+d x)^2 \tanh (e+f x)}{f}+\frac {(c+d x)^3}{3 d}\)

\(\Big \downarrow \) 4201

\(\displaystyle -\frac {2 i d \left (2 i \int \frac {e^{2 (e+f x)} (c+d x)}{1+e^{2 (e+f x)}}dx-\frac {i (c+d x)^2}{2 d}\right )}{f}-\frac {(c+d x)^2 \tanh (e+f x)}{f}+\frac {(c+d x)^3}{3 d}\)

\(\Big \downarrow \) 2620

\(\displaystyle -\frac {2 i d \left (2 i \left (\frac {(c+d x) \log \left (e^{2 (e+f x)}+1\right )}{2 f}-\frac {d \int \log \left (1+e^{2 (e+f x)}\right )dx}{2 f}\right )-\frac {i (c+d x)^2}{2 d}\right )}{f}-\frac {(c+d x)^2 \tanh (e+f x)}{f}+\frac {(c+d x)^3}{3 d}\)

\(\Big \downarrow \) 2715

\(\displaystyle -\frac {2 i d \left (2 i \left (\frac {(c+d x) \log \left (e^{2 (e+f x)}+1\right )}{2 f}-\frac {d \int e^{-2 (e+f x)} \log \left (1+e^{2 (e+f x)}\right )de^{2 (e+f x)}}{4 f^2}\right )-\frac {i (c+d x)^2}{2 d}\right )}{f}-\frac {(c+d x)^2 \tanh (e+f x)}{f}+\frac {(c+d x)^3}{3 d}\)

\(\Big \downarrow \) 2838

\(\displaystyle -\frac {2 i d \left (2 i \left (\frac {(c+d x) \log \left (e^{2 (e+f x)}+1\right )}{2 f}+\frac {d \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{4 f^2}\right )-\frac {i (c+d x)^2}{2 d}\right )}{f}-\frac {(c+d x)^2 \tanh (e+f x)}{f}+\frac {(c+d x)^3}{3 d}\)

Input:

Int[(c + d*x)^2*Tanh[e + f*x]^2,x]
 

Output:

(c + d*x)^3/(3*d) - ((2*I)*d*(((-1/2*I)*(c + d*x)^2)/d + (2*I)*(((c + d*x) 
*Log[1 + E^(2*(e + f*x))])/(2*f) + (d*PolyLog[2, -E^(2*(e + f*x))])/(4*f^2 
))))/f - ((c + d*x)^2*Tanh[e + f*x])/f
 

Defintions of rubi rules used

rule 17

Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4201

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x 
_Symbol] :> Simp[(-I)*((c + d*x)^(m + 1)/(d*(m + 1))), x] + Simp[2*I   Int[ 
(c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x], x] 
 /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
 

rule 4203

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symb 
ol] :> Simp[b*(c + d*x)^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] + (-Si 
mp[b*d*(m/(f*(n - 1)))   Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1), x] 
, x] - Simp[b^2   Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; Free 
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(184\) vs. \(2(86)=172\).

Time = 0.40 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.10

method	result	size
risch	\(\frac {d^{2} x^{3}}{3}+d c \,x^{2}+c^{2} x +\frac {c^{3}}{3 d}+\frac {2 x^{2} d^{2}+4 c d x +2 c^{2}}{f \left (1+{\mathrm e}^{2 f x +2 e}\right )}-\frac {4 d c \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}+\frac {2 d c \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f^{2}}-\frac {2 d^{2} x^{2}}{f}-\frac {4 d^{2} e x}{f^{2}}-\frac {2 d^{2} e^{2}}{f^{3}}+\frac {2 d^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}+\frac {d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{f^{3}}+\frac {4 d^{2} e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}\)	\(185\)

Input:

int((d*x+c)^2*tanh(f*x+e)^2,x,method=_RETURNVERBOSE)
 

Output:

1/3*d^2*x^3+d*c*x^2+c^2*x+1/3/d*c^3+2*(d^2*x^2+2*c*d*x+c^2)/f/(1+exp(2*f*x 
+2*e))-4/f^2*d*c*ln(exp(f*x+e))+2/f^2*d*c*ln(1+exp(2*f*x+2*e))-2/f*d^2*x^2 
-4/f^2*d^2*e*x-2/f^3*d^2*e^2+2/f^2*d^2*ln(1+exp(2*f*x+2*e))*x+d^2*polylog( 
2,-exp(2*f*x+2*e))/f^3+4/f^3*d^2*e*ln(exp(f*x+e))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 840, normalized size of antiderivative = 9.55 \[ \int (c+d x)^2 \tanh ^2(e+f x) \, dx =\text {Too large to display} \] Input:

integrate((d*x+c)^2*tanh(f*x+e)^2,x, algorithm="fricas")
 

Output:

1/3*(d^2*f^3*x^3 + 3*c*d*f^3*x^2 + 3*c^2*f^3*x + 6*d^2*e^2 - 12*c*d*e*f + 
6*c^2*f^2 + (d^2*f^3*x^3 + 6*d^2*e^2 - 12*c*d*e*f + 3*(c*d*f^3 - 2*d^2*f^2 
)*x^2 + 3*(c^2*f^3 - 4*c*d*f^2)*x)*cosh(f*x + e)^2 + 2*(d^2*f^3*x^3 + 6*d^ 
2*e^2 - 12*c*d*e*f + 3*(c*d*f^3 - 2*d^2*f^2)*x^2 + 3*(c^2*f^3 - 4*c*d*f^2) 
*x)*cosh(f*x + e)*sinh(f*x + e) + (d^2*f^3*x^3 + 6*d^2*e^2 - 12*c*d*e*f + 
3*(c*d*f^3 - 2*d^2*f^2)*x^2 + 3*(c^2*f^3 - 4*c*d*f^2)*x)*sinh(f*x + e)^2 + 
 6*(d^2*cosh(f*x + e)^2 + 2*d^2*cosh(f*x + e)*sinh(f*x + e) + d^2*sinh(f*x 
 + e)^2 + d^2)*dilog(I*cosh(f*x + e) + I*sinh(f*x + e)) + 6*(d^2*cosh(f*x 
+ e)^2 + 2*d^2*cosh(f*x + e)*sinh(f*x + e) + d^2*sinh(f*x + e)^2 + d^2)*di 
log(-I*cosh(f*x + e) - I*sinh(f*x + e)) - 6*(d^2*e - c*d*f + (d^2*e - c*d* 
f)*cosh(f*x + e)^2 + 2*(d^2*e - c*d*f)*cosh(f*x + e)*sinh(f*x + e) + (d^2* 
e - c*d*f)*sinh(f*x + e)^2)*log(cosh(f*x + e) + sinh(f*x + e) + I) - 6*(d^ 
2*e - c*d*f + (d^2*e - c*d*f)*cosh(f*x + e)^2 + 2*(d^2*e - c*d*f)*cosh(f*x 
 + e)*sinh(f*x + e) + (d^2*e - c*d*f)*sinh(f*x + e)^2)*log(cosh(f*x + e) + 
 sinh(f*x + e) - I) + 6*(d^2*f*x + d^2*e + (d^2*f*x + d^2*e)*cosh(f*x + e) 
^2 + 2*(d^2*f*x + d^2*e)*cosh(f*x + e)*sinh(f*x + e) + (d^2*f*x + d^2*e)*s 
inh(f*x + e)^2)*log(I*cosh(f*x + e) + I*sinh(f*x + e) + 1) + 6*(d^2*f*x + 
d^2*e + (d^2*f*x + d^2*e)*cosh(f*x + e)^2 + 2*(d^2*f*x + d^2*e)*cosh(f*x + 
 e)*sinh(f*x + e) + (d^2*f*x + d^2*e)*sinh(f*x + e)^2)*log(-I*cosh(f*x + e 
) - I*sinh(f*x + e) + 1))/(f^3*cosh(f*x + e)^2 + 2*f^3*cosh(f*x + e)*si...
 

Sympy [F]

\[ \int (c+d x)^2 \tanh ^2(e+f x) \, dx=\int \left (c + d x\right )^{2} \tanh ^{2}{\left (e + f x \right )}\, dx \] Input:

integrate((d*x+c)**2*tanh(f*x+e)**2,x)
 

Output:

Integral((c + d*x)**2*tanh(e + f*x)**2, x)
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int (c+d x)^2 \tanh ^2(e+f x) \, dx=\int { {\left (d x + c\right )}^{2} \tanh \left (f x + e\right )^{2} \,d x } \] Input:

integrate((d*x+c)^2*tanh(f*x+e)^2,x, algorithm="maxima")
 

Output:

c^2*(x + e/f - 2/(f*(e^(-2*f*x - 2*e) + 1))) - c*d*(2*x*e^(2*f*x + 2*e)/(f 
*e^(2*f*x + 2*e) + f) - (f*x^2 + (f*x^2*e^(2*e) - 2*x*e^(2*e))*e^(2*f*x))/ 
(f*e^(2*f*x + 2*e) + f) - 2*log((e^(2*f*x + 2*e) + 1)*e^(-2*e))/f^2) + 1/3 
*d^2*((f*x^3*e^(2*f*x + 2*e) + f*x^3 + 6*x^2)/(f*e^(2*f*x + 2*e) + f) - 12 
*integrate(x/(f*e^(2*f*x + 2*e) + f), x))
 

Giac [F]

\[ \int (c+d x)^2 \tanh ^2(e+f x) \, dx=\int { {\left (d x + c\right )}^{2} \tanh \left (f x + e\right )^{2} \,d x } \] Input:

integrate((d*x+c)^2*tanh(f*x+e)^2,x, algorithm="giac")
 

Output:

integrate((d*x + c)^2*tanh(f*x + e)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^2 \tanh ^2(e+f x) \, dx=\int {\mathrm {tanh}\left (e+f\,x\right )}^2\,{\left (c+d\,x\right )}^2 \,d x \] Input:

int(tanh(e + f*x)^2*(c + d*x)^2,x)
 

Output:

int(tanh(e + f*x)^2*(c + d*x)^2, x)
 

Reduce [F]

\[ \int (c+d x)^2 \tanh ^2(e+f x) \, dx=\frac {-12 e^{2 f x +2 e} \left (\int \frac {x}{e^{4 f x +4 e}+2 e^{2 f x +2 e}+1}d x \right ) d^{2} f^{2}+6 e^{2 f x +2 e} \mathrm {log}\left (e^{2 f x +2 e}+1\right ) c d f +3 e^{2 f x +2 e} \mathrm {log}\left (e^{2 f x +2 e}+1\right ) d^{2}+3 e^{2 f x +2 e} c^{2} f^{3} x -6 e^{2 f x +2 e} c^{2} f^{2}+3 e^{2 f x +2 e} c d \,f^{3} x^{2}-12 e^{2 f x +2 e} c d \,f^{2} x +e^{2 f x +2 e} d^{2} f^{3} x^{3}-6 e^{2 f x +2 e} d^{2} f x -12 \left (\int \frac {x}{e^{4 f x +4 e}+2 e^{2 f x +2 e}+1}d x \right ) d^{2} f^{2}+6 \,\mathrm {log}\left (e^{2 f x +2 e}+1\right ) c d f +3 \,\mathrm {log}\left (e^{2 f x +2 e}+1\right ) d^{2}+3 c^{2} f^{3} x +3 c d \,f^{3} x^{2}+d^{2} f^{3} x^{3}+6 d^{2} f^{2} x^{2}}{3 f^{3} \left (e^{2 f x +2 e}+1\right )} \] Input:

int((d*x+c)^2*tanh(f*x+e)^2,x)
 

Output:

( - 12*e**(2*e + 2*f*x)*int(x/(e**(4*e + 4*f*x) + 2*e**(2*e + 2*f*x) + 1), 
x)*d**2*f**2 + 6*e**(2*e + 2*f*x)*log(e**(2*e + 2*f*x) + 1)*c*d*f + 3*e**( 
2*e + 2*f*x)*log(e**(2*e + 2*f*x) + 1)*d**2 + 3*e**(2*e + 2*f*x)*c**2*f**3 
*x - 6*e**(2*e + 2*f*x)*c**2*f**2 + 3*e**(2*e + 2*f*x)*c*d*f**3*x**2 - 12* 
e**(2*e + 2*f*x)*c*d*f**2*x + e**(2*e + 2*f*x)*d**2*f**3*x**3 - 6*e**(2*e 
+ 2*f*x)*d**2*f*x - 12*int(x/(e**(4*e + 4*f*x) + 2*e**(2*e + 2*f*x) + 1),x 
)*d**2*f**2 + 6*log(e**(2*e + 2*f*x) + 1)*c*d*f + 3*log(e**(2*e + 2*f*x) + 
 1)*d**2 + 3*c**2*f**3*x + 3*c*d*f**3*x**2 + d**2*f**3*x**3 + 6*d**2*f**2* 
x**2)/(3*f**3*(e**(2*e + 2*f*x) + 1))