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

Optimal. Leaf size=88 \[ \frac {2 d (c+d x) \log \left (e^{2 (e+f x)}+1\right )}{f^2}-\frac {(c+d x)^2 \tanh (e+f x)}{f}-\frac {(c+d x)^2}{f}+\frac {(c+d x)^3}{3 d}+\frac {d^2 \text {Li}_2\left (-e^{2 (e+f x)}\right )}{f^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3720, 3718, 2190, 2279, 2391, 32} \[ \frac {d^2 \text {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3}+\frac {2 d (c+d x) \log \left (e^{2 (e+f x)}+1\right )}{f^2}-\frac {(c+d x)^2 \tanh (e+f x)}{f}-\frac {(c+d x)^2}{f}+\frac {(c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Tanh[e + f*x]^2,x]

[Out]

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

Rule 32

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

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2391

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 3718

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] + Dist[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 3720

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

Rubi steps

\begin {align*} \int (c+d x)^2 \tanh ^2(e+f x) \, dx &=-\frac {(c+d x)^2 \tanh (e+f x)}{f}+\frac {(2 d) \int (c+d x) \tanh (e+f x) \, dx}{f}+\int (c+d x)^2 \, dx\\ &=-\frac {(c+d x)^2}{f}+\frac {(c+d x)^3}{3 d}-\frac {(c+d x)^2 \tanh (e+f x)}{f}+\frac {(4 d) \int \frac {e^{2 (e+f x)} (c+d x)}{1+e^{2 (e+f x)}} \, dx}{f}\\ &=-\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 {(c+d x)^2 \tanh (e+f x)}{f}-\frac {\left (2 d^2\right ) \int \log \left (1+e^{2 (e+f x)}\right ) \, dx}{f^2}\\ &=-\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 {(c+d x)^2 \tanh (e+f x)}{f}-\frac {d^2 \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 (e+f x)}\right )}{f^3}\\ &=-\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 \text {Li}_2\left (-e^{2 (e+f x)}\right )}{f^3}-\frac {(c+d x)^2 \tanh (e+f x)}{f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 6.34, size = 301, normalized size = 3.42 \[ \frac {\text {sech}(e) \text {sech}(e+f x) \left (c^2 (-\sinh (f x))-2 c d x \sinh (f x)-d^2 x^2 \sinh (f x)\right )}{f}+\frac {1}{3} x \left (3 c^2+3 c d x+d^2 x^2\right )+\frac {2 c d \text {sech}(e) (\cosh (e) \log (\sinh (e) \sinh (f x)+\cosh (e) \cosh (f x))-f x \sinh (e))}{f^2 \left (\cosh ^2(e)-\sinh ^2(e)\right )}+\frac {d^2 \text {csch}(e) \text {sech}(e) \left (f^2 x^2 e^{-\tanh ^{-1}(\coth (e))}-\frac {i \coth (e) \left (i \text {Li}_2\left (e^{2 i \left (i f x+i \tanh ^{-1}(\coth (e))\right )}\right )-f x \left (-\pi +2 i \tanh ^{-1}(\coth (e))\right )-2 \left (i \tanh ^{-1}(\coth (e))+i f x\right ) \log \left (1-e^{2 i \left (i \tanh ^{-1}(\coth (e))+i f x\right )}\right )+2 i \tanh ^{-1}(\coth (e)) \log \left (i \sinh \left (\tanh ^{-1}(\coth (e))+f x\right )\right )-\pi \log \left (e^{2 f x}+1\right )+\pi \log (\cosh (f x))\right )}{\sqrt {1-\coth ^2(e)}}\right )}{f^3 \sqrt {\text {csch}^2(e) \left (\sinh ^2(e)-\cosh ^2(e)\right )}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(3*c^2 + 3*c*d*x + d^2*x^2))/3 + (2*c*d*Sech[e]*(Cosh[e]*Log[Cosh[e]*Cosh[f*x] + Sinh[e]*Sinh[f*x]] - f*x*S
inh[e]))/(f^2*(Cosh[e]^2 - Sinh[e]^2)) + (d^2*Csch[e]*((f^2*x^2)/E^ArcTanh[Coth[e]] - (I*Coth[e]*(-(f*x*(-Pi +
(2*I)*ArcTanh[Coth[e]])) - Pi*Log[1 + E^(2*f*x)] - 2*(I*f*x + I*ArcTanh[Coth[e]])*Log[1 - E^((2*I)*(I*f*x + I
*ArcTanh[Coth[e]]))] + Pi*Log[Cosh[f*x]] + (2*I)*ArcTanh[Coth[e]]*Log[I*Sinh[f*x + ArcTanh[Coth[e]]]] + I*Poly
Log[2, E^((2*I)*(I*f*x + I*ArcTanh[Coth[e]]))]))/Sqrt[1 - Coth[e]^2])*Sech[e])/(f^3*Sqrt[Csch[e]^2*(-Cosh[e]^2
+ Sinh[e]^2)]) + (Sech[e]*Sech[e + f*x]*(-(c^2*Sinh[f*x]) - 2*c*d*x*Sinh[f*x] - d^2*x^2*Sinh[f*x]))/f

________________________________________________________________________________________

fricas [C]  time = 0.58, size = 840, normalized size = 9.55 \[ \frac {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} + {\left (d^{2} f^{3} x^{3} + 6 \, d^{2} e^{2} - 12 \, c d e f + 3 \, {\left (c d f^{3} - 2 \, d^{2} f^{2}\right )} x^{2} + 3 \, {\left (c^{2} f^{3} - 4 \, c d f^{2}\right )} x\right )} \cosh \left (f x + e\right )^{2} + 2 \, {\left (d^{2} f^{3} x^{3} + 6 \, d^{2} e^{2} - 12 \, c d e f + 3 \, {\left (c d f^{3} - 2 \, d^{2} f^{2}\right )} x^{2} + 3 \, {\left (c^{2} f^{3} - 4 \, c d f^{2}\right )} x\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + {\left (d^{2} f^{3} x^{3} + 6 \, d^{2} e^{2} - 12 \, c d e f + 3 \, {\left (c d f^{3} - 2 \, d^{2} f^{2}\right )} x^{2} + 3 \, {\left (c^{2} f^{3} - 4 \, c d f^{2}\right )} x\right )} \sinh \left (f x + e\right )^{2} + 6 \, {\left (d^{2} \cosh \left (f x + e\right )^{2} + 2 \, d^{2} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + d^{2} \sinh \left (f x + e\right )^{2} + d^{2}\right )} {\rm Li}_2\left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 6 \, {\left (d^{2} \cosh \left (f x + e\right )^{2} + 2 \, d^{2} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + d^{2} \sinh \left (f x + e\right )^{2} + d^{2}\right )} {\rm Li}_2\left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) - 6 \, {\left (d^{2} e - c d f + {\left (d^{2} e - c d f\right )} \cosh \left (f x + e\right )^{2} + 2 \, {\left (d^{2} e - c d f\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + {\left (d^{2} e - c d f\right )} \sinh \left (f x + e\right )^{2}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + i\right ) - 6 \, {\left (d^{2} e - c d f + {\left (d^{2} e - c d f\right )} \cosh \left (f x + e\right )^{2} + 2 \, {\left (d^{2} e - c d f\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + {\left (d^{2} e - c d f\right )} \sinh \left (f x + e\right )^{2}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - i\right ) + 6 \, {\left (d^{2} f x + d^{2} e + {\left (d^{2} f x + d^{2} e\right )} \cosh \left (f x + e\right )^{2} + 2 \, {\left (d^{2} f x + d^{2} e\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + {\left (d^{2} f x + d^{2} e\right )} \sinh \left (f x + e\right )^{2}\right )} \log \left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right ) + 1\right ) + 6 \, {\left (d^{2} f x + d^{2} e + {\left (d^{2} f x + d^{2} e\right )} \cosh \left (f x + e\right )^{2} + 2 \, {\left (d^{2} f x + d^{2} e\right )} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + {\left (d^{2} f x + d^{2} e\right )} \sinh \left (f x + e\right )^{2}\right )} \log \left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right ) + 1\right )}{3 \, {\left (f^{3} \cosh \left (f x + e\right )^{2} + 2 \, f^{3} \cosh \left (f x + e\right ) \sinh \left (f x + e\right ) + f^{3} \sinh \left (f x + e\right )^{2} + f^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)*dilog(-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)*cos
h(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) + 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)*sinh(f*x + e) + f^3*sinh(f*x + e)^2 + f^3)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} \tanh \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.16, size = 177, normalized size = 2.01 \[ \frac {d^{2} x^{3}}{3}+c d \,x^{2}+c^{2} x +\frac {2 d^{2} x^{2}+4 c d x +2 c^{2}}{f \left ({\mathrm e}^{2 f x +2 e}+1\right )}+\frac {2 d c \ln \left ({\mathrm e}^{2 f x +2 e}+1\right )}{f^{2}}-\frac {4 d c \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}-\frac {2 d^{2} x^{2}}{f}-\frac {4 d^{2} e x}{f^{2}}-\frac {2 d^{2} e^{2}}{f^{3}}+\frac {2 d^{2} \ln \left ({\mathrm e}^{2 f x +2 e}+1\right ) x}{f^{2}}+\frac {d^{2} \polylog \left (2, -{\mathrm e}^{2 f x +2 e}\right )}{f^{3}}+\frac {4 d^{2} e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ c^{2} {\left (x + \frac {e}{f} - \frac {2}{f {\left (e^{\left (-2 \, f x - 2 \, e\right )} + 1\right )}}\right )} - c d {\left (\frac {2 \, x e^{\left (2 \, f x + 2 \, e\right )}}{f e^{\left (2 \, f x + 2 \, e\right )} + f} - \frac {f x^{2} + {\left (f x^{2} e^{\left (2 \, e\right )} - 2 \, x e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{f e^{\left (2 \, f x + 2 \, e\right )} + f} - \frac {2 \, \log \left ({\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )} e^{\left (-2 \, e\right )}\right )}{f^{2}}\right )} + \frac {1}{3} \, d^{2} {\left (\frac {f x^{3} e^{\left (2 \, f x + 2 \, e\right )} + f x^{3} + 6 \, x^{2}}{f e^{\left (2 \, f x + 2 \, e\right )} + f} - 12 \, \int \frac {x}{f e^{\left (2 \, f x + 2 \, e\right )} + f}\,{d x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tanh}\left (e+f\,x\right )}^2\,{\left (c+d\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(e + f*x)^2*(c + d*x)^2,x)

[Out]

int(tanh(e + f*x)^2*(c + d*x)^2, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{2} \tanh ^{2}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________