\(\int (c+d x)^2 (a+b \tanh (e+f x))^2 \, dx\) [59]

Optimal result

Integrand size = 20, antiderivative size = 211 \[ \int (c+d x)^2 (a+b \tanh (e+f x))^2 \, dx=-\frac {b^2 (c+d x)^2}{f}+\frac {a^2 (c+d x)^3}{3 d}-\frac {2 a b (c+d x)^3}{3 d}+\frac {b^2 (c+d x)^3}{3 d}+\frac {2 b^2 d (c+d x) \log \left (1+e^{2 (e+f x)}\right )}{f^2}+\frac {2 a b (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {b^2 d^2 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3}+\frac {2 a b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}-\frac {a b d^2 \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{f^3}-\frac {b^2 (c+d x)^2 \tanh (e+f x)}{f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3803, 3799, 2221, 2611, 2320, 6724, 3801, 2317, 2438, 32} \[ \int (c+d x)^2 (a+b \tanh (e+f x))^2 \, dx=\frac {a^2 (c+d x)^3}{3 d}+\frac {2 a b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}+\frac {2 a b (c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {2 a b (c+d x)^3}{3 d}-\frac {a b d^2 \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{f^3}+\frac {2 b^2 d (c+d x) \log \left (e^{2 (e+f x)}+1\right )}{f^2}-\frac {b^2 (c+d x)^2 \tanh (e+f x)}{f}-\frac {b^2 (c+d x)^2}{f}+\frac {b^2 (c+d x)^3}{3 d}+\frac {b^2 d^2 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3} \]

[In]

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

[Out]

-((b^2*(c + d*x)^2)/f) + (a^2*(c + d*x)^3)/(3*d) - (2*a*b*(c + d*x)^3)/(3*d) + (b^2*(c + d*x)^3)/(3*d) + (2*b^
2*d*(c + d*x)*Log[1 + E^(2*(e + f*x))])/f^2 + (2*a*b*(c + d*x)^2*Log[1 + E^(2*(e + f*x))])/f + (b^2*d^2*PolyLo
g[2, -E^(2*(e + f*x))])/f^3 + (2*a*b*d*(c + d*x)*PolyLog[2, -E^(2*(e + f*x))])/f^2 - (a*b*d^2*PolyLog[3, -E^(2
*(e + f*x))])/f^3 - (b^2*(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 2221

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]
- 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 2317

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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2438

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 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3799

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 3801

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]

Rule 3803

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

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

Mathematica [A] (verified)

Time = 1.81 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.12 \[ \int (c+d x)^2 (a+b \tanh (e+f x))^2 \, dx=\frac {1}{3} \left (\frac {2 b \left (\frac {f x \left (3 b d \left (-2 c e^{2 e}+d x\right )+2 a f \left (-3 c^2 e^{2 e}+3 c d x+d^2 x^2\right )\right )}{1+e^{2 e}}+3 d x (b d+a f (2 c+d x)) \log \left (1+e^{-2 (e+f x)}\right )+3 c (b d+a c f) \log \left (1+e^{2 (e+f x)}\right )\right )}{f^2}-\frac {3 b d (b d+2 a f (c+d x)) \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )}{f^3}-\frac {3 a b d^2 \operatorname {PolyLog}\left (3,-e^{-2 (e+f x)}\right )}{f^3}-\frac {3 b^2 (c+d x)^2 \text {sech}(e) \text {sech}(e+f x) \sinh (f x)}{f}+x \left (3 c^2+3 c d x+d^2 x^2\right ) \left (a^2+b^2+2 a b \tanh (e)\right )\right ) \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(541\) vs. \(2(205)=410\).

Time = 0.36 (sec) , antiderivative size = 542, normalized size of antiderivative = 2.57

method	result	size
risch	\(\frac {4 b a \,d^{2} e^{2} x}{f^{2}}-\frac {4 b c d a \,e^{2}}{f^{2}}-\frac {4 b \,e^{2} a \,d^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {2 b a \,d^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{f}+\frac {2 b^{2} \left (x^{2} d^{2}+2 c d x +c^{2}\right )}{f \left (1+{\mathrm e}^{2 f x +2 e}\right )}-\frac {4 b a \,c^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f}+\frac {2 b^{2} c d \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f^{2}}-\frac {4 b^{2} c d \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}+d \,b^{2} c \,x^{2}+b^{2} c^{2} x +a^{2} d c \,x^{2}+a^{2} c^{2} x -\frac {a b \,d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right )}{f^{3}}+\frac {d^{2} b^{2} x^{3}}{3}+\frac {b^{2} c^{3}}{3 d}+\frac {a^{2} d^{2} x^{3}}{3}+\frac {a^{2} c^{3}}{3 d}-\frac {2 b^{2} d^{2} x^{2}}{f}-\frac {2 b^{2} d^{2} e^{2}}{f^{3}}+\frac {2 b a \,d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}+\frac {2 b c d a \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{f^{2}}-\frac {4 b^{2} d^{2} e x}{f^{2}}+\frac {b^{2} d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{f^{3}}+\frac {8 b a \,d^{2} e^{3}}{3 f^{3}}+\frac {4 b^{2} e \,d^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {2 b^{2} d^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}+\frac {2 b a \,c^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f}-\frac {2 d^{2} a b \,x^{3}}{3}+\frac {2 c^{3} a b}{3 d}-2 d a b c \,x^{2}+2 a b \,c^{2} x -\frac {8 b c d a e x}{f}+\frac {4 b c d a \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f}+\frac {8 b e c d a \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}\)	\(542\)

[In]

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

[Out]

4/f^2*b*a*d^2*e^2*x-4/f^2*b*c*d*a*e^2-4/f^3*b*e^2*a*d^2*ln(exp(f*x+e))+2/f*b*a*d^2*ln(1+exp(2*f*x+2*e))*x^2+2/
f*b^2*(d^2*x^2+2*c*d*x+c^2)/(1+exp(2*f*x+2*e))-4/f*b*a*c^2*ln(exp(f*x+e))+2/f^2*b^2*c*d*ln(1+exp(2*f*x+2*e))-4
/f^2*b^2*c*d*ln(exp(f*x+e))+d*b^2*c*x^2+b^2*c^2*x+a^2*d*c*x^2+a^2*c^2*x-a*b*d^2*polylog(3,-exp(2*f*x+2*e))/f^3
+1/3*d^2*b^2*x^3+1/3/d*b^2*c^3+1/3*a^2*d^2*x^3+1/3*a^2/d*c^3-2/f*b^2*d^2*x^2-2/f^3*b^2*d^2*e^2+2/f^2*b*a*d^2*p
olylog(2,-exp(2*f*x+2*e))*x+2/f^2*b*c*d*a*polylog(2,-exp(2*f*x+2*e))-4/f^2*b^2*d^2*e*x+b^2*d^2*polylog(2,-exp(
2*f*x+2*e))/f^3+8/3/f^3*b*a*d^2*e^3+4/f^3*b^2*e*d^2*ln(exp(f*x+e))+2/f^2*b^2*d^2*ln(1+exp(2*f*x+2*e))*x+2/f*b*
a*c^2*ln(1+exp(2*f*x+2*e))-2/3*d^2*a*b*x^3+2/3/d*c^3*a*b-2*d*a*b*c*x^2+2*a*b*c^2*x-8/f*b*c*d*a*e*x+4/f*b*c*d*a
*ln(1+exp(2*f*x+2*e))*x+8/f^2*b*e*c*d*a*ln(exp(f*x+e))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 2123, normalized size of antiderivative = 10.06 \[ \int (c+d x)^2 (a+b \tanh (e+f x))^2 \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/3*((a^2 - 2*a*b + b^2)*d^2*f^3*x^3 + 3*(a^2 - 2*a*b + b^2)*c*d*f^3*x^2 - 4*a*b*d^2*e^3 + 3*(a^2 - 2*a*b + b^
2)*c^2*f^3*x + 6*b^2*d^2*e^2 - 6*(2*a*b*c^2*e - b^2*c^2)*f^2 + ((a^2 - 2*a*b + b^2)*d^2*f^3*x^3 - 4*a*b*d^2*e^
3 - 12*a*b*c^2*e*f^2 + 6*b^2*d^2*e^2 - 3*(2*b^2*d^2*f^2 - (a^2 - 2*a*b + b^2)*c*d*f^3)*x^2 + 12*(a*b*c*d*e^2 -
b^2*c*d*e)*f - 3*(4*b^2*c*d*f^2 - (a^2 - 2*a*b + b^2)*c^2*f^3)*x)*cosh(f*x + e)^2 + 2*((a^2 - 2*a*b + b^2)*d^
2*f^3*x^3 - 4*a*b*d^2*e^3 - 12*a*b*c^2*e*f^2 + 6*b^2*d^2*e^2 - 3*(2*b^2*d^2*f^2 - (a^2 - 2*a*b + b^2)*c*d*f^3)
*x^2 + 12*(a*b*c*d*e^2 - b^2*c*d*e)*f - 3*(4*b^2*c*d*f^2 - (a^2 - 2*a*b + b^2)*c^2*f^3)*x)*cosh(f*x + e)*sinh(
f*x + e) + ((a^2 - 2*a*b + b^2)*d^2*f^3*x^3 - 4*a*b*d^2*e^3 - 12*a*b*c^2*e*f^2 + 6*b^2*d^2*e^2 - 3*(2*b^2*d^2*
f^2 - (a^2 - 2*a*b + b^2)*c*d*f^3)*x^2 + 12*(a*b*c*d*e^2 - b^2*c*d*e)*f - 3*(4*b^2*c*d*f^2 - (a^2 - 2*a*b + b^
2)*c^2*f^3)*x)*sinh(f*x + e)^2 + 12*(a*b*c*d*e^2 - b^2*c*d*e)*f + 6*(2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2 + (
2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*cosh(f*x + e)^2 + 2*(2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*cosh(f*x +
e)*sinh(f*x + e) + (2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*sinh(f*x + e)^2)*dilog(I*cosh(f*x + e) + I*sinh(f*x
+ e)) + 6*(2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2 + (2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*cosh(f*x + e)^2 +
2*(2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2)*cosh(f*x + e)*sinh(f*x + e) + (2*a*b*d^2*f*x + 2*a*b*c*d*f + b^2*d^2
)*sinh(f*x + e)^2)*dilog(-I*cosh(f*x + e) - I*sinh(f*x + e)) + 6*(a*b*d^2*e^2 + a*b*c^2*f^2 - b^2*d^2*e + (a*b
*d^2*e^2 + a*b*c^2*f^2 - b^2*d^2*e - (2*a*b*c*d*e - b^2*c*d)*f)*cosh(f*x + e)^2 + 2*(a*b*d^2*e^2 + a*b*c^2*f^2
- b^2*d^2*e - (2*a*b*c*d*e - b^2*c*d)*f)*cosh(f*x + e)*sinh(f*x + e) + (a*b*d^2*e^2 + a*b*c^2*f^2 - b^2*d^2*e
- (2*a*b*c*d*e - b^2*c*d)*f)*sinh(f*x + e)^2 - (2*a*b*c*d*e - b^2*c*d)*f)*log(cosh(f*x + e) + sinh(f*x + e) +
I) + 6*(a*b*d^2*e^2 + a*b*c^2*f^2 - b^2*d^2*e + (a*b*d^2*e^2 + a*b*c^2*f^2 - b^2*d^2*e - (2*a*b*c*d*e - b^2*c
*d)*f)*cosh(f*x + e)^2 + 2*(a*b*d^2*e^2 + a*b*c^2*f^2 - b^2*d^2*e - (2*a*b*c*d*e - b^2*c*d)*f)*cosh(f*x + e)*s
inh(f*x + e) + (a*b*d^2*e^2 + a*b*c^2*f^2 - b^2*d^2*e - (2*a*b*c*d*e - b^2*c*d)*f)*sinh(f*x + e)^2 - (2*a*b*c*
d*e - b^2*c*d)*f)*log(cosh(f*x + e) + sinh(f*x + e) - I) + 6*(a*b*d^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*d*e*f +
b^2*d^2*e + (a*b*d^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*d*e*f + b^2*d^2*e + (2*a*b*c*d*f^2 + b^2*d^2*f)*x)*cosh(f
*x + e)^2 + 2*(a*b*d^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*d*e*f + b^2*d^2*e + (2*a*b*c*d*f^2 + b^2*d^2*f)*x)*cosh
(f*x + e)*sinh(f*x + e) + (a*b*d^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*d*e*f + b^2*d^2*e + (2*a*b*c*d*f^2 + b^2*d^
2*f)*x)*sinh(f*x + e)^2 + (2*a*b*c*d*f^2 + b^2*d^2*f)*x)*log(I*cosh(f*x + e) + I*sinh(f*x + e) + 1) + 6*(a*b*d
^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*d*e*f + b^2*d^2*e + (a*b*d^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*d*e*f + b^2*d^
2*e + (2*a*b*c*d*f^2 + b^2*d^2*f)*x)*cosh(f*x + e)^2 + 2*(a*b*d^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*d*e*f + b^2*
d^2*e + (2*a*b*c*d*f^2 + b^2*d^2*f)*x)*cosh(f*x + e)*sinh(f*x + e) + (a*b*d^2*f^2*x^2 - a*b*d^2*e^2 + 2*a*b*c*
d*e*f + b^2*d^2*e + (2*a*b*c*d*f^2 + b^2*d^2*f)*x)*sinh(f*x + e)^2 + (2*a*b*c*d*f^2 + b^2*d^2*f)*x)*log(-I*cos
h(f*x + e) - I*sinh(f*x + e) + 1) - 12*(a*b*d^2*cosh(f*x + e)^2 + 2*a*b*d^2*cosh(f*x + e)*sinh(f*x + e) + a*b*
d^2*sinh(f*x + e)^2 + a*b*d^2)*polylog(3, I*cosh(f*x + e) + I*sinh(f*x + e)) - 12*(a*b*d^2*cosh(f*x + e)^2 + 2
*a*b*d^2*cosh(f*x + e)*sinh(f*x + e) + a*b*d^2*sinh(f*x + e)^2 + a*b*d^2)*polylog(3, -I*cosh(f*x + e) - I*sinh
(f*x + e)))/(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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 411 vs. \(2 (203) = 406\).

Time = 0.28 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.95 \[ \int (c+d x)^2 (a+b \tanh (e+f x))^2 \, dx=\frac {1}{3} \, a^{2} d^{2} x^{3} + a^{2} c d x^{2} + b^{2} c^{2} {\left (x + \frac {e}{f} - \frac {2}{f {\left (e^{\left (-2 \, f x - 2 \, e\right )} + 1\right )}}\right )} + a^{2} c^{2} x + b^{2} c d {\left (\frac {f x^{2} + {\left (f x^{2} e^{\left (2 \, e\right )} - 4 \, x e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{f e^{\left (2 \, f x + 2 \, e\right )} + f} + \frac {2 \, \log \left ({\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )} e^{\left (-2 \, e\right )}\right )}{f^{2}}\right )} + \frac {2 \, a b c^{2} \log \left (\cosh \left (f x + e\right )\right )}{f} + \frac {{\left (2 \, f^{2} x^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} a b d^{2}}{f^{3}} + \frac {{\left (2 \, a b d^{2} f + b^{2} d^{2} f\right )} x^{3} + 6 \, {\left (a b c d f + b^{2} d^{2}\right )} x^{2} + {\left (6 \, a b c d f x^{2} e^{\left (2 \, e\right )} + {\left (2 \, a b d^{2} f e^{\left (2 \, e\right )} + b^{2} d^{2} f e^{\left (2 \, e\right )}\right )} x^{3}\right )} e^{\left (2 \, f x\right )}}{3 \, {\left (f e^{\left (2 \, f x + 2 \, e\right )} + f\right )}} + \frac {{\left (2 \, a b c d f + b^{2} d^{2}\right )} {\left (2 \, f x \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right )\right )}}{f^{3}} - \frac {2 \, {\left (2 \, a b d^{2} f^{3} x^{3} + 3 \, {\left (2 \, a b c d f + b^{2} d^{2}\right )} f^{2} x^{2}\right )}}{3 \, f^{3}} \]

[In]

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

[Out]

1/3*a^2*d^2*x^3 + a^2*c*d*x^2 + b^2*c^2*(x + e/f - 2/(f*(e^(-2*f*x - 2*e) + 1))) + a^2*c^2*x + b^2*c*d*((f*x^2
+ (f*x^2*e^(2*e) - 4*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) + 2*a*b*c^2*log(cosh(f*x + e))/f + (2*f^2*x^2*log(e^(2*f*x + 2*e) + 1) + 2*f*x*dilog(-e^(2*f*x + 2*e)) - po
lylog(3, -e^(2*f*x + 2*e)))*a*b*d^2/f^3 + 1/3*((2*a*b*d^2*f + b^2*d^2*f)*x^3 + 6*(a*b*c*d*f + b^2*d^2)*x^2 + (
6*a*b*c*d*f*x^2*e^(2*e) + (2*a*b*d^2*f*e^(2*e) + b^2*d^2*f*e^(2*e))*x^3)*e^(2*f*x))/(f*e^(2*f*x + 2*e) + f) +
(2*a*b*c*d*f + b^2*d^2)*(2*f*x*log(e^(2*f*x + 2*e) + 1) + dilog(-e^(2*f*x + 2*e)))/f^3 - 2/3*(2*a*b*d^2*f^3*x^
3 + 3*(2*a*b*c*d*f + b^2*d^2)*f^2*x^2)/f^3

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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