3.53 \(\int (c+d x)^3 (a+b \tanh (e+f x)) \, dx\)

Optimal. Leaf size=137 \[ \frac {a (c+d x)^4}{4 d}-\frac {3 b d^2 (c+d x) \text {Li}_3\left (-e^{2 (e+f x)}\right )}{2 f^3}+\frac {3 b d (c+d x)^2 \text {Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2}+\frac {b (c+d x)^3 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {b (c+d x)^4}{4 d}+\frac {3 b d^3 \text {Li}_4\left (-e^{2 (e+f x)}\right )}{4 f^4} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3722, 3718, 2190, 2531, 6609, 2282, 6589} \[ -\frac {3 b d^2 (c+d x) \text {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac {3 b d (c+d x)^2 \text {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}+\frac {3 b d^3 \text {PolyLog}\left (4,-e^{2 (e+f x)}\right )}{4 f^4}+\frac {a (c+d x)^4}{4 d}+\frac {b (c+d x)^3 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {b (c+d x)^4}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2282

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 2531

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 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 3722

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 6589

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]

Rule 6609

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

Rubi steps

\begin {align*} \int (c+d x)^3 (a+b \tanh (e+f x)) \, dx &=\int \left (a (c+d x)^3+b (c+d x)^3 \tanh (e+f x)\right ) \, dx\\ &=\frac {a (c+d x)^4}{4 d}+b \int (c+d x)^3 \tanh (e+f x) \, dx\\ &=\frac {a (c+d x)^4}{4 d}-\frac {b (c+d x)^4}{4 d}+(2 b) \int \frac {e^{2 (e+f x)} (c+d x)^3}{1+e^{2 (e+f x)}} \, dx\\ &=\frac {a (c+d x)^4}{4 d}-\frac {b (c+d x)^4}{4 d}+\frac {b (c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}-\frac {(3 b d) \int (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right ) \, dx}{f}\\ &=\frac {a (c+d x)^4}{4 d}-\frac {b (c+d x)^4}{4 d}+\frac {b (c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {3 b d (c+d x)^2 \text {Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2}-\frac {\left (3 b d^2\right ) \int (c+d x) \text {Li}_2\left (-e^{2 (e+f x)}\right ) \, dx}{f^2}\\ &=\frac {a (c+d x)^4}{4 d}-\frac {b (c+d x)^4}{4 d}+\frac {b (c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {3 b d (c+d x)^2 \text {Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2}-\frac {3 b d^2 (c+d x) \text {Li}_3\left (-e^{2 (e+f x)}\right )}{2 f^3}+\frac {\left (3 b d^3\right ) \int \text {Li}_3\left (-e^{2 (e+f x)}\right ) \, dx}{2 f^3}\\ &=\frac {a (c+d x)^4}{4 d}-\frac {b (c+d x)^4}{4 d}+\frac {b (c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {3 b d (c+d x)^2 \text {Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2}-\frac {3 b d^2 (c+d x) \text {Li}_3\left (-e^{2 (e+f x)}\right )}{2 f^3}+\frac {\left (3 b d^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{2 (e+f x)}\right )}{4 f^4}\\ &=\frac {a (c+d x)^4}{4 d}-\frac {b (c+d x)^4}{4 d}+\frac {b (c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {3 b d (c+d x)^2 \text {Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2}-\frac {3 b d^2 (c+d x) \text {Li}_3\left (-e^{2 (e+f x)}\right )}{2 f^3}+\frac {3 b d^3 \text {Li}_4\left (-e^{2 (e+f x)}\right )}{4 f^4}\\ \end {align*}

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Mathematica [A]  time = 2.17, size = 161, normalized size = 1.18 \[ \frac {1}{4} \left (x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) (a+b \tanh (e))+b \left (-\frac {3 d \left (2 f^2 (c+d x)^2 \text {Li}_2\left (-e^{-2 (e+f x)}\right )+d \left (2 f (c+d x) \text {Li}_3\left (-e^{-2 (e+f x)}\right )+d \text {Li}_4\left (-e^{-2 (e+f x)}\right )\right )\right )}{f^4}+\frac {4 (c+d x)^3 \log \left (e^{-2 (e+f x)}+1\right )}{f}+\frac {2 (c+d x)^4}{d \left (e^{2 e}+1\right )}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*((2*(c + d*x)^4)/(d*(1 + E^(2*e))) + (4*(c + d*x)^3*Log[1 + E^(-2*(e + f*x))])/f - (3*d*(2*f^2*(c + d*x)^2*
PolyLog[2, -E^(-2*(e + f*x))] + d*(2*f*(c + d*x)*PolyLog[3, -E^(-2*(e + f*x))] + d*PolyLog[4, -E^(-2*(e + f*x)
)])))/f^4) + x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3)*(a + b*Tanh[e]))/4

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fricas [C]  time = 0.56, size = 583, normalized size = 4.26 \[ \frac {{\left (a - b\right )} d^{3} f^{4} x^{4} + 4 \, {\left (a - b\right )} c d^{2} f^{4} x^{3} + 6 \, {\left (a - b\right )} c^{2} d f^{4} x^{2} + 4 \, {\left (a - b\right )} c^{3} f^{4} x + 24 \, b d^{3} {\rm polylog}\left (4, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 24 \, b d^{3} {\rm polylog}\left (4, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) + 12 \, {\left (b d^{3} f^{2} x^{2} + 2 \, b c d^{2} f^{2} x + b c^{2} d f^{2}\right )} {\rm Li}_2\left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 12 \, {\left (b d^{3} f^{2} x^{2} + 2 \, b c d^{2} f^{2} x + b c^{2} d f^{2}\right )} {\rm Li}_2\left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) - 4 \, {\left (b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b c^{2} d e f^{2} - b c^{3} f^{3}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + i\right ) - 4 \, {\left (b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b c^{2} d e f^{2} - b c^{3} f^{3}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - i\right ) + 4 \, {\left (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + 3 \, b c^{2} d f^{3} x + b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b c^{2} d e f^{2}\right )} \log \left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right ) + 1\right ) + 4 \, {\left (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + 3 \, b c^{2} d f^{3} x + b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b c^{2} d e f^{2}\right )} \log \left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right ) + 1\right ) - 24 \, {\left (b d^{3} f x + b c d^{2} f\right )} {\rm polylog}\left (3, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) - 24 \, {\left (b d^{3} f x + b c d^{2} f\right )} {\rm polylog}\left (3, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right )}{4 \, f^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((a - b)*d^3*f^4*x^4 + 4*(a - b)*c*d^2*f^4*x^3 + 6*(a - b)*c^2*d*f^4*x^2 + 4*(a - b)*c^3*f^4*x + 24*b*d^3*
polylog(4, I*cosh(f*x + e) + I*sinh(f*x + e)) + 24*b*d^3*polylog(4, -I*cosh(f*x + e) - I*sinh(f*x + e)) + 12*(
b*d^3*f^2*x^2 + 2*b*c*d^2*f^2*x + b*c^2*d*f^2)*dilog(I*cosh(f*x + e) + I*sinh(f*x + e)) + 12*(b*d^3*f^2*x^2 +
2*b*c*d^2*f^2*x + b*c^2*d*f^2)*dilog(-I*cosh(f*x + e) - I*sinh(f*x + e)) - 4*(b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*
b*c^2*d*e*f^2 - b*c^3*f^3)*log(cosh(f*x + e) + sinh(f*x + e) + I) - 4*(b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d
*e*f^2 - b*c^3*f^3)*log(cosh(f*x + e) + sinh(f*x + e) - I) + 4*(b*d^3*f^3*x^3 + 3*b*c*d^2*f^3*x^2 + 3*b*c^2*d*
f^3*x + b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2)*log(I*cosh(f*x + e) + I*sinh(f*x + e) + 1) + 4*(b*d^3*f
^3*x^3 + 3*b*c*d^2*f^3*x^2 + 3*b*c^2*d*f^3*x + b*d^3*e^3 - 3*b*c*d^2*e^2*f + 3*b*c^2*d*e*f^2)*log(-I*cosh(f*x
+ e) - I*sinh(f*x + e) + 1) - 24*(b*d^3*f*x + b*c*d^2*f)*polylog(3, I*cosh(f*x + e) + I*sinh(f*x + e)) - 24*(b
*d^3*f*x + b*c*d^2*f)*polylog(3, -I*cosh(f*x + e) - I*sinh(f*x + e)))/f^4

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{3} {\left (b \tanh \left (f x + e\right ) + a\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)^3*(b*tanh(f*x + e) + a), x)

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maple [B]  time = 0.30, size = 452, normalized size = 3.30 \[ a c \,d^{2} x^{3}-b c \,d^{2} x^{3}+\frac {3 a \,c^{2} d \,x^{2}}{2}-\frac {3 b \,c^{2} d \,x^{2}}{2}-\frac {3 b \,e^{4} d^{3}}{2 f^{4}}+\frac {b \,c^{3} \ln \left ({\mathrm e}^{2 f x +2 e}+1\right )}{f}-\frac {2 b \,c^{3} \ln \left ({\mathrm e}^{f x +e}\right )}{f}+\frac {3 b \,d^{3} \polylog \left (4, -{\mathrm e}^{2 f x +2 e}\right )}{4 f^{4}}+\frac {a \,d^{3} x^{4}}{4}-\frac {b \,d^{3} x^{4}}{4}+c^{3} a x +b \,c^{3} x -\frac {6 b \,c^{2} d e x}{f}+\frac {6 b c \,d^{2} e^{2} x}{f^{2}}+\frac {4 b c \,d^{2} e^{3}}{f^{3}}-\frac {2 b \,e^{3} d^{3} x}{f^{3}}-\frac {3 b \,c^{2} d \,e^{2}}{f^{2}}+\frac {b \,d^{3} \ln \left ({\mathrm e}^{2 f x +2 e}+1\right ) x^{3}}{f}+\frac {2 b \,d^{3} e^{3} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{4}}-\frac {3 b \,d^{3} \polylog \left (3, -{\mathrm e}^{2 f x +2 e}\right ) x}{2 f^{3}}+\frac {3 b \,c^{2} d \polylog \left (2, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{2}}-\frac {3 b c \,d^{2} \polylog \left (3, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{3}}+\frac {3 b \,d^{3} \polylog \left (2, -{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{2 f^{2}}+\frac {3 b \,c^{2} d \ln \left ({\mathrm e}^{2 f x +2 e}+1\right ) x}{f}+\frac {3 b c \,d^{2} \polylog \left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}-\frac {6 b c \,d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {3 b c \,d^{2} \ln \left ({\mathrm e}^{2 f x +2 e}+1\right ) x^{2}}{f}+\frac {6 b \,c^{2} d e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*(a+b*tanh(f*x+e)),x)

[Out]

a*c*d^2*x^3-b*c*d^2*x^3+3/2*a*c^2*d*x^2-3/2*b*c^2*d*x^2-3/2/f^4*b*e^4*d^3+1/f*b*c^3*ln(exp(2*f*x+2*e)+1)-2/f*b
*c^3*ln(exp(f*x+e))+3/4*b*d^3*polylog(4,-exp(2*f*x+2*e))/f^4+1/4*a*d^3*x^4-1/4*b*d^3*x^4+c^3*a*x+b*c^3*x-6/f*b
*c^2*d*e*x+6/f^2*b*c*d^2*e^2*x+3/f^2*b*c*d^2*polylog(2,-exp(2*f*x+2*e))*x+4/f^3*b*c*d^2*e^3-2/f^3*b*e^3*d^3*x-
3/f^2*b*c^2*d*e^2-3/2/f^3*b*d^3*polylog(3,-exp(2*f*x+2*e))*x+1/f*b*d^3*ln(exp(2*f*x+2*e)+1)*x^3+3/2/f^2*b*c^2*
d*polylog(2,-exp(2*f*x+2*e))+2/f^4*b*d^3*e^3*ln(exp(f*x+e))-3/2/f^3*b*c*d^2*polylog(3,-exp(2*f*x+2*e))+3/2/f^2
*b*d^3*polylog(2,-exp(2*f*x+2*e))*x^2+3/f*b*c^2*d*ln(exp(2*f*x+2*e)+1)*x-6/f^3*b*c*d^2*e^2*ln(exp(f*x+e))+3/f*
b*c*d^2*ln(exp(2*f*x+2*e)+1)*x^2+6/f^2*b*c^2*d*e*ln(exp(f*x+e))

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maxima [B]  time = 0.39, size = 304, normalized size = 2.22 \[ \frac {1}{4} \, a d^{3} x^{4} + \frac {1}{4} \, b d^{3} x^{4} + a c d^{2} x^{3} + b c d^{2} x^{3} + \frac {3}{2} \, a c^{2} d x^{2} + \frac {3}{2} \, b c^{2} d x^{2} + a c^{3} x + \frac {b c^{3} \log \left (\cosh \left (f x + e\right )\right )}{f} + \frac {3 \, {\left (2 \, f x \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right )\right )} b c^{2} d}{2 \, f^{2}} + \frac {3 \, {\left (2 \, f^{2} x^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} b c d^{2}}{2 \, f^{3}} + \frac {{\left (4 \, f^{3} x^{3} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 6 \, f^{2} x^{2} {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - 6 \, f x {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )}) + 3 \, {\rm Li}_{4}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} b d^{3}}{3 \, f^{4}} - \frac {b d^{3} f^{4} x^{4} + 4 \, b c d^{2} f^{4} x^{3} + 6 \, b c^{2} d f^{4} x^{2}}{2 \, f^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (a+b\,\mathrm {tanh}\left (e+f\,x\right )\right )\,{\left (c+d\,x\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tanh(e + f*x))*(c + d*x)^3,x)

[Out]

int((a + b*tanh(e + f*x))*(c + d*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tanh {\left (e + f x \right )}\right ) \left (c + d x\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3*(a+b*tanh(f*x+e)),x)

[Out]

Integral((a + b*tanh(e + f*x))*(c + d*x)**3, x)

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