\(\int (c+d x)^2 (a+b \tanh (e+f x)) \, dx\) [54]

Optimal result

Integrand size = 18, antiderivative size = 103 \[ \int (c+d x)^2 (a+b \tanh (e+f x)) \, dx=\frac {a (c+d x)^3}{3 d}-\frac {b (c+d x)^3}{3 d}+\frac {b (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}-\frac {b d^2 \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3803, 3799, 2221, 2611, 2320, 6724} \[ \int (c+d x)^2 (a+b \tanh (e+f x)) \, dx=\frac {a (c+d x)^3}{3 d}+\frac {b d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}+\frac {b (c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {b (c+d x)^3}{3 d}-\frac {b d^2 \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.43 \[ \int (c+d x)^2 (a+b \tanh (e+f x)) \, dx=\frac {1}{6} \left (\frac {b e^{2 e} \left (\frac {4 e^{-2 e} (c+d x)^3}{d}+\frac {6 \left (1+e^{-2 e}\right ) (c+d x)^2 \log \left (1+e^{-2 (e+f x)}\right )}{f}-\frac {3 d \left (1+e^{-2 e}\right ) \left (2 f (c+d x) \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )+d \operatorname {PolyLog}\left (3,-e^{-2 (e+f x)}\right )\right )}{f^3}\right )}{1+e^{2 e}}+2 x \left (3 c^2+3 c d x+d^2 x^2\right ) (a+b \tanh (e))\right ) \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(289\) vs. \(2(97)=194\).

Time = 0.24 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.82

method	result	size
risch	\(\frac {a \,d^{2} x^{3}}{3}-\frac {d^{2} b \,x^{3}}{3}+a d c \,x^{2}-d b c \,x^{2}+a \,c^{2} x +b \,c^{2} x +\frac {a \,c^{3}}{3 d}+\frac {b \,c^{3}}{3 d}+\frac {b \,c^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f}-\frac {2 b \,c^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f}+\frac {4 b \,d^{2} e^{3}}{3 f^{3}}+\frac {b \,d^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{f}-\frac {b \,d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{3}}-\frac {4 b d c e x}{f}+\frac {4 b d c e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}-\frac {2 b \,d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {2 b \,d^{2} e^{2} x}{f^{2}}+\frac {b \,d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}-\frac {2 b d c \,e^{2}}{f^{2}}+\frac {2 b d c \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f}+\frac {b d c \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{f^{2}}\)	\(290\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 364, normalized size of antiderivative = 3.53 \[ \int (c+d x)^2 (a+b \tanh (e+f x)) \, dx=\frac {{\left (a - b\right )} d^{2} f^{3} x^{3} + 3 \, {\left (a - b\right )} c d f^{3} x^{2} + 3 \, {\left (a - b\right )} c^{2} f^{3} x - 6 \, b d^{2} {\rm polylog}\left (3, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) - 6 \, b d^{2} {\rm polylog}\left (3, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) + 6 \, {\left (b d^{2} f x + b c d f\right )} {\rm Li}_2\left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 6 \, {\left (b d^{2} f x + b c d f\right )} {\rm Li}_2\left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) + 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + i\right ) + 3 \, {\left (b d^{2} e^{2} - 2 \, b c d e f + b c^{2} f^{2}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - i\right ) + 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right ) + 1\right ) + 3 \, {\left (b d^{2} f^{2} x^{2} + 2 \, b c d f^{2} x - b d^{2} e^{2} + 2 \, b c d e f\right )} \log \left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right ) + 1\right )}{3 \, f^{3}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.74 \[ \int (c+d x)^2 (a+b \tanh (e+f x)) \, dx=\frac {1}{3} \, a d^{2} x^{3} + \frac {1}{3} \, b d^{2} x^{3} + a c d x^{2} + b c d x^{2} + a c^{2} x + \frac {b c^{2} \log \left (\cosh \left (f x + e\right )\right )}{f} + \frac {{\left (2 \, f x \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right )\right )} b c d}{f^{2}} + \frac {{\left (2 \, f^{2} x^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} b d^{2}}{2 \, f^{3}} - \frac {2 \, {\left (b d^{2} f^{3} x^{3} + 3 \, b c d f^{3} x^{2}\right )}}{3 \, f^{3}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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