3.43 \(\int (c+d x)^2 (a+b \coth (e+f x))^2 \, dx\)

Optimal. Leaf size=209 \[ \frac {a^2 (c+d x)^3}{3 d}+\frac {2 a b d (c+d x) \text {Li}_2\left (e^{2 (e+f x)}\right )}{f^2}+\frac {2 a b (c+d x)^2 \log \left (1-e^{2 (e+f x)}\right )}{f}-\frac {2 a b (c+d x)^3}{3 d}-\frac {a b d^2 \text {Li}_3\left (e^{2 (e+f x)}\right )}{f^3}+\frac {2 b^2 d (c+d x) \log \left (1-e^{2 (e+f x)}\right )}{f^2}-\frac {b^2 (c+d x)^2 \coth (e+f x)}{f}-\frac {b^2 (c+d x)^2}{f}+\frac {b^2 (c+d x)^3}{3 d}+\frac {b^2 d^2 \text {Li}_2\left (e^{2 (e+f x)}\right )}{f^3} \]

[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-b^2*(d*x+c)^2*coth(f*x+e)/f+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

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Rubi [A]  time = 0.40, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3722, 3716, 2190, 2531, 2282, 6589, 3720, 2279, 2391, 32} \[ \frac {2 a b d (c+d x) \text {PolyLog}\left (2,e^{2 (e+f x)}\right )}{f^2}-\frac {a b d^2 \text {PolyLog}\left (3,e^{2 (e+f x)}\right )}{f^3}+\frac {b^2 d^2 \text {PolyLog}\left (2,e^{2 (e+f x)}\right )}{f^3}+\frac {a^2 (c+d x)^3}{3 d}+\frac {2 a b (c+d x)^2 \log \left (1-e^{2 (e+f x)}\right )}{f}-\frac {2 a b (c+d x)^3}{3 d}+\frac {2 b^2 d (c+d x) \log \left (1-e^{2 (e+f x)}\right )}{f^2}-\frac {b^2 (c+d x)^2 \coth (e+f x)}{f}-\frac {b^2 (c+d x)^2}{f}+\frac {b^2 (c+d x)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*(a + b*Coth[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) - (b^2*
(c + d*x)^2*Coth[e + f*x])/f + (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*PolyLog[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

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 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 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 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 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && 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]

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]

Rubi steps

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

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Mathematica [B]  time = 10.33, size = 478, normalized size = 2.29 \[ \frac {1}{3} x \text {csch}(e) \left (3 c^2+3 c d x+d^2 x^2\right ) \left (a^2 \sinh (e)+2 a b \cosh (e)+b^2 \sinh (e)\right )+\frac {2}{3} b \left (-\frac {3 d \text {Li}_2\left (-e^{-e-f x}\right ) (2 a c f+b d)}{f^3}-\frac {3 d \text {Li}_2\left (e^{-e-f x}\right ) (2 a c f+b d)}{f^3}+\frac {3 d x \log \left (1-e^{-e-f x}\right ) (2 a c f+b d)}{f^2}+\frac {3 d x \log \left (e^{-e-f x}+1\right ) (2 a c f+b d)}{f^2}-\frac {3 c \left (f x-\log \left (1-e^{e+f x}\right )\right ) (a c f+b d)}{f^2}-\frac {3 c \left (f x-\log \left (e^{e+f x}+1\right )\right ) (a c f+b d)}{f^2}-\frac {3 d x^2 (2 a c f+b d)}{\left (e^{2 e}-1\right ) f}-\frac {6 c x (a c f+b d)}{\left (e^{2 e}-1\right ) f}-\frac {6 a d^2 \left (f x \text {Li}_2\left (-e^{-e-f x}\right )+\text {Li}_3\left (-e^{-e-f x}\right )\right )}{f^3}-\frac {6 a d^2 \left (f x \text {Li}_2\left (e^{-e-f x}\right )+\text {Li}_3\left (e^{-e-f x}\right )\right )}{f^3}+\frac {3 a d^2 x^2 \log \left (1-e^{-e-f x}\right )}{f}+\frac {3 a d^2 x^2 \log \left (e^{-e-f x}+1\right )}{f}-\frac {2 a d^2 x^3}{e^{2 e}-1}\right )+\frac {\text {csch}(e) \text {csch}(e+f x) \left (b^2 c^2 \sinh (f x)+2 b^2 c d x \sinh (f x)+b^2 d^2 x^2 \sinh (f x)\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.45, size = 1854, normalized size = 8.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*(a+b*coth(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(cosh(f*x + e) + 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)*s
inh(f*x + e)^2)*dilog(-cosh(f*x + e) - sinh(f*x + e)) + 6*(a*b*d^2*f^2*x^2 + a*b*c^2*f^2 + b^2*c*d*f - (a*b*d^
2*f^2*x^2 + a*b*c^2*f^2 + b^2*c*d*f + (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*c^2*f^2 + b^2*c*d*f + (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*c^
2*f^2 + b^2*c*d*f + (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(cosh(f
*x + e) + sinh(f*x + e) + 1) + 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)*si
nh(f*x + e)^2 - (2*a*b*c*d*e - b^2*c*d)*f)*log(cosh(f*x + e) + 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(-cosh(f*x + e) - 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, cosh(f*x + e) + sinh(f*x + e)) + 12*(a*b*d^2*cosh(f*x + e)^2 + 2*a*b*d^2*cosh(f*x + e)*sin
h(f*x + e) + a*b*d^2*sinh(f*x + e)^2 - a*b*d^2)*polylog(3, -cosh(f*x + e) - 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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.59, size = 793, normalized size = 3.79 \[ -\frac {8 b a c d e x}{f}+b^{2} c d \,x^{2}+a^{2} c d \,x^{2}+\frac {2 b a \,c^{2} \ln \left ({\mathrm e}^{f x +e}-1\right )}{f}+\frac {2 b a \,c^{2} \ln \left ({\mathrm e}^{f x +e}+1\right )}{f}+\frac {2 b^{2} c d \ln \left ({\mathrm e}^{f x +e}-1\right )}{f^{2}}+\frac {2 b^{2} c d \ln \left ({\mathrm e}^{f x +e}+1\right )}{f^{2}}+\frac {2 b^{2} d^{2} \ln \left (1-{\mathrm e}^{f x +e}\right ) x}{f^{2}}+\frac {2 b^{2} d^{2} \ln \left (1-{\mathrm e}^{f x +e}\right ) e}{f^{3}}+\frac {2 b^{2} d^{2} \ln \left ({\mathrm e}^{f x +e}+1\right ) x}{f^{2}}-\frac {2 b^{2} d^{2} e \ln \left ({\mathrm e}^{f x +e}-1\right )}{f^{3}}+\frac {4 b^{2} d^{2} e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}-\frac {4 b a \,c^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f}-\frac {4 b^{2} c d \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}-\frac {4 b a \,d^{2} \polylog \left (3, {\mathrm e}^{f x +e}\right )}{f^{3}}-\frac {4 b a \,d^{2} \polylog \left (3, -{\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {4 b \ln \left (1-{\mathrm e}^{f x +e}\right ) a c d e}{f^{2}}-\frac {4 b a c d e \ln \left ({\mathrm e}^{f x +e}-1\right )}{f^{2}}+\frac {4 b \ln \left (1-{\mathrm e}^{f x +e}\right ) a c d x}{f}+\frac {4 b \ln \left ({\mathrm e}^{f x +e}+1\right ) a c d x}{f}+\frac {8 b a c d e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}-\frac {2 b^{2} \left (d^{2} x^{2}+2 c d x +c^{2}\right )}{f \left ({\mathrm e}^{2 f x +2 e}-1\right )}+\frac {a^{2} d^{2} x^{3}}{3}+\frac {b^{2} d^{2} x^{3}}{3}+c^{2} a^{2} x +c^{2} b^{2} x -\frac {2 b^{2} d^{2} x^{2}}{f}-\frac {2 b^{2} d^{2} e^{2}}{f^{3}}+\frac {2 b^{2} d^{2} \polylog \left (2, {\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {2 b^{2} d^{2} \polylog \left (2, -{\mathrm e}^{f x +e}\right )}{f^{3}}-\frac {2 a b \,d^{2} x^{3}}{3}+2 c^{2} a b x -2 a b c d \,x^{2}-\frac {4 b^{2} d^{2} e x}{f^{2}}+\frac {8 b a \,d^{2} e^{3}}{3 f^{3}}+\frac {4 b a \,d^{2} e^{2} x}{f^{2}}-\frac {4 b a c d \,e^{2}}{f^{2}}-\frac {2 b a \,d^{2} e^{2} \ln \left (1-{\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {2 b a \,d^{2} \ln \left (1-{\mathrm e}^{f x +e}\right ) x^{2}}{f}+\frac {2 b a \,d^{2} \ln \left ({\mathrm e}^{f x +e}+1\right ) x^{2}}{f}+\frac {2 b a \,d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}-1\right )}{f^{3}}-\frac {4 b a \,d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {4 b a c d \polylog \left (2, {\mathrm e}^{f x +e}\right )}{f^{2}}+\frac {4 b a c d \polylog \left (2, -{\mathrm e}^{f x +e}\right )}{f^{2}}+\frac {4 b a \,d^{2} \polylog \left (2, {\mathrm e}^{f x +e}\right ) x}{f^{2}}+\frac {4 b a \,d^{2} \polylog \left (2, -{\mathrm e}^{f x +e}\right ) x}{f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.48, size = 494, normalized size = 2.36 \[ \frac {1}{3} \, a^{2} d^{2} x^{3} + a^{2} c d x^{2} + a^{2} c^{2} x - \frac {4 \, b^{2} c d x}{f} + \frac {2 \, a b c^{2} \log \left (\sinh \left (f x + e\right )\right )}{f} + \frac {2 \, b^{2} c d \log \left (e^{\left (f x + e\right )} + 1\right )}{f^{2}} + \frac {2 \, b^{2} c d \log \left (e^{\left (f x + e\right )} - 1\right )}{f^{2}} + \frac {2 \, {\left (f^{2} x^{2} \log \left (e^{\left (f x + e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (-e^{\left (f x + e\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (f x + e\right )})\right )} a b d^{2}}{f^{3}} + \frac {2 \, {\left (f^{2} x^{2} \log \left (-e^{\left (f x + e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (e^{\left (f x + e\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (f x + e\right )})\right )} a b d^{2}}{f^{3}} - \frac {6 \, b^{2} c^{2} + 3 \, {\left (c^{2} f + 4 \, c d\right )} b^{2} x + {\left (2 \, a b d^{2} f + b^{2} d^{2} f\right )} x^{3} + 3 \, {\left (2 \, a b c d f + {\left (c d f + 2 \, d^{2}\right )} b^{2}\right )} x^{2} - {\left (3 \, b^{2} c^{2} f x e^{\left (2 \, e\right )} + {\left (2 \, a b d^{2} f e^{\left (2 \, e\right )} + b^{2} d^{2} f e^{\left (2 \, e\right )}\right )} x^{3} + 3 \, {\left (2 \, a b c d f e^{\left (2 \, e\right )} + b^{2} c d f e^{\left (2 \, e\right )}\right )} x^{2}\right )} e^{\left (2 \, f x\right )}}{3 \, {\left (f e^{\left (2 \, f x + 2 \, e\right )} - f\right )}} + \frac {2 \, {\left (2 \, a b c d f + b^{2} d^{2}\right )} {\left (f x \log \left (e^{\left (f x + e\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (f x + e\right )}\right )\right )}}{f^{3}} + \frac {2 \, {\left (2 \, a b c d f + b^{2} d^{2}\right )} {\left (f x \log \left (-e^{\left (f x + e\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (f x + e\right )}\right )\right )}}{f^{3}} - \frac {2 \, {\left (2 \, a b d^{2} f^{3} x^{3} + 3 \, {\left (2 \, a b c d f + b^{2} d^{2}\right )} f^{2} x^{2}\right )}}{3 \, f^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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