3.34 \(\int (c+d x)^2 \text {csch}^3(a+b x) \, dx\)

Optimal. Leaf size=154 \[ -\frac {d^2 \text {Li}_3\left (-e^{a+b x}\right )}{b^3}+\frac {d^2 \text {Li}_3\left (e^{a+b x}\right )}{b^3}-\frac {d^2 \tanh ^{-1}(\cosh (a+b x))}{b^3}+\frac {d (c+d x) \text {Li}_2\left (-e^{a+b x}\right )}{b^2}-\frac {d (c+d x) \text {Li}_2\left (e^{a+b x}\right )}{b^2}-\frac {d (c+d x) \text {csch}(a+b x)}{b^2}+\frac {(c+d x)^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {(c+d x)^2 \coth (a+b x) \text {csch}(a+b x)}{2 b} \]

[Out]

(d*x+c)^2*arctanh(exp(b*x+a))/b-d^2*arctanh(cosh(b*x+a))/b^3-d*(d*x+c)*csch(b*x+a)/b^2-1/2*(d*x+c)^2*coth(b*x+
a)*csch(b*x+a)/b+d*(d*x+c)*polylog(2,-exp(b*x+a))/b^2-d*(d*x+c)*polylog(2,exp(b*x+a))/b^2-d^2*polylog(3,-exp(b
*x+a))/b^3+d^2*polylog(3,exp(b*x+a))/b^3

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Rubi [A]  time = 0.17, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {4186, 3770, 4182, 2531, 2282, 6589} \[ \frac {d (c+d x) \text {PolyLog}\left (2,-e^{a+b x}\right )}{b^2}-\frac {d (c+d x) \text {PolyLog}\left (2,e^{a+b x}\right )}{b^2}-\frac {d^2 \text {PolyLog}\left (3,-e^{a+b x}\right )}{b^3}+\frac {d^2 \text {PolyLog}\left (3,e^{a+b x}\right )}{b^3}-\frac {d (c+d x) \text {csch}(a+b x)}{b^2}-\frac {d^2 \tanh ^{-1}(\cosh (a+b x))}{b^3}+\frac {(c+d x)^2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}-\frac {(c+d x)^2 \coth (a+b x) \text {csch}(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^2*Csch[a + b*x]^3,x]

[Out]

((c + d*x)^2*ArcTanh[E^(a + b*x)])/b - (d^2*ArcTanh[Cosh[a + b*x]])/b^3 - (d*(c + d*x)*Csch[a + b*x])/b^2 - ((
c + d*x)^2*Coth[a + b*x]*Csch[a + b*x])/(2*b) + (d*(c + d*x)*PolyLog[2, -E^(a + b*x)])/b^2 - (d*(c + d*x)*Poly
Log[2, E^(a + b*x)])/b^2 - (d^2*PolyLog[3, -E^(a + b*x)])/b^3 + (d^2*PolyLog[3, E^(a + b*x)])/b^3

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 4186

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

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

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Mathematica [B]  time = 10.27, size = 420, normalized size = 2.73 \[ \frac {\text {csch}\left (\frac {a}{2}\right ) \text {csch}\left (\frac {a}{2}+\frac {b x}{2}\right ) \left (c d \sinh \left (\frac {b x}{2}\right )+d^2 x \sinh \left (\frac {b x}{2}\right )\right )}{2 b^2}+\frac {\text {sech}\left (\frac {a}{2}\right ) \text {sech}\left (\frac {a}{2}+\frac {b x}{2}\right ) \left (c d \sinh \left (\frac {b x}{2}\right )+d^2 x \sinh \left (\frac {b x}{2}\right )\right )}{2 b^2}-\frac {d \text {csch}(a) (c+d x)}{b^2}+\frac {b^2 \left (-c^2\right ) \log \left (1-e^{a+b x}\right )+b^2 c^2 \log \left (e^{a+b x}+1\right )-2 b^2 c d x \log \left (1-e^{a+b x}\right )+2 b^2 c d x \log \left (e^{a+b x}+1\right )-b^2 d^2 x^2 \log \left (1-e^{a+b x}\right )+b^2 d^2 x^2 \log \left (e^{a+b x}+1\right )+2 b d (c+d x) \text {Li}_2\left (-e^{a+b x}\right )-2 b d (c+d x) \text {Li}_2\left (e^{a+b x}\right )-2 d^2 \text {Li}_3\left (-e^{a+b x}\right )+2 d^2 \text {Li}_3\left (e^{a+b x}\right )+2 d^2 \log \left (1-e^{a+b x}\right )-2 d^2 \log \left (e^{a+b x}+1\right )}{2 b^3}+\frac {\left (-c^2-2 c d x-d^2 x^2\right ) \text {csch}^2\left (\frac {a}{2}+\frac {b x}{2}\right )}{8 b}+\frac {\left (-c^2-2 c d x-d^2 x^2\right ) \text {sech}^2\left (\frac {a}{2}+\frac {b x}{2}\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^2*Csch[a + b*x]^3,x]

[Out]

-((d*(c + d*x)*Csch[a])/b^2) + ((-c^2 - 2*c*d*x - d^2*x^2)*Csch[a/2 + (b*x)/2]^2)/(8*b) + (-(b^2*c^2*Log[1 - E
^(a + b*x)]) + 2*d^2*Log[1 - E^(a + b*x)] - 2*b^2*c*d*x*Log[1 - E^(a + b*x)] - b^2*d^2*x^2*Log[1 - E^(a + b*x)
] + b^2*c^2*Log[1 + E^(a + b*x)] - 2*d^2*Log[1 + E^(a + b*x)] + 2*b^2*c*d*x*Log[1 + E^(a + b*x)] + b^2*d^2*x^2
*Log[1 + E^(a + b*x)] + 2*b*d*(c + d*x)*PolyLog[2, -E^(a + b*x)] - 2*b*d*(c + d*x)*PolyLog[2, E^(a + b*x)] - 2
*d^2*PolyLog[3, -E^(a + b*x)] + 2*d^2*PolyLog[3, E^(a + b*x)])/(2*b^3) + ((-c^2 - 2*c*d*x - d^2*x^2)*Sech[a/2
+ (b*x)/2]^2)/(8*b) + (Csch[a/2]*Csch[a/2 + (b*x)/2]*(c*d*Sinh[(b*x)/2] + d^2*x*Sinh[(b*x)/2]))/(2*b^2) + (Sec
h[a/2]*Sech[a/2 + (b*x)/2]*(c*d*Sinh[(b*x)/2] + d^2*x*Sinh[(b*x)/2]))/(2*b^2)

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fricas [C]  time = 0.55, size = 2218, normalized size = 14.40 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*csch(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/2*(2*(b^2*d^2*x^2 + b^2*c^2 + 2*b*c*d + 2*(b^2*c*d + b*d^2)*x)*cosh(b*x + a)^3 + 6*(b^2*d^2*x^2 + b^2*c^2 +
2*b*c*d + 2*(b^2*c*d + b*d^2)*x)*cosh(b*x + a)*sinh(b*x + a)^2 + 2*(b^2*d^2*x^2 + b^2*c^2 + 2*b*c*d + 2*(b^2*
c*d + b*d^2)*x)*sinh(b*x + a)^3 + 2*(b^2*d^2*x^2 + b^2*c^2 - 2*b*c*d + 2*(b^2*c*d - b*d^2)*x)*cosh(b*x + a) +
2*((b*d^2*x + b*c*d)*cosh(b*x + a)^4 + 4*(b*d^2*x + b*c*d)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*d^2*x + b*c*d)*s
inh(b*x + a)^4 + b*d^2*x + b*c*d - 2*(b*d^2*x + b*c*d)*cosh(b*x + a)^2 - 2*(b*d^2*x + b*c*d - 3*(b*d^2*x + b*c
*d)*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 4*((b*d^2*x + b*c*d)*cosh(b*x + a)^3 - (b*d^2*x + b*c*d)*cosh(b*x + a))
*sinh(b*x + a))*dilog(cosh(b*x + a) + sinh(b*x + a)) - 2*((b*d^2*x + b*c*d)*cosh(b*x + a)^4 + 4*(b*d^2*x + b*c
*d)*cosh(b*x + a)*sinh(b*x + a)^3 + (b*d^2*x + b*c*d)*sinh(b*x + a)^4 + b*d^2*x + b*c*d - 2*(b*d^2*x + b*c*d)*
cosh(b*x + a)^2 - 2*(b*d^2*x + b*c*d - 3*(b*d^2*x + b*c*d)*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 4*((b*d^2*x + b*
c*d)*cosh(b*x + a)^3 - (b*d^2*x + b*c*d)*cosh(b*x + a))*sinh(b*x + a))*dilog(-cosh(b*x + a) - sinh(b*x + a)) -
(b^2*d^2*x^2 + 2*b^2*c*d*x + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*cosh(b*x + a)^4 + 4*(b^2*d^2*x^2 +
2*b^2*c*d*x + b^2*c^2 - 2*d^2)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*
sinh(b*x + a)^4 + b^2*c^2 - 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*cosh(b*x + a)^2 - 2*(b^2*d^2*x^2 +
2*b^2*c*d*x + b^2*c^2 - 3*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*cosh(b*x + a)^2 - 2*d^2)*sinh(b*x + a
)^2 - 2*d^2 + 4*((b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*cosh(b*x + a)^3 - (b^2*d^2*x^2 + 2*b^2*c*d*x +
b^2*c^2 - 2*d^2)*cosh(b*x + a))*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) + 1) + ((b^2*c^2 - 2*a*b*c*d
+ (a^2 - 2)*d^2)*cosh(b*x + a)^4 + 4*(b^2*c^2 - 2*a*b*c*d + (a^2 - 2)*d^2)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^
2*c^2 - 2*a*b*c*d + (a^2 - 2)*d^2)*sinh(b*x + a)^4 + b^2*c^2 - 2*a*b*c*d + (a^2 - 2)*d^2 - 2*(b^2*c^2 - 2*a*b*
c*d + (a^2 - 2)*d^2)*cosh(b*x + a)^2 - 2*(b^2*c^2 - 2*a*b*c*d + (a^2 - 2)*d^2 - 3*(b^2*c^2 - 2*a*b*c*d + (a^2
- 2)*d^2)*cosh(b*x + a)^2)*sinh(b*x + a)^2 + 4*((b^2*c^2 - 2*a*b*c*d + (a^2 - 2)*d^2)*cosh(b*x + a)^3 - (b^2*c
^2 - 2*a*b*c*d + (a^2 - 2)*d^2)*cosh(b*x + a))*sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) - 1) + (b^2*d^
2*x^2 + 2*b^2*c*d*x + (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*cosh(b*x + a)^4 + 4*(b^2*d^2*x^2 + 2*b
^2*c*d*x + 2*a*b*c*d - a^2*d^2)*cosh(b*x + a)*sinh(b*x + a)^3 + (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d
^2)*sinh(b*x + a)^4 + 2*a*b*c*d - a^2*d^2 - 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*cosh(b*x + a)^
2 - 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2 - 3*(b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*c
osh(b*x + a)^2)*sinh(b*x + a)^2 + 4*((b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*cosh(b*x + a)^3 - (b^2*
d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*cosh(b*x + a))*sinh(b*x + a))*log(-cosh(b*x + a) - sinh(b*x + a)
+ 1) - 2*(d^2*cosh(b*x + a)^4 + 4*d^2*cosh(b*x + a)*sinh(b*x + a)^3 + d^2*sinh(b*x + a)^4 - 2*d^2*cosh(b*x + a
)^2 + 2*(3*d^2*cosh(b*x + a)^2 - d^2)*sinh(b*x + a)^2 + d^2 + 4*(d^2*cosh(b*x + a)^3 - d^2*cosh(b*x + a))*sinh
(b*x + a))*polylog(3, cosh(b*x + a) + sinh(b*x + a)) + 2*(d^2*cosh(b*x + a)^4 + 4*d^2*cosh(b*x + a)*sinh(b*x +
a)^3 + d^2*sinh(b*x + a)^4 - 2*d^2*cosh(b*x + a)^2 + 2*(3*d^2*cosh(b*x + a)^2 - d^2)*sinh(b*x + a)^2 + d^2 +
4*(d^2*cosh(b*x + a)^3 - d^2*cosh(b*x + a))*sinh(b*x + a))*polylog(3, -cosh(b*x + a) - sinh(b*x + a)) + 2*(b^2
*d^2*x^2 + b^2*c^2 - 2*b*c*d + 3*(b^2*d^2*x^2 + b^2*c^2 + 2*b*c*d + 2*(b^2*c*d + b*d^2)*x)*cosh(b*x + a)^2 + 2
*(b^2*c*d - b*d^2)*x)*sinh(b*x + a))/(b^3*cosh(b*x + a)^4 + 4*b^3*cosh(b*x + a)*sinh(b*x + a)^3 + b^3*sinh(b*x
+ a)^4 - 2*b^3*cosh(b*x + a)^2 + b^3 + 2*(3*b^3*cosh(b*x + a)^2 - b^3)*sinh(b*x + a)^2 + 4*(b^3*cosh(b*x + a)
^3 - b^3*cosh(b*x + a))*sinh(b*x + a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} \operatorname {csch}\left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*csch(b*x+a)^3,x, algorithm="giac")

[Out]

integrate((d*x + c)^2*csch(b*x + a)^3, x)

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maple [B]  time = 0.07, size = 444, normalized size = 2.88 \[ -\frac {{\mathrm e}^{b x +a} \left (b \,d^{2} x^{2} {\mathrm e}^{2 b x +2 a}+2 b c d x \,{\mathrm e}^{2 b x +2 a}+b \,c^{2} {\mathrm e}^{2 b x +2 a}+b \,d^{2} x^{2}+2 d^{2} x \,{\mathrm e}^{2 b x +2 a}+2 b c d x +2 c d \,{\mathrm e}^{2 b x +2 a}+b \,c^{2}-2 d^{2} x -2 c d \right )}{b^{2} \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}-\frac {d^{2} \polylog \left (3, -{\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {d^{2} \polylog \left (3, {\mathrm e}^{b x +a}\right )}{b^{3}}-\frac {2 d^{2} \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {d^{2} \ln \left (1+{\mathrm e}^{b x +a}\right ) x^{2}}{2 b}+\frac {d^{2} \polylog \left (2, -{\mathrm e}^{b x +a}\right ) x}{b^{2}}-\frac {d^{2} \ln \left (1-{\mathrm e}^{b x +a}\right ) x^{2}}{2 b}-\frac {d^{2} \polylog \left (2, {\mathrm e}^{b x +a}\right ) x}{b^{2}}+\frac {c d \polylog \left (2, -{\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {c d \polylog \left (2, {\mathrm e}^{b x +a}\right )}{b^{2}}-\frac {2 c d a \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{2}}+\frac {c d \ln \left (1+{\mathrm e}^{b x +a}\right ) x}{b}+\frac {c d \ln \left (1+{\mathrm e}^{b x +a}\right ) a}{b^{2}}-\frac {c d \ln \left (1-{\mathrm e}^{b x +a}\right ) x}{b}-\frac {c d \ln \left (1-{\mathrm e}^{b x +a}\right ) a}{b^{2}}+\frac {d^{2} a^{2} \arctanh \left ({\mathrm e}^{b x +a}\right )}{b^{3}}+\frac {c^{2} \arctanh \left ({\mathrm e}^{b x +a}\right )}{b}-\frac {d^{2} \ln \left (1+{\mathrm e}^{b x +a}\right ) a^{2}}{2 b^{3}}+\frac {d^{2} \ln \left (1-{\mathrm e}^{b x +a}\right ) a^{2}}{2 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^2*csch(b*x+a)^3,x)

[Out]

-exp(b*x+a)*(b*d^2*x^2*exp(2*b*x+2*a)+2*b*c*d*x*exp(2*b*x+2*a)+b*c^2*exp(2*b*x+2*a)+b*d^2*x^2+2*d^2*x*exp(2*b*
x+2*a)+2*b*c*d*x+2*c*d*exp(2*b*x+2*a)+b*c^2-2*d^2*x-2*c*d)/b^2/(exp(2*b*x+2*a)-1)^2-d^2*polylog(3,-exp(b*x+a))
/b^3+d^2*polylog(3,exp(b*x+a))/b^3-2/b^3*d^2*arctanh(exp(b*x+a))+1/2/b*d^2*ln(1+exp(b*x+a))*x^2+1/b^2*d^2*poly
log(2,-exp(b*x+a))*x-1/2/b*d^2*ln(1-exp(b*x+a))*x^2-1/b^2*d^2*polylog(2,exp(b*x+a))*x+1/b^2*c*d*polylog(2,-exp
(b*x+a))-1/b^2*c*d*polylog(2,exp(b*x+a))-2/b^2*c*d*a*arctanh(exp(b*x+a))+1/b*c*d*ln(1+exp(b*x+a))*x+1/b^2*c*d*
ln(1+exp(b*x+a))*a-1/b*c*d*ln(1-exp(b*x+a))*x-1/b^2*c*d*ln(1-exp(b*x+a))*a+1/b^3*d^2*a^2*arctanh(exp(b*x+a))+1
/b*c^2*arctanh(exp(b*x+a))-1/2/b^3*d^2*ln(1+exp(b*x+a))*a^2+1/2/b^3*d^2*ln(1-exp(b*x+a))*a^2

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maxima [B]  time = 1.04, size = 393, normalized size = 2.55 \[ \frac {1}{2} \, c^{2} {\left (\frac {\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} - \frac {\log \left (e^{\left (-b x - a\right )} - 1\right )}{b} + \frac {2 \, {\left (e^{\left (-b x - a\right )} + e^{\left (-3 \, b x - 3 \, a\right )}\right )}}{b {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )}}\right )} + \frac {{\left (b x \log \left (e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (b x + a\right )}\right )\right )} c d}{b^{2}} - \frac {{\left (b x \log \left (-e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (b x + a\right )}\right )\right )} c d}{b^{2}} - \frac {{\left (b d^{2} x^{2} e^{\left (3 \, a\right )} + 2 \, c d e^{\left (3 \, a\right )} + 2 \, {\left (b c d + d^{2}\right )} x e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )} + {\left (b d^{2} x^{2} e^{a} - 2 \, c d e^{a} + 2 \, {\left (b c d - d^{2}\right )} x e^{a}\right )} e^{\left (b x\right )}}{b^{2} e^{\left (4 \, b x + 4 \, a\right )} - 2 \, b^{2} e^{\left (2 \, b x + 2 \, a\right )} + b^{2}} + \frac {{\left (b^{2} x^{2} \log \left (e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (b x + a\right )})\right )} d^{2}}{2 \, b^{3}} - \frac {{\left (b^{2} x^{2} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (b x + a\right )})\right )} d^{2}}{2 \, b^{3}} - \frac {d^{2} \log \left (e^{\left (b x + a\right )} + 1\right )}{b^{3}} + \frac {d^{2} \log \left (e^{\left (b x + a\right )} - 1\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^2*csch(b*x+a)^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,x\right )}^2}{{\mathrm {sinh}\left (a+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^2/sinh(a + b*x)^3,x)

[Out]

int((c + d*x)^2/sinh(a + b*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{2} \operatorname {csch}^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**2*csch(b*x+a)**3,x)

[Out]

Integral((c + d*x)**2*csch(a + b*x)**3, x)

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