3.239 \(\int \frac {(e+f x)^2 \text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx\)

Optimal. Leaf size=433 \[ \frac {2 b f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a d^3 \sqrt {a^2+b^2}}-\frac {2 b f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a d^3 \sqrt {a^2+b^2}}-\frac {2 b f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a d^2 \sqrt {a^2+b^2}}+\frac {2 b f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a d^2 \sqrt {a^2+b^2}}-\frac {b (e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{a d \sqrt {a^2+b^2}}+\frac {b (e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{a d \sqrt {a^2+b^2}}+\frac {2 f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a d^3}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d} \]

[Out]

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

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Rubi [A]  time = 0.81, antiderivative size = 433, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5575, 4182, 2531, 2282, 6589, 3322, 2264, 2190} \[ -\frac {2 b f (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a d^2 \sqrt {a^2+b^2}}+\frac {2 b f (e+f x) \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{a d^2 \sqrt {a^2+b^2}}+\frac {2 b f^2 \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a d^3 \sqrt {a^2+b^2}}-\frac {2 b f^2 \text {PolyLog}\left (3,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{a d^3 \sqrt {a^2+b^2}}-\frac {2 f (e+f x) \text {PolyLog}\left (2,-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {PolyLog}\left (2,e^{c+d x}\right )}{a d^2}+\frac {2 f^2 \text {PolyLog}\left (3,-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {PolyLog}\left (3,e^{c+d x}\right )}{a d^3}-\frac {b (e+f x)^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{a d \sqrt {a^2+b^2}}+\frac {b (e+f x)^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{a d \sqrt {a^2+b^2}}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && 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 3322

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol] :> Dist[2,
Int[((c + d*x)^m*E^(-(I*e) + f*fz*x))/(-(I*b) + 2*a*E^(-(I*e) + f*fz*x) + I*b*E^(2*(-(I*e) + f*fz*x))), x], x]
/; FreeQ[{a, b, c, d, e, f, fz}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

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 5575

Int[(Csch[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/a, Int[(e + f*x)^m*Csch[c + d*x]^n, x], x] - Dist[b/a, Int[((e + f*x)^m*Csch[c + d*x]^(n - 1))/
(a + b*Sinh[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, 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 \frac {(e+f x)^2 \text {csch}(c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac {\int (e+f x)^2 \text {csch}(c+d x) \, dx}{a}-\frac {b \int \frac {(e+f x)^2}{a+b \sinh (c+d x)} \, dx}{a}\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {(2 b) \int \frac {e^{c+d x} (e+f x)^2}{-b+2 a e^{c+d x}+b e^{2 (c+d x)}} \, dx}{a}-\frac {(2 f) \int (e+f x) \log \left (1-e^{c+d x}\right ) \, dx}{a d}+\frac {(2 f) \int (e+f x) \log \left (1+e^{c+d x}\right ) \, dx}{a d}\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}-\frac {\left (2 b^2\right ) \int \frac {e^{c+d x} (e+f x)^2}{2 a-2 \sqrt {a^2+b^2}+2 b e^{c+d x}} \, dx}{a \sqrt {a^2+b^2}}+\frac {\left (2 b^2\right ) \int \frac {e^{c+d x} (e+f x)^2}{2 a+2 \sqrt {a^2+b^2}+2 b e^{c+d x}} \, dx}{a \sqrt {a^2+b^2}}+\frac {\left (2 f^2\right ) \int \text {Li}_2\left (-e^{c+d x}\right ) \, dx}{a d^2}-\frac {\left (2 f^2\right ) \int \text {Li}_2\left (e^{c+d x}\right ) \, dx}{a d^2}\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {b (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}+\frac {b (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}+\frac {(2 b f) \int (e+f x) \log \left (1+\frac {2 b e^{c+d x}}{2 a-2 \sqrt {a^2+b^2}}\right ) \, dx}{a \sqrt {a^2+b^2} d}-\frac {(2 b f) \int (e+f x) \log \left (1+\frac {2 b e^{c+d x}}{2 a+2 \sqrt {a^2+b^2}}\right ) \, dx}{a \sqrt {a^2+b^2} d}+\frac {\left (2 f^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{c+d x}\right )}{a d^3}-\frac {\left (2 f^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{c+d x}\right )}{a d^3}\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {b (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}+\frac {b (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}-\frac {2 b f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {2 b f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {2 f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a d^3}+\frac {\left (2 b f^2\right ) \int \text {Li}_2\left (-\frac {2 b e^{c+d x}}{2 a-2 \sqrt {a^2+b^2}}\right ) \, dx}{a \sqrt {a^2+b^2} d^2}-\frac {\left (2 b f^2\right ) \int \text {Li}_2\left (-\frac {2 b e^{c+d x}}{2 a+2 \sqrt {a^2+b^2}}\right ) \, dx}{a \sqrt {a^2+b^2} d^2}\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {b (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}+\frac {b (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}-\frac {2 b f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {2 b f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {2 f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a d^3}+\frac {\left (2 b f^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {b x}{-a+\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{a \sqrt {a^2+b^2} d^3}-\frac {\left (2 b f^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-\frac {b x}{a+\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{a \sqrt {a^2+b^2} d^3}\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{c+d x}\right )}{a d}-\frac {b (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}+\frac {b (e+f x)^2 \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}-\frac {2 f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )}{a d^2}+\frac {2 f (e+f x) \text {Li}_2\left (e^{c+d x}\right )}{a d^2}-\frac {2 b f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {2 b f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {2 f^2 \text {Li}_3\left (-e^{c+d x}\right )}{a d^3}-\frac {2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a d^3}+\frac {2 b f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^3}-\frac {2 b f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^3}\\ \end {align*}

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Mathematica [A]  time = 1.90, size = 454, normalized size = 1.05 \[ \frac {\frac {b \left (2 d^2 e^2 \tanh ^{-1}\left (\frac {a+b e^{c+d x}}{\sqrt {a^2+b^2}}\right )-2 d^2 e f x \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+2 d^2 e f x \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )-d^2 f^2 x^2 \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+d^2 f^2 x^2 \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )-2 d f (e+f x) \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )+2 d f (e+f x) \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+2 f^2 \text {Li}_3\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )-2 f^2 \text {Li}_3\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )\right )}{\sqrt {a^2+b^2}}+d^2 (e+f x)^2 \log \left (1-e^{c+d x}\right )-d^2 (e+f x)^2 \log \left (e^{c+d x}+1\right )-2 d f (e+f x) \text {Li}_2\left (-e^{c+d x}\right )+2 d f (e+f x) \text {Li}_2\left (e^{c+d x}\right )+2 f^2 \text {Li}_3\left (-e^{c+d x}\right )-2 f^2 \text {Li}_3\left (e^{c+d x}\right )}{a d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [C]  time = 0.58, size = 1096, normalized size = 2.53 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 0.47, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{2} \mathrm {csch}\left (d x +c \right )}{a +b \sinh \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^2*csch(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

int((f*x+e)^2*csch(d*x+c)/(a+b*sinh(d*x+c)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -e^{2} {\left (\frac {b \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a d} + \frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} - \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d}\right )} - \frac {2 \, {\left (d x \log \left (e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} + \frac {2 \, {\left (d x \log \left (-e^{\left (d x + c\right )} + 1\right ) + {\rm Li}_2\left (e^{\left (d x + c\right )}\right )\right )} e f}{a d^{2}} - \frac {{\left (d^{2} x^{2} \log \left (e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (-e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(-e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} + \frac {{\left (d^{2} x^{2} \log \left (-e^{\left (d x + c\right )} + 1\right ) + 2 \, d x {\rm Li}_2\left (e^{\left (d x + c\right )}\right ) - 2 \, {\rm Li}_{3}(e^{\left (d x + c\right )})\right )} f^{2}}{a d^{3}} - \int \frac {2 \, {\left (b f^{2} x^{2} e^{c} + 2 \, b e f x e^{c}\right )} e^{\left (d x\right )}}{a b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{2} e^{\left (d x + c\right )} - a b}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**2*csch(d*x+c)/(a+b*sinh(d*x+c)),x)

[Out]

Integral((e + f*x)**2*csch(c + d*x)/(a + b*sinh(c + d*x)), x)

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