3.43 \(\int \frac {x}{a+b \text {csch}(c+d \sqrt {x})} \, dx\)

Optimal. Leaf size=449 \[ -\frac {12 b \text {Li}_4\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a d^4 \sqrt {a^2+b^2}}+\frac {12 b \text {Li}_4\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a d^4 \sqrt {a^2+b^2}}+\frac {12 b \sqrt {x} \text {Li}_3\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a d^3 \sqrt {a^2+b^2}}-\frac {12 b \sqrt {x} \text {Li}_3\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a d^3 \sqrt {a^2+b^2}}-\frac {6 b x \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a d^2 \sqrt {a^2+b^2}}+\frac {6 b x \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a d^2 \sqrt {a^2+b^2}}-\frac {2 b x^{3/2} \log \left (\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}+1\right )}{a d \sqrt {a^2+b^2}}+\frac {2 b x^{3/2} \log \left (\frac {a e^{c+d \sqrt {x}}}{\sqrt {a^2+b^2}+b}+1\right )}{a d \sqrt {a^2+b^2}}+\frac {x^2}{2 a} \]

[Out]

1/2*x^2/a-2*b*x^(3/2)*ln(1+a*exp(c+d*x^(1/2))/(b-(a^2+b^2)^(1/2)))/a/d/(a^2+b^2)^(1/2)+2*b*x^(3/2)*ln(1+a*exp(
c+d*x^(1/2))/(b+(a^2+b^2)^(1/2)))/a/d/(a^2+b^2)^(1/2)-6*b*x*polylog(2,-a*exp(c+d*x^(1/2))/(b-(a^2+b^2)^(1/2)))
/a/d^2/(a^2+b^2)^(1/2)+6*b*x*polylog(2,-a*exp(c+d*x^(1/2))/(b+(a^2+b^2)^(1/2)))/a/d^2/(a^2+b^2)^(1/2)-12*b*pol
ylog(4,-a*exp(c+d*x^(1/2))/(b-(a^2+b^2)^(1/2)))/a/d^4/(a^2+b^2)^(1/2)+12*b*polylog(4,-a*exp(c+d*x^(1/2))/(b+(a
^2+b^2)^(1/2)))/a/d^4/(a^2+b^2)^(1/2)+12*b*polylog(3,-a*exp(c+d*x^(1/2))/(b-(a^2+b^2)^(1/2)))*x^(1/2)/a/d^3/(a
^2+b^2)^(1/2)-12*b*polylog(3,-a*exp(c+d*x^(1/2))/(b+(a^2+b^2)^(1/2)))*x^(1/2)/a/d^3/(a^2+b^2)^(1/2)

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Rubi [A]  time = 0.90, antiderivative size = 449, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5437, 4191, 3322, 2264, 2190, 2531, 6609, 2282, 6589} \[ -\frac {6 b x \text {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a d^2 \sqrt {a^2+b^2}}+\frac {6 b x \text {PolyLog}\left (2,-\frac {a e^{c+d \sqrt {x}}}{\sqrt {a^2+b^2}+b}\right )}{a d^2 \sqrt {a^2+b^2}}+\frac {12 b \sqrt {x} \text {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a d^3 \sqrt {a^2+b^2}}-\frac {12 b \sqrt {x} \text {PolyLog}\left (3,-\frac {a e^{c+d \sqrt {x}}}{\sqrt {a^2+b^2}+b}\right )}{a d^3 \sqrt {a^2+b^2}}-\frac {12 b \text {PolyLog}\left (4,-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a d^4 \sqrt {a^2+b^2}}+\frac {12 b \text {PolyLog}\left (4,-\frac {a e^{c+d \sqrt {x}}}{\sqrt {a^2+b^2}+b}\right )}{a d^4 \sqrt {a^2+b^2}}-\frac {2 b x^{3/2} \log \left (\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}+1\right )}{a d \sqrt {a^2+b^2}}+\frac {2 b x^{3/2} \log \left (\frac {a e^{c+d \sqrt {x}}}{\sqrt {a^2+b^2}+b}+1\right )}{a d \sqrt {a^2+b^2}}+\frac {x^2}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Csch[c + d*Sqrt[x]]),x]

[Out]

x^2/(2*a) - (2*b*x^(3/2)*Log[1 + (a*E^(c + d*Sqrt[x]))/(b - Sqrt[a^2 + b^2])])/(a*Sqrt[a^2 + b^2]*d) + (2*b*x^
(3/2)*Log[1 + (a*E^(c + d*Sqrt[x]))/(b + Sqrt[a^2 + b^2])])/(a*Sqrt[a^2 + b^2]*d) - (6*b*x*PolyLog[2, -((a*E^(
c + d*Sqrt[x]))/(b - Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 + b^2]*d^2) + (6*b*x*PolyLog[2, -((a*E^(c + d*Sqrt[x]))/(
b + Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 + b^2]*d^2) + (12*b*Sqrt[x]*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b - Sqrt[a
^2 + b^2]))])/(a*Sqrt[a^2 + b^2]*d^3) - (12*b*Sqrt[x]*PolyLog[3, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[a^2 + b^2])
)])/(a*Sqrt[a^2 + b^2]*d^3) - (12*b*PolyLog[4, -((a*E^(c + d*Sqrt[x]))/(b - Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 +
b^2]*d^4) + (12*b*PolyLog[4, -((a*E^(c + d*Sqrt[x]))/(b + Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 + b^2]*d^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 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 4191

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] &
& IGtQ[m, 0]

Rule 5437

Int[((a_.) + Csch[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Csch[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplif
y[(m + 1)/n], 0] && IntegerQ[p]

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 \frac {x}{a+b \text {csch}\left (c+d \sqrt {x}\right )} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^3}{a+b \text {csch}(c+d x)} \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (\frac {x^3}{a}-\frac {b x^3}{a (b+a \sinh (c+d x))}\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {x^2}{2 a}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {x^3}{b+a \sinh (c+d x)} \, dx,x,\sqrt {x}\right )}{a}\\ &=\frac {x^2}{2 a}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {e^{c+d x} x^3}{-a+2 b e^{c+d x}+a e^{2 (c+d x)}} \, dx,x,\sqrt {x}\right )}{a}\\ &=\frac {x^2}{2 a}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {e^{c+d x} x^3}{2 b-2 \sqrt {a^2+b^2}+2 a e^{c+d x}} \, dx,x,\sqrt {x}\right )}{\sqrt {a^2+b^2}}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {e^{c+d x} x^3}{2 b+2 \sqrt {a^2+b^2}+2 a e^{c+d x}} \, dx,x,\sqrt {x}\right )}{\sqrt {a^2+b^2}}\\ &=\frac {x^2}{2 a}-\frac {2 b x^{3/2} \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}+\frac {2 b x^{3/2} \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}+\frac {(6 b) \operatorname {Subst}\left (\int x^2 \log \left (1+\frac {2 a e^{c+d x}}{2 b-2 \sqrt {a^2+b^2}}\right ) \, dx,x,\sqrt {x}\right )}{a \sqrt {a^2+b^2} d}-\frac {(6 b) \operatorname {Subst}\left (\int x^2 \log \left (1+\frac {2 a e^{c+d x}}{2 b+2 \sqrt {a^2+b^2}}\right ) \, dx,x,\sqrt {x}\right )}{a \sqrt {a^2+b^2} d}\\ &=\frac {x^2}{2 a}-\frac {2 b x^{3/2} \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}+\frac {2 b x^{3/2} \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}-\frac {6 b x \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {6 b x \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {(12 b) \operatorname {Subst}\left (\int x \text {Li}_2\left (-\frac {2 a e^{c+d x}}{2 b-2 \sqrt {a^2+b^2}}\right ) \, dx,x,\sqrt {x}\right )}{a \sqrt {a^2+b^2} d^2}-\frac {(12 b) \operatorname {Subst}\left (\int x \text {Li}_2\left (-\frac {2 a e^{c+d x}}{2 b+2 \sqrt {a^2+b^2}}\right ) \, dx,x,\sqrt {x}\right )}{a \sqrt {a^2+b^2} d^2}\\ &=\frac {x^2}{2 a}-\frac {2 b x^{3/2} \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}+\frac {2 b x^{3/2} \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}-\frac {6 b x \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {6 b x \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {12 b \sqrt {x} \text {Li}_3\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^3}-\frac {12 b \sqrt {x} \text {Li}_3\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^3}-\frac {(12 b) \operatorname {Subst}\left (\int \text {Li}_3\left (-\frac {2 a e^{c+d x}}{2 b-2 \sqrt {a^2+b^2}}\right ) \, dx,x,\sqrt {x}\right )}{a \sqrt {a^2+b^2} d^3}+\frac {(12 b) \operatorname {Subst}\left (\int \text {Li}_3\left (-\frac {2 a e^{c+d x}}{2 b+2 \sqrt {a^2+b^2}}\right ) \, dx,x,\sqrt {x}\right )}{a \sqrt {a^2+b^2} d^3}\\ &=\frac {x^2}{2 a}-\frac {2 b x^{3/2} \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}+\frac {2 b x^{3/2} \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}-\frac {6 b x \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {6 b x \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {12 b \sqrt {x} \text {Li}_3\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^3}-\frac {12 b \sqrt {x} \text {Li}_3\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^3}-\frac {(12 b) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {a x}{-b+\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{a \sqrt {a^2+b^2} d^4}+\frac {(12 b) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {a x}{b+\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{a \sqrt {a^2+b^2} d^4}\\ &=\frac {x^2}{2 a}-\frac {2 b x^{3/2} \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}+\frac {2 b x^{3/2} \log \left (1+\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d}-\frac {6 b x \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {6 b x \text {Li}_2\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2}+\frac {12 b \sqrt {x} \text {Li}_3\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^3}-\frac {12 b \sqrt {x} \text {Li}_3\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^3}-\frac {12 b \text {Li}_4\left (-\frac {a e^{c+d \sqrt {x}}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^4}+\frac {12 b \text {Li}_4\left (-\frac {a e^{c+d \sqrt {x}}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^4}\\ \end {align*}

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Mathematica [A]  time = 2.04, size = 488, normalized size = 1.09 \[ \frac {d^4 x^2 \sqrt {e^{2 c} \left (a^2+b^2\right )}-4 b e^c d^3 x^{3/2} \log \left (\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {e^{2 c} \left (a^2+b^2\right )}}+1\right )+4 b e^c d^3 x^{3/2} \log \left (\frac {a e^{2 c+d \sqrt {x}}}{\sqrt {e^{2 c} \left (a^2+b^2\right )}+b e^c}+1\right )-12 b e^c d^2 x \text {Li}_2\left (-\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+12 b e^c d^2 x \text {Li}_2\left (-\frac {a e^{2 c+d \sqrt {x}}}{e^c b+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+24 b e^c d \sqrt {x} \text {Li}_3\left (-\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-24 b e^c d \sqrt {x} \text {Li}_3\left (-\frac {a e^{2 c+d \sqrt {x}}}{e^c b+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )-24 b e^c \text {Li}_4\left (-\frac {a e^{2 c+d \sqrt {x}}}{b e^c-\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )+24 b e^c \text {Li}_4\left (-\frac {a e^{2 c+d \sqrt {x}}}{e^c b+\sqrt {\left (a^2+b^2\right ) e^{2 c}}}\right )}{2 a d^4 \sqrt {e^{2 c} \left (a^2+b^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Csch[c + d*Sqrt[x]]),x]

[Out]

(d^4*Sqrt[(a^2 + b^2)*E^(2*c)]*x^2 - 4*b*d^3*E^c*x^(3/2)*Log[1 + (a*E^(2*c + d*Sqrt[x]))/(b*E^c - Sqrt[(a^2 +
b^2)*E^(2*c)])] + 4*b*d^3*E^c*x^(3/2)*Log[1 + (a*E^(2*c + d*Sqrt[x]))/(b*E^c + Sqrt[(a^2 + b^2)*E^(2*c)])] - 1
2*b*d^2*E^c*x*PolyLog[2, -((a*E^(2*c + d*Sqrt[x]))/(b*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))] + 12*b*d^2*E^c*x*Poly
Log[2, -((a*E^(2*c + d*Sqrt[x]))/(b*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))] + 24*b*d*E^c*Sqrt[x]*PolyLog[3, -((a*E^
(2*c + d*Sqrt[x]))/(b*E^c - Sqrt[(a^2 + b^2)*E^(2*c)]))] - 24*b*d*E^c*Sqrt[x]*PolyLog[3, -((a*E^(2*c + d*Sqrt[
x]))/(b*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))] - 24*b*E^c*PolyLog[4, -((a*E^(2*c + d*Sqrt[x]))/(b*E^c - Sqrt[(a^2
+ b^2)*E^(2*c)]))] + 24*b*E^c*PolyLog[4, -((a*E^(2*c + d*Sqrt[x]))/(b*E^c + Sqrt[(a^2 + b^2)*E^(2*c)]))])/(2*a
*d^4*Sqrt[(a^2 + b^2)*E^(2*c)])

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x}{b \operatorname {csch}\left (d \sqrt {x} + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*csch(c+d*x^(1/2))),x, algorithm="fricas")

[Out]

integral(x/(b*csch(d*sqrt(x) + c) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*csch(c+d*x^(1/2))),x, algorithm="giac")

[Out]

integrate(x/(b*csch(d*sqrt(x) + c) + a), x)

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maple [F]  time = 0.79, size = 0, normalized size = 0.00 \[ \int \frac {x}{a +b \,\mathrm {csch}\left (c +d \sqrt {x}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a+b*csch(c+d*x^(1/2))),x)

[Out]

int(x/(a+b*csch(c+d*x^(1/2))),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -2 \, b \int \frac {x e^{\left (d \sqrt {x} + c\right )}}{a^{2} e^{\left (2 \, d \sqrt {x} + 2 \, c\right )} + 2 \, a b e^{\left (d \sqrt {x} + c\right )} - a^{2}}\,{d x} + \frac {x^{2}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*csch(c+d*x^(1/2))),x, algorithm="maxima")

[Out]

-2*b*integrate(x*e^(d*sqrt(x) + c)/(a^2*e^(2*d*sqrt(x) + 2*c) + 2*a*b*e^(d*sqrt(x) + c) - a^2), x) + 1/2*x^2/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b/sinh(c + d*x^(1/2))),x)

[Out]

int(x/(a + b/sinh(c + d*x^(1/2))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*csch(c+d*x**(1/2))),x)

[Out]

Integral(x/(a + b*csch(c + d*sqrt(x))), x)

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