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

Optimal. Leaf size=164 \[ \frac {a x^2}{2}-\frac {12 b \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {12 b \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {12 b \sqrt {x} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 b \sqrt {x} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {6 b x \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {6 b x \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}-\frac {4 b x^{3/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d} \]

[Out]

1/2*a*x^2-4*b*x^(3/2)*arctanh(exp(c+d*x^(1/2)))/d-6*b*x*polylog(2,-exp(c+d*x^(1/2)))/d^2+6*b*x*polylog(2,exp(c
+d*x^(1/2)))/d^2-12*b*polylog(4,-exp(c+d*x^(1/2)))/d^4+12*b*polylog(4,exp(c+d*x^(1/2)))/d^4+12*b*polylog(3,-ex
p(c+d*x^(1/2)))*x^(1/2)/d^3-12*b*polylog(3,exp(c+d*x^(1/2)))*x^(1/2)/d^3

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Rubi [A]  time = 0.17, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {14, 5437, 4182, 2531, 6609, 2282, 6589} \[ -\frac {6 b x \text {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {6 b x \text {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 b \sqrt {x} \text {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 b \sqrt {x} \text {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 b \text {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {12 b \text {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {a x^2}{2}-\frac {4 b x^{3/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*Csch[c + d*Sqrt[x]]),x]

[Out]

(a*x^2)/2 - (4*b*x^(3/2)*ArcTanh[E^(c + d*Sqrt[x])])/d - (6*b*x*PolyLog[2, -E^(c + d*Sqrt[x])])/d^2 + (6*b*x*P
olyLog[2, E^(c + d*Sqrt[x])])/d^2 + (12*b*Sqrt[x]*PolyLog[3, -E^(c + d*Sqrt[x])])/d^3 - (12*b*Sqrt[x]*PolyLog[
3, E^(c + d*Sqrt[x])])/d^3 - (12*b*PolyLog[4, -E^(c + d*Sqrt[x])])/d^4 + (12*b*PolyLog[4, E^(c + d*Sqrt[x])])/
d^4

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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

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Mathematica [A]  time = 2.77, size = 181, normalized size = 1.10 \[ \frac {a x^2}{2}+\frac {2 b \left (d^3 x^{3/2} \log \left (1-e^{c+d \sqrt {x}}\right )-d^3 x^{3/2} \log \left (e^{c+d \sqrt {x}}+1\right )-3 d^2 x \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )+3 d^2 x \text {Li}_2\left (e^{c+d \sqrt {x}}\right )+6 d \sqrt {x} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )-6 d \sqrt {x} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )-6 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )+6 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )\right )}{d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b x \operatorname {csch}\left (d \sqrt {x} + c\right ) + a x, x\right ) \]

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(b*x*csch(d*sqrt(x) + c) + a*x, x)

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

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((b*csch(d*sqrt(x) + c) + a)*x, x)

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maple [F]  time = 0.63, size = 0, normalized size = 0.00 \[ \int x \left (a +b \,\mathrm {csch}\left (c +d \sqrt {x}\right )\right )\, dx \]

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 [A]  time = 0.58, size = 173, normalized size = 1.05 \[ \frac {1}{2} \, a x^{2} - \frac {2 \, {\left (\log \left (e^{\left (d \sqrt {x} + c\right )} + 1\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{3} + 3 \, {\rm Li}_2\left (-e^{\left (d \sqrt {x} + c\right )}\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{2} - 6 \, \log \left (e^{\left (d \sqrt {x}\right )}\right ) {\rm Li}_{3}(-e^{\left (d \sqrt {x} + c\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (d \sqrt {x} + c\right )})\right )} b}{d^{4}} + \frac {2 \, {\left (\log \left (-e^{\left (d \sqrt {x} + c\right )} + 1\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{3} + 3 \, {\rm Li}_2\left (e^{\left (d \sqrt {x} + c\right )}\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{2} - 6 \, \log \left (e^{\left (d \sqrt {x}\right )}\right ) {\rm Li}_{3}(e^{\left (d \sqrt {x} + c\right )}) + 6 \, {\rm Li}_{4}(e^{\left (d \sqrt {x} + c\right )})\right )} b}{d^{4}} \]

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]

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

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

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 \[ \int x \left (a + b \operatorname {csch}{\left (c + d \sqrt {x} \right )}\right )\, dx \]

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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