\(\int x (a+b \text {csch}(c+d \sqrt {x})) \, dx\) [33]

Optimal result

Integrand size = 16, antiderivative size = 164 \[ \int x \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^2}{2}-\frac {4 b x^{3/2} \text {arctanh}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {6 b x \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {6 b x \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 b \sqrt {x} \operatorname {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 b \operatorname {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {12 b \operatorname {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4} \]

[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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {14, 5545, 4267, 2611, 6744, 2320, 6724} \[ \int x \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^2}{2}-\frac {4 b x^{3/2} \text {arctanh}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {12 b \operatorname {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {12 b \operatorname {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {12 b \sqrt {x} \operatorname {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 b \sqrt {x} \operatorname {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {6 b x \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {6 b x \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2} \]

[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 2320

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 2611

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 4267

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 5545

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 6724

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 6744

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.10 \[ \int x \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {a x^2}{2}+\frac {2 b \left (d^3 x^{3/2} \log \left (1-e^{c+d \sqrt {x}}\right )-d^3 x^{3/2} \log \left (1+e^{c+d \sqrt {x}}\right )-3 d^2 x \operatorname {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )+3 d^2 x \operatorname {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )+6 d \sqrt {x} \operatorname {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )-6 d \sqrt {x} \operatorname {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )-6 \operatorname {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )+6 \operatorname {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )\right )}{d^4} \]

[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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.05 \[ \int x \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\frac {1}{2} \, a x^{2} - \frac {2 \, {\left (\log \left (e^{\left (d \sqrt {x} + c\right )} + 1\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{3} + 3 \, {\rm Li}_2\left (-e^{\left (d \sqrt {x} + c\right )}\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{2} - 6 \, \log \left (e^{\left (d \sqrt {x}\right )}\right ) {\rm Li}_{3}(-e^{\left (d \sqrt {x} + c\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (d \sqrt {x} + c\right )})\right )} b}{d^{4}} + \frac {2 \, {\left (\log \left (-e^{\left (d \sqrt {x} + c\right )} + 1\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{3} + 3 \, {\rm Li}_2\left (e^{\left (d \sqrt {x} + c\right )}\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{2} - 6 \, \log \left (e^{\left (d \sqrt {x}\right )}\right ) {\rm Li}_{3}(e^{\left (d \sqrt {x} + c\right )}) + 6 \, {\rm Li}_{4}(e^{\left (d \sqrt {x} + c\right )})\right )} b}{d^{4}} \]

[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

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int x \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx=\int x\,\left (a+\frac {b}{\mathrm {sinh}\left (c+d\,\sqrt {x}\right )}\right ) \,d x \]

[In]

int(x*(a + b/sinh(c + d*x^(1/2))),x)

[Out]

int(x*(a + b/sinh(c + d*x^(1/2))), x)