3.1.37 \(\int x^2 (a+b \text {csch}(c+d \sqrt {x}))^2 \, dx\) [37]

Optimal. Leaf size=441 \[ -\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 a b x^2 \text {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 b^2 x^{3/2} \text {PolyLog}\left (2,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {30 b^2 x \text {PolyLog}\left (3,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 a b x \text {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 a b x \text {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {30 b^2 \sqrt {x} \text {PolyLog}\left (4,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 a b \sqrt {x} \text {PolyLog}\left (5,-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {PolyLog}\left (5,e^{c+d \sqrt {x}}\right )}{d^5}-\frac {15 b^2 \text {PolyLog}\left (5,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {480 a b \text {PolyLog}\left (6,-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {480 a b \text {PolyLog}\left (6,e^{c+d \sqrt {x}}\right )}{d^6} \]

[Out]

-2*b^2*x^(5/2)/d+1/3*a^2*x^3-8*a*b*x^(5/2)*arctanh(exp(c+d*x^(1/2)))/d-2*b^2*x^(5/2)*coth(c+d*x^(1/2))/d+10*b^
2*x^2*ln(1-exp(2*c+2*d*x^(1/2)))/d^2-20*a*b*x^2*polylog(2,-exp(c+d*x^(1/2)))/d^2+20*a*b*x^2*polylog(2,exp(c+d*
x^(1/2)))/d^2+20*b^2*x^(3/2)*polylog(2,exp(2*c+2*d*x^(1/2)))/d^3+80*a*b*x^(3/2)*polylog(3,-exp(c+d*x^(1/2)))/d
^3-80*a*b*x^(3/2)*polylog(3,exp(c+d*x^(1/2)))/d^3-30*b^2*x*polylog(3,exp(2*c+2*d*x^(1/2)))/d^4-240*a*b*x*polyl
og(4,-exp(c+d*x^(1/2)))/d^4+240*a*b*x*polylog(4,exp(c+d*x^(1/2)))/d^4-15*b^2*polylog(5,exp(2*c+2*d*x^(1/2)))/d
^6-480*a*b*polylog(6,-exp(c+d*x^(1/2)))/d^6+480*a*b*polylog(6,exp(c+d*x^(1/2)))/d^6+30*b^2*polylog(4,exp(2*c+2
*d*x^(1/2)))*x^(1/2)/d^5+480*a*b*polylog(5,-exp(c+d*x^(1/2)))*x^(1/2)/d^5-480*a*b*polylog(5,exp(c+d*x^(1/2)))*
x^(1/2)/d^5

________________________________________________________________________________________

Rubi [A]
time = 0.44, antiderivative size = 441, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5545, 4275, 4267, 2611, 6744, 2320, 6724, 4269, 3797, 2221} \begin {gather*} \frac {a^2 x^3}{3}-\frac {480 a b \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {480 a b \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {240 a b x \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 a b x \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {15 b^2 \text {Li}_5\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {30 b^2 \sqrt {x} \text {Li}_4\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {30 b^2 x \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {20 b^2 x^{3/2} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}-\frac {2 b^2 x^{5/2}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*Csch[c + d*Sqrt[x]])^2,x]

[Out]

(-2*b^2*x^(5/2))/d + (a^2*x^3)/3 - (8*a*b*x^(5/2)*ArcTanh[E^(c + d*Sqrt[x])])/d - (2*b^2*x^(5/2)*Coth[c + d*Sq
rt[x]])/d + (10*b^2*x^2*Log[1 - E^(2*(c + d*Sqrt[x]))])/d^2 - (20*a*b*x^2*PolyLog[2, -E^(c + d*Sqrt[x])])/d^2
+ (20*a*b*x^2*PolyLog[2, E^(c + d*Sqrt[x])])/d^2 + (20*b^2*x^(3/2)*PolyLog[2, E^(2*(c + d*Sqrt[x]))])/d^3 + (8
0*a*b*x^(3/2)*PolyLog[3, -E^(c + d*Sqrt[x])])/d^3 - (80*a*b*x^(3/2)*PolyLog[3, E^(c + d*Sqrt[x])])/d^3 - (30*b
^2*x*PolyLog[3, E^(2*(c + d*Sqrt[x]))])/d^4 - (240*a*b*x*PolyLog[4, -E^(c + d*Sqrt[x])])/d^4 + (240*a*b*x*Poly
Log[4, E^(c + d*Sqrt[x])])/d^4 + (30*b^2*Sqrt[x]*PolyLog[4, E^(2*(c + d*Sqrt[x]))])/d^5 + (480*a*b*Sqrt[x]*Pol
yLog[5, -E^(c + d*Sqrt[x])])/d^5 - (480*a*b*Sqrt[x]*PolyLog[5, E^(c + d*Sqrt[x])])/d^5 - (15*b^2*PolyLog[5, E^
(2*(c + d*Sqrt[x]))])/d^6 - (480*a*b*PolyLog[6, -E^(c + d*Sqrt[x])])/d^6 + (480*a*b*PolyLog[6, E^(c + d*Sqrt[x
])])/d^6

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 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 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((
c + d*x)^(m + 1)/(d*(m + 1))), x] + Dist[2*I, Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[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 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4275

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 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*} \int x^2 \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )^2 \, dx &=2 \text {Subst}\left (\int x^5 (a+b \text {csch}(c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 \text {Subst}\left (\int \left (a^2 x^5+2 a b x^5 \text {csch}(c+d x)+b^2 x^5 \text {csch}^2(c+d x)\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {a^2 x^3}{3}+(4 a b) \text {Subst}\left (\int x^5 \text {csch}(c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \text {Subst}\left (\int x^5 \text {csch}^2(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}-\frac {(20 a b) \text {Subst}\left (\int x^4 \log \left (1-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(20 a b) \text {Subst}\left (\int x^4 \log \left (1+e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {\left (10 b^2\right ) \text {Subst}\left (\int x^4 \coth (c+d x) \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {(80 a b) \text {Subst}\left (\int x^3 \text {Li}_2\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {(80 a b) \text {Subst}\left (\int x^3 \text {Li}_2\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {\left (20 b^2\right ) \text {Subst}\left (\int \frac {e^{2 (c+d x)} x^4}{1-e^{2 (c+d x)}} \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {(240 a b) \text {Subst}\left (\int x^2 \text {Li}_3\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {(240 a b) \text {Subst}\left (\int x^2 \text {Li}_3\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {\left (40 b^2\right ) \text {Subst}\left (\int x^3 \log \left (1-e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=-\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 b^2 x^{3/2} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {240 a b x \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 a b x \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {(480 a b) \text {Subst}\left (\int x \text {Li}_4\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(480 a b) \text {Subst}\left (\int x \text {Li}_4\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {\left (60 b^2\right ) \text {Subst}\left (\int x^2 \text {Li}_2\left (e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=-\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 b^2 x^{3/2} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {30 b^2 x \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 a b x \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 a b x \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(480 a b) \text {Subst}\left (\int \text {Li}_5\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5}+\frac {(480 a b) \text {Subst}\left (\int \text {Li}_5\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5}+\frac {\left (60 b^2\right ) \text {Subst}\left (\int x \text {Li}_3\left (e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=-\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 b^2 x^{3/2} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {30 b^2 x \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 a b x \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 a b x \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {30 b^2 \sqrt {x} \text {Li}_4\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(480 a b) \text {Subst}\left (\int \frac {\text {Li}_5(-x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^6}+\frac {(480 a b) \text {Subst}\left (\int \frac {\text {Li}_5(x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^6}-\frac {\left (30 b^2\right ) \text {Subst}\left (\int \text {Li}_4\left (e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^5}\\ &=-\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 b^2 x^{3/2} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {30 b^2 x \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 a b x \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 a b x \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {30 b^2 \sqrt {x} \text {Li}_4\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {480 a b \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}-\frac {\left (15 b^2\right ) \text {Subst}\left (\int \frac {\text {Li}_4(x)}{x} \, dx,x,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^6}\\ &=-\frac {2 b^2 x^{5/2}}{d}+\frac {a^2 x^3}{3}-\frac {8 a b x^{5/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^{5/2} \coth \left (c+d \sqrt {x}\right )}{d}+\frac {10 b^2 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {20 a b x^2 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 a b x^2 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {20 b^2 x^{3/2} \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {80 a b x^{3/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {80 a b x^{3/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {30 b^2 x \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {240 a b x \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {240 a b x \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {30 b^2 \sqrt {x} \text {Li}_4\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {480 a b \sqrt {x} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {480 a b \sqrt {x} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {15 b^2 \text {Li}_5\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {480 a b \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {480 a b \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 8.57, size = 516, normalized size = 1.17 \begin {gather*} \frac {a^2 x^3}{3}+\frac {b \left (-\frac {4 b d^5 e^{2 c} x^{5/2}}{-1+e^{2 c}}+4 a d^5 x^{5/2} \log \left (1-e^{c+d \sqrt {x}}\right )-4 a d^5 x^{5/2} \log \left (1+e^{c+d \sqrt {x}}\right )+10 b d^4 x^2 \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )-20 a d^4 x^2 \text {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )+20 a d^4 x^2 \text {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )+20 b d^3 x^{3/2} \text {PolyLog}\left (2,e^{2 \left (c+d \sqrt {x}\right )}\right )+80 a d^3 x^{3/2} \text {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )-80 a d^3 x^{3/2} \text {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )-30 b d^2 x \text {PolyLog}\left (3,e^{2 \left (c+d \sqrt {x}\right )}\right )-240 a d^2 x \text {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )+240 a d^2 x \text {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )+30 b d \sqrt {x} \text {PolyLog}\left (4,e^{2 \left (c+d \sqrt {x}\right )}\right )+480 a d \sqrt {x} \text {PolyLog}\left (5,-e^{c+d \sqrt {x}}\right )-480 a d \sqrt {x} \text {PolyLog}\left (5,e^{c+d \sqrt {x}}\right )-15 b \text {PolyLog}\left (5,e^{2 \left (c+d \sqrt {x}\right )}\right )-480 a \text {PolyLog}\left (6,-e^{c+d \sqrt {x}}\right )+480 a \text {PolyLog}\left (6,e^{c+d \sqrt {x}}\right )\right )}{d^6}+\frac {b^2 x^{5/2} \text {csch}\left (\frac {c}{2}\right ) \text {csch}\left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right ) \sinh \left (\frac {d \sqrt {x}}{2}\right )}{d}-\frac {b^2 x^{5/2} \text {sech}\left (\frac {c}{2}\right ) \text {sech}\left (\frac {1}{2} \left (c+d \sqrt {x}\right )\right ) \sinh \left (\frac {d \sqrt {x}}{2}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x^3)/3 + (b*((-4*b*d^5*E^(2*c)*x^(5/2))/(-1 + E^(2*c)) + 4*a*d^5*x^(5/2)*Log[1 - E^(c + d*Sqrt[x])] - 4*a
*d^5*x^(5/2)*Log[1 + E^(c + d*Sqrt[x])] + 10*b*d^4*x^2*Log[1 - E^(2*(c + d*Sqrt[x]))] - 20*a*d^4*x^2*PolyLog[2
, -E^(c + d*Sqrt[x])] + 20*a*d^4*x^2*PolyLog[2, E^(c + d*Sqrt[x])] + 20*b*d^3*x^(3/2)*PolyLog[2, E^(2*(c + d*S
qrt[x]))] + 80*a*d^3*x^(3/2)*PolyLog[3, -E^(c + d*Sqrt[x])] - 80*a*d^3*x^(3/2)*PolyLog[3, E^(c + d*Sqrt[x])] -
30*b*d^2*x*PolyLog[3, E^(2*(c + d*Sqrt[x]))] - 240*a*d^2*x*PolyLog[4, -E^(c + d*Sqrt[x])] + 240*a*d^2*x*PolyL
og[4, E^(c + d*Sqrt[x])] + 30*b*d*Sqrt[x]*PolyLog[4, E^(2*(c + d*Sqrt[x]))] + 480*a*d*Sqrt[x]*PolyLog[5, -E^(c
+ d*Sqrt[x])] - 480*a*d*Sqrt[x]*PolyLog[5, E^(c + d*Sqrt[x])] - 15*b*PolyLog[5, E^(2*(c + d*Sqrt[x]))] - 480*
a*PolyLog[6, -E^(c + d*Sqrt[x])] + 480*a*PolyLog[6, E^(c + d*Sqrt[x])]))/d^6 + (b^2*x^(5/2)*Csch[c/2]*Csch[(c
+ d*Sqrt[x])/2]*Sinh[(d*Sqrt[x])/2])/d - (b^2*x^(5/2)*Sech[c/2]*Sech[(c + d*Sqrt[x])/2]*Sinh[(d*Sqrt[x])/2])/d

________________________________________________________________________________________

Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{2} \left (a +b \,\mathrm {csch}\left (c +d \sqrt {x}\right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.42, size = 496, normalized size = 1.12 \begin {gather*} \frac {1}{3} \, a^{2} x^{3} - \frac {4 \, b^{2} x^{\frac {5}{2}}}{d e^{\left (2 \, d \sqrt {x} + 2 \, c\right )} - d} - \frac {4 \, {\left (d^{5} x^{\frac {5}{2}} \log \left (e^{\left (d \sqrt {x} + c\right )} + 1\right ) + 5 \, d^{4} x^{2} {\rm Li}_2\left (-e^{\left (d \sqrt {x} + c\right )}\right ) - 20 \, d^{3} x^{\frac {3}{2}} {\rm Li}_{3}(-e^{\left (d \sqrt {x} + c\right )}) + 60 \, d^{2} x {\rm Li}_{4}(-e^{\left (d \sqrt {x} + c\right )}) - 120 \, d \sqrt {x} {\rm Li}_{5}(-e^{\left (d \sqrt {x} + c\right )}) + 120 \, {\rm Li}_{6}(-e^{\left (d \sqrt {x} + c\right )})\right )} a b}{d^{6}} + \frac {4 \, {\left (d^{5} x^{\frac {5}{2}} \log \left (-e^{\left (d \sqrt {x} + c\right )} + 1\right ) + 5 \, d^{4} x^{2} {\rm Li}_2\left (e^{\left (d \sqrt {x} + c\right )}\right ) - 20 \, d^{3} x^{\frac {3}{2}} {\rm Li}_{3}(e^{\left (d \sqrt {x} + c\right )}) + 60 \, d^{2} x {\rm Li}_{4}(e^{\left (d \sqrt {x} + c\right )}) - 120 \, d \sqrt {x} {\rm Li}_{5}(e^{\left (d \sqrt {x} + c\right )}) + 120 \, {\rm Li}_{6}(e^{\left (d \sqrt {x} + c\right )})\right )} a b}{d^{6}} + \frac {10 \, {\left (d^{4} x^{2} \log \left (e^{\left (d \sqrt {x} + c\right )} + 1\right ) + 4 \, d^{3} x^{\frac {3}{2}} {\rm Li}_2\left (-e^{\left (d \sqrt {x} + c\right )}\right ) - 12 \, d^{2} x {\rm Li}_{3}(-e^{\left (d \sqrt {x} + c\right )}) + 24 \, d \sqrt {x} {\rm Li}_{4}(-e^{\left (d \sqrt {x} + c\right )}) - 24 \, {\rm Li}_{5}(-e^{\left (d \sqrt {x} + c\right )})\right )} b^{2}}{d^{6}} + \frac {10 \, {\left (d^{4} x^{2} \log \left (-e^{\left (d \sqrt {x} + c\right )} + 1\right ) + 4 \, d^{3} x^{\frac {3}{2}} {\rm Li}_2\left (e^{\left (d \sqrt {x} + c\right )}\right ) - 12 \, d^{2} x {\rm Li}_{3}(e^{\left (d \sqrt {x} + c\right )}) + 24 \, d \sqrt {x} {\rm Li}_{4}(e^{\left (d \sqrt {x} + c\right )}) - 24 \, {\rm Li}_{5}(e^{\left (d \sqrt {x} + c\right )})\right )} b^{2}}{d^{6}} - \frac {2 \, {\left (a b d^{6} x^{3} + 3 \, b^{2} d^{5} x^{\frac {5}{2}}\right )}}{3 \, d^{6}} + \frac {2 \, {\left (a b d^{6} x^{3} - 3 \, b^{2} d^{5} x^{\frac {5}{2}}\right )}}{3 \, d^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a^2*x^3 - 4*b^2*x^(5/2)/(d*e^(2*d*sqrt(x) + 2*c) - d) - 4*(d^5*x^(5/2)*log(e^(d*sqrt(x) + c) + 1) + 5*d^4*
x^2*dilog(-e^(d*sqrt(x) + c)) - 20*d^3*x^(3/2)*polylog(3, -e^(d*sqrt(x) + c)) + 60*d^2*x*polylog(4, -e^(d*sqrt
(x) + c)) - 120*d*sqrt(x)*polylog(5, -e^(d*sqrt(x) + c)) + 120*polylog(6, -e^(d*sqrt(x) + c)))*a*b/d^6 + 4*(d^
5*x^(5/2)*log(-e^(d*sqrt(x) + c) + 1) + 5*d^4*x^2*dilog(e^(d*sqrt(x) + c)) - 20*d^3*x^(3/2)*polylog(3, e^(d*sq
rt(x) + c)) + 60*d^2*x*polylog(4, e^(d*sqrt(x) + c)) - 120*d*sqrt(x)*polylog(5, e^(d*sqrt(x) + c)) + 120*polyl
og(6, e^(d*sqrt(x) + c)))*a*b/d^6 + 10*(d^4*x^2*log(e^(d*sqrt(x) + c) + 1) + 4*d^3*x^(3/2)*dilog(-e^(d*sqrt(x)
+ c)) - 12*d^2*x*polylog(3, -e^(d*sqrt(x) + c)) + 24*d*sqrt(x)*polylog(4, -e^(d*sqrt(x) + c)) - 24*polylog(5,
-e^(d*sqrt(x) + c)))*b^2/d^6 + 10*(d^4*x^2*log(-e^(d*sqrt(x) + c) + 1) + 4*d^3*x^(3/2)*dilog(e^(d*sqrt(x) + c
)) - 12*d^2*x*polylog(3, e^(d*sqrt(x) + c)) + 24*d*sqrt(x)*polylog(4, e^(d*sqrt(x) + c)) - 24*polylog(5, e^(d*
sqrt(x) + c)))*b^2/d^6 - 2/3*(a*b*d^6*x^3 + 3*b^2*d^5*x^(5/2))/d^6 + 2/3*(a*b*d^6*x^3 - 3*b^2*d^5*x^(5/2))/d^6

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b^2*x^2*csch(d*sqrt(x) + c)^2 + 2*a*b*x^2*csch(d*sqrt(x) + c) + a^2*x^2, x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \left (a + b \operatorname {csch}{\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,{\left (a+\frac {b}{\mathrm {sinh}\left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________