3.1.31 \(\int x^3 (a+b \text {csch}(c+d \sqrt {x})) \, dx\) [31]

Optimal. Leaf size=356 \[ \frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 b x^2 \text {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 b x^{3/2} \text {PolyLog}\left (5,-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 b x^{3/2} \text {PolyLog}\left (5,e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 b x \text {PolyLog}\left (6,-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 b x \text {PolyLog}\left (6,e^{c+d \sqrt {x}}\right )}{d^6}+\frac {10080 b \sqrt {x} \text {PolyLog}\left (7,-e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 b \sqrt {x} \text {PolyLog}\left (7,e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 b \text {PolyLog}\left (8,-e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 b \text {PolyLog}\left (8,e^{c+d \sqrt {x}}\right )}{d^8} \]

[Out]

1/4*a*x^4-4*b*x^(7/2)*arctanh(exp(c+d*x^(1/2)))/d-14*b*x^3*polylog(2,-exp(c+d*x^(1/2)))/d^2+14*b*x^3*polylog(2
,exp(c+d*x^(1/2)))/d^2+84*b*x^(5/2)*polylog(3,-exp(c+d*x^(1/2)))/d^3-84*b*x^(5/2)*polylog(3,exp(c+d*x^(1/2)))/
d^3-420*b*x^2*polylog(4,-exp(c+d*x^(1/2)))/d^4+420*b*x^2*polylog(4,exp(c+d*x^(1/2)))/d^4+1680*b*x^(3/2)*polylo
g(5,-exp(c+d*x^(1/2)))/d^5-1680*b*x^(3/2)*polylog(5,exp(c+d*x^(1/2)))/d^5-5040*b*x*polylog(6,-exp(c+d*x^(1/2))
)/d^6+5040*b*x*polylog(6,exp(c+d*x^(1/2)))/d^6-10080*b*polylog(8,-exp(c+d*x^(1/2)))/d^8+10080*b*polylog(8,exp(
c+d*x^(1/2)))/d^8+10080*b*polylog(7,-exp(c+d*x^(1/2)))*x^(1/2)/d^7-10080*b*polylog(7,exp(c+d*x^(1/2)))*x^(1/2)
/d^7

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Rubi [A]
time = 0.28, antiderivative size = 356, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {14, 5545, 4267, 2611, 6744, 2320, 6724} \begin {gather*} \frac {a x^4}{4}-\frac {10080 b \text {Li}_8\left (-e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 b \text {Li}_8\left (e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 b \sqrt {x} \text {Li}_7\left (-e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 b \sqrt {x} \text {Li}_7\left (e^{c+d \sqrt {x}}\right )}{d^7}-\frac {5040 b x \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 b x \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}+\frac {1680 b x^{3/2} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {420 b x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Csch[c + d*Sqrt[x]]),x]

[Out]

(a*x^4)/4 - (4*b*x^(7/2)*ArcTanh[E^(c + d*Sqrt[x])])/d - (14*b*x^3*PolyLog[2, -E^(c + d*Sqrt[x])])/d^2 + (14*b
*x^3*PolyLog[2, E^(c + d*Sqrt[x])])/d^2 + (84*b*x^(5/2)*PolyLog[3, -E^(c + d*Sqrt[x])])/d^3 - (84*b*x^(5/2)*Po
lyLog[3, E^(c + d*Sqrt[x])])/d^3 - (420*b*x^2*PolyLog[4, -E^(c + d*Sqrt[x])])/d^4 + (420*b*x^2*PolyLog[4, E^(c
+ d*Sqrt[x])])/d^4 + (1680*b*x^(3/2)*PolyLog[5, -E^(c + d*Sqrt[x])])/d^5 - (1680*b*x^(3/2)*PolyLog[5, E^(c +
d*Sqrt[x])])/d^5 - (5040*b*x*PolyLog[6, -E^(c + d*Sqrt[x])])/d^6 + (5040*b*x*PolyLog[6, E^(c + d*Sqrt[x])])/d^
6 + (10080*b*Sqrt[x]*PolyLog[7, -E^(c + d*Sqrt[x])])/d^7 - (10080*b*Sqrt[x]*PolyLog[7, E^(c + d*Sqrt[x])])/d^7
- (10080*b*PolyLog[8, -E^(c + d*Sqrt[x])])/d^8 + (10080*b*PolyLog[8, E^(c + d*Sqrt[x])])/d^8

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*} \int x^3 \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx &=\int \left (a x^3+b x^3 \text {csch}\left (c+d \sqrt {x}\right )\right ) \, dx\\ &=\frac {a x^4}{4}+b \int x^3 \text {csch}\left (c+d \sqrt {x}\right ) \, dx\\ &=\frac {a x^4}{4}+(2 b) \text {Subst}\left (\int x^7 \text {csch}(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {(14 b) \text {Subst}\left (\int x^6 \log \left (1-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(14 b) \text {Subst}\left (\int x^6 \log \left (1+e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {(84 b) \text {Subst}\left (\int x^5 \text {Li}_2\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {(84 b) \text {Subst}\left (\int x^5 \text {Li}_2\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {(420 b) \text {Subst}\left (\int x^4 \text {Li}_3\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {(420 b) \text {Subst}\left (\int x^4 \text {Li}_3\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 b x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {(1680 b) \text {Subst}\left (\int x^3 \text {Li}_4\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(1680 b) \text {Subst}\left (\int x^3 \text {Li}_4\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 b x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(5040 b) \text {Subst}\left (\int x^2 \text {Li}_5\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5}+\frac {(5040 b) \text {Subst}\left (\int x^2 \text {Li}_5\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^5}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 b x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 b x \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 b x \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}+\frac {(10080 b) \text {Subst}\left (\int x \text {Li}_6\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^6}-\frac {(10080 b) \text {Subst}\left (\int x \text {Li}_6\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^6}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 b x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 b x \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 b x \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}+\frac {10080 b \sqrt {x} \text {Li}_7\left (-e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 b \sqrt {x} \text {Li}_7\left (e^{c+d \sqrt {x}}\right )}{d^7}-\frac {(10080 b) \text {Subst}\left (\int \text {Li}_7\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^7}+\frac {(10080 b) \text {Subst}\left (\int \text {Li}_7\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^7}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 b x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 b x \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 b x \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}+\frac {10080 b \sqrt {x} \text {Li}_7\left (-e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 b \sqrt {x} \text {Li}_7\left (e^{c+d \sqrt {x}}\right )}{d^7}-\frac {(10080 b) \text {Subst}\left (\int \frac {\text {Li}_7(-x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^8}+\frac {(10080 b) \text {Subst}\left (\int \frac {\text {Li}_7(x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^8}\\ &=\frac {a x^4}{4}-\frac {4 b x^{7/2} \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {14 b x^3 \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {14 b x^3 \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {84 b x^{5/2} \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {84 b x^{5/2} \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {420 b x^2 \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {420 b x^2 \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {1680 b x^{3/2} \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {1680 b x^{3/2} \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {5040 b x \text {Li}_6\left (-e^{c+d \sqrt {x}}\right )}{d^6}+\frac {5040 b x \text {Li}_6\left (e^{c+d \sqrt {x}}\right )}{d^6}+\frac {10080 b \sqrt {x} \text {Li}_7\left (-e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 b \sqrt {x} \text {Li}_7\left (e^{c+d \sqrt {x}}\right )}{d^7}-\frac {10080 b \text {Li}_8\left (-e^{c+d \sqrt {x}}\right )}{d^8}+\frac {10080 b \text {Li}_8\left (e^{c+d \sqrt {x}}\right )}{d^8}\\ \end {align*}

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Mathematica [A]
time = 1.97, size = 365, normalized size = 1.03 \begin {gather*} \frac {a x^4}{4}+\frac {2 b \left (d^7 x^{7/2} \log \left (1-e^{c+d \sqrt {x}}\right )-d^7 x^{7/2} \log \left (1+e^{c+d \sqrt {x}}\right )-7 d^6 x^3 \text {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )+7 d^6 x^3 \text {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )+42 d^5 x^{5/2} \text {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )-42 d^5 x^{5/2} \text {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )-210 d^4 x^2 \text {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )+210 d^4 x^2 \text {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )+840 d^3 x^{3/2} \text {PolyLog}\left (5,-e^{c+d \sqrt {x}}\right )-840 d^3 x^{3/2} \text {PolyLog}\left (5,e^{c+d \sqrt {x}}\right )-2520 d^2 x \text {PolyLog}\left (6,-e^{c+d \sqrt {x}}\right )+2520 d^2 x \text {PolyLog}\left (6,e^{c+d \sqrt {x}}\right )+5040 d \sqrt {x} \text {PolyLog}\left (7,-e^{c+d \sqrt {x}}\right )-5040 d \sqrt {x} \text {PolyLog}\left (7,e^{c+d \sqrt {x}}\right )-5040 \text {PolyLog}\left (8,-e^{c+d \sqrt {x}}\right )+5040 \text {PolyLog}\left (8,e^{c+d \sqrt {x}}\right )\right )}{d^8} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*Csch[c + d*Sqrt[x]]),x]

[Out]

(a*x^4)/4 + (2*b*(d^7*x^(7/2)*Log[1 - E^(c + d*Sqrt[x])] - d^7*x^(7/2)*Log[1 + E^(c + d*Sqrt[x])] - 7*d^6*x^3*
PolyLog[2, -E^(c + d*Sqrt[x])] + 7*d^6*x^3*PolyLog[2, E^(c + d*Sqrt[x])] + 42*d^5*x^(5/2)*PolyLog[3, -E^(c + d
*Sqrt[x])] - 42*d^5*x^(5/2)*PolyLog[3, E^(c + d*Sqrt[x])] - 210*d^4*x^2*PolyLog[4, -E^(c + d*Sqrt[x])] + 210*d
^4*x^2*PolyLog[4, E^(c + d*Sqrt[x])] + 840*d^3*x^(3/2)*PolyLog[5, -E^(c + d*Sqrt[x])] - 840*d^3*x^(3/2)*PolyLo
g[5, E^(c + d*Sqrt[x])] - 2520*d^2*x*PolyLog[6, -E^(c + d*Sqrt[x])] + 2520*d^2*x*PolyLog[6, E^(c + d*Sqrt[x])]
+ 5040*d*Sqrt[x]*PolyLog[7, -E^(c + d*Sqrt[x])] - 5040*d*Sqrt[x]*PolyLog[7, E^(c + d*Sqrt[x])] - 5040*PolyLog
[8, -E^(c + d*Sqrt[x])] + 5040*PolyLog[8, E^(c + d*Sqrt[x])]))/d^8

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Maple [F]
time = 2.43, size = 0, normalized size = 0.00 \[\int x^{3} \left (a +b \,\mathrm {csch}\left (c +d \sqrt {x}\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*csch(c+d*x^(1/2))),x)

[Out]

int(x^3*(a+b*csch(c+d*x^(1/2))),x)

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Maxima [A]
time = 0.46, size = 349, normalized size = 0.98 \begin {gather*} \frac {1}{4} \, a x^{4} - \frac {2 \, {\left (\log \left (e^{\left (d \sqrt {x} + c\right )} + 1\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{7} + 7 \, {\rm Li}_2\left (-e^{\left (d \sqrt {x} + c\right )}\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{6} - 42 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{5} {\rm Li}_{3}(-e^{\left (d \sqrt {x} + c\right )}) + 210 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{4} {\rm Li}_{4}(-e^{\left (d \sqrt {x} + c\right )}) - 840 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{3} {\rm Li}_{5}(-e^{\left (d \sqrt {x} + c\right )}) + 2520 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{2} {\rm Li}_{6}(-e^{\left (d \sqrt {x} + c\right )}) - 5040 \, \log \left (e^{\left (d \sqrt {x}\right )}\right ) {\rm Li}_{7}(-e^{\left (d \sqrt {x} + c\right )}) + 5040 \, {\rm Li}_{8}(-e^{\left (d \sqrt {x} + c\right )})\right )} b}{d^{8}} + \frac {2 \, {\left (\log \left (-e^{\left (d \sqrt {x} + c\right )} + 1\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{7} + 7 \, {\rm Li}_2\left (e^{\left (d \sqrt {x} + c\right )}\right ) \log \left (e^{\left (d \sqrt {x}\right )}\right )^{6} - 42 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{5} {\rm Li}_{3}(e^{\left (d \sqrt {x} + c\right )}) + 210 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{4} {\rm Li}_{4}(e^{\left (d \sqrt {x} + c\right )}) - 840 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{3} {\rm Li}_{5}(e^{\left (d \sqrt {x} + c\right )}) + 2520 \, \log \left (e^{\left (d \sqrt {x}\right )}\right )^{2} {\rm Li}_{6}(e^{\left (d \sqrt {x} + c\right )}) - 5040 \, \log \left (e^{\left (d \sqrt {x}\right )}\right ) {\rm Li}_{7}(e^{\left (d \sqrt {x} + c\right )}) + 5040 \, {\rm Li}_{8}(e^{\left (d \sqrt {x} + c\right )})\right )} b}{d^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*csch(c+d*x^(1/2))),x, algorithm="maxima")

[Out]

1/4*a*x^4 - 2*(log(e^(d*sqrt(x) + c) + 1)*log(e^(d*sqrt(x)))^7 + 7*dilog(-e^(d*sqrt(x) + c))*log(e^(d*sqrt(x))
)^6 - 42*log(e^(d*sqrt(x)))^5*polylog(3, -e^(d*sqrt(x) + c)) + 210*log(e^(d*sqrt(x)))^4*polylog(4, -e^(d*sqrt(
x) + c)) - 840*log(e^(d*sqrt(x)))^3*polylog(5, -e^(d*sqrt(x) + c)) + 2520*log(e^(d*sqrt(x)))^2*polylog(6, -e^(
d*sqrt(x) + c)) - 5040*log(e^(d*sqrt(x)))*polylog(7, -e^(d*sqrt(x) + c)) + 5040*polylog(8, -e^(d*sqrt(x) + c))
)*b/d^8 + 2*(log(-e^(d*sqrt(x) + c) + 1)*log(e^(d*sqrt(x)))^7 + 7*dilog(e^(d*sqrt(x) + c))*log(e^(d*sqrt(x)))^
6 - 42*log(e^(d*sqrt(x)))^5*polylog(3, e^(d*sqrt(x) + c)) + 210*log(e^(d*sqrt(x)))^4*polylog(4, e^(d*sqrt(x) +
c)) - 840*log(e^(d*sqrt(x)))^3*polylog(5, e^(d*sqrt(x) + c)) + 2520*log(e^(d*sqrt(x)))^2*polylog(6, e^(d*sqrt
(x) + c)) - 5040*log(e^(d*sqrt(x)))*polylog(7, e^(d*sqrt(x) + c)) + 5040*polylog(8, e^(d*sqrt(x) + c)))*b/d^8

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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^3*(a+b*csch(c+d*x^(1/2))),x, algorithm="fricas")

[Out]

integral(b*x^3*csch(d*sqrt(x) + c) + a*x^3, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \left (a + b \operatorname {csch}{\left (c + d \sqrt {x} \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*csch(c+d*x**(1/2))),x)

[Out]

Integral(x**3*(a + b*csch(c + d*sqrt(x))), x)

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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^3*(a+b*csch(c+d*x^(1/2))),x, algorithm="giac")

[Out]

integrate((b*csch(d*sqrt(x) + c) + a)*x^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^3\,\left (a+\frac {b}{\mathrm {sinh}\left (c+d\,\sqrt {x}\right )}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b/sinh(c + d*x^(1/2))),x)

[Out]

int(x^3*(a + b/sinh(c + d*x^(1/2))), x)

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