3.1.56 \(\int x^{3/2} (a+b \text {csch}(c+d \sqrt {x}))^2 \, dx\) [56]

Optimal. Leaf size=363 \[ -\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^2 \coth \left (c+d \sqrt {x}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 a b x^{3/2} \text {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 a b x^{3/2} \text {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 b^2 x \text {PolyLog}\left (2,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 a b x \text {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 b^2 \sqrt {x} \text {PolyLog}\left (3,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 a b \sqrt {x} \text {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 a b \sqrt {x} \text {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )}{d^4}+\frac {6 b^2 \text {PolyLog}\left (4,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {96 a b \text {PolyLog}\left (5,-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {96 a b \text {PolyLog}\left (5,e^{c+d \sqrt {x}}\right )}{d^5} \]

[Out]

-2*b^2*x^2/d+2/5*a^2*x^(5/2)-8*a*b*x^2*arctanh(exp(c+d*x^(1/2)))/d-2*b^2*x^2*coth(c+d*x^(1/2))/d+8*b^2*x^(3/2)
*ln(1-exp(2*c+2*d*x^(1/2)))/d^2-16*a*b*x^(3/2)*polylog(2,-exp(c+d*x^(1/2)))/d^2+16*a*b*x^(3/2)*polylog(2,exp(c
+d*x^(1/2)))/d^2+12*b^2*x*polylog(2,exp(2*c+2*d*x^(1/2)))/d^3+48*a*b*x*polylog(3,-exp(c+d*x^(1/2)))/d^3-48*a*b
*x*polylog(3,exp(c+d*x^(1/2)))/d^3+6*b^2*polylog(4,exp(2*c+2*d*x^(1/2)))/d^5+96*a*b*polylog(5,-exp(c+d*x^(1/2)
))/d^5-96*a*b*polylog(5,exp(c+d*x^(1/2)))/d^5-12*b^2*polylog(3,exp(2*c+2*d*x^(1/2)))*x^(1/2)/d^4-96*a*b*polylo
g(4,-exp(c+d*x^(1/2)))*x^(1/2)/d^4+96*a*b*polylog(4,exp(c+d*x^(1/2)))*x^(1/2)/d^4

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Rubi [A]
time = 0.38, antiderivative size = 363, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 10, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {5545, 4275, 4267, 2611, 6744, 2320, 6724, 4269, 3797, 2221} \begin {gather*} \frac {2}{5} a^2 x^{5/2}+\frac {96 a b \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {96 a b \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}-\frac {96 a b \sqrt {x} \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 a b \sqrt {x} \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {48 a b x \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 a b x \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {16 a b x^{3/2} \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 a b x^{3/2} \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}+\frac {6 b^2 \text {Li}_4\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {12 b^2 \sqrt {x} \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {12 b^2 x \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {8 b^2 x^{3/2} \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {2 b^2 x^2 \coth \left (c+d \sqrt {x}\right )}{d}-\frac {2 b^2 x^2}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(a + b*Csch[c + d*Sqrt[x]])^2,x]

[Out]

(-2*b^2*x^2)/d + (2*a^2*x^(5/2))/5 - (8*a*b*x^2*ArcTanh[E^(c + d*Sqrt[x])])/d - (2*b^2*x^2*Coth[c + d*Sqrt[x]]
)/d + (8*b^2*x^(3/2)*Log[1 - E^(2*(c + d*Sqrt[x]))])/d^2 - (16*a*b*x^(3/2)*PolyLog[2, -E^(c + d*Sqrt[x])])/d^2
+ (16*a*b*x^(3/2)*PolyLog[2, E^(c + d*Sqrt[x])])/d^2 + (12*b^2*x*PolyLog[2, E^(2*(c + d*Sqrt[x]))])/d^3 + (48
*a*b*x*PolyLog[3, -E^(c + d*Sqrt[x])])/d^3 - (48*a*b*x*PolyLog[3, E^(c + d*Sqrt[x])])/d^3 - (12*b^2*Sqrt[x]*Po
lyLog[3, E^(2*(c + d*Sqrt[x]))])/d^4 - (96*a*b*Sqrt[x]*PolyLog[4, -E^(c + d*Sqrt[x])])/d^4 + (96*a*b*Sqrt[x]*P
olyLog[4, E^(c + d*Sqrt[x])])/d^4 + (6*b^2*PolyLog[4, E^(2*(c + d*Sqrt[x]))])/d^5 + (96*a*b*PolyLog[5, -E^(c +
d*Sqrt[x])])/d^5 - (96*a*b*PolyLog[5, E^(c + d*Sqrt[x])])/d^5

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^{3/2} \left (a+b \text {csch}\left (c+d \sqrt {x}\right )\right )^2 \, dx &=2 \text {Subst}\left (\int x^4 (a+b \text {csch}(c+d x))^2 \, dx,x,\sqrt {x}\right )\\ &=2 \text {Subst}\left (\int \left (a^2 x^4+2 a b x^4 \text {csch}(c+d x)+b^2 x^4 \text {csch}^2(c+d x)\right ) \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{5} a^2 x^{5/2}+(4 a b) \text {Subst}\left (\int x^4 \text {csch}(c+d x) \, dx,x,\sqrt {x}\right )+\left (2 b^2\right ) \text {Subst}\left (\int x^4 \text {csch}^2(c+d x) \, dx,x,\sqrt {x}\right )\\ &=\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^2 \coth \left (c+d \sqrt {x}\right )}{d}-\frac {(16 a b) \text {Subst}\left (\int x^3 \log \left (1-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {(16 a b) \text {Subst}\left (\int x^3 \log \left (1+e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d}+\frac {\left (8 b^2\right ) \text {Subst}\left (\int x^3 \coth (c+d x) \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^2 \coth \left (c+d \sqrt {x}\right )}{d}-\frac {16 a b x^{3/2} \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 a b x^{3/2} \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {(48 a b) \text {Subst}\left (\int x^2 \text {Li}_2\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {(48 a b) \text {Subst}\left (\int x^2 \text {Li}_2\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^2}-\frac {\left (16 b^2\right ) \text {Subst}\left (\int \frac {e^{2 (c+d x)} x^3}{1-e^{2 (c+d x)}} \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^2 \coth \left (c+d \sqrt {x}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 a b x^{3/2} \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 a b x^{3/2} \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {48 a b x \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 a b x \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {(96 a b) \text {Subst}\left (\int x \text {Li}_3\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}+\frac {(96 a b) \text {Subst}\left (\int x \text {Li}_3\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^3}-\frac {\left (24 b^2\right ) \text {Subst}\left (\int x^2 \log \left (1-e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^2}\\ &=-\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^2 \coth \left (c+d \sqrt {x}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 a b x^{3/2} \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 a b x^{3/2} \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 b^2 x \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 a b x \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {96 a b \sqrt {x} \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 a b \sqrt {x} \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {(96 a b) \text {Subst}\left (\int \text {Li}_4\left (-e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {(96 a b) \text {Subst}\left (\int \text {Li}_4\left (e^{c+d x}\right ) \, dx,x,\sqrt {x}\right )}{d^4}-\frac {\left (24 b^2\right ) \text {Subst}\left (\int x \text {Li}_2\left (e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^3}\\ &=-\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^2 \coth \left (c+d \sqrt {x}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 a b x^{3/2} \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 a b x^{3/2} \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 b^2 x \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 a b x \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 b^2 \sqrt {x} \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 a b \sqrt {x} \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 a b \sqrt {x} \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {(96 a b) \text {Subst}\left (\int \frac {\text {Li}_4(-x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^5}-\frac {(96 a b) \text {Subst}\left (\int \frac {\text {Li}_4(x)}{x} \, dx,x,e^{c+d \sqrt {x}}\right )}{d^5}+\frac {\left (12 b^2\right ) \text {Subst}\left (\int \text {Li}_3\left (e^{2 (c+d x)}\right ) \, dx,x,\sqrt {x}\right )}{d^4}\\ &=-\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^2 \coth \left (c+d \sqrt {x}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 a b x^{3/2} \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 a b x^{3/2} \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 b^2 x \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 a b x \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 b^2 \sqrt {x} \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 a b \sqrt {x} \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 a b \sqrt {x} \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 a b \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {96 a b \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}+\frac {\left (6 b^2\right ) \text {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}\\ &=-\frac {2 b^2 x^2}{d}+\frac {2}{5} a^2 x^{5/2}-\frac {8 a b x^2 \tanh ^{-1}\left (e^{c+d \sqrt {x}}\right )}{d}-\frac {2 b^2 x^2 \coth \left (c+d \sqrt {x}\right )}{d}+\frac {8 b^2 x^{3/2} \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {16 a b x^{3/2} \text {Li}_2\left (-e^{c+d \sqrt {x}}\right )}{d^2}+\frac {16 a b x^{3/2} \text {Li}_2\left (e^{c+d \sqrt {x}}\right )}{d^2}+\frac {12 b^2 x \text {Li}_2\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {48 a b x \text {Li}_3\left (-e^{c+d \sqrt {x}}\right )}{d^3}-\frac {48 a b x \text {Li}_3\left (e^{c+d \sqrt {x}}\right )}{d^3}-\frac {12 b^2 \sqrt {x} \text {Li}_3\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {96 a b \sqrt {x} \text {Li}_4\left (-e^{c+d \sqrt {x}}\right )}{d^4}+\frac {96 a b \sqrt {x} \text {Li}_4\left (e^{c+d \sqrt {x}}\right )}{d^4}+\frac {6 b^2 \text {Li}_4\left (e^{2 \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {96 a b \text {Li}_5\left (-e^{c+d \sqrt {x}}\right )}{d^5}-\frac {96 a b \text {Li}_5\left (e^{c+d \sqrt {x}}\right )}{d^5}\\ \end {align*}

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Mathematica [A]
time = 6.85, size = 439, normalized size = 1.21 \begin {gather*} \frac {2}{5} a^2 x^{5/2}+\frac {2 b \left (-\frac {2 b d^4 e^{2 c} x^2}{-1+e^{2 c}}+2 a d^4 x^2 \log \left (1-e^{c+d \sqrt {x}}\right )-2 a d^4 x^2 \log \left (1+e^{c+d \sqrt {x}}\right )+4 b d^3 x^{3/2} \log \left (1-e^{2 \left (c+d \sqrt {x}\right )}\right )-8 a d^3 x^{3/2} \text {PolyLog}\left (2,-e^{c+d \sqrt {x}}\right )+8 a d^3 x^{3/2} \text {PolyLog}\left (2,e^{c+d \sqrt {x}}\right )+6 b d^2 x \text {PolyLog}\left (2,e^{2 \left (c+d \sqrt {x}\right )}\right )+24 a d^2 x \text {PolyLog}\left (3,-e^{c+d \sqrt {x}}\right )-24 a d^2 x \text {PolyLog}\left (3,e^{c+d \sqrt {x}}\right )-6 b d \sqrt {x} \text {PolyLog}\left (3,e^{2 \left (c+d \sqrt {x}\right )}\right )-48 a d \sqrt {x} \text {PolyLog}\left (4,-e^{c+d \sqrt {x}}\right )+48 a d \sqrt {x} \text {PolyLog}\left (4,e^{c+d \sqrt {x}}\right )+3 b \text {PolyLog}\left (4,e^{2 \left (c+d \sqrt {x}\right )}\right )+48 a \text {PolyLog}\left (5,-e^{c+d \sqrt {x}}\right )-48 a \text {PolyLog}\left (5,e^{c+d \sqrt {x}}\right )\right )}{d^5}+\frac {b^2 x^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^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^(3/2)*(a + b*Csch[c + d*Sqrt[x]])^2,x]

[Out]

(2*a^2*x^(5/2))/5 + (2*b*((-2*b*d^4*E^(2*c)*x^2)/(-1 + E^(2*c)) + 2*a*d^4*x^2*Log[1 - E^(c + d*Sqrt[x])] - 2*a
*d^4*x^2*Log[1 + E^(c + d*Sqrt[x])] + 4*b*d^3*x^(3/2)*Log[1 - E^(2*(c + d*Sqrt[x]))] - 8*a*d^3*x^(3/2)*PolyLog
[2, -E^(c + d*Sqrt[x])] + 8*a*d^3*x^(3/2)*PolyLog[2, E^(c + d*Sqrt[x])] + 6*b*d^2*x*PolyLog[2, E^(2*(c + d*Sqr
t[x]))] + 24*a*d^2*x*PolyLog[3, -E^(c + d*Sqrt[x])] - 24*a*d^2*x*PolyLog[3, E^(c + d*Sqrt[x])] - 6*b*d*Sqrt[x]
*PolyLog[3, E^(2*(c + d*Sqrt[x]))] - 48*a*d*Sqrt[x]*PolyLog[4, -E^(c + d*Sqrt[x])] + 48*a*d*Sqrt[x]*PolyLog[4,
E^(c + d*Sqrt[x])] + 3*b*PolyLog[4, E^(2*(c + d*Sqrt[x]))] + 48*a*PolyLog[5, -E^(c + d*Sqrt[x])] - 48*a*PolyL
og[5, E^(c + d*Sqrt[x])]))/d^5 + (b^2*x^2*Csch[c/2]*Csch[(c + d*Sqrt[x])/2]*Sinh[(d*Sqrt[x])/2])/d - (b^2*x^2*
Sech[c/2]*Sech[(c + d*Sqrt[x])/2]*Sinh[(d*Sqrt[x])/2])/d

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{\frac {3}{2}} \left (a +b \,\mathrm {csch}\left (c +d \sqrt {x}\right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.43, size = 422, normalized size = 1.16 \begin {gather*} \frac {2}{5} \, a^{2} x^{\frac {5}{2}} - \frac {4 \, b^{2} x^{2}}{d e^{\left (2 \, d \sqrt {x} + 2 \, c\right )} - d} - \frac {4 \, {\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 )} a b}{d^{5}} + \frac {4 \, {\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 )} a b}{d^{5}} + \frac {8 \, {\left (d^{3} x^{\frac {3}{2}} \log \left (e^{\left (d \sqrt {x} + c\right )} + 1\right ) + 3 \, d^{2} x {\rm Li}_2\left (-e^{\left (d \sqrt {x} + c\right )}\right ) - 6 \, d \sqrt {x} {\rm Li}_{3}(-e^{\left (d \sqrt {x} + c\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (d \sqrt {x} + c\right )})\right )} b^{2}}{d^{5}} + \frac {8 \, {\left (d^{3} x^{\frac {3}{2}} \log \left (-e^{\left (d \sqrt {x} + c\right )} + 1\right ) + 3 \, d^{2} x {\rm Li}_2\left (e^{\left (d \sqrt {x} + c\right )}\right ) - 6 \, d \sqrt {x} {\rm Li}_{3}(e^{\left (d \sqrt {x} + c\right )}) + 6 \, {\rm Li}_{4}(e^{\left (d \sqrt {x} + c\right )})\right )} b^{2}}{d^{5}} - \frac {2 \, {\left (2 \, a b d^{5} x^{\frac {5}{2}} + 5 \, b^{2} d^{4} x^{2}\right )}}{5 \, d^{5}} + \frac {2 \, {\left (2 \, a b d^{5} x^{\frac {5}{2}} - 5 \, b^{2} d^{4} x^{2}\right )}}{5 \, d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integral(b^2*x^(3/2)*csch(d*sqrt(x) + c)^2 + 2*a*b*x^(3/2)*csch(d*sqrt(x) + c) + a^2*x^(3/2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{\frac {3}{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**(3/2)*(a+b*csch(c+d*x**(1/2)))**2,x)

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^{3/2}\,{\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^(3/2)*(a + b/sinh(c + d*x^(1/2)))^2,x)

[Out]

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

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