3.3.25 \(\int \frac {x^4 (a+b \sinh ^{-1}(c x))^2}{d+c^2 d x^2} \, dx\) [225]

Optimal. Leaf size=277 \[ -\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d}+\frac {22 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {2 i b^2 \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {2 i b^2 \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{c^5 d} \]

[Out]

-22/9*b^2*x/c^4/d+2/27*b^2*x^3/c^2/d-x*(a+b*arcsinh(c*x))^2/c^4/d+1/3*x^3*(a+b*arcsinh(c*x))^2/c^2/d+2*(a+b*ar
csinh(c*x))^2*arctan(c*x+(c^2*x^2+1)^(1/2))/c^5/d-2*I*b*(a+b*arcsinh(c*x))*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)
))/c^5/d+2*I*b*(a+b*arcsinh(c*x))*polylog(2,I*(c*x+(c^2*x^2+1)^(1/2)))/c^5/d+2*I*b^2*polylog(3,-I*(c*x+(c^2*x^
2+1)^(1/2)))/c^5/d-2*I*b^2*polylog(3,I*(c*x+(c^2*x^2+1)^(1/2)))/c^5/d+22/9*b*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1
/2)/c^5/d-2/9*b*x^2*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/c^3/d

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Rubi [A]
time = 0.41, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {5812, 5789, 4265, 2611, 2320, 6724, 5798, 8, 30} \begin {gather*} \frac {2 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^5 d}-\frac {2 i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d}+\frac {2 i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {22 b \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}+\frac {2 i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {2 i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2),x]

[Out]

(-22*b^2*x)/(9*c^4*d) + (2*b^2*x^3)/(27*c^2*d) + (22*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(9*c^5*d) - (2*
b*x^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(9*c^3*d) - (x*(a + b*ArcSinh[c*x])^2)/(c^4*d) + (x^3*(a + b*Arc
Sinh[c*x])^2)/(3*c^2*d) + (2*(a + b*ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/(c^5*d) - ((2*I)*b*(a + b*ArcSinh[
c*x])*PolyLog[2, (-I)*E^ArcSinh[c*x]])/(c^5*d) + ((2*I)*b*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^ArcSinh[c*x]])/(
c^5*d) + ((2*I)*b^2*PolyLog[3, (-I)*E^ArcSinh[c*x]])/(c^5*d) - ((2*I)*b^2*PolyLog[3, I*E^ArcSinh[c*x]])/(c^5*d
)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 4265

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

Rule 5789

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/(c*d), Subst[Int[(a +
b*x)^n*Sech[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0]

Rule 5798

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^
(p + 1)*((a + b*ArcSinh[c*x])^n/(2*e*(p + 1))), x] - Dist[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^p], Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e
, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

Rule 5812

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(e*(m + 2*p + 1))), x] + (-Dist[f^2*((m - 1)/(c^2*
(m + 2*p + 1))), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[b*f*(n/(c*(m + 2*p + 1)
))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1)
, x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && IGtQ[m, 1] && NeQ[m + 2*p + 1, 0
]

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]

Rubi steps

\begin {align*} \int \frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx &=\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}-\frac {\int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx}{c^2}-\frac {(2 b) \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{3 c d}\\ &=-\frac {2 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {\int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx}{c^4}+\frac {(4 b) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{9 c^3 d}+\frac {(2 b) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{c^3 d}+\frac {\left (2 b^2\right ) \int x^2 \, dx}{9 c^2 d}\\ &=\frac {2 b^2 x^3}{27 c^2 d}+\frac {22 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {\text {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d}-\frac {\left (4 b^2\right ) \int 1 \, dx}{9 c^4 d}-\frac {\left (2 b^2\right ) \int 1 \, dx}{c^4 d}\\ &=-\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d}+\frac {22 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {(2 i b) \text {Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d}+\frac {(2 i b) \text {Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d}\\ &=-\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d}+\frac {22 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {\left (2 i b^2\right ) \text {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d}-\frac {\left (2 i b^2\right ) \text {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d}\\ &=-\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d}+\frac {22 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {\left (2 i b^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {\left (2 i b^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c^5 d}\\ &=-\frac {22 b^2 x}{9 c^4 d}+\frac {2 b^2 x^3}{27 c^2 d}+\frac {22 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^5 d}-\frac {2 b x^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{9 c^3 d}-\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c^2 d}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}+\frac {2 i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}-\frac {2 i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{c^5 d}\\ \end {align*}

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Mathematica [A]
time = 0.85, size = 365, normalized size = 1.32 \begin {gather*} \frac {-3 a^2 c x+a^2 c^3 x^3+3 a^2 \text {ArcTan}(c x)-\frac {2}{3} a b \left (-11 \sqrt {1+c^2 x^2}+c^2 x^2 \sqrt {1+c^2 x^2}+9 c x \sinh ^{-1}(c x)-3 c^3 x^3 \sinh ^{-1}(c x)-9 i \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )+9 i \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+9 i \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )-9 i \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )\right )+3 b^2 \left (\frac {5}{2} \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)-\frac {5}{4} c x \left (2+\sinh ^{-1}(c x)^2\right )-\frac {1}{18} \sinh ^{-1}(c x) \cosh \left (3 \sinh ^{-1}(c x)\right )+i \left (-\sinh ^{-1}(c x)^2 \left (\log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-\log \left (1+i e^{-\sinh ^{-1}(c x)}\right )\right )-2 \sinh ^{-1}(c x) \left (\text {PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )-\text {PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )\right )-2 \text {PolyLog}\left (3,-i e^{-\sinh ^{-1}(c x)}\right )+2 \text {PolyLog}\left (3,i e^{-\sinh ^{-1}(c x)}\right )\right )+\frac {1}{108} \left (2+9 \sinh ^{-1}(c x)^2\right ) \sinh \left (3 \sinh ^{-1}(c x)\right )\right )}{3 c^5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2),x]

[Out]

(-3*a^2*c*x + a^2*c^3*x^3 + 3*a^2*ArcTan[c*x] - (2*a*b*(-11*Sqrt[1 + c^2*x^2] + c^2*x^2*Sqrt[1 + c^2*x^2] + 9*
c*x*ArcSinh[c*x] - 3*c^3*x^3*ArcSinh[c*x] - (9*I)*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] + (9*I)*ArcSinh[c*x]*
Log[1 + I*E^ArcSinh[c*x]] + (9*I)*PolyLog[2, (-I)*E^ArcSinh[c*x]] - (9*I)*PolyLog[2, I*E^ArcSinh[c*x]]))/3 + 3
*b^2*((5*Sqrt[1 + c^2*x^2]*ArcSinh[c*x])/2 - (5*c*x*(2 + ArcSinh[c*x]^2))/4 - (ArcSinh[c*x]*Cosh[3*ArcSinh[c*x
]])/18 + I*(-(ArcSinh[c*x]^2*(Log[1 - I/E^ArcSinh[c*x]] - Log[1 + I/E^ArcSinh[c*x]])) - 2*ArcSinh[c*x]*(PolyLo
g[2, (-I)/E^ArcSinh[c*x]] - PolyLog[2, I/E^ArcSinh[c*x]]) - 2*PolyLog[3, (-I)/E^ArcSinh[c*x]] + 2*PolyLog[3, I
/E^ArcSinh[c*x]]) + ((2 + 9*ArcSinh[c*x]^2)*Sinh[3*ArcSinh[c*x]])/108))/(3*c^5*d)

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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {x^{4} \left (a +b \arcsinh \left (c x \right )\right )^{2}}{c^{2} d \,x^{2}+d}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d),x)

[Out]

int(x^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d),x, algorithm="maxima")

[Out]

1/3*a^2*((c^2*x^3 - 3*x)/(c^4*d) + 3*arctan(c*x)/(c^5*d)) + integrate(b^2*x^4*log(c*x + sqrt(c^2*x^2 + 1))^2/(
c^2*d*x^2 + d) + 2*a*b*x^4*log(c*x + sqrt(c^2*x^2 + 1))/(c^2*d*x^2 + d), x)

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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^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d),x, algorithm="fricas")

[Out]

integral((b^2*x^4*arcsinh(c*x)^2 + 2*a*b*x^4*arcsinh(c*x) + a^2*x^4)/(c^2*d*x^2 + d), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{2} x^{4}}{c^{2} x^{2} + 1}\, dx + \int \frac {b^{2} x^{4} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx + \int \frac {2 a b x^{4} \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*asinh(c*x))**2/(c**2*d*x**2+d),x)

[Out]

(Integral(a**2*x**4/(c**2*x**2 + 1), x) + Integral(b**2*x**4*asinh(c*x)**2/(c**2*x**2 + 1), x) + Integral(2*a*
b*x**4*asinh(c*x)/(c**2*x**2 + 1), x))/d

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{d\,c^2\,x^2+d} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*asinh(c*x))^2)/(d + c^2*d*x^2),x)

[Out]

int((x^4*(a + b*asinh(c*x))^2)/(d + c^2*d*x^2), x)

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