3.226 \(\int \frac {x^3 (a+b \sinh ^{-1}(c x))^2}{d+c^2 d x^2} \, dx\)

Optimal. Leaf size=199 \[ -\frac {b \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^4 d}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^4 d}-\frac {\log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}-\frac {b x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac {b^2 \text {Li}_3\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d}+\frac {b^2 x^2}{4 c^2 d} \]

[Out]

1/4*b^2*x^2/c^2/d+1/4*(a+b*arcsinh(c*x))^2/c^4/d+1/2*x^2*(a+b*arcsinh(c*x))^2/c^2/d+1/3*(a+b*arcsinh(c*x))^3/b
/c^4/d-(a+b*arcsinh(c*x))^2*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)/c^4/d-b*(a+b*arcsinh(c*x))*polylog(2,-(c*x+(c^2*x^
2+1)^(1/2))^2)/c^4/d+1/2*b^2*polylog(3,-(c*x+(c^2*x^2+1)^(1/2))^2)/c^4/d-1/2*b*x*(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 = 199, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {5767, 5714, 3718, 2190, 2531, 2282, 6589, 5758, 5675, 30} \[ -\frac {b \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^4 d}+\frac {b^2 \text {PolyLog}\left (3,-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d}+\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}-\frac {b x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^4 d}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^4 d}-\frac {\log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^4 d}+\frac {b^2 x^2}{4 c^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*x^2)/(4*c^2*d) - (b*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(2*c^3*d) + (a + b*ArcSinh[c*x])^2/(4*c^4*d
) + (x^2*(a + b*ArcSinh[c*x])^2)/(2*c^2*d) + (a + b*ArcSinh[c*x])^3/(3*b*c^4*d) - ((a + b*ArcSinh[c*x])^2*Log[
1 + E^(2*ArcSinh[c*x])])/(c^4*d) - (b*(a + b*ArcSinh[c*x])*PolyLog[2, -E^(2*ArcSinh[c*x])])/(c^4*d) + (b^2*Pol
yLog[3, -E^(2*ArcSinh[c*x])])/(2*c^4*d)

Rule 30

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

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2282

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 2531

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 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5675

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

Rule 5714

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

Rule 5758

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

Rule 5767

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*d^IntPart[p]*(d
+ e*x^2)^FracPart[p])/(c*(m + 2*p + 1)*(1 + c^2*x^2)^FracPart[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] && GtQ[m
, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[m]

Rule 6589

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^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx &=\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}-\frac {\int \frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx}{c^2}-\frac {b \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{c d}\\ &=-\frac {b x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}-\frac {\operatorname {Subst}\left (\int (a+b x)^2 \tanh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d}+\frac {b \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{2 c^3 d}+\frac {b^2 \int x \, dx}{2 c^2 d}\\ &=\frac {b^2 x^2}{4 c^2 d}-\frac {b x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^4 d}+\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac {2 \operatorname {Subst}\left (\int \frac {e^{2 x} (a+b x)^2}{1+e^{2 x}} \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d}\\ &=\frac {b^2 x^2}{4 c^2 d}-\frac {b x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^4 d}+\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}+\frac {(2 b) \operatorname {Subst}\left (\int (a+b x) \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d}\\ &=\frac {b^2 x^2}{4 c^2 d}-\frac {b x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^4 d}+\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}-\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}+\frac {b^2 \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^4 d}\\ &=\frac {b^2 x^2}{4 c^2 d}-\frac {b x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^4 d}+\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}-\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d}\\ &=\frac {b^2 x^2}{4 c^2 d}-\frac {b x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^4 d}+\frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c^4 d}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}-\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{c^4 d}+\frac {b^2 \text {Li}_3\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 c^4 d}\\ \end {align*}

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Mathematica [C]  time = 0.46, size = 279, normalized size = 1.40 \[ \frac {12 a^2 c^2 x^2-12 a^2 \log \left (c^2 x^2+1\right )-12 a b c x \sqrt {c^2 x^2+1}+24 a b c^2 x^2 \sinh ^{-1}(c x)-48 a b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )-48 a b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )+24 a b \sinh ^{-1}(c x)^2+12 a b \sinh ^{-1}(c x)-48 a b \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-48 a b \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+24 b^2 \sinh ^{-1}(c x) \text {Li}_2\left (-e^{-2 \sinh ^{-1}(c x)}\right )+12 b^2 \text {Li}_3\left (-e^{-2 \sinh ^{-1}(c x)}\right )-8 b^2 \sinh ^{-1}(c x)^3-6 b^2 \sinh \left (2 \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)-24 b^2 \sinh ^{-1}(c x)^2 \log \left (e^{-2 \sinh ^{-1}(c x)}+1\right )+6 b^2 \sinh ^{-1}(c x)^2 \cosh \left (2 \sinh ^{-1}(c x)\right )+3 b^2 \cosh \left (2 \sinh ^{-1}(c x)\right )}{24 c^4 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^3*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2),x]

[Out]

(12*a^2*c^2*x^2 - 12*a*b*c*x*Sqrt[1 + c^2*x^2] + 12*a*b*ArcSinh[c*x] + 24*a*b*c^2*x^2*ArcSinh[c*x] + 24*a*b*Ar
cSinh[c*x]^2 - 8*b^2*ArcSinh[c*x]^3 + 3*b^2*Cosh[2*ArcSinh[c*x]] + 6*b^2*ArcSinh[c*x]^2*Cosh[2*ArcSinh[c*x]] -
24*b^2*ArcSinh[c*x]^2*Log[1 + E^(-2*ArcSinh[c*x])] - 48*a*b*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] - 48*a*b*A
rcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] - 12*a^2*Log[1 + c^2*x^2] + 24*b^2*ArcSinh[c*x]*PolyLog[2, -E^(-2*ArcSin
h[c*x])] - 48*a*b*PolyLog[2, (-I)*E^ArcSinh[c*x]] - 48*a*b*PolyLog[2, I*E^ArcSinh[c*x]] + 12*b^2*PolyLog[3, -E
^(-2*ArcSinh[c*x])] - 6*b^2*ArcSinh[c*x]*Sinh[2*ArcSinh[c*x]])/(24*c^4*d)

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{3} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b x^{3} \operatorname {arsinh}\left (c x\right ) + a^{2} x^{3}}{c^{2} d x^{2} + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.20, size = 380, normalized size = 1.91 \[ \frac {a^{2} x^{2}}{2 c^{2} d}-\frac {a^{2} \ln \left (c^{2} x^{2}+1\right )}{2 c^{4} d}+\frac {b^{2} \arcsinh \left (c x \right )^{3}}{3 c^{4} d}+\frac {b^{2} \arcsinh \left (c x \right )^{2} x^{2}}{2 c^{2} d}-\frac {b^{2} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x}{2 c^{3} d}+\frac {b^{2} \arcsinh \left (c x \right )^{2}}{4 c^{4} d}+\frac {b^{2} x^{2}}{4 c^{2} d}+\frac {b^{2}}{8 c^{4} d}-\frac {b^{2} \arcsinh \left (c x \right )^{2} \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{c^{4} d}-\frac {b^{2} \arcsinh \left (c x \right ) \polylog \left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{c^{4} d}+\frac {b^{2} \polylog \left (3, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 c^{4} d}+\frac {a b \arcsinh \left (c x \right )^{2}}{c^{4} d}+\frac {a b \arcsinh \left (c x \right ) x^{2}}{c^{2} d}-\frac {a b x \sqrt {c^{2} x^{2}+1}}{2 c^{3} d}+\frac {a b \arcsinh \left (c x \right )}{2 c^{4} d}-\frac {2 a b \arcsinh \left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{c^{4} d}-\frac {a b \polylog \left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{c^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/c^2*a^2/d*x^2-1/2/c^4*a^2/d*ln(c^2*x^2+1)+1/3/c^4*b^2/d*arcsinh(c*x)^3+1/2/c^2*b^2/d*arcsinh(c*x)^2*x^2-1/
2/c^3*b^2/d*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x+1/4/c^4*b^2/d*arcsinh(c*x)^2+1/4*b^2*x^2/c^2/d+1/8/c^4*b^2/d-1/c^
4*b^2/d*arcsinh(c*x)^2*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)-1/c^4*b^2/d*arcsinh(c*x)*polylog(2,-(c*x+(c^2*x^2+1)^(1
/2))^2)+1/2*b^2*polylog(3,-(c*x+(c^2*x^2+1)^(1/2))^2)/c^4/d+1/c^4*a*b/d*arcsinh(c*x)^2+1/c^2*a*b/d*arcsinh(c*x
)*x^2-1/2/c^3*a*b/d*x*(c^2*x^2+1)^(1/2)+1/2/c^4*a*b/d*arcsinh(c*x)-2/c^4*a*b/d*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2
+1)^(1/2))^2)-1/c^4*a*b/d*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a^{2} {\left (\frac {x^{2}}{c^{2} d} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4} d}\right )} + \frac {{\left (b^{2} c^{2} x^{2} - b^{2} \log \left (c^{2} x^{2} + 1\right )\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{2 \, c^{4} d} + \int -\frac {{\left (b^{2} c^{2} x^{2} - {\left (2 \, a b c^{4} - b^{2} c^{4}\right )} x^{4} - {\left (b^{2} c^{2} x^{2} + b^{2}\right )} \log \left (c^{2} x^{2} + 1\right ) - {\left (b^{2} c x \log \left (c^{2} x^{2} + 1\right ) + {\left (2 \, a b c^{3} - b^{2} c^{3}\right )} x^{3}\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{6} d x^{3} + c^{4} d x + {\left (c^{5} d x^{2} + c^{3} d\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{d\,c^2\,x^2+d} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{2} x^{3}}{c^{2} x^{2} + 1}\, dx + \int \frac {b^{2} x^{3} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx + \int \frac {2 a b x^{3} \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2*x**3/(c**2*x**2 + 1), x) + Integral(b**2*x**3*asinh(c*x)**2/(c**2*x**2 + 1), x) + Integral(2*a*
b*x**3*asinh(c*x)/(c**2*x**2 + 1), x))/d

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