3.3.27 \(\int \frac {x^2 (a+b \sinh ^{-1}(c x))^2}{d+c^2 d x^2} \, dx\) [227]

Optimal. Leaf size=198 \[ \frac {2 b^2 x}{c^2 d}-\frac {2 b \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}-\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}-\frac {2 i b^2 \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac {2 i b^2 \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{c^3 d} \]

[Out]

2*b^2*x/c^2/d+x*(a+b*arcsinh(c*x))^2/c^2/d-2*(a+b*arcsinh(c*x))^2*arctan(c*x+(c^2*x^2+1)^(1/2))/c^3/d+2*I*b*(a
+b*arcsinh(c*x))*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)))/c^3/d-2*I*b*(a+b*arcsinh(c*x))*polylog(2,I*(c*x+(c^2*x^
2+1)^(1/2)))/c^3/d-2*I*b^2*polylog(3,-I*(c*x+(c^2*x^2+1)^(1/2)))/c^3/d+2*I*b^2*polylog(3,I*(c*x+(c^2*x^2+1)^(1
/2)))/c^3/d-2*b*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/c^3/d

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Rubi [A]
time = 0.20, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5812, 5789, 4265, 2611, 2320, 6724, 5798, 8} \begin {gather*} -\frac {2 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^3 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^3 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^3 d}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d}-\frac {2 b \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d}-\frac {2 i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac {2 i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{c^3 d}+\frac {2 b^2 x}{c^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*x)/(c^2*d) - (2*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(c^3*d) + (x*(a + b*ArcSinh[c*x])^2)/(c^2*d)
- (2*(a + b*ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/(c^3*d) + ((2*I)*b*(a + b*ArcSinh[c*x])*PolyLog[2, (-I)*E^
ArcSinh[c*x]])/(c^3*d) - ((2*I)*b*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^ArcSinh[c*x]])/(c^3*d) - ((2*I)*b^2*Poly
Log[3, (-I)*E^ArcSinh[c*x]])/(c^3*d) + ((2*I)*b^2*PolyLog[3, I*E^ArcSinh[c*x]])/(c^3*d)

Rule 8

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

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

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Mathematica [A]
time = 0.40, size = 293, normalized size = 1.48 \begin {gather*} \frac {a^2 x}{c^2 d}-\frac {a^2 \text {ArcTan}(c x)}{c^3 d}+\frac {2 a b \left (-\sqrt {1+c^2 x^2}+c x \sinh ^{-1}(c x)+i \sinh ^{-1}(c x) \left (\log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-\log \left (1+i e^{-\sinh ^{-1}(c x)}\right )\right )+i \left (\text {PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )-\text {PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )\right )\right )}{c^3 d}+\frac {b^2 \left (-2 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)+c x \left (2+\sinh ^{-1}(c x)^2\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 \left (\text {PolyLog}\left (3,-i e^{-\sinh ^{-1}(c x)}\right )-\text {PolyLog}\left (3,i e^{-\sinh ^{-1}(c x)}\right )\right )\right )\right )}{c^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*x)/(c^2*d) - (a^2*ArcTan[c*x])/(c^3*d) + (2*a*b*(-Sqrt[1 + c^2*x^2] + c*x*ArcSinh[c*x] + I*ArcSinh[c*x]*(
Log[1 - I/E^ArcSinh[c*x]] - Log[1 + I/E^ArcSinh[c*x]]) + I*(PolyLog[2, (-I)/E^ArcSinh[c*x]] - PolyLog[2, I/E^A
rcSinh[c*x]])))/(c^3*d) + (b^2*(-2*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + c*x*(2 + ArcSinh[c*x]^2) - I*(-(ArcSinh[c*
x]^2*(Log[1 - I/E^ArcSinh[c*x]] - Log[1 + I/E^ArcSinh[c*x]])) - 2*ArcSinh[c*x]*(PolyLog[2, (-I)/E^ArcSinh[c*x]
] - PolyLog[2, I/E^ArcSinh[c*x]]) - 2*(PolyLog[3, (-I)/E^ArcSinh[c*x]] - PolyLog[3, I/E^ArcSinh[c*x]]))))/(c^3
*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

a^2*(x/(c^2*d) - arctan(c*x)/(c^3*d)) + integrate(b^2*x^2*log(c*x + sqrt(c^2*x^2 + 1))^2/(c^2*d*x^2 + d) + 2*a
*b*x^2*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^2*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d),x, algorithm="fricas")

[Out]

integral((b^2*x^2*arcsinh(c*x)^2 + 2*a*b*x^2*arcsinh(c*x) + a^2*x^2)/(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^{2}}{c^{2} x^{2} + 1}\, dx + \int \frac {b^{2} x^{2} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx + \int \frac {2 a b x^{2} \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**2*(a+b*asinh(c*x))**2/(c**2*d*x**2+d),x)

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,{\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^2*(a + b*asinh(c*x))^2)/(d + c^2*d*x^2),x)

[Out]

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

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