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

Optimal. Leaf size=210 \[ \frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt {c^2 x^2+1}}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (c^2 x^2+1\right )}-\frac {i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c d^2}+\frac {i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c d^2}+\frac {\tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c d^2}+\frac {i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{c d^2}-\frac {i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{c d^2}-\frac {b^2 \tan ^{-1}(c x)}{c d^2} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {5690, 5693, 4180, 2531, 2282, 6589, 5717, 203} \[ -\frac {i b \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c d^2}+\frac {i b \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c d^2}+\frac {i b^2 \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{c d^2}-\frac {i b^2 \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{c d^2}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt {c^2 x^2+1}}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (c^2 x^2+1\right )}+\frac {\tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c d^2}-\frac {b^2 \tan ^{-1}(c x)}{c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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 4180

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 5690

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSinh[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a +
b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 + c^2*x^2)^FracPar
t[p]), Int[x*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ
[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 5693

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 5717

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[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 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 {\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^2} \, dx &=\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}-\frac {(b c) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{d^2}+\frac {\int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx}{2 d}\\ &=\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt {1+c^2 x^2}}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}-\frac {b^2 \int \frac {1}{1+c^2 x^2} \, dx}{d^2}+\frac {\operatorname {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 c d^2}\\ &=\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt {1+c^2 x^2}}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c d^2}-\frac {b^2 \tan ^{-1}(c x)}{c d^2}-\frac {(i b) \operatorname {Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c d^2}+\frac {(i b) \operatorname {Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c d^2}\\ &=\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt {1+c^2 x^2}}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c d^2}-\frac {b^2 \tan ^{-1}(c x)}{c d^2}-\frac {i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c d^2}+\frac {i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c d^2}+\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c d^2}-\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c d^2}\\ &=\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt {1+c^2 x^2}}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c d^2}-\frac {b^2 \tan ^{-1}(c x)}{c d^2}-\frac {i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c d^2}+\frac {i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c d^2}+\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c d^2}-\frac {\left (i b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{c d^2}\\ &=\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c d^2 \sqrt {1+c^2 x^2}}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{c d^2}-\frac {b^2 \tan ^{-1}(c x)}{c d^2}-\frac {i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c d^2}+\frac {i b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c d^2}+\frac {i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{c d^2}-\frac {i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{c d^2}\\ \end {align*}

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Mathematica [A]  time = 1.42, size = 403, normalized size = 1.92 \[ \frac {\frac {a^2 x}{c^2 x^2+1}+\frac {a^2 \tan ^{-1}(c x)}{c}+\frac {2 a b \left (-i \left (c^2 x^2+1\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )+i \left (c^2 x^2+1\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )+\sqrt {c^2 x^2+1}+i c^2 x^2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-i c^2 x^2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )+c x \sinh ^{-1}(c x)+i \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-i \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )\right )}{c^3 x^2+c}+\frac {2 b^2 \left (\frac {c x \sinh ^{-1}(c x)^2}{2 c^2 x^2+2}+\frac {\sinh ^{-1}(c x)}{\sqrt {c^2 x^2+1}}-\frac {1}{2} i \left (2 \sinh ^{-1}(c x) \text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )-2 \sinh ^{-1}(c x) \text {Li}_2\left (i e^{-\sinh ^{-1}(c x)}\right )+2 \text {Li}_3\left (-i e^{-\sinh ^{-1}(c x)}\right )-2 \text {Li}_3\left (i e^{-\sinh ^{-1}(c x)}\right )+\sinh ^{-1}(c x)^2 \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-\sinh ^{-1}(c x)^2 \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )-4 i \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )\right )}{c}}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a^2*x)/(1 + c^2*x^2) + (a^2*ArcTan[c*x])/c + (2*a*b*(Sqrt[1 + c^2*x^2] + c*x*ArcSinh[c*x] + I*ArcSinh[c*x]*L
og[1 - I*E^ArcSinh[c*x]] + I*c^2*x^2*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] - I*ArcSinh[c*x]*Log[1 + I*E^ArcSi
nh[c*x]] - I*c^2*x^2*ArcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] - I*(1 + c^2*x^2)*PolyLog[2, (-I)*E^ArcSinh[c*x]]
+ I*(1 + c^2*x^2)*PolyLog[2, I*E^ArcSinh[c*x]]))/(c + c^3*x^2) + (2*b^2*(ArcSinh[c*x]/Sqrt[1 + c^2*x^2] + (c*x
*ArcSinh[c*x]^2)/(2 + 2*c^2*x^2) - (I/2)*((-4*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + ArcSinh[c*x]^2*Log[1 - I/E^Arc
Sinh[c*x]] - ArcSinh[c*x]^2*Log[1 + I/E^ArcSinh[c*x]] + 2*ArcSinh[c*x]*PolyLog[2, (-I)/E^ArcSinh[c*x]] - 2*Arc
Sinh[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]])))
/c)/(2*d^2)

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}}{c^{4} d^{2} x^{4} + 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsinh \left (c x \right )\right )^{2}}{\left (c^{2} d \,x^{2}+d \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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