3.3.29 \(\int \frac {(a+b \sinh ^{-1}(c x))^2}{d+c^2 d x^2} \, dx\) [229]

Optimal. Leaf size=138 \[ \frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{c d}-\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c d}+\frac {2 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c d}+\frac {2 i b^2 \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{c d}-\frac {2 i b^2 \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{c d} \]

[Out]

2*(a+b*arcsinh(c*x))^2*arctan(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*(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

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Rubi [A]
time = 0.08, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {5789, 4265, 2611, 2320, 6724} \begin {gather*} \frac {2 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c 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 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 d}+\frac {2 i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{c d}-\frac {2 i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/(c*d) - ((2*I)*b*(a + b*ArcSinh[c*x])*PolyLog[2, (-I)*E^ArcS
inh[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)

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

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Mathematica [A]
time = 0.17, size = 274, normalized size = 1.99 \begin {gather*} -\frac {c \left (a^2 \sqrt {-c^2} \text {ArcTan}(c x)-2 a b c \sinh ^{-1}(c x) \log \left (1+\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )-b^2 c \sinh ^{-1}(c x)^2 \log \left (1+\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )+2 a b c \sinh ^{-1}(c x) \log \left (1+\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+b^2 c \sinh ^{-1}(c x)^2 \log \left (1+\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+2 b c \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )-2 b c \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )-2 b^2 c \text {PolyLog}\left (3,\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )+2 b^2 c \text {PolyLog}\left (3,\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )\right )}{\left (-c^2\right )^{3/2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

(Integral(a**2/(c**2*x**2 + 1), x) + Integral(b**2*asinh(c*x)**2/(c**2*x**2 + 1), x) + Integral(2*a*b*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((a+b*arcsinh(c*x))^2/(c^2*d*x^2+d),x, algorithm="giac")

[Out]

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

[Out]

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

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