3.3.30 \(\int \frac {(a+b \sinh ^{-1}(c x))^2}{x (d+c^2 d x^2)} \, dx\) [230]

Optimal. Leaf size=116 \[ -\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}-\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{d}+\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{d}+\frac {b^2 \text {PolyLog}\left (3,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d}-\frac {b^2 \text {PolyLog}\left (3,e^{2 \sinh ^{-1}(c x)}\right )}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5799, 5569, 4267, 2611, 2320, 6724} \begin {gather*} -\frac {b \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}+\frac {b \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac {2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d}+\frac {b^2 \text {Li}_3\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d}-\frac {b^2 \text {Li}_3\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcSinh[c*x])^2*ArcTanh[E^(2*ArcSinh[c*x])])/d - (b*(a + b*ArcSinh[c*x])*PolyLog[2, -E^(2*ArcSinh[c
*x])])/d + (b*(a + b*ArcSinh[c*x])*PolyLog[2, E^(2*ArcSinh[c*x])])/d + (b^2*PolyLog[3, -E^(2*ArcSinh[c*x])])/(
2*d) - (b^2*PolyLog[3, E^(2*ArcSinh[c*x])])/(2*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 4267

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

Rule 5569

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rule 5799

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Dist[1/d, Subst[Int[(
a + b*x)^n/(Cosh[x]*Sinh[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}{x \left (d+c^2 d x^2\right )} \, dx &=\frac {\text {Subst}\left (\int (a+b x)^2 \text {csch}(x) \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=\frac {2 \text {Subst}\left (\int (a+b x)^2 \text {csch}(2 x) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}-\frac {(2 b) \text {Subst}\left (\int (a+b x) \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}+\frac {(2 b) \text {Subst}\left (\int (a+b x) \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}-\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d}+\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}+\frac {b^2 \text {Subst}\left (\int \text {Li}_2\left (-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}-\frac {b^2 \text {Subst}\left (\int \text {Li}_2\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}-\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d}+\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}+\frac {b^2 \text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d}-\frac {b^2 \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d}\\ &=-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}-\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d}+\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}+\frac {b^2 \text {Li}_3\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d}-\frac {b^2 \text {Li}_3\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(400\) vs. \(2(116)=232\).
time = 0.26, size = 400, normalized size = 3.45 \begin {gather*} -\frac {2 a^3+6 a^2 b \sinh ^{-1}(c x)+12 a b^2 \sinh ^{-1}(c x) \log \left (1+\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )+6 b^3 \sinh ^{-1}(c x)^2 \log \left (1+\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )+12 a b^2 \sinh ^{-1}(c x) \log \left (1+\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+6 b^3 \sinh ^{-1}(c x)^2 \log \left (1+\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )-6 a^2 b \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-12 a b^2 \sinh ^{-1}(c x) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-6 b^3 \sinh ^{-1}(c x)^2 \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+3 a^2 b \log \left (1+c^2 x^2\right )+12 b^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )+12 b^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )-6 a b^2 \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )-6 b^3 \sinh ^{-1}(c x) \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )-12 b^3 \text {PolyLog}\left (3,\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )-12 b^3 \text {PolyLog}\left (3,\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+3 b^3 \text {PolyLog}\left (3,e^{2 \sinh ^{-1}(c x)}\right )}{6 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(2*a^3 + 6*a^2*b*ArcSinh[c*x] + 12*a*b^2*ArcSinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 6*b^3*ArcS
inh[c*x]^2*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 12*a*b^2*ArcSinh[c*x]*Log[1 + (Sqrt[-c^2]*E^ArcSinh[c*x])/
c] + 6*b^3*ArcSinh[c*x]^2*Log[1 + (Sqrt[-c^2]*E^ArcSinh[c*x])/c] - 6*a^2*b*Log[1 - E^(2*ArcSinh[c*x])] - 12*a*
b^2*ArcSinh[c*x]*Log[1 - E^(2*ArcSinh[c*x])] - 6*b^3*ArcSinh[c*x]^2*Log[1 - E^(2*ArcSinh[c*x])] + 3*a^2*b*Log[
1 + c^2*x^2] + 12*b^2*(a + b*ArcSinh[c*x])*PolyLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 12*b^2*(a + b*ArcSinh[c
*x])*PolyLog[2, (Sqrt[-c^2]*E^ArcSinh[c*x])/c] - 6*a*b^2*PolyLog[2, E^(2*ArcSinh[c*x])] - 6*b^3*ArcSinh[c*x]*P
olyLog[2, E^(2*ArcSinh[c*x])] - 12*b^3*PolyLog[3, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] - 12*b^3*PolyLog[3, (Sqrt[-c^
2]*E^ArcSinh[c*x])/c] + 3*b^3*PolyLog[3, E^(2*ArcSinh[c*x])])/(b*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(349\) vs. \(2(157)=314\).
time = 3.11, size = 350, normalized size = 3.02

method	result	size
derivativedivides	\(-\frac {a^{2} \ln \left (c^{2} x^{2}+1\right )}{2 d}+\frac {a^{2} \ln \left (c x \right )}{d}-\frac {b^{2} \arcsinh \left (c x \right )^{2} \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d}-\frac {b^{2} \arcsinh \left (c x \right ) \polylog \left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {b^{2} \polylog \left (3, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 d}+\frac {b^{2} \arcsinh \left (c x \right )^{2} \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )}{d}+\frac {2 b^{2} \arcsinh \left (c x \right ) \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )}{d}-\frac {2 b^{2} \polylog \left (3, c x +\sqrt {c^{2} x^{2}+1}\right )}{d}+\frac {b^{2} \arcsinh \left (c x \right )^{2} \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{d}+\frac {2 b^{2} \arcsinh \left (c x \right ) \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )}{d}-\frac {2 b^{2} \polylog \left (3, -c x -\sqrt {c^{2} x^{2}+1}\right )}{d}+\frac {2 a b \left (\dilog \left (\frac {1}{\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}}\right )-\frac {\dilog \left (\frac {1}{\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{4}}\right )}{4}\right )}{d}\)	\(350\)
default	\(-\frac {a^{2} \ln \left (c^{2} x^{2}+1\right )}{2 d}+\frac {a^{2} \ln \left (c x \right )}{d}-\frac {b^{2} \arcsinh \left (c x \right )^{2} \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d}-\frac {b^{2} \arcsinh \left (c x \right ) \polylog \left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d}+\frac {b^{2} \polylog \left (3, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 d}+\frac {b^{2} \arcsinh \left (c x \right )^{2} \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )}{d}+\frac {2 b^{2} \arcsinh \left (c x \right ) \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )}{d}-\frac {2 b^{2} \polylog \left (3, c x +\sqrt {c^{2} x^{2}+1}\right )}{d}+\frac {b^{2} \arcsinh \left (c x \right )^{2} \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{d}+\frac {2 b^{2} \arcsinh \left (c x \right ) \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )}{d}-\frac {2 b^{2} \polylog \left (3, -c x -\sqrt {c^{2} x^{2}+1}\right )}{d}+\frac {2 a b \left (\dilog \left (\frac {1}{\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}}\right )-\frac {\dilog \left (\frac {1}{\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{4}}\right )}{4}\right )}{d}\)	\(350\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/x/(c^2*d*x^2+d),x, algorithm="maxima")

[Out]

-1/2*a^2*(log(c^2*x^2 + 1)/d - 2*log(x)/d) + integrate(b^2*log(c*x + sqrt(c^2*x^2 + 1))^2/(c^2*d*x^3 + d*x) +
2*a*b*log(c*x + sqrt(c^2*x^2 + 1))/(c^2*d*x^3 + d*x), 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/x/(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^3 + d*x), 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^{3} + x}\, dx + \int \frac {b^{2} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{2} x^{3} + x}\, dx + \int \frac {2 a b \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{3} + x}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c**2*x**3 + x), x) + Integral(b**2*asinh(c*x)**2/(c**2*x**3 + x), x) + Integral(2*a*b*asinh(c*
x)/(c**2*x**3 + x), 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/x/(c^2*d*x^2+d),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^2/((c^2*d*x^2 + d)*x), 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}{x\,\left (d\,c^2\,x^2+d\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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