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

Optimal. Leaf size=113 \[ \frac{b c^2 \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d}-\frac{b c^2 \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{2 d}+\frac{2 c^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac{a+b \sinh ^{-1}(c x)}{2 d x^2}-\frac{b c \sqrt{c^2 x^2+1}}{2 d x} \]

[Out]

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

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Rubi [A]  time = 0.198218, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {5747, 5720, 5461, 4182, 2279, 2391, 264} \[ \frac{b c^2 \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d}-\frac{b c^2 \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{2 d}+\frac{2 c^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac{a+b \sinh ^{-1}(c x)}{2 d x^2}-\frac{b c \sqrt{c^2 x^2+1}}{2 d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5747

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

Rule 5720

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 5461

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 4182

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 2279

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

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

Mathematica [B]  time = 0.272453, size = 240, normalized size = 2.12 \[ \frac{-c^2 \left (b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )+2 \log \left (1-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )\right )+2 b c^2 \text{PolyLog}\left (2,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right )+2 b c^2 \text{PolyLog}\left (2,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+\frac{c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{b}-\frac{a+b \sinh ^{-1}(c x)}{x^2}+a c^2 \log \left (c^2 x^2+1\right )-\frac{b c \sqrt{c^2 x^2+1}}{x}-b c^2 \sinh ^{-1}(c x)^2+2 b c^2 \sinh ^{-1}(c x) \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )+2 b c^2 \sinh ^{-1}(c x) \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-((b*c*Sqrt[1 + c^2*x^2])/x) - b*c^2*ArcSinh[c*x]^2 - (a + b*ArcSinh[c*x])/x^2 + (c^2*(a + b*ArcSinh[c*x])^2)
/b + 2*b*c^2*ArcSinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 2*b*c^2*ArcSinh[c*x]*Log[1 + (Sqrt[-c^2]*E^
ArcSinh[c*x])/c] + a*c^2*Log[1 + c^2*x^2] + 2*b*c^2*PolyLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 2*b*c^2*PolyLo
g[2, (Sqrt[-c^2]*E^ArcSinh[c*x])/c] - c^2*(2*(a + b*ArcSinh[c*x])*Log[1 - E^(2*ArcSinh[c*x])] + b*PolyLog[2, E
^(2*ArcSinh[c*x])]))/(2*d)

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Maple [A]  time = 0.097, size = 266, normalized size = 2.4 \begin{align*} -{\frac{a}{2\,d{x}^{2}}}-{\frac{{c}^{2}a\ln \left ( cx \right ) }{d}}+{\frac{{c}^{2}a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,d}}-{\frac{bc}{2\,dx}\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{{c}^{2}b}{2\,d}}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{2\,d{x}^{2}}}+{\frac{{c}^{2}b{\it Arcsinh} \left ( cx \right ) }{d}\ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{{c}^{2}b}{2\,d}{\it polylog} \left ( 2,- \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-{\frac{{c}^{2}b{\it Arcsinh} \left ( cx \right ) }{d}\ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }-{\frac{{c}^{2}b}{d}{\it polylog} \left ( 2,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }-{\frac{{c}^{2}b{\it Arcsinh} \left ( cx \right ) }{d}\ln \left ( 1-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }-{\frac{{c}^{2}b}{d}{\it polylog} \left ( 2,cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a/d/x^2-c^2*a/d*ln(c*x)+1/2*c^2*a/d*ln(c^2*x^2+1)-1/2*b*c*(c^2*x^2+1)^(1/2)/d/x+1/2*c^2*b/d-1/2*b/d*arcsi
nh(c*x)/x^2+c^2*b/d*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)+1/2*b*c^2*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^
2)/d-c^2*b/d*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))-c^2*b/d*polylog(2,-c*x-(c^2*x^2+1)^(1/2))-c^2*b/d*arcsin
h(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))-c^2*b/d*polylog(2,c*x+(c^2*x^2+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \,{\left (\frac{c^{2} \log \left (c^{2} x^{2} + 1\right )}{d} - \frac{2 \, c^{2} \log \left (x\right )}{d} - \frac{1}{d x^{2}}\right )} a + b \int \frac{\log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{2} d x^{5} + d x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{arsinh}\left (c x\right ) + a}{c^{2} d x^{5} + d x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arcsinh(c*x) + a)/(c^2*d*x^5 + d*x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{c^{2} x^{5} + x^{3}}\, dx + \int \frac{b \operatorname{asinh}{\left (c x \right )}}{c^{2} x^{5} + x^{3}}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**2*x**5 + x**3), x) + Integral(b*asinh(c*x)/(c**2*x**5 + x**3), x))/d

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arsinh}\left (c x\right ) + a}{{\left (c^{2} d x^{2} + d\right )} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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