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

Optimal. Leaf size=159 \[ -\frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac{b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac{a+b \sinh ^{-1}(c x)}{2 d^3 \left (c^2 x^2+1\right )}+\frac{a+b \sinh ^{-1}(c x)}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac{2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}-\frac{2 b c x}{3 d^3 \sqrt{c^2 x^2+1}}-\frac{b c x}{12 d^3 \left (c^2 x^2+1\right )^{3/2}} \]

[Out]

-(b*c*x)/(12*d^3*(1 + c^2*x^2)^(3/2)) - (2*b*c*x)/(3*d^3*Sqrt[1 + c^2*x^2]) + (a + b*ArcSinh[c*x])/(4*d^3*(1 +
c^2*x^2)^2) + (a + b*ArcSinh[c*x])/(2*d^3*(1 + c^2*x^2)) - (2*(a + b*ArcSinh[c*x])*ArcTanh[E^(2*ArcSinh[c*x])
])/d^3 - (b*PolyLog[2, -E^(2*ArcSinh[c*x])])/(2*d^3) + (b*PolyLog[2, E^(2*ArcSinh[c*x])])/(2*d^3)

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Rubi [A]  time = 0.25158, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5755, 5720, 5461, 4182, 2279, 2391, 191, 192} \[ -\frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac{b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac{a+b \sinh ^{-1}(c x)}{2 d^3 \left (c^2 x^2+1\right )}+\frac{a+b \sinh ^{-1}(c x)}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac{2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}-\frac{2 b c x}{3 d^3 \sqrt{c^2 x^2+1}}-\frac{b c x}{12 d^3 \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*x)/(12*d^3*(1 + c^2*x^2)^(3/2)) - (2*b*c*x)/(3*d^3*Sqrt[1 + c^2*x^2]) + (a + b*ArcSinh[c*x])/(4*d^3*(1 +
c^2*x^2)^2) + (a + b*ArcSinh[c*x])/(2*d^3*(1 + c^2*x^2)) - (2*(a + b*ArcSinh[c*x])*ArcTanh[E^(2*ArcSinh[c*x])
])/d^3 - (b*PolyLog[2, -E^(2*ArcSinh[c*x])])/(2*d^3) + (b*PolyLog[2, E^(2*ArcSinh[c*x])])/(2*d^3)

Rule 5755

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)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p
+ 1)), Int[(f*x)^m*(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)^F
racPart[p])/(2*f*(p + 1)*(1 + c^2*x^2)^FracPart[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, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])

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 191

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

Rule 192

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

Rubi steps

\begin{align*} \int \frac{a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )^3} \, dx &=\frac{a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac{(b c) \int \frac{1}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{4 d^3}+\frac{\int \frac{a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )^2} \, dx}{d}\\ &=-\frac{b c x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac{a+b \sinh ^{-1}(c x)}{2 d^3 \left (1+c^2 x^2\right )}-\frac{(b c) \int \frac{1}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{6 d^3}-\frac{(b c) \int \frac{1}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{2 d^3}+\frac{\int \frac{a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )} \, dx}{d^2}\\ &=-\frac{b c x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{2 b c x}{3 d^3 \sqrt{1+c^2 x^2}}+\frac{a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac{a+b \sinh ^{-1}(c x)}{2 d^3 \left (1+c^2 x^2\right )}+\frac{\operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=-\frac{b c x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{2 b c x}{3 d^3 \sqrt{1+c^2 x^2}}+\frac{a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac{a+b \sinh ^{-1}(c x)}{2 d^3 \left (1+c^2 x^2\right )}+\frac{2 \operatorname{Subst}\left (\int (a+b x) \text{csch}(2 x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=-\frac{b c x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{2 b c x}{3 d^3 \sqrt{1+c^2 x^2}}+\frac{a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac{a+b \sinh ^{-1}(c x)}{2 d^3 \left (1+c^2 x^2\right )}-\frac{2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac{b \operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}+\frac{b \operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=-\frac{b c x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{2 b c x}{3 d^3 \sqrt{1+c^2 x^2}}+\frac{a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac{a+b \sinh ^{-1}(c x)}{2 d^3 \left (1+c^2 x^2\right )}-\frac{2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac{b \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac{b \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}\\ &=-\frac{b c x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{2 b c x}{3 d^3 \sqrt{1+c^2 x^2}}+\frac{a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac{a+b \sinh ^{-1}(c x)}{2 d^3 \left (1+c^2 x^2\right )}-\frac{2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac{b \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac{b \text{Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}\\ \end{align*}

Mathematica [A]  time = 0.632703, size = 289, normalized size = 1.82 \[ \frac{-4 b \text{PolyLog}\left (2,\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}\right )-4 b \text{PolyLog}\left (2,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+2 b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )-\frac{2 a^2}{b}+\frac{2 a}{c^2 x^2+1}+\frac{a}{\left (c^2 x^2+1\right )^2}-2 a \log \left (c^2 x^2+1\right )-4 a \sinh ^{-1}(c x)+4 a \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-\frac{8 b c x}{3 \sqrt{c^2 x^2+1}}-\frac{b c x}{3 \left (c^2 x^2+1\right )^{3/2}}+\frac{2 b \sinh ^{-1}(c x)}{c^2 x^2+1}+\frac{b \sinh ^{-1}(c x)}{\left (c^2 x^2+1\right )^2}-4 b \sinh ^{-1}(c x) \log \left (\frac{c e^{\sinh ^{-1}(c x)}}{\sqrt{-c^2}}+1\right )-4 b \sinh ^{-1}(c x) \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )+4 b \sinh ^{-1}(c x) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )}{4 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-2*a^2)/b + a/(1 + c^2*x^2)^2 - (b*c*x)/(3*(1 + c^2*x^2)^(3/2)) + (2*a)/(1 + c^2*x^2) - (8*b*c*x)/(3*Sqrt[1
+ c^2*x^2]) - 4*a*ArcSinh[c*x] + (b*ArcSinh[c*x])/(1 + c^2*x^2)^2 + (2*b*ArcSinh[c*x])/(1 + c^2*x^2) - 4*b*Arc
Sinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] - 4*b*ArcSinh[c*x]*Log[1 + (Sqrt[-c^2]*E^ArcSinh[c*x])/c] + 4
*a*Log[1 - E^(2*ArcSinh[c*x])] + 4*b*ArcSinh[c*x]*Log[1 - E^(2*ArcSinh[c*x])] - 2*a*Log[1 + c^2*x^2] - 4*b*Pol
yLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] - 4*b*PolyLog[2, (Sqrt[-c^2]*E^ArcSinh[c*x])/c] + 2*b*PolyLog[2, E^(2*A
rcSinh[c*x])])/(4*d^3)

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Maple [B]  time = 0.156, size = 451, normalized size = 2.8 \begin{align*}{\frac{a\ln \left ( cx \right ) }{{d}^{3}}}+{\frac{a}{4\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}}}+{\frac{a}{2\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) }}-{\frac{a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,{d}^{3}}}-{\frac{2\,b{c}^{3}{x}^{3}}{3\,{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{2\,b{c}^{4}{x}^{4}}{3\,{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }}+{\frac{b{\it Arcsinh} \left ( cx \right ){c}^{2}{x}^{2}}{2\,{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }}-{\frac{3\,bcx}{4\,{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }\sqrt{{c}^{2}{x}^{2}+1}}+{\frac{4\,b{c}^{2}{x}^{2}}{3\,{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }}+{\frac{3\,b{\it Arcsinh} \left ( cx \right ) }{4\,{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }}+{\frac{2\,b}{3\,{d}^{3} \left ({c}^{4}{x}^{4}+2\,{c}^{2}{x}^{2}+1 \right ) }}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{{d}^{3}}\ln \left ( 1+ \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-{\frac{b}{2\,{d}^{3}}{\it polylog} \left ( 2,- \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+{\frac{b{\it Arcsinh} \left ( cx \right ) }{{d}^{3}}\ln \left ( 1+cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) }+{\frac{b}{{d}^{3}}{\it polylog} \left ( 2,-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }+{\frac{b{\it Arcsinh} \left ( cx \right ) }{{d}^{3}}\ln \left ( 1-cx-\sqrt{{c}^{2}{x}^{2}+1} \right ) }+{\frac{b}{{d}^{3}}{\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/(c^2*d*x^2+d)^3,x)

[Out]

a/d^3*ln(c*x)+1/4*a/d^3/(c^2*x^2+1)^2+1/2*a/d^3/(c^2*x^2+1)-1/2*a/d^3*ln(c^2*x^2+1)-2/3*b/d^3/(c^4*x^4+2*c^2*x
^2+1)*c^3*x^3*(c^2*x^2+1)^(1/2)+2/3*b/d^3/(c^4*x^4+2*c^2*x^2+1)*c^4*x^4+1/2*b/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsin
h(c*x)*c^2*x^2-3/4*b/d^3/(c^4*x^4+2*c^2*x^2+1)*c*x*(c^2*x^2+1)^(1/2)+4/3*b/d^3/(c^4*x^4+2*c^2*x^2+1)*c^2*x^2+3
/4*b/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)+2/3*b/d^3/(c^4*x^4+2*c^2*x^2+1)-b/d^3*arcsinh(c*x)*ln(1+(c*x+(c^2*
x^2+1)^(1/2))^2)-1/2*b*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/d^3+b/d^3*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2)
)+b/d^3*polylog(2,-c*x-(c^2*x^2+1)^(1/2))+b/d^3*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))+b/d^3*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}{4} \, a{\left (\frac{2 \, c^{2} x^{2} + 3}{c^{4} d^{3} x^{4} + 2 \, c^{2} d^{3} x^{2} + d^{3}} - \frac{2 \, \log \left (c^{2} x^{2} + 1\right )}{d^{3}} + \frac{4 \, \log \left (x\right )}{d^{3}}\right )} + b \int \frac{\log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{6} d^{3} x^{7} + 3 \, c^{4} d^{3} x^{5} + 3 \, c^{2} d^{3} x^{3} + d^{3} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*((2*c^2*x^2 + 3)/(c^4*d^3*x^4 + 2*c^2*d^3*x^2 + d^3) - 2*log(c^2*x^2 + 1)/d^3 + 4*log(x)/d^3) + b*integr
ate(log(c*x + sqrt(c^2*x^2 + 1))/(c^6*d^3*x^7 + 3*c^4*d^3*x^5 + 3*c^2*d^3*x^3 + d^3*x), 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^{6} d^{3} x^{7} + 3 \, c^{4} d^{3} x^{5} + 3 \, c^{2} d^{3} x^{3} + d^{3} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arcsinh(c*x) + a)/(c^6*d^3*x^7 + 3*c^4*d^3*x^5 + 3*c^2*d^3*x^3 + d^3*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a}{c^{6} x^{7} + 3 c^{4} x^{5} + 3 c^{2} x^{3} + x}\, dx + \int \frac{b \operatorname{asinh}{\left (c x \right )}}{c^{6} x^{7} + 3 c^{4} x^{5} + 3 c^{2} x^{3} + x}\, dx}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**6*x**7 + 3*c**4*x**5 + 3*c**2*x**3 + x), x) + Integral(b*asinh(c*x)/(c**6*x**7 + 3*c**4*x**5 +
3*c**2*x**3 + x), x))/d**3

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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 )}^{3} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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