3.1.8 \(\int \frac {(d+c^2 d x^2) (a+b \sinh ^{-1}(c x))}{x^3} \, dx\) [8]

Optimal. Leaf size=128 \[ -\frac {b c d \sqrt {1+c^2 x^2}}{2 x}+\frac {1}{2} b c^2 d \sinh ^{-1}(c x)-\frac {d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac {c^2 d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+c^2 d \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right )-\frac {1}{2} b c^2 d \text {PolyLog}\left (2,e^{-2 \sinh ^{-1}(c x)}\right ) \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5802, 283, 221, 5775, 3797, 2221, 2317, 2438} \begin {gather*} -\frac {d \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 x^2}+\frac {c^2 d \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+c^2 d \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac {1}{2} b c^2 d \text {Li}_2\left (e^{-2 \sinh ^{-1}(c x)}\right )-\frac {b c d \sqrt {c^2 x^2+1}}{2 x}+\frac {1}{2} b c^2 d \sinh ^{-1}(c x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 221

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

Rule 283

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

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

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 2438

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 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((
c + d*x)^(m + 1)/(d*(m + 1))), x] + Dist[2*I, Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5775

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Dist[1/b, Subst[Int[x^n*Coth[-a/b + x/b], x],
x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 5802

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(f*x)
^(m + 1)*(d + e*x^2)^p*((a + b*ArcSinh[c*x])/(f*(m + 1))), x] + (-Dist[b*c*(d^p/(f*(m + 1))), Int[(f*x)^(m + 1
)*(1 + c^2*x^2)^(p - 1/2), x], x] - Dist[2*e*(p/(f^2*(m + 1))), Int[(f*x)^(m + 2)*(d + e*x^2)^(p - 1)*(a + b*A
rcSinh[c*x]), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && IGtQ[p, 0] && ILtQ[(m + 1)/2, 0]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 111, normalized size = 0.87 \begin {gather*} -\frac {a d}{2 x^2}-\frac {b c d \sqrt {1+c^2 x^2}}{2 x}-\frac {b d \sinh ^{-1}(c x)}{2 x^2}-\frac {1}{2} b c^2 d \sinh ^{-1}(c x)^2+b c^2 d \sinh ^{-1}(c x) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+a c^2 d \log (x)+\frac {1}{2} b c^2 d \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 4.84, size = 166, normalized size = 1.30

method	result	size
derivativedivides	\(c^{2} \left (a d \ln \left (c x \right )-\frac {a d}{2 c^{2} x^{2}}-\frac {b d \arcsinh \left (c x \right )^{2}}{2}-\frac {b d \sqrt {c^{2} x^{2}+1}}{2 c x}+\frac {b d}{2}-\frac {b d \arcsinh \left (c x \right )}{2 c^{2} x^{2}}+b d \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+b d \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )+b d \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+b d \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )\right )\)	\(166\)
default	\(c^{2} \left (a d \ln \left (c x \right )-\frac {a d}{2 c^{2} x^{2}}-\frac {b d \arcsinh \left (c x \right )^{2}}{2}-\frac {b d \sqrt {c^{2} x^{2}+1}}{2 c x}+\frac {b d}{2}-\frac {b d \arcsinh \left (c x \right )}{2 c^{2} x^{2}}+b d \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )+b d \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )+b d \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )+b d \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )\right )\)	\(166\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} d \left (\int \frac {a}{x^{3}}\, dx + \int \frac {a c^{2}}{x}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{x^{3}}\, dx + \int \frac {b c^{2} \operatorname {asinh}{\left (c x \right )}}{x}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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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 )\,\left (d\,c^2\,x^2+d\right )}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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