3.2.51 \(\int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {d+c^2 d x^2}} \, dx\) [151]

Optimal. Leaf size=122 \[ -\frac {2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {d+c^2 d x^2}}-\frac {b \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {d+c^2 d x^2}}+\frac {b \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{\sqrt {d+c^2 d x^2}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {5816, 4267, 2317, 2438} \begin {gather*} -\frac {2 \sqrt {c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {c^2 d x^2+d}}-\frac {b \sqrt {c^2 x^2+1} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{\sqrt {c^2 d x^2+d}}+\frac {b \sqrt {c^2 x^2+1} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{\sqrt {c^2 d x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 5816

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[(1/c^(m
+ 1))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], Subst[Int[(a + b*x)^n*Sinh[x]^m, x], x, ArcSinh[c*x]], x] /; F
reeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]
time = 0.24, size = 129, normalized size = 1.06 \begin {gather*} \frac {a \log (x)}{\sqrt {d}}-\frac {a \log \left (d+\sqrt {d} \sqrt {d \left (1+c^2 x^2\right )}\right )}{\sqrt {d}}+\frac {b \sqrt {1+c^2 x^2} \left (\sinh ^{-1}(c x) \left (\log \left (1-e^{-\sinh ^{-1}(c x)}\right )-\log \left (1+e^{-\sinh ^{-1}(c x)}\right )\right )+\text {PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-\text {PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )\right )}{\sqrt {d \left (1+c^2 x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.89, size = 234, normalized size = 1.92

method	result	size
default	\(-\frac {a \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {c^{2} d \,x^{2}+d}}{x}\right )}{\sqrt {d}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )}{\sqrt {c^{2} x^{2}+1}\, d}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )}{\sqrt {c^{2} x^{2}+1}\, d}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{\sqrt {c^{2} x^{2}+1}\, d}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )}{\sqrt {c^{2} x^{2}+1}\, d}\)	\(234\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/x/(c^2*d*x^2+d)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

[Out]

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

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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))/x/(c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/x/(c**2*d*x**2+d)**(1/2),x)

[Out]

Integral((a + b*asinh(c*x))/(x*sqrt(d*(c**2*x**2 + 1))), x)

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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))/x/(c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/(sqrt(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 {a+b\,\mathrm {asinh}\left (c\,x\right )}{x\,\sqrt {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))/(x*(d + c^2*d*x^2)^(1/2)),x)

[Out]

int((a + b*asinh(c*x))/(x*(d + c^2*d*x^2)^(1/2)), x)

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