3.1.32 \(\int \frac {a+b \sinh ^{-1}(c x)}{d+c^2 d x^2} \, dx\) [32]

Optimal. Leaf size=70 \[ \frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{c d}-\frac {i b \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c d}+\frac {i b \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c d} \]

[Out]

2*(a+b*arcsinh(c*x))*arctan(c*x+(c^2*x^2+1)^(1/2))/c/d-I*b*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)))/c/d+I*b*polyl
og(2,I*(c*x+(c^2*x^2+1)^(1/2)))/c/d

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Rubi [A]
time = 0.05, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5789, 4265, 2317, 2438} \begin {gather*} \frac {2 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c d}-\frac {i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c d}+\frac {i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/(c*d) - (I*b*PolyLog[2, (-I)*E^ArcSinh[c*x]])/(c*d) + (I*b*Pol
yLog[2, I*E^ArcSinh[c*x]])/(c*d)

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 4265

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

Rule 5789

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

Rubi steps

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

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Mathematica [A]
time = 0.07, size = 135, normalized size = 1.93 \begin {gather*} -\frac {c \left (a \sqrt {-c^2} \text {ArcTan}(c x)-b c \sinh ^{-1}(c x) \log \left (1+\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )+b c \sinh ^{-1}(c x) \log \left (1+\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+b c \text {PolyLog}\left (2,\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )-b c \text {PolyLog}\left (2,\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )\right )}{\left (-c^2\right )^{3/2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*(a*Sqrt[-c^2]*ArcTan[c*x] - b*c*ArcSinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + b*c*ArcSinh[c*x]*Lo
g[1 + (Sqrt[-c^2]*E^ArcSinh[c*x])/c] + b*c*PolyLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] - b*c*PolyLog[2, (Sqrt[-c
^2]*E^ArcSinh[c*x])/c]))/((-c^2)^(3/2)*d))

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Maple [A]
time = 2.54, size = 157, normalized size = 2.24

method	result	size
derivativedivides	\(\frac {\frac {a \arctan \left (c x \right )}{d}+\frac {i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}+\frac {b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{d}-\frac {i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}-\frac {b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}+\frac {b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}}{c}\)	\(157\)
default	\(\frac {\frac {a \arctan \left (c x \right )}{d}+\frac {i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}+\frac {b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{d}-\frac {i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}-\frac {b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}+\frac {b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}}{c}\)	\(157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/(c^2*d*x^2 + d), x) + a*arctan(c*x)/(c*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))/(c^2*d*x^2+d),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a}{c^{2} x^{2} + 1}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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