3.1.34 \(\int \frac {a+b \sinh ^{-1}(c x)}{x^2 (d+c^2 d x^2)} \, dx\) [34]

Optimal. Leaf size=101 \[ -\frac {a+b \sinh ^{-1}(c x)}{d x}-\frac {2 c \left (a+b \sinh ^{-1}(c x)\right ) \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {b c \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )}{d}+\frac {i b c \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {i b c \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{d} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5809, 5789, 4265, 2317, 2438, 272, 65, 214} \begin {gather*} -\frac {2 c \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac {a+b \sinh ^{-1}(c x)}{d x}-\frac {b c \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )}{d}+\frac {i b c \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac {i b c \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

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

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]

Rule 5809

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/(f*(m + 1)))*Simp[(d +
e*x^2)^p/(1 + c^2*x^2)^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, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && ILtQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 182, normalized size = 1.80 \begin {gather*} -\frac {a+b \sinh ^{-1}(c x)+a c x \text {ArcTan}(c x)+b c x \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )+b \sqrt {-c^2} x \sinh ^{-1}(c x) \log \left (1+\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )-b \sqrt {-c^2} x \sinh ^{-1}(c x) \log \left (1+\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )-b \sqrt {-c^2} x \text {PolyLog}\left (2,\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )+b \sqrt {-c^2} x \text {PolyLog}\left (2,\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )}{d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.70, size = 203, normalized size = 2.01

method	result	size
derivativedivides	\(c \left (-\frac {a \arctan \left (c x \right )}{d}-\frac {a}{d c x}-\frac {b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{d}-\frac {b \arcsinh \left (c x \right )}{d c x}-\frac {b \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{d}-\frac {b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}+\frac {b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}+\frac {i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}-\frac {i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}\right )\)	\(203\)
default	\(c \left (-\frac {a \arctan \left (c x \right )}{d}-\frac {a}{d c x}-\frac {b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{d}-\frac {b \arcsinh \left (c x \right )}{d c x}-\frac {b \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{d}-\frac {b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}+\frac {b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}+\frac {i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}-\frac {i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{d}\right )\)	\(203\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

-a*(c*arctan(c*x)/d + 1/(d*x)) + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/(c^2*d*x^4 + d*x^2), x)

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

[Out]

integral((b*arcsinh(c*x) + a)/(c^2*d*x^4 + d*x^2), 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^{4} + x^{2}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{4} + x^{2}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**2*x**4 + x**2), x) + Integral(b*asinh(c*x)/(c**2*x**4 + x**2), 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))/x^2/(c^2*d*x^2+d),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/((c^2*d*x^2 + d)*x^2), 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^2\,\left (d\,c^2\,x^2+d\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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