3.1.41 \(\int \frac {a+b \sinh ^{-1}(c x)}{(d+c^2 d x^2)^2} \, dx\) [41]

Optimal. Leaf size=124 \[ \frac {b}{2 c d^2 \sqrt {1+c^2 x^2}}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (1+c^2 x^2\right )}+\frac {\left (a+b \sinh ^{-1}(c x)\right ) \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{c d^2}-\frac {i b \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{2 c d^2}+\frac {i b \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{2 c d^2} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5788, 5789, 4265, 2317, 2438, 267} \begin {gather*} \frac {\text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c d^2}+\frac {x \left (a+b \sinh ^{-1}(c x)\right )}{2 d^2 \left (c^2 x^2+1\right )}+\frac {b}{2 c d^2 \sqrt {c^2 x^2+1}}-\frac {i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{2 c d^2}+\frac {i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{2 c d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 267

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

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 5788

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*(d + e*x^2)^(
p + 1)*((a + b*ArcSinh[c*x])^n/(2*d*(p + 1))), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a
+ b*ArcSinh[c*x])^n, x], x] + Dist[b*c*(n/(2*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[x*(1 + c^2*x^2
)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0] &
& LtQ[p, -1] && NeQ[p, -3/2]

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

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Mathematica [A]
time = 0.07, size = 216, normalized size = 1.74 \begin {gather*} \frac {a c x+b \sqrt {1+c^2 x^2}+b c x \sinh ^{-1}(c x)+a \text {ArcTan}(c x)+a c^2 x^2 \text {ArcTan}(c x)+i b \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )+i b c^2 x^2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-i b \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )-i b c^2 x^2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )-i b \left (1+c^2 x^2\right ) \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )+i b \left (1+c^2 x^2\right ) \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{2 d^2 \left (c+c^3 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*c*x + b*Sqrt[1 + c^2*x^2] + b*c*x*ArcSinh[c*x] + a*ArcTan[c*x] + a*c^2*x^2*ArcTan[c*x] + I*b*ArcSinh[c*x]*L
og[1 - I*E^ArcSinh[c*x]] + I*b*c^2*x^2*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] - I*b*ArcSinh[c*x]*Log[1 + I*E^A
rcSinh[c*x]] - I*b*c^2*x^2*ArcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] - I*b*(1 + c^2*x^2)*PolyLog[2, (-I)*E^ArcSin
h[c*x]] + I*b*(1 + c^2*x^2)*PolyLog[2, I*E^ArcSinh[c*x]])/(2*d^2*(c + c^3*x^2))

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Maple [A]
time = 0.64, size = 219, normalized size = 1.77

method	result	size
derivativedivides	\(\frac {\frac {a c x}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {a \arctan \left (c x \right )}{2 d^{2}}+\frac {b \arcsinh \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{2 d^{2}}+\frac {b}{2 d^{2} \sqrt {c^{2} x^{2}+1}}+\frac {b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}-\frac {b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}-\frac {i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}+\frac {i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}}{c}\)	\(219\)
default	\(\frac {\frac {a c x}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {a \arctan \left (c x \right )}{2 d^{2}}+\frac {b \arcsinh \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}+1\right )}+\frac {b \arcsinh \left (c x \right ) \arctan \left (c x \right )}{2 d^{2}}+\frac {b}{2 d^{2} \sqrt {c^{2} x^{2}+1}}+\frac {b \arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}-\frac {b \arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}-\frac {i b \dilog \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}+\frac {i b \dilog \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )}{2 d^{2}}}{c}\)	\(219\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

1/2*a*(x/(c^2*d^2*x^2 + d^2) + arctan(c*x)/(c*d^2)) + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/(c^4*d^2*x^4 +
2*c^2*d^2*x^2 + d^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))/(c^2*d*x^2+d)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \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)**2,x)

[Out]

(Integral(a/(c**4*x**4 + 2*c**2*x**2 + 1), x) + Integral(b*asinh(c*x)/(c**4*x**4 + 2*c**2*x**2 + 1), x))/d**2

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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)^2,x, algorithm="giac")

[Out]

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

[Out]

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

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