3.1.44 \(\int \frac {a+b \sinh ^{-1}(c x)}{x^3 (d+c^2 d x^2)^2} \, dx\) [44]

Optimal. Leaf size=146 \[ -\frac {b c}{2 d^2 x \sqrt {1+c^2 x^2}}-\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \left (1+c^2 x^2\right )}-\frac {a+b \sinh ^{-1}(c x)}{2 d^2 x^2 \left (1+c^2 x^2\right )}+\frac {4 c^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^2}+\frac {b c^2 \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{d^2}-\frac {b c^2 \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{d^2} \]

[Out]

-c^2*(a+b*arcsinh(c*x))/d^2/(c^2*x^2+1)+1/2*(-a-b*arcsinh(c*x))/d^2/x^2/(c^2*x^2+1)+4*c^2*(a+b*arcsinh(c*x))*a
rctanh((c*x+(c^2*x^2+1)^(1/2))^2)/d^2+b*c^2*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/d^2-b*c^2*polylog(2,(c*x+(c^
2*x^2+1)^(1/2))^2)/d^2-1/2*b*c/d^2/x/(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.18, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5809, 5811, 5799, 5569, 4267, 2317, 2438, 197, 277} \begin {gather*} -\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \left (c^2 x^2+1\right )}-\frac {a+b \sinh ^{-1}(c x)}{2 d^2 x^2 \left (c^2 x^2+1\right )}+\frac {4 c^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2}+\frac {b c^2 \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d^2}-\frac {b c^2 \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^2}-\frac {b c}{2 d^2 x \sqrt {c^2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*c)/(d^2*x*Sqrt[1 + c^2*x^2]) - (c^2*(a + b*ArcSinh[c*x]))/(d^2*(1 + c^2*x^2)) - (a + b*ArcSinh[c*x])/(
2*d^2*x^2*(1 + c^2*x^2)) + (4*c^2*(a + b*ArcSinh[c*x])*ArcTanh[E^(2*ArcSinh[c*x])])/d^2 + (b*c^2*PolyLog[2, -E
^(2*ArcSinh[c*x])])/d^2 - (b*c^2*PolyLog[2, E^(2*ArcSinh[c*x])])/d^2

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -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 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 5569

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rule 5799

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Dist[1/d, Subst[Int[(
a + b*x)^n/(Cosh[x]*Sinh[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]

Rule 5811

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/(2*d*f*(p + 1))), x] + (Dist[(m + 2*p + 3)/(2*d*(
p + 1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] + Dist[b*c*(n/(2*f*(p + 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, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && (IntegerQ[m] ||
IntegerQ[p] || EqQ[n, 1])

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{x^3 \left (d+c^2 d x^2\right )^2} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{2 d^2 x^2 \left (1+c^2 x^2\right )}-\left (2 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )^2} \, dx+\frac {(b c) \int \frac {1}{x^2 \left (1+c^2 x^2\right )^{3/2}} \, dx}{2 d^2}\\ &=-\frac {b c}{2 d^2 x \sqrt {1+c^2 x^2}}-\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \left (1+c^2 x^2\right )}-\frac {a+b \sinh ^{-1}(c x)}{2 d^2 x^2 \left (1+c^2 x^2\right )}-\frac {\left (2 c^2\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )} \, dx}{d}\\ &=-\frac {b c}{2 d^2 x \sqrt {1+c^2 x^2}}-\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \left (1+c^2 x^2\right )}-\frac {a+b \sinh ^{-1}(c x)}{2 d^2 x^2 \left (1+c^2 x^2\right )}-\frac {\left (2 c^2\right ) \text {Subst}\left (\int (a+b x) \text {csch}(x) \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2}\\ &=-\frac {b c}{2 d^2 x \sqrt {1+c^2 x^2}}-\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \left (1+c^2 x^2\right )}-\frac {a+b \sinh ^{-1}(c x)}{2 d^2 x^2 \left (1+c^2 x^2\right )}-\frac {\left (4 c^2\right ) \text {Subst}\left (\int (a+b x) \text {csch}(2 x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2}\\ &=-\frac {b c}{2 d^2 x \sqrt {1+c^2 x^2}}-\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \left (1+c^2 x^2\right )}-\frac {a+b \sinh ^{-1}(c x)}{2 d^2 x^2 \left (1+c^2 x^2\right )}+\frac {4 c^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^2}+\frac {\left (2 b c^2\right ) \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2}-\frac {\left (2 b c^2\right ) \text {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2}\\ &=-\frac {b c}{2 d^2 x \sqrt {1+c^2 x^2}}-\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \left (1+c^2 x^2\right )}-\frac {a+b \sinh ^{-1}(c x)}{2 d^2 x^2 \left (1+c^2 x^2\right )}+\frac {4 c^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^2}+\frac {\left (b c^2\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{d^2}-\frac {\left (b c^2\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{d^2}\\ &=-\frac {b c}{2 d^2 x \sqrt {1+c^2 x^2}}-\frac {c^2 \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \left (1+c^2 x^2\right )}-\frac {a+b \sinh ^{-1}(c x)}{2 d^2 x^2 \left (1+c^2 x^2\right )}+\frac {4 c^2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^2}+\frac {b c^2 \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d^2}-\frac {b c^2 \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^2}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(326\) vs. \(2(146)=292\).
time = 0.32, size = 326, normalized size = 2.23 \begin {gather*} \frac {\frac {2 a^2 c^2}{b}-\frac {2 a}{x^2}+\frac {b c}{x \sqrt {1+c^2 x^2}}+\frac {2 b c^3 x}{\sqrt {1+c^2 x^2}}-\frac {2 b c \sqrt {1+c^2 x^2}}{x}+\frac {a}{x^2+c^2 x^4}+4 a c^2 \sinh ^{-1}(c x)-\frac {2 b \sinh ^{-1}(c x)}{x^2}+\frac {b \sinh ^{-1}(c x)}{x^2+c^2 x^4}+4 b c^2 \sinh ^{-1}(c x) \log \left (1+\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )+4 b c^2 \sinh ^{-1}(c x) \log \left (1+\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )-4 a c^2 \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-4 b c^2 \sinh ^{-1}(c x) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+2 a c^2 \log \left (1+c^2 x^2\right )+4 b c^2 \text {PolyLog}\left (2,\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )+4 b c^2 \text {PolyLog}\left (2,\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )-2 b c^2 \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2*c^2)/b - (2*a)/x^2 + (b*c)/(x*Sqrt[1 + c^2*x^2]) + (2*b*c^3*x)/Sqrt[1 + c^2*x^2] - (2*b*c*Sqrt[1 + c^2
*x^2])/x + a/(x^2 + c^2*x^4) + 4*a*c^2*ArcSinh[c*x] - (2*b*ArcSinh[c*x])/x^2 + (b*ArcSinh[c*x])/(x^2 + c^2*x^4
) + 4*b*c^2*ArcSinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 4*b*c^2*ArcSinh[c*x]*Log[1 + (Sqrt[-c^2]*E^A
rcSinh[c*x])/c] - 4*a*c^2*Log[1 - E^(2*ArcSinh[c*x])] - 4*b*c^2*ArcSinh[c*x]*Log[1 - E^(2*ArcSinh[c*x])] + 2*a
*c^2*Log[1 + c^2*x^2] + 4*b*c^2*PolyLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] + 4*b*c^2*PolyLog[2, (Sqrt[-c^2]*E^A
rcSinh[c*x])/c] - 2*b*c^2*PolyLog[2, E^(2*ArcSinh[c*x])])/(2*d^2)

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Maple [A]
time = 3.70, size = 293, normalized size = 2.01

method	result	size
derivativedivides	\(c^{2} \left (-\frac {a}{2 d^{2} c^{2} x^{2}}-\frac {2 a \ln \left (c x \right )}{d^{2}}+\frac {a \ln \left (c^{2} x^{2}+1\right )}{d^{2}}-\frac {a}{2 d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b \arcsinh \left (c x \right )}{d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b}{2 d^{2} c x \sqrt {c^{2} x^{2}+1}}-\frac {b \arcsinh \left (c x \right )}{2 d^{2} c^{2} x^{2} \left (c^{2} x^{2}+1\right )}-\frac {2 b \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{d^{2}}-\frac {2 b \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )}{d^{2}}+\frac {2 b \arcsinh \left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d^{2}}+\frac {b \polylog \left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d^{2}}-\frac {2 b \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )}{d^{2}}-\frac {2 b \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )}{d^{2}}\right )\)	\(293\)
default	\(c^{2} \left (-\frac {a}{2 d^{2} c^{2} x^{2}}-\frac {2 a \ln \left (c x \right )}{d^{2}}+\frac {a \ln \left (c^{2} x^{2}+1\right )}{d^{2}}-\frac {a}{2 d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b \arcsinh \left (c x \right )}{d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b}{2 d^{2} c x \sqrt {c^{2} x^{2}+1}}-\frac {b \arcsinh \left (c x \right )}{2 d^{2} c^{2} x^{2} \left (c^{2} x^{2}+1\right )}-\frac {2 b \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{d^{2}}-\frac {2 b \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )}{d^{2}}+\frac {2 b \arcsinh \left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d^{2}}+\frac {b \polylog \left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d^{2}}-\frac {2 b \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )}{d^{2}}-\frac {2 b \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )}{d^{2}}\right )\)	\(293\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*(-1/2*a/d^2/c^2/x^2-2*a/d^2*ln(c*x)+a/d^2*ln(c^2*x^2+1)-1/2*a/d^2/(c^2*x^2+1)-b/d^2*arcsinh(c*x)/(c^2*x^2+
1)-1/2*b/d^2/c/x/(c^2*x^2+1)^(1/2)-1/2*b/d^2/c^2/x^2/(c^2*x^2+1)*arcsinh(c*x)-2*b/d^2*arcsinh(c*x)*ln(1+c*x+(c
^2*x^2+1)^(1/2))-2*b/d^2*polylog(2,-c*x-(c^2*x^2+1)^(1/2))+2*b/d^2*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2
)+b*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/d^2-2*b/d^2*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))-2*b/d^2*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^3/(c^2*d*x^2+d)^2,x, algorithm="maxima")

[Out]

1/2*a*(2*c^2*log(c^2*x^2 + 1)/d^2 - 4*c^2*log(x)/d^2 - (2*c^2*x^2 + 1)/(c^2*d^2*x^4 + d^2*x^2)) + b*integrate(
log(c*x + sqrt(c^2*x^2 + 1))/(c^4*d^2*x^7 + 2*c^2*d^2*x^5 + d^2*x^3), 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^3/(c^2*d*x^2+d)^2,x, algorithm="fricas")

[Out]

integral((b*arcsinh(c*x) + a)/(c^4*d^2*x^7 + 2*c^2*d^2*x^5 + d^2*x^3), 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^{7} + 2 c^{2} x^{5} + x^{3}}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{7} + 2 c^{2} x^{5} + x^{3}}\, dx}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**4*x**7 + 2*c**2*x**5 + x**3), x) + Integral(b*asinh(c*x)/(c**4*x**7 + 2*c**2*x**5 + x**3), 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))/x^3/(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^3), 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^3\,{\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))/(x^3*(d + c^2*d*x^2)^2),x)

[Out]

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

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