3.1.51 \(\int \frac {a+b \sinh ^{-1}(c x)}{x (d+c^2 d x^2)^3} \, dx\) [51]

Optimal. Leaf size=159 \[ -\frac {b c x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {2 b c x}{3 d^3 \sqrt {1+c^2 x^2}}+\frac {a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac {a+b \sinh ^{-1}(c x)}{2 d^3 \left (1+c^2 x^2\right )}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {b \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {b \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3} \]

[Out]

-1/12*b*c*x/d^3/(c^2*x^2+1)^(3/2)+1/4*(a+b*arcsinh(c*x))/d^3/(c^2*x^2+1)^2+1/2*(a+b*arcsinh(c*x))/d^3/(c^2*x^2
+1)-2*(a+b*arcsinh(c*x))*arctanh((c*x+(c^2*x^2+1)^(1/2))^2)/d^3-1/2*b*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/d^
3+1/2*b*polylog(2,(c*x+(c^2*x^2+1)^(1/2))^2)/d^3-2/3*b*c*x/d^3/(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.21, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5811, 5799, 5569, 4267, 2317, 2438, 197, 198} \begin {gather*} \frac {a+b \sinh ^{-1}(c x)}{2 d^3 \left (c^2 x^2+1\right )}+\frac {a+b \sinh ^{-1}(c x)}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac {2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}-\frac {2 b c x}{3 d^3 \sqrt {c^2 x^2+1}}-\frac {b c x}{12 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac {b \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {b \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(b*c*x)/(d^3*(1 + c^2*x^2)^(3/2)) - (2*b*c*x)/(3*d^3*Sqrt[1 + c^2*x^2]) + (a + b*ArcSinh[c*x])/(4*d^3*(1
+ c^2*x^2)^2) + (a + b*ArcSinh[c*x])/(2*d^3*(1 + c^2*x^2)) - (2*(a + b*ArcSinh[c*x])*ArcTanh[E^(2*ArcSinh[c*x
])])/d^3 - (b*PolyLog[2, -E^(2*ArcSinh[c*x])])/(2*d^3) + (b*PolyLog[2, E^(2*ArcSinh[c*x])])/(2*d^3)

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 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && 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 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 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 \left (d+c^2 d x^2\right )^3} \, dx &=\frac {a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {(b c) \int \frac {1}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{4 d^3}+\frac {\int \frac {a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )^2} \, dx}{d}\\ &=-\frac {b c x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac {a+b \sinh ^{-1}(c x)}{2 d^3 \left (1+c^2 x^2\right )}-\frac {(b c) \int \frac {1}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{6 d^3}-\frac {(b c) \int \frac {1}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{2 d^3}+\frac {\int \frac {a+b \sinh ^{-1}(c x)}{x \left (d+c^2 d x^2\right )} \, dx}{d^2}\\ &=-\frac {b c x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {2 b c x}{3 d^3 \sqrt {1+c^2 x^2}}+\frac {a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac {a+b \sinh ^{-1}(c x)}{2 d^3 \left (1+c^2 x^2\right )}+\frac {\text {Subst}\left (\int (a+b x) \text {csch}(x) \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=-\frac {b c x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {2 b c x}{3 d^3 \sqrt {1+c^2 x^2}}+\frac {a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac {a+b \sinh ^{-1}(c x)}{2 d^3 \left (1+c^2 x^2\right )}+\frac {2 \text {Subst}\left (\int (a+b x) \text {csch}(2 x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=-\frac {b c x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {2 b c x}{3 d^3 \sqrt {1+c^2 x^2}}+\frac {a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac {a+b \sinh ^{-1}(c x)}{2 d^3 \left (1+c^2 x^2\right )}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {b \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}+\frac {b \text {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=-\frac {b c x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {2 b c x}{3 d^3 \sqrt {1+c^2 x^2}}+\frac {a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac {a+b \sinh ^{-1}(c x)}{2 d^3 \left (1+c^2 x^2\right )}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {b \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {b \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}\\ &=-\frac {b c x}{12 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {2 b c x}{3 d^3 \sqrt {1+c^2 x^2}}+\frac {a+b \sinh ^{-1}(c x)}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac {a+b \sinh ^{-1}(c x)}{2 d^3 \left (1+c^2 x^2\right )}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {b \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {b \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}\\ \end {align*}

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Mathematica [A]
time = 0.43, size = 289, normalized size = 1.82 \begin {gather*} \frac {-\frac {2 a^2}{b}+\frac {a}{\left (1+c^2 x^2\right )^2}-\frac {b c x}{3 \left (1+c^2 x^2\right )^{3/2}}+\frac {2 a}{1+c^2 x^2}-\frac {8 b c x}{3 \sqrt {1+c^2 x^2}}-4 a \sinh ^{-1}(c x)+\frac {b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^2}+\frac {2 b \sinh ^{-1}(c x)}{1+c^2 x^2}-4 b \sinh ^{-1}(c x) \log \left (1+\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )-4 b \sinh ^{-1}(c x) \log \left (1+\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+4 a \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+4 b \sinh ^{-1}(c x) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-2 a \log \left (1+c^2 x^2\right )-4 b \text {PolyLog}\left (2,\frac {c e^{\sinh ^{-1}(c x)}}{\sqrt {-c^2}}\right )-4 b \text {PolyLog}\left (2,\frac {\sqrt {-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )+2 b \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{4 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a^2)/b + a/(1 + c^2*x^2)^2 - (b*c*x)/(3*(1 + c^2*x^2)^(3/2)) + (2*a)/(1 + c^2*x^2) - (8*b*c*x)/(3*Sqrt[1
+ c^2*x^2]) - 4*a*ArcSinh[c*x] + (b*ArcSinh[c*x])/(1 + c^2*x^2)^2 + (2*b*ArcSinh[c*x])/(1 + c^2*x^2) - 4*b*Arc
Sinh[c*x]*Log[1 + (c*E^ArcSinh[c*x])/Sqrt[-c^2]] - 4*b*ArcSinh[c*x]*Log[1 + (Sqrt[-c^2]*E^ArcSinh[c*x])/c] + 4
*a*Log[1 - E^(2*ArcSinh[c*x])] + 4*b*ArcSinh[c*x]*Log[1 - E^(2*ArcSinh[c*x])] - 2*a*Log[1 + c^2*x^2] - 4*b*Pol
yLog[2, (c*E^ArcSinh[c*x])/Sqrt[-c^2]] - 4*b*PolyLog[2, (Sqrt[-c^2]*E^ArcSinh[c*x])/c] + 2*b*PolyLog[2, E^(2*A
rcSinh[c*x])])/(4*d^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(450\) vs. \(2(170)=340\).
time = 4.46, size = 451, normalized size = 2.84

method	result	size
derivativedivides	\(-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d^{3}}+\frac {a}{4 d^{3} \left (c^{2} x^{2}+1\right )^{2}}+\frac {a}{2 d^{3} \left (c^{2} x^{2}+1\right )}+\frac {a \ln \left (c x \right )}{d^{3}}-\frac {2 b \sqrt {c^{2} x^{2}+1}\, c^{3} x^{3}}{3 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}+\frac {2 b \,c^{4} x^{4}}{3 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}+\frac {b \arcsinh \left (c x \right ) c^{2} x^{2}}{2 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}-\frac {3 b c x \sqrt {c^{2} x^{2}+1}}{4 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}+\frac {4 b \,c^{2} x^{2}}{3 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}+\frac {3 b \arcsinh \left (c x \right )}{4 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}+\frac {2 b}{3 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}+\frac {b \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{d^{3}}+\frac {b \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )}{d^{3}}-\frac {b \arcsinh \left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d^{3}}-\frac {b \polylog \left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 d^{3}}+\frac {b \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )}{d^{3}}+\frac {b \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )}{d^{3}}\)	\(451\)
default	\(-\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 d^{3}}+\frac {a}{4 d^{3} \left (c^{2} x^{2}+1\right )^{2}}+\frac {a}{2 d^{3} \left (c^{2} x^{2}+1\right )}+\frac {a \ln \left (c x \right )}{d^{3}}-\frac {2 b \sqrt {c^{2} x^{2}+1}\, c^{3} x^{3}}{3 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}+\frac {2 b \,c^{4} x^{4}}{3 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}+\frac {b \arcsinh \left (c x \right ) c^{2} x^{2}}{2 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}-\frac {3 b c x \sqrt {c^{2} x^{2}+1}}{4 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}+\frac {4 b \,c^{2} x^{2}}{3 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}+\frac {3 b \arcsinh \left (c x \right )}{4 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}+\frac {2 b}{3 d^{3} \left (c^{4} x^{4}+2 c^{2} x^{2}+1\right )}+\frac {b \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{d^{3}}+\frac {b \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )}{d^{3}}-\frac {b \arcsinh \left (c x \right ) \ln \left (1+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{d^{3}}-\frac {b \polylog \left (2, -\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}\right )}{2 d^{3}}+\frac {b \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )}{d^{3}}+\frac {b \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )}{d^{3}}\)	\(451\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a/d^3*ln(c^2*x^2+1)+1/4*a/d^3/(c^2*x^2+1)^2+1/2*a/d^3/(c^2*x^2+1)+a/d^3*ln(c*x)-2/3*b/d^3/(c^4*x^4+2*c^2*
x^2+1)*(c^2*x^2+1)^(1/2)*c^3*x^3+2/3*b/d^3/(c^4*x^4+2*c^2*x^2+1)*c^4*x^4+1/2*b/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsi
nh(c*x)*c^2*x^2-3/4*b/d^3/(c^4*x^4+2*c^2*x^2+1)*c*x*(c^2*x^2+1)^(1/2)+4/3*b/d^3/(c^4*x^4+2*c^2*x^2+1)*c^2*x^2+
3/4*b/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)+2/3*b/d^3/(c^4*x^4+2*c^2*x^2+1)+b/d^3*arcsinh(c*x)*ln(1+c*x+(c^2*
x^2+1)^(1/2))+b/d^3*polylog(2,-c*x-(c^2*x^2+1)^(1/2))-b/d^3*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)-1/2*b
*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/d^3+b/d^3*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))+b/d^3*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)^3,x, algorithm="maxima")

[Out]

1/4*a*((2*c^2*x^2 + 3)/(c^4*d^3*x^4 + 2*c^2*d^3*x^2 + d^3) - 2*log(c^2*x^2 + 1)/d^3 + 4*log(x)/d^3) + b*integr
ate(log(c*x + sqrt(c^2*x^2 + 1))/(c^6*d^3*x^7 + 3*c^4*d^3*x^5 + 3*c^2*d^3*x^3 + d^3*x), 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/(c^2*d*x^2+d)^3,x, algorithm="fricas")

[Out]

integral((b*arcsinh(c*x) + a)/(c^6*d^3*x^7 + 3*c^4*d^3*x^5 + 3*c^2*d^3*x^3 + d^3*x), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a}{c^{6} x^{7} + 3 c^{4} x^{5} + 3 c^{2} x^{3} + x}\, dx + \int \frac {b \operatorname {asinh}{\left (c x \right )}}{c^{6} x^{7} + 3 c^{4} x^{5} + 3 c^{2} x^{3} + x}\, dx}{d^{3}} \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)**3,x)

[Out]

(Integral(a/(c**6*x**7 + 3*c**4*x**5 + 3*c**2*x**3 + x), x) + Integral(b*asinh(c*x)/(c**6*x**7 + 3*c**4*x**5 +
3*c**2*x**3 + x), x))/d**3

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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)^3,x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)/((c^2*d*x^2 + d)^3*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\,{\left (d\,c^2\,x^2+d\right )}^3} \,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)^3),x)

[Out]

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

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