3.3.6 \(\int \frac {(d+c^2 d x^2) (a+b \sinh ^{-1}(c x))^2}{x^4} \, dx\) [206]

Optimal. Leaf size=158 \[ -\frac {b^2 c^2 d}{3 x}-\frac {b c d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2}-\frac {2 c^2 d \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x}-\frac {d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x^3}-\frac {10}{3} b c^3 d \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-\frac {5}{3} b^2 c^3 d \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )+\frac {5}{3} b^2 c^3 d \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right ) \]

[Out]

-1/3*b^2*c^2*d/x-2/3*c^2*d*(a+b*arcsinh(c*x))^2/x-1/3*d*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2/x^3-10/3*b*c^3*d*(a+b
*arcsinh(c*x))*arctanh(c*x+(c^2*x^2+1)^(1/2))-5/3*b^2*c^3*d*polylog(2,-c*x-(c^2*x^2+1)^(1/2))+5/3*b^2*c^3*d*po
lylog(2,c*x+(c^2*x^2+1)^(1/2))-1/3*b*c*d*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/x^2

________________________________________________________________________________________

Rubi [A]
time = 0.26, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5807, 5776, 5816, 4267, 2317, 2438, 5805, 30} \begin {gather*} -\frac {10}{3} b c^3 d \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )-\frac {b c d \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2}-\frac {d \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x^3}-\frac {2 c^2 d \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x}-\frac {5}{3} b^2 c^3 d \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )+\frac {5}{3} b^2 c^3 d \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )-\frac {b^2 c^2 d}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + c^2*d*x^2)*(a + b*ArcSinh[c*x])^2)/x^4,x]

[Out]

-1/3*(b^2*c^2*d)/x - (b*c*d*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(3*x^2) - (2*c^2*d*(a + b*ArcSinh[c*x])^2)
/(3*x) - (d*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/(3*x^3) - (10*b*c^3*d*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh
[c*x]])/3 - (5*b^2*c^3*d*PolyLog[2, -E^ArcSinh[c*x]])/3 + (5*b^2*c^3*d*PolyLog[2, E^ArcSinh[c*x]])/3

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && 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 5776

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcS
inh[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[
1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5805

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

Rule 5807

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

Rule 5816

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

Rubi steps

\begin {align*} \int \frac {\left (d+c^2 d x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{x^4} \, dx &=-\frac {d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x^3}+\frac {1}{3} (2 b c d) \int \frac {\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x^3} \, dx+\frac {1}{3} \left (2 c^2 d\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x^2} \, dx\\ &=-\frac {b c d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2}-\frac {2 c^2 d \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x}-\frac {d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x^3}+\frac {1}{3} \left (b^2 c^2 d\right ) \int \frac {1}{x^2} \, dx+\frac {1}{3} \left (b c^3 d\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx+\frac {1}{3} \left (4 b c^3 d\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx\\ &=-\frac {b^2 c^2 d}{3 x}-\frac {b c d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2}-\frac {2 c^2 d \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x}-\frac {d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x^3}+\frac {1}{3} \left (b c^3 d\right ) \text {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )+\frac {1}{3} \left (4 b c^3 d\right ) \text {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac {b^2 c^2 d}{3 x}-\frac {b c d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2}-\frac {2 c^2 d \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x}-\frac {d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x^3}-\frac {10}{3} b c^3 d \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-\frac {1}{3} \left (b^2 c^3 d\right ) \text {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )+\frac {1}{3} \left (b^2 c^3 d\right ) \text {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )-\frac {1}{3} \left (4 b^2 c^3 d\right ) \text {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )+\frac {1}{3} \left (4 b^2 c^3 d\right ) \text {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )\\ &=-\frac {b^2 c^2 d}{3 x}-\frac {b c d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2}-\frac {2 c^2 d \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x}-\frac {d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x^3}-\frac {10}{3} b c^3 d \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-\frac {1}{3} \left (b^2 c^3 d\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )+\frac {1}{3} \left (b^2 c^3 d\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )-\frac {1}{3} \left (4 b^2 c^3 d\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )+\frac {1}{3} \left (4 b^2 c^3 d\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )\\ &=-\frac {b^2 c^2 d}{3 x}-\frac {b c d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2}-\frac {2 c^2 d \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x}-\frac {d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 x^3}-\frac {10}{3} b c^3 d \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )-\frac {5}{3} b^2 c^3 d \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )+\frac {5}{3} b^2 c^3 d \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.49, size = 245, normalized size = 1.55 \begin {gather*} -\frac {d \left (a^2+3 a^2 c^2 x^2+b^2 c^2 x^2+a b c x \sqrt {1+c^2 x^2}+2 a b \sinh ^{-1}(c x)+6 a b c^2 x^2 \sinh ^{-1}(c x)+b^2 c x \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)+b^2 \sinh ^{-1}(c x)^2+3 b^2 c^2 x^2 \sinh ^{-1}(c x)^2+5 a b c^3 x^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )-5 b^2 c^3 x^3 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )+5 b^2 c^3 x^3 \sinh ^{-1}(c x) \log \left (1+e^{-\sinh ^{-1}(c x)}\right )-5 b^2 c^3 x^3 \text {PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )+5 b^2 c^3 x^3 \text {PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )\right )}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + c^2*d*x^2)*(a + b*ArcSinh[c*x])^2)/x^4,x]

[Out]

-1/3*(d*(a^2 + 3*a^2*c^2*x^2 + b^2*c^2*x^2 + a*b*c*x*Sqrt[1 + c^2*x^2] + 2*a*b*ArcSinh[c*x] + 6*a*b*c^2*x^2*Ar
cSinh[c*x] + b^2*c*x*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + b^2*ArcSinh[c*x]^2 + 3*b^2*c^2*x^2*ArcSinh[c*x]^2 + 5*a*
b*c^3*x^3*ArcTanh[Sqrt[1 + c^2*x^2]] - 5*b^2*c^3*x^3*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] + 5*b^2*c^3*x^3*A
rcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] - 5*b^2*c^3*x^3*PolyLog[2, -E^(-ArcSinh[c*x])] + 5*b^2*c^3*x^3*PolyLog[
2, E^(-ArcSinh[c*x])]))/x^3

________________________________________________________________________________________

Maple [A]
time = 5.94, size = 272, normalized size = 1.72

method	result	size
derivativedivides	\(c^{3} \left (a^{2} d \left (-\frac {1}{c x}-\frac {1}{3 c^{3} x^{3}}\right )-\frac {b^{2} d \arcsinh \left (c x \right )^{2}}{c x}-\frac {b^{2} d \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}}{3 c^{2} x^{2}}-\frac {b^{2} d \arcsinh \left (c x \right )^{2}}{3 c^{3} x^{3}}-\frac {b^{2} d}{3 c x}-\frac {5 b^{2} d \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{3}-\frac {5 b^{2} d \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )}{3}+\frac {5 b^{2} d \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )}{3}+\frac {5 b^{2} d \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )}{3}+2 b d a \left (-\frac {\arcsinh \left (c x \right )}{c x}-\frac {\arcsinh \left (c x \right )}{3 c^{3} x^{3}}-\frac {5 \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}-\frac {\sqrt {c^{2} x^{2}+1}}{6 c^{2} x^{2}}\right )\right )\)	\(272\)
default	\(c^{3} \left (a^{2} d \left (-\frac {1}{c x}-\frac {1}{3 c^{3} x^{3}}\right )-\frac {b^{2} d \arcsinh \left (c x \right )^{2}}{c x}-\frac {b^{2} d \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}}{3 c^{2} x^{2}}-\frac {b^{2} d \arcsinh \left (c x \right )^{2}}{3 c^{3} x^{3}}-\frac {b^{2} d}{3 c x}-\frac {5 b^{2} d \arcsinh \left (c x \right ) \ln \left (1+c x +\sqrt {c^{2} x^{2}+1}\right )}{3}-\frac {5 b^{2} d \polylog \left (2, -c x -\sqrt {c^{2} x^{2}+1}\right )}{3}+\frac {5 b^{2} d \arcsinh \left (c x \right ) \ln \left (1-c x -\sqrt {c^{2} x^{2}+1}\right )}{3}+\frac {5 b^{2} d \polylog \left (2, c x +\sqrt {c^{2} x^{2}+1}\right )}{3}+2 b d a \left (-\frac {\arcsinh \left (c x \right )}{c x}-\frac {\arcsinh \left (c x \right )}{3 c^{3} x^{3}}-\frac {5 \arctanh \left (\frac {1}{\sqrt {c^{2} x^{2}+1}}\right )}{6}-\frac {\sqrt {c^{2} x^{2}+1}}{6 c^{2} x^{2}}\right )\right )\)	\(272\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c^2*d*x^2+d)*(a+b*arcsinh(c*x))^2/x^4,x,method=_RETURNVERBOSE)

[Out]

c^3*(a^2*d*(-1/c/x-1/3/c^3/x^3)-b^2*d*arcsinh(c*x)^2/c/x-1/3*b^2*d/c^2/x^2*arcsinh(c*x)*(c^2*x^2+1)^(1/2)-1/3*
b^2*d/c^3/x^3*arcsinh(c*x)^2-1/3*b^2*d/c/x-5/3*b^2*d*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))-5/3*b^2*d*polylo
g(2,-c*x-(c^2*x^2+1)^(1/2))+5/3*b^2*d*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))+5/3*b^2*d*polylog(2,c*x+(c^2*x^
2+1)^(1/2))+2*b*d*a*(-arcsinh(c*x)/c/x-1/3*arcsinh(c*x)/c^3/x^3-5/6*arctanh(1/(c^2*x^2+1)^(1/2))-1/6/c^2/x^2*(
c^2*x^2+1)^(1/2)))

________________________________________________________________________________________

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((c^2*d*x^2+d)*(a+b*arcsinh(c*x))^2/x^4,x, algorithm="maxima")

[Out]

-2*(c*arcsinh(1/(c*abs(x))) + arcsinh(c*x)/x)*a*b*c^2*d + 1/3*((c^2*arcsinh(1/(c*abs(x))) - sqrt(c^2*x^2 + 1)/
x^2)*c - 2*arcsinh(c*x)/x^3)*a*b*d - a^2*c^2*d/x - 1/3*a^2*d/x^3 - 1/3*(3*b^2*c^2*d*x^2 + b^2*d)*log(c*x + sqr
t(c^2*x^2 + 1))^2/x^3 + integrate(2/3*(3*b^2*c^5*d*x^4 + 4*b^2*c^3*d*x^2 + b^2*c*d + (3*b^2*c^4*d*x^3 + b^2*c^
2*d*x)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1))/(c^3*x^6 + c*x^4 + (c^2*x^5 + x^3)*sqrt(c^2*x^2 + 1)),
x)

________________________________________________________________________________________

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((c^2*d*x^2+d)*(a+b*arcsinh(c*x))^2/x^4,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,\left (d\,c^2\,x^2+d\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asinh(c*x))^2*(d + c^2*d*x^2))/x^4,x)

[Out]

int(((a + b*asinh(c*x))^2*(d + c^2*d*x^2))/x^4, x)

________________________________________________________________________________________