3.3.99 \(\int \frac {(a+b \sinh ^{-1}(c x))^2}{x^4 \sqrt {d+c^2 d x^2}} \, dx\) [299]

Optimal. Leaf size=299 \[ -\frac {b^2 c^2 \left (1+c^2 x^2\right )}{3 x \sqrt {d+c^2 d x^2}}-\frac {b c \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2 \sqrt {d+c^2 d x^2}}-\frac {2 c^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \sqrt {d+c^2 d x^2}}-\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {2 c^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x}-\frac {4 b c^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right )}{3 \sqrt {d+c^2 d x^2}}+\frac {2 b^2 c^3 \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,e^{-2 \sinh ^{-1}(c x)}\right )}{3 \sqrt {d+c^2 d x^2}} \]

[Out]

-1/3*b^2*c^2*(c^2*x^2+1)/x/(c^2*d*x^2+d)^(1/2)-1/3*b*c*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/x^2/(c^2*d*x^2+d)^
(1/2)-2/3*c^3*(a+b*arcsinh(c*x))^2*(c^2*x^2+1)^(1/2)/(c^2*d*x^2+d)^(1/2)-4/3*b*c^3*(a+b*arcsinh(c*x))*ln(1-1/(
c*x+(c^2*x^2+1)^(1/2))^2)*(c^2*x^2+1)^(1/2)/(c^2*d*x^2+d)^(1/2)+2/3*b^2*c^3*polylog(2,1/(c*x+(c^2*x^2+1)^(1/2)
)^2)*(c^2*x^2+1)^(1/2)/(c^2*d*x^2+d)^(1/2)-1/3*(a+b*arcsinh(c*x))^2*(c^2*d*x^2+d)^(1/2)/d/x^3+2/3*c^2*(a+b*arc
sinh(c*x))^2*(c^2*d*x^2+d)^(1/2)/d/x

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Rubi [A]
time = 0.32, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {5809, 5800, 5775, 3797, 2221, 2317, 2438, 5776, 270} \begin {gather*} \frac {2 c^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x}-\frac {b c \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2 \sqrt {c^2 d x^2+d}}-\frac {\sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}-\frac {2 c^3 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \sqrt {c^2 d x^2+d}}-\frac {4 b c^3 \sqrt {c^2 x^2+1} \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 \sqrt {c^2 d x^2+d}}-\frac {b^2 c^2 \left (c^2 x^2+1\right )}{3 x \sqrt {c^2 d x^2+d}}+\frac {2 b^2 c^3 \sqrt {c^2 x^2+1} \text {Li}_2\left (e^{-2 \sinh ^{-1}(c x)}\right )}{3 \sqrt {c^2 d x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 270

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

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

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 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((
c + d*x)^(m + 1)/(d*(m + 1))), x] + Dist[2*I, Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5775

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Dist[1/b, Subst[Int[x^n*Coth[-a/b + x/b], x],
x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 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 5800

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[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, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]

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 {\left (a+b \sinh ^{-1}(c x)\right )^2}{x^4 \sqrt {d+c^2 d x^2}} \, dx &=-\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}-\frac {1}{3} \left (2 c^2\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x^2 \sqrt {d+c^2 d x^2}} \, dx+\frac {\left (2 b c \sqrt {1+c^2 x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x^3} \, dx}{3 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b c \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2 \sqrt {d+c^2 d x^2}}-\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {2 c^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x}+\frac {\left (b^2 c^2 \sqrt {1+c^2 x^2}\right ) \int \frac {1}{x^2 \sqrt {1+c^2 x^2}} \, dx}{3 \sqrt {d+c^2 d x^2}}-\frac {\left (4 b c^3 \sqrt {1+c^2 x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x} \, dx}{3 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b^2 c^2 \left (1+c^2 x^2\right )}{3 x \sqrt {d+c^2 d x^2}}-\frac {b c \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2 \sqrt {d+c^2 d x^2}}-\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {2 c^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x}-\frac {\left (4 b c^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int (a+b x) \coth (x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b^2 c^2 \left (1+c^2 x^2\right )}{3 x \sqrt {d+c^2 d x^2}}-\frac {b c \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2 \sqrt {d+c^2 d x^2}}+\frac {2 c^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \sqrt {d+c^2 d x^2}}-\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {2 c^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x}+\frac {\left (8 b c^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \frac {e^{2 x} (a+b x)}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(c x)\right )}{3 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b^2 c^2 \left (1+c^2 x^2\right )}{3 x \sqrt {d+c^2 d x^2}}-\frac {b c \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2 \sqrt {d+c^2 d x^2}}+\frac {2 c^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \sqrt {d+c^2 d x^2}}-\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {2 c^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x}-\frac {4 b c^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )}{3 \sqrt {d+c^2 d x^2}}+\frac {\left (4 b^2 c^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b^2 c^2 \left (1+c^2 x^2\right )}{3 x \sqrt {d+c^2 d x^2}}-\frac {b c \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2 \sqrt {d+c^2 d x^2}}+\frac {2 c^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \sqrt {d+c^2 d x^2}}-\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {2 c^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x}-\frac {4 b c^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )}{3 \sqrt {d+c^2 d x^2}}+\frac {\left (2 b^2 c^3 \sqrt {1+c^2 x^2}\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{3 \sqrt {d+c^2 d x^2}}\\ &=-\frac {b^2 c^2 \left (1+c^2 x^2\right )}{3 x \sqrt {d+c^2 d x^2}}-\frac {b c \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^2 \sqrt {d+c^2 d x^2}}+\frac {2 c^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 \sqrt {d+c^2 d x^2}}-\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x^3}+\frac {2 c^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d x}-\frac {4 b c^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )}{3 \sqrt {d+c^2 d x^2}}-\frac {2 b^2 c^3 \sqrt {1+c^2 x^2} \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{3 \sqrt {d+c^2 d x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.48, size = 278, normalized size = 0.93 \begin {gather*} \frac {-a^2+a^2 c^2 x^2-b^2 c^2 x^2+2 a^2 c^4 x^4-b^2 c^4 x^4-a b c x \sqrt {1+c^2 x^2}+b^2 \left (-1+c^2 x^2+2 c^4 x^4-2 c^3 x^3 \sqrt {1+c^2 x^2}\right ) \sinh ^{-1}(c x)^2-b \sinh ^{-1}(c x) \left (b c x \sqrt {1+c^2 x^2}-2 a \left (-1+c^2 x^2+2 c^4 x^4\right )+4 b c^3 x^3 \sqrt {1+c^2 x^2} \log \left (1-e^{-2 \sinh ^{-1}(c x)}\right )\right )-4 a b c^3 x^3 \sqrt {1+c^2 x^2} \log (c x)+2 b^2 c^3 x^3 \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,e^{-2 \sinh ^{-1}(c x)}\right )}{3 x^3 \sqrt {d+c^2 d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2145\) vs. \(2(281)=562\).
time = 3.98, size = 2146, normalized size = 7.18

method	result	size
default	\(\text {Expression too large to display}\)	\(2146\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))^2/x^4/(c^2*d*x^2+d)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/3*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/x^3/d*arcsinh(c*x)-4/3*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^
2+1)^(1/2)/d*ln((c*x+(c^2*x^2+1)^(1/2))^2-1)*c^3-4/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x^3/d*a
rcsinh(c*x)*(c^2*x^2+1)*c^6-b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d*arcsinh(c*x)*(c^2*x^2+1)^(1/2)
*c^3+1/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x/d*arcsinh(c*x)^2*c^4-4/3*b^2*(d*(c^2*x^2+1))^(1/2
)/(c^2*x^2+1)^(1/2)/d*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))*c^3+4/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*
c^2*x^2-1)*x^5/d*arcsinh(c*x)*c^8-2/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x/d*c^4+1/3*b^2*(d*(c^
2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/x/d*c^2+1/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/x^3/d*ar
csinh(c*x)^2-4/3*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d*polylog(2,-c*x-(c^2*x^2+1)^(1/2))*c^3-1/3*b^2*(
d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d*(c^2*x^2+1)^(1/2)*c^3+4/3*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)
^(1/2)/d*arcsinh(c*x)^2*c^3-4/3*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d*polylog(2,c*x+(c^2*x^2+1)^(1/2))
*c^3+2/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x^5/d*c^8-1/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+
2*c^2*x^2-1)*x^3/d*c^6-4/3*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/
2))*c^3+2/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d*arcsinh(c*x)^2*(c^2*x^2+1)^(1/2)*c^3+b^2*(d*(c
^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x^2/d*(c^2*x^2+1)^(1/2)*c^5-2/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+
2*c^2*x^2-1)*x^3/d*(c^2*x^2+1)*c^6-2*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x^2/d*arcsinh(c*x)^2*(c
^2*x^2+1)^(1/2)*c^5+2/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x/d*arcsinh(c*x)*(c^2*x^2+1)*c^4+1/3
*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/x^2/d*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c-4/3*a*b*(d*(c^2*x^2+
1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x^3/d*(c^2*x^2+1)*c^6+4*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x^
3/d*arcsinh(c*x)*c^6+2/3*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x/d*(c^2*x^2+1)*c^4+2/3*a*b*(d*(c^2
*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x/d*arcsinh(c*x)*c^4+2*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1
)*x^3/d*arcsinh(c*x)^2*c^6-4*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x^2/d*arcsinh(c*x)*(c^2*x^2+1)^
(1/2)*c^5+4/3*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/d*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^3-8/3*a*b*(
d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)/x/d*arcsinh(c*x)*c^2+1/3*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c
^2*x^2-1)/x^2/d*c*(c^2*x^2+1)^(1/2)-2/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x/d*arcsinh(c*x)*c^4
+2/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x^3/d*arcsinh(c*x)*c^6-4/3*b^2*(d*(c^2*x^2+1))^(1/2)/(3
*c^4*x^4+2*c^2*x^2-1)/x/d*arcsinh(c*x)^2*c^2+8/3*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/d*arcsinh(c*x)*c^
3+4/3*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x^5/d*c^8+2/3*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c
^2*x^2-1)*x^3/d*c^6-2/3*a*b*(d*(c^2*x^2+1))^(1/2)/(3*c^4*x^4+2*c^2*x^2-1)*x/d*c^4-a*b*(d*(c^2*x^2+1))^(1/2)/(3
*c^4*x^4+2*c^2*x^2-1)/d*(c^2*x^2+1)^(1/2)*c^3+a^2*(-1/3/d/x^3*(c^2*d*x^2+d)^(1/2)+2/3*c^2/d/x*(c^2*d*x^2+d)^(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))^2/x^4/(c^2*d*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

-1/3*(4*c^2*log(x)/sqrt(d) + 1/(sqrt(d)*x^2))*a*b*c + 2/3*a*b*(2*sqrt(c^2*d*x^2 + d)*c^2/(d*x) - sqrt(c^2*d*x^
2 + d)/(d*x^3))*arcsinh(c*x) + 1/3*a^2*(2*sqrt(c^2*d*x^2 + d)*c^2/(d*x) - sqrt(c^2*d*x^2 + d)/(d*x^3)) + b^2*i
ntegrate(log(c*x + sqrt(c^2*x^2 + 1))^2/(sqrt(c^2*d*x^2 + d)*x^4), 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))^2/x^4/(c^2*d*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))^2/x^4/(c^2*d*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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