3.625 \(\int \frac {d+e x^2}{(a+b \sinh ^{-1}(c x))^2} \, dx\)

Optimal. Leaf size=247 \[ \frac {e \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{4 b^2 c^3}-\frac {3 e \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b^2 c^3}-\frac {e \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{4 b^2 c^3}+\frac {3 e \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b^2 c^3}-\frac {d \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b^2 c}+\frac {d \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b^2 c}-\frac {d \sqrt {c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {e x^2 \sqrt {c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

[Out]

d*cosh(a/b)*Shi((a+b*arcsinh(c*x))/b)/b^2/c-1/4*e*cosh(a/b)*Shi((a+b*arcsinh(c*x))/b)/b^2/c^3+3/4*e*cosh(3*a/b
)*Shi(3*(a+b*arcsinh(c*x))/b)/b^2/c^3-d*Chi((a+b*arcsinh(c*x))/b)*sinh(a/b)/b^2/c+1/4*e*Chi((a+b*arcsinh(c*x))
/b)*sinh(a/b)/b^2/c^3-3/4*e*Chi(3*(a+b*arcsinh(c*x))/b)*sinh(3*a/b)/b^2/c^3-d*(c^2*x^2+1)^(1/2)/b/c/(a+b*arcsi
nh(c*x))-e*x^2*(c^2*x^2+1)^(1/2)/b/c/(a+b*arcsinh(c*x))

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Rubi [A]  time = 0.48, antiderivative size = 239, normalized size of antiderivative = 0.97, number of steps used = 15, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {5706, 5655, 5779, 3303, 3298, 3301, 5665} \[ \frac {e \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{4 b^2 c^3}-\frac {3 e \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b^2 c^3}-\frac {e \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{4 b^2 c^3}+\frac {3 e \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b^2 c^3}-\frac {d \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{b^2 c}+\frac {d \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{b^2 c}-\frac {d \sqrt {c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {e x^2 \sqrt {c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*Sqrt[1 + c^2*x^2])/(b*c*(a + b*ArcSinh[c*x]))) - (e*x^2*Sqrt[1 + c^2*x^2])/(b*c*(a + b*ArcSinh[c*x])) - (
d*CoshIntegral[a/b + ArcSinh[c*x]]*Sinh[a/b])/(b^2*c) + (e*CoshIntegral[a/b + ArcSinh[c*x]]*Sinh[a/b])/(4*b^2*
c^3) - (3*e*CoshIntegral[(3*a)/b + 3*ArcSinh[c*x]]*Sinh[(3*a)/b])/(4*b^2*c^3) + (d*Cosh[a/b]*SinhIntegral[a/b
+ ArcSinh[c*x]])/(b^2*c) - (e*Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]])/(4*b^2*c^3) + (3*e*Cosh[(3*a)/b]*Sin
hIntegral[(3*a)/b + 3*ArcSinh[c*x]])/(4*b^2*c^3)

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 5655

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^(n + 1
))/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcSinh[c*x])^(n + 1))/Sqrt[1 + c^2*x^2], x], x] /; F
reeQ[{a, b, c}, x] && LtQ[n, -1]

Rule 5665

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 + c^2*x^2]*(a + b*ArcSi
nh[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0]
&& GeQ[n, -2] && LtQ[n, -1]

Rule 5706

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a
+ b*ArcSinh[c*x])^n, (d + e*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, n}, x] && NeQ[e, c^2*d] && IntegerQ[p] &&
(p > 0 || IGtQ[n, 0])

Rule 5779

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

Rubi steps

\begin {align*} \int \frac {d+e x^2}{\left (a+b \sinh ^{-1}(c x)\right )^2} \, dx &=\int \left (\frac {d}{\left (a+b \sinh ^{-1}(c x)\right )^2}+\frac {e x^2}{\left (a+b \sinh ^{-1}(c x)\right )^2}\right ) \, dx\\ &=d \int \frac {1}{\left (a+b \sinh ^{-1}(c x)\right )^2} \, dx+e \int \frac {x^2}{\left (a+b \sinh ^{-1}(c x)\right )^2} \, dx\\ &=-\frac {d \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {e x^2 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {(c d) \int \frac {x}{\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )} \, dx}{b}+\frac {e \operatorname {Subst}\left (\int \left (-\frac {\sinh (x)}{4 (a+b x)}+\frac {3 \sinh (3 x)}{4 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}\\ &=-\frac {d \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {e x^2 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {d \operatorname {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c}-\frac {e \operatorname {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^3}+\frac {(3 e) \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^3}\\ &=-\frac {d \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {e x^2 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {\left (d \cosh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c}-\frac {\left (e \cosh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^3}+\frac {\left (3 e \cosh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^3}-\frac {\left (d \sinh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c}+\frac {\left (e \sinh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^3}-\frac {\left (3 e \sinh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^3}\\ &=-\frac {d \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {e x^2 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {d \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )}{b^2 c}+\frac {e \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )}{4 b^2 c^3}-\frac {3 e \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {3 a}{b}\right )}{4 b^2 c^3}+\frac {d \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{b^2 c}-\frac {e \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{4 b^2 c^3}+\frac {3 e \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{4 b^2 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.91, size = 190, normalized size = 0.77 \[ -\frac {\sinh \left (\frac {a}{b}\right ) \left (4 c^2 d-e\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )-4 c^2 d \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )+\frac {4 b c^2 d \sqrt {c^2 x^2+1}}{a+b \sinh ^{-1}(c x)}+\frac {4 b c^2 e x^2 \sqrt {c^2 x^2+1}}{a+b \sinh ^{-1}(c x)}+3 e \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+e \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )-3 e \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )}{4 b^2 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((4*b*c^2*d*Sqrt[1 + c^2*x^2])/(a + b*ArcSinh[c*x]) + (4*b*c^2*e*x^2*Sqrt[1 + c^2*x^2])/(a + b*ArcSinh[c*
x]) + (4*c^2*d - e)*CoshIntegral[a/b + ArcSinh[c*x]]*Sinh[a/b] + 3*e*CoshIntegral[3*(a/b + ArcSinh[c*x])]*Sinh
[(3*a)/b] - 4*c^2*d*Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]] + e*Cosh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]]
- 3*e*Cosh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c*x])])/(b^2*c^3)

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e x^{2} + d}{b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e x^{2} + d}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.25, size = 438, normalized size = 1.77 \[ \frac {\frac {\left (4 c^{3} x^{3}-4 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+3 c x -\sqrt {c^{2} x^{2}+1}\right ) e}{8 c^{2} b \left (a +b \arcsinh \left (c x \right )\right )}+\frac {3 e \,{\mathrm e}^{\frac {3 a}{b}} \Ei \left (1, 3 \arcsinh \left (c x \right )+\frac {3 a}{b}\right )}{8 c^{2} b^{2}}-\frac {e \left (4 c^{3} x^{3}+3 c x +4 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+\sqrt {c^{2} x^{2}+1}\right )}{8 b \,c^{2} \left (a +b \arcsinh \left (c x \right )\right )}-\frac {3 e \,{\mathrm e}^{-\frac {3 a}{b}} \Ei \left (1, -3 \arcsinh \left (c x \right )-\frac {3 a}{b}\right )}{8 b^{2} c^{2}}+\frac {\left (c x -\sqrt {c^{2} x^{2}+1}\right ) d}{2 b \left (a +b \arcsinh \left (c x \right )\right )}+\frac {d \,{\mathrm e}^{\frac {a}{b}} \Ei \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right )}{2 b^{2}}-\frac {\left (c x -\sqrt {c^{2} x^{2}+1}\right ) e}{8 c^{2} b \left (a +b \arcsinh \left (c x \right )\right )}-\frac {e \,{\mathrm e}^{\frac {a}{b}} \Ei \left (1, \arcsinh \left (c x \right )+\frac {a}{b}\right )}{8 c^{2} b^{2}}-\frac {d \left (c x +\sqrt {c^{2} x^{2}+1}\right )}{2 b \left (a +b \arcsinh \left (c x \right )\right )}-\frac {d \,{\mathrm e}^{-\frac {a}{b}} \Ei \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right )}{2 b^{2}}+\frac {e \left (c x +\sqrt {c^{2} x^{2}+1}\right )}{8 c^{2} b \left (a +b \arcsinh \left (c x \right )\right )}+\frac {e \,{\mathrm e}^{-\frac {a}{b}} \Ei \left (1, -\arcsinh \left (c x \right )-\frac {a}{b}\right )}{8 c^{2} b^{2}}}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(1/8*(4*c^3*x^3-4*c^2*x^2*(c^2*x^2+1)^(1/2)+3*c*x-(c^2*x^2+1)^(1/2))*e/c^2/b/(a+b*arcsinh(c*x))+3/8*e/c^2/
b^2*exp(3*a/b)*Ei(1,3*arcsinh(c*x)+3*a/b)-1/8/b*e/c^2*(4*c^3*x^3+3*c*x+4*c^2*x^2*(c^2*x^2+1)^(1/2)+(c^2*x^2+1)
^(1/2))/(a+b*arcsinh(c*x))-3/8/b^2*e/c^2*exp(-3*a/b)*Ei(1,-3*arcsinh(c*x)-3*a/b)+1/2*(c*x-(c^2*x^2+1)^(1/2))*d
/b/(a+b*arcsinh(c*x))+1/2*d/b^2*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)-1/8*(c*x-(c^2*x^2+1)^(1/2))*e/c^2/b/(a+b*arcsi
nh(c*x))-1/8/c^2*e/b^2*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)-1/2/b*d*(c*x+(c^2*x^2+1)^(1/2))/(a+b*arcsinh(c*x))-1/2/
b^2*d*exp(-a/b)*Ei(1,-arcsinh(c*x)-a/b)+1/8/c^2/b*e*(c*x+(c^2*x^2+1)^(1/2))/(a+b*arcsinh(c*x))+1/8/c^2/b^2*e*e
xp(-a/b)*Ei(1,-arcsinh(c*x)-a/b))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {c^{3} e x^{5} + {\left (c^{3} d + c e\right )} x^{3} + c d x + {\left (c^{2} e x^{4} + {\left (c^{2} d + e\right )} x^{2} + d\right )} \sqrt {c^{2} x^{2} + 1}}{a b c^{3} x^{2} + \sqrt {c^{2} x^{2} + 1} a b c^{2} x + a b c + {\left (b^{2} c^{3} x^{2} + \sqrt {c^{2} x^{2} + 1} b^{2} c^{2} x + b^{2} c\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )} + \int \frac {3 \, c^{5} e x^{6} + {\left (c^{5} d + 6 \, c^{3} e\right )} x^{4} + {\left (2 \, c^{3} d + 3 \, c e\right )} x^{2} + {\left (3 \, c^{3} e x^{4} + {\left (c^{3} d + c e\right )} x^{2} - c d\right )} {\left (c^{2} x^{2} + 1\right )} + c d + {\left (6 \, c^{4} e x^{5} + {\left (2 \, c^{4} d + 7 \, c^{2} e\right )} x^{3} + {\left (c^{2} d + 2 \, e\right )} x\right )} \sqrt {c^{2} x^{2} + 1}}{a b c^{5} x^{4} + {\left (c^{2} x^{2} + 1\right )} a b c^{3} x^{2} + 2 \, a b c^{3} x^{2} + a b c + {\left (b^{2} c^{5} x^{4} + {\left (c^{2} x^{2} + 1\right )} b^{2} c^{3} x^{2} + 2 \, b^{2} c^{3} x^{2} + b^{2} c + 2 \, {\left (b^{2} c^{4} x^{3} + b^{2} c^{2} x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (a b c^{4} x^{3} + a b c^{2} x\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c^3*e*x^5 + (c^3*d + c*e)*x^3 + c*d*x + (c^2*e*x^4 + (c^2*d + e)*x^2 + d)*sqrt(c^2*x^2 + 1))/(a*b*c^3*x^2 +
sqrt(c^2*x^2 + 1)*a*b*c^2*x + a*b*c + (b^2*c^3*x^2 + sqrt(c^2*x^2 + 1)*b^2*c^2*x + b^2*c)*log(c*x + sqrt(c^2*x
^2 + 1))) + integrate((3*c^5*e*x^6 + (c^5*d + 6*c^3*e)*x^4 + (2*c^3*d + 3*c*e)*x^2 + (3*c^3*e*x^4 + (c^3*d + c
*e)*x^2 - c*d)*(c^2*x^2 + 1) + c*d + (6*c^4*e*x^5 + (2*c^4*d + 7*c^2*e)*x^3 + (c^2*d + 2*e)*x)*sqrt(c^2*x^2 +
1))/(a*b*c^5*x^4 + (c^2*x^2 + 1)*a*b*c^3*x^2 + 2*a*b*c^3*x^2 + a*b*c + (b^2*c^5*x^4 + (c^2*x^2 + 1)*b^2*c^3*x^
2 + 2*b^2*c^3*x^2 + b^2*c + 2*(b^2*c^4*x^3 + b^2*c^2*x)*sqrt(c^2*x^2 + 1))*log(c*x + sqrt(c^2*x^2 + 1)) + 2*(a
*b*c^4*x^3 + a*b*c^2*x)*sqrt(c^2*x^2 + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {e\,x^2+d}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {d + e x^{2}}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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