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

Optimal. Leaf size=249 \[ \frac {e \sin \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}(c x)}{b}\right )}{4 b^2 c^3}-\frac {3 e \sin \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{4 b^2 c^3}-\frac {e \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c x)}{b}\right )}{4 b^2 c^3}+\frac {3 e \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 \left (a+b \sin ^{-1}(c x)\right )}{b}\right )}{4 b^2 c^3}+\frac {d \sin \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}(c x)}{b}\right )}{b^2 c}-\frac {d \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c x)}{b}\right )}{b^2 c}-\frac {d \sqrt {1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {e x^2 \sqrt {1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

[Out]

-d*cos(a/b)*Si((a+b*arcsin(c*x))/b)/b^2/c-1/4*e*cos(a/b)*Si((a+b*arcsin(c*x))/b)/b^2/c^3+3/4*e*cos(3*a/b)*Si(3
*(a+b*arcsin(c*x))/b)/b^2/c^3+d*Ci((a+b*arcsin(c*x))/b)*sin(a/b)/b^2/c+1/4*e*Ci((a+b*arcsin(c*x))/b)*sin(a/b)/
b^2/c^3-3/4*e*Ci(3*(a+b*arcsin(c*x))/b)*sin(3*a/b)/b^2/c^3-d*(-c^2*x^2+1)^(1/2)/b/c/(a+b*arcsin(c*x))-e*x^2*(-
c^2*x^2+1)^(1/2)/b/c/(a+b*arcsin(c*x))

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Rubi [A]  time = 0.42, antiderivative size = 241, 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 = {4667, 4621, 4723, 3303, 3299, 3302, 4631} \[ \frac {e \sin \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{4 b^2 c^3}-\frac {3 e \sin \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b^2 c^3}-\frac {e \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{4 b^2 c^3}+\frac {3 e \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b^2 c^3}+\frac {d \sin \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{b^2 c}-\frac {d \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{b^2 c}-\frac {d \sqrt {1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {e x^2 \sqrt {1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*Sqrt[1 - c^2*x^2])/(b*c*(a + b*ArcSin[c*x]))) - (e*x^2*Sqrt[1 - c^2*x^2])/(b*c*(a + b*ArcSin[c*x])) + (d*
CosIntegral[a/b + ArcSin[c*x]]*Sin[a/b])/(b^2*c) + (e*CosIntegral[a/b + ArcSin[c*x]]*Sin[a/b])/(4*b^2*c^3) - (
3*e*CosIntegral[(3*a)/b + 3*ArcSin[c*x]]*Sin[(3*a)/b])/(4*b^2*c^3) - (d*Cos[a/b]*SinIntegral[a/b + ArcSin[c*x]
])/(b^2*c) - (e*Cos[a/b]*SinIntegral[a/b + ArcSin[c*x]])/(4*b^2*c^3) + (3*e*Cos[(3*a)/b]*SinIntegral[(3*a)/b +
 3*ArcSin[c*x]])/(4*b^2*c^3)

Rule 3299

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

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 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 4621

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

Rule 4631

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 - c^2*x^2]*(a + b*ArcSin
[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)
, Sin[x]^(m - 1)*(m - (m + 1)*Sin[x]^2), x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && G
eQ[n, -2] && LtQ[n, -1]

Rule 4667

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

Rule 4723

Int[((a_.) + ArcSin[(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*Sin[x]^m*Cos[x]^(2*p + 1), x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n},
x] && EqQ[c^2*d + e, 0] && 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 \sin ^{-1}(c x)\right )^2} \, dx &=\int \left (\frac {d}{\left (a+b \sin ^{-1}(c x)\right )^2}+\frac {e x^2}{\left (a+b \sin ^{-1}(c x)\right )^2}\right ) \, dx\\ &=d \int \frac {1}{\left (a+b \sin ^{-1}(c x)\right )^2} \, dx+e \int \frac {x^2}{\left (a+b \sin ^{-1}(c x)\right )^2} \, dx\\ &=-\frac {d \sqrt {1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {e x^2 \sqrt {1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {(c d) \int \frac {x}{\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )} \, dx}{b}+\frac {e \operatorname {Subst}\left (\int \left (-\frac {\sin (x)}{4 (a+b x)}+\frac {3 \sin (3 x)}{4 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{b c^3}\\ &=-\frac {d \sqrt {1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {e x^2 \sqrt {1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {d \operatorname {Subst}\left (\int \frac {\sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c}-\frac {e \operatorname {Subst}\left (\int \frac {\sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^3}+\frac {(3 e) \operatorname {Subst}\left (\int \frac {\sin (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^3}\\ &=-\frac {d \sqrt {1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {e x^2 \sqrt {1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {\left (d \cos \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c}-\frac {\left (e \cos \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^3}+\frac {\left (3 e \cos \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^3}+\frac {\left (d \sin \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{b c}+\frac {\left (e \sin \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^3}-\frac {\left (3 e \sin \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c x)\right )}{4 b c^3}\\ &=-\frac {d \sqrt {1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )}-\frac {e x^2 \sqrt {1-c^2 x^2}}{b c \left (a+b \sin ^{-1}(c x)\right )}+\frac {d \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c x)\right ) \sin \left (\frac {a}{b}\right )}{b^2 c}+\frac {e \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c x)\right ) \sin \left (\frac {a}{b}\right )}{4 b^2 c^3}-\frac {3 e \text {Ci}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right ) \sin \left (\frac {3 a}{b}\right )}{4 b^2 c^3}-\frac {d \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{b^2 c}-\frac {e \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )}{4 b^2 c^3}+\frac {3 e \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c x)\right )}{4 b^2 c^3}\\ \end {align*}

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Mathematica [A]  time = 1.04, size = 191, normalized size = 0.77 \[ -\frac {-\sin \left (\frac {a}{b}\right ) \left (4 c^2 d+e\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )+4 c^2 d \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )+\frac {4 b c^2 d \sqrt {1-c^2 x^2}}{a+b \sin ^{-1}(c x)}+\frac {4 b c^2 e x^2 \sqrt {1-c^2 x^2}}{a+b \sin ^{-1}(c x)}+3 e \sin \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )+e \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c x)\right )-3 e \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c x)\right )\right )}{4 b^2 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((4*b*c^2*d*Sqrt[1 - c^2*x^2])/(a + b*ArcSin[c*x]) + (4*b*c^2*e*x^2*Sqrt[1 - c^2*x^2])/(a + b*ArcSin[c*x]
) - (4*c^2*d + e)*CosIntegral[a/b + ArcSin[c*x]]*Sin[a/b] + 3*e*CosIntegral[3*(a/b + ArcSin[c*x])]*Sin[(3*a)/b
] + 4*c^2*d*Cos[a/b]*SinIntegral[a/b + ArcSin[c*x]] + e*Cos[a/b]*SinIntegral[a/b + ArcSin[c*x]] - 3*e*Cos[(3*a
)/b]*SinIntegral[3*(a/b + ArcSin[c*x])])/(b^2*c^3)

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e x^{2} + d}{b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.82, size = 905, normalized size = 3.63 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*c^2*d*arcsin(c*x)*cos_integral(a/b + arcsin(c*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 3*b*arcsin(c*
x)*cos(a/b)^2*cos_integral(3*a/b + 3*arcsin(c*x))*e*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 3*b*arcsin(c*
x)*cos(a/b)^3*e*sin_integral(3*a/b + 3*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - b*c^2*d*arcsin(c*x)*co
s(a/b)*sin_integral(a/b + arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + a*c^2*d*cos_integral(a/b + arcsin(c
*x))*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 3*a*cos(a/b)^2*cos_integral(3*a/b + 3*arcsin(c*x))*e*sin(a/b
)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 3*a*cos(a/b)^3*e*sin_integral(3*a/b + 3*arcsin(c*x))/(b^3*c^3*arcsin(c*x
) + a*b^2*c^3) - a*c^2*d*cos(a/b)*sin_integral(a/b + arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 3/4*b*ar
csin(c*x)*cos_integral(3*a/b + 3*arcsin(c*x))*e*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 1/4*b*arcsin(c*x)
*cos_integral(a/b + arcsin(c*x))*e*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 9/4*b*arcsin(c*x)*cos(a/b)*e*s
in_integral(3*a/b + 3*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 1/4*b*arcsin(c*x)*cos(a/b)*e*sin_integr
al(a/b + arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - sqrt(-c^2*x^2 + 1)*b*c^2*d/(b^3*c^3*arcsin(c*x) + a*
b^2*c^3) + 3/4*a*cos_integral(3*a/b + 3*arcsin(c*x))*e*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) + 1/4*a*cos_
integral(a/b + arcsin(c*x))*e*sin(a/b)/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 9/4*a*cos(a/b)*e*sin_integral(3*a/b
 + 3*arcsin(c*x))/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - 1/4*a*cos(a/b)*e*sin_integral(a/b + arcsin(c*x))/(b^3*c^
3*arcsin(c*x) + a*b^2*c^3) + (-c^2*x^2 + 1)^(3/2)*b*e/(b^3*c^3*arcsin(c*x) + a*b^2*c^3) - sqrt(-c^2*x^2 + 1)*b
*e/(b^3*c^3*arcsin(c*x) + a*b^2*c^3)

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maple [A]  time = 0.19, size = 367, normalized size = 1.47 \[ \frac {-4 \arcsin \left (c x \right ) \Si \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b \,c^{2} d +4 \arcsin \left (c x \right ) \Ci \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b \,c^{2} d -4 \Si \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a \,c^{2} d +4 \Ci \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a \,c^{2} d +3 \arcsin \left (c x \right ) \Si \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) b e -3 \arcsin \left (c x \right ) \Ci \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) b e -\arcsin \left (c x \right ) \Si \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b e +\arcsin \left (c x \right ) \Ci \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b e -4 \sqrt {-c^{2} x^{2}+1}\, b \,c^{2} d +3 \Si \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) a e -3 \Ci \left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) a e -\Si \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a e +\Ci \left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a e +\cos \left (3 \arcsin \left (c x \right )\right ) b e -\sqrt {-c^{2} x^{2}+1}\, b e}{4 c^{3} \left (a +b \arcsin \left (c x \right )\right ) b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/c^3*(-4*arcsin(c*x)*Si(arcsin(c*x)+a/b)*cos(a/b)*b*c^2*d+4*arcsin(c*x)*Ci(arcsin(c*x)+a/b)*sin(a/b)*b*c^2*
d-4*Si(arcsin(c*x)+a/b)*cos(a/b)*a*c^2*d+4*Ci(arcsin(c*x)+a/b)*sin(a/b)*a*c^2*d+3*arcsin(c*x)*Si(3*arcsin(c*x)
+3*a/b)*cos(3*a/b)*b*e-3*arcsin(c*x)*Ci(3*arcsin(c*x)+3*a/b)*sin(3*a/b)*b*e-arcsin(c*x)*Si(arcsin(c*x)+a/b)*co
s(a/b)*b*e+arcsin(c*x)*Ci(arcsin(c*x)+a/b)*sin(a/b)*b*e-4*(-c^2*x^2+1)^(1/2)*b*c^2*d+3*Si(3*arcsin(c*x)+3*a/b)
*cos(3*a/b)*a*e-3*Ci(3*arcsin(c*x)+3*a/b)*sin(3*a/b)*a*e-Si(arcsin(c*x)+a/b)*cos(a/b)*a*e+Ci(arcsin(c*x)+a/b)*
sin(a/b)*a*e+cos(3*arcsin(c*x))*b*e-(-c^2*x^2+1)^(1/2)*b*e)/(a+b*arcsin(c*x))/b^2

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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