3.1.48 \(\int \frac {x^2 (a+b \text {ArcSin}(c x))}{(d-c^2 d x^2)^3} \, dx\) [48]

Optimal. Leaf size=202 \[ -\frac {b}{12 c^3 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {b}{8 c^3 d^3 \sqrt {1-c^2 x^2}}+\frac {x (a+b \text {ArcSin}(c x))}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {x (a+b \text {ArcSin}(c x))}{8 c^2 d^3 \left (1-c^2 x^2\right )}+\frac {i (a+b \text {ArcSin}(c x)) \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )}{4 c^3 d^3}-\frac {i b \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{8 c^3 d^3}+\frac {i b \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{8 c^3 d^3} \]

[Out]

-1/12*b/c^3/d^3/(-c^2*x^2+1)^(3/2)+1/4*x*(a+b*arcsin(c*x))/c^2/d^3/(-c^2*x^2+1)^2-1/8*x*(a+b*arcsin(c*x))/c^2/
d^3/(-c^2*x^2+1)+1/4*I*(a+b*arcsin(c*x))*arctan(I*c*x+(-c^2*x^2+1)^(1/2))/c^3/d^3-1/8*I*b*polylog(2,-I*(I*c*x+
(-c^2*x^2+1)^(1/2)))/c^3/d^3+1/8*I*b*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c^3/d^3+1/8*b/c^3/d^3/(-c^2*x^2+1
)^(1/2)

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Rubi [A]
time = 0.12, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4791, 4747, 4749, 4266, 2317, 2438, 267} \begin {gather*} \frac {i \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{4 c^3 d^3}-\frac {x (a+b \text {ArcSin}(c x))}{8 c^2 d^3 \left (1-c^2 x^2\right )}+\frac {x (a+b \text {ArcSin}(c x))}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {i b \text {Li}_2\left (-i e^{i \text {ArcSin}(c x)}\right )}{8 c^3 d^3}+\frac {i b \text {Li}_2\left (i e^{i \text {ArcSin}(c x)}\right )}{8 c^3 d^3}+\frac {b}{8 c^3 d^3 \sqrt {1-c^2 x^2}}-\frac {b}{12 c^3 d^3 \left (1-c^2 x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*b/(c^3*d^3*(1 - c^2*x^2)^(3/2)) + b/(8*c^3*d^3*Sqrt[1 - c^2*x^2]) + (x*(a + b*ArcSin[c*x]))/(4*c^2*d^3*(
1 - c^2*x^2)^2) - (x*(a + b*ArcSin[c*x]))/(8*c^2*d^3*(1 - c^2*x^2)) + ((I/4)*(a + b*ArcSin[c*x])*ArcTan[E^(I*A
rcSin[c*x])])/(c^3*d^3) - ((I/8)*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c^3*d^3) + ((I/8)*b*PolyLog[2, I*E^(I*
ArcSin[c*x])])/(c^3*d^3)

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && 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 4266

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

Rule 4747

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*(d + e*x^2)^(p
+ 1)*((a + b*ArcSin[c*x])^n/(2*d*(p + 1))), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a +
b*ArcSin[c*x])^n, x], x] + Dist[b*c*(n/(2*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p], Int[x*(1 - c^2*x^2)^(
p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] &
& LtQ[p, -1] && NeQ[p, -3/2]

Rule 4749

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/(c*d), Subst[Int[(a +
b*x)^n*Sec[x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]

Rule 4791

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

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^3} \, dx &=\frac {x \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {b \int \frac {x}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{4 c d^3}-\frac {\int \frac {a+b \sin ^{-1}(c x)}{\left (d-c^2 d x^2\right )^2} \, dx}{4 c^2 d}\\ &=-\frac {b}{12 c^3 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{8 c^2 d^3 \left (1-c^2 x^2\right )}+\frac {b \int \frac {x}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{8 c d^3}-\frac {\int \frac {a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx}{8 c^2 d^2}\\ &=-\frac {b}{12 c^3 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {b}{8 c^3 d^3 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{8 c^2 d^3 \left (1-c^2 x^2\right )}-\frac {\text {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{8 c^3 d^3}\\ &=-\frac {b}{12 c^3 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {b}{8 c^3 d^3 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{8 c^2 d^3 \left (1-c^2 x^2\right )}+\frac {i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c^3 d^3}+\frac {b \text {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{8 c^3 d^3}-\frac {b \text {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{8 c^3 d^3}\\ &=-\frac {b}{12 c^3 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {b}{8 c^3 d^3 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{8 c^2 d^3 \left (1-c^2 x^2\right )}+\frac {i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c^3 d^3}-\frac {(i b) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{8 c^3 d^3}+\frac {(i b) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{8 c^3 d^3}\\ &=-\frac {b}{12 c^3 d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac {b}{8 c^3 d^3 \sqrt {1-c^2 x^2}}+\frac {x \left (a+b \sin ^{-1}(c x)\right )}{4 c^2 d^3 \left (1-c^2 x^2\right )^2}-\frac {x \left (a+b \sin ^{-1}(c x)\right )}{8 c^2 d^3 \left (1-c^2 x^2\right )}+\frac {i \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c^3 d^3}-\frac {i b \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{8 c^3 d^3}+\frac {i b \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{8 c^3 d^3}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(445\) vs. \(2(202)=404\).
time = 0.46, size = 445, normalized size = 2.20 \begin {gather*} \frac {-\frac {2 b \sqrt {1-c^2 x^2}}{(-1+c x)^2}+\frac {b c x \sqrt {1-c^2 x^2}}{(-1+c x)^2}-\frac {3 b \sqrt {1-c^2 x^2}}{-1+c x}-\frac {2 b \sqrt {1-c^2 x^2}}{(1+c x)^2}-\frac {b c x \sqrt {1-c^2 x^2}}{(1+c x)^2}+\frac {3 b \sqrt {1-c^2 x^2}}{1+c x}+\frac {12 a c x}{\left (-1+c^2 x^2\right )^2}+\frac {6 a c x}{-1+c^2 x^2}+3 i b \pi \text {ArcSin}(c x)+\frac {3 b \text {ArcSin}(c x)}{(-1+c x)^2}+\frac {3 b \text {ArcSin}(c x)}{-1+c x}-\frac {3 b \text {ArcSin}(c x)}{(1+c x)^2}+\frac {3 b \text {ArcSin}(c x)}{1+c x}-3 b \pi \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-6 b \text {ArcSin}(c x) \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-3 b \pi \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+6 b \text {ArcSin}(c x) \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+3 a \log (1-c x)-3 a \log (1+c x)+3 b \pi \log \left (-\cos \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )+3 b \pi \log \left (\sin \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )-6 i b \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )+6 i b \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{48 c^3 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*b*Sqrt[1 - c^2*x^2])/(-1 + c*x)^2 + (b*c*x*Sqrt[1 - c^2*x^2])/(-1 + c*x)^2 - (3*b*Sqrt[1 - c^2*x^2])/(-1
+ c*x) - (2*b*Sqrt[1 - c^2*x^2])/(1 + c*x)^2 - (b*c*x*Sqrt[1 - c^2*x^2])/(1 + c*x)^2 + (3*b*Sqrt[1 - c^2*x^2])
/(1 + c*x) + (12*a*c*x)/(-1 + c^2*x^2)^2 + (6*a*c*x)/(-1 + c^2*x^2) + (3*I)*b*Pi*ArcSin[c*x] + (3*b*ArcSin[c*x
])/(-1 + c*x)^2 + (3*b*ArcSin[c*x])/(-1 + c*x) - (3*b*ArcSin[c*x])/(1 + c*x)^2 + (3*b*ArcSin[c*x])/(1 + c*x) -
3*b*Pi*Log[1 - I*E^(I*ArcSin[c*x])] - 6*b*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - 3*b*Pi*Log[1 + I*E^(I*Ar
cSin[c*x])] + 6*b*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] + 3*a*Log[1 - c*x] - 3*a*Log[1 + c*x] + 3*b*Pi*Log[
-Cos[(Pi + 2*ArcSin[c*x])/4]] + 3*b*Pi*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - (6*I)*b*PolyLog[2, (-I)*E^(I*ArcSin[
c*x])] + (6*I)*b*PolyLog[2, I*E^(I*ArcSin[c*x])])/(48*c^3*d^3)

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Maple [A]
time = 0.30, size = 358, normalized size = 1.77

method	result	size
derivativedivides	\(\frac {-\frac {a}{16 d^{3} \left (c x +1\right )^{2}}+\frac {a}{16 d^{3} \left (c x +1\right )}-\frac {a \ln \left (c x +1\right )}{16 d^{3}}+\frac {a}{16 d^{3} \left (c x -1\right )^{2}}+\frac {a}{16 d^{3} \left (c x -1\right )}+\frac {a \ln \left (c x -1\right )}{16 d^{3}}+\frac {b \arcsin \left (c x \right ) c^{3} x^{3}}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {b \,c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {b \arcsin \left (c x \right ) c x}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {b \sqrt {-c^{2} x^{2}+1}}{24 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 d^{3}}-\frac {b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 d^{3}}-\frac {i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 d^{3}}+\frac {i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 d^{3}}}{c^{3}}\)	\(358\)
default	\(\frac {-\frac {a}{16 d^{3} \left (c x +1\right )^{2}}+\frac {a}{16 d^{3} \left (c x +1\right )}-\frac {a \ln \left (c x +1\right )}{16 d^{3}}+\frac {a}{16 d^{3} \left (c x -1\right )^{2}}+\frac {a}{16 d^{3} \left (c x -1\right )}+\frac {a \ln \left (c x -1\right )}{16 d^{3}}+\frac {b \arcsin \left (c x \right ) c^{3} x^{3}}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}-\frac {b \,c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {b \arcsin \left (c x \right ) c x}{8 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {b \sqrt {-c^{2} x^{2}+1}}{24 d^{3} \left (c^{4} x^{4}-2 c^{2} x^{2}+1\right )}+\frac {b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 d^{3}}-\frac {b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 d^{3}}-\frac {i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 d^{3}}+\frac {i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{8 d^{3}}}{c^{3}}\)	\(358\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(-1/16*a/d^3/(c*x+1)^2+1/16*a/d^3/(c*x+1)-1/16*a/d^3*ln(c*x+1)+1/16*a/d^3/(c*x-1)^2+1/16*a/d^3/(c*x-1)+1
/16*a/d^3*ln(c*x-1)+1/8*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*c^3*x^3-1/8*b/d^3/(c^4*x^4-2*c^2*x^2+1)*c^2*x^
2*(-c^2*x^2+1)^(1/2)+1/8*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*c*x+1/24*b/d^3/(c^4*x^4-2*c^2*x^2+1)*(-c^2*x^
2+1)^(1/2)+1/8*b/d^3*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))-1/8*b/d^3*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*
x^2+1)^(1/2)))-1/8*I*b/d^3*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+1/8*I*b/d^3*dilog(1-I*(I*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(x^2*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^3,x, algorithm="maxima")

[Out]

1/16*a*(2*(c^2*x^3 + x)/(c^6*d^3*x^4 - 2*c^4*d^3*x^2 + c^2*d^3) - log(c*x + 1)/(c^3*d^3) + log(c*x - 1)/(c^3*d
^3)) - 1/16*((c^4*x^4 - 2*c^2*x^2 + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - (c^4*x^4 - 2*
c^2*x^2 + 1)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1) - 2*(c^3*x^3 + c*x)*arctan2(c*x, sqrt(c*
x + 1)*sqrt(-c*x + 1)) + 16*(c^7*d^3*x^4 - 2*c^5*d^3*x^2 + c^3*d^3)*integrate(-1/16*(2*c^3*x^3 + 2*c*x - (c^4*
x^4 - 2*c^2*x^2 + 1)*log(c*x + 1) + (c^4*x^4 - 2*c^2*x^2 + 1)*log(-c*x + 1))*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^8
*d^3*x^6 - 3*c^6*d^3*x^4 + 3*c^4*d^3*x^2 - c^2*d^3), x))*b/(c^7*d^3*x^4 - 2*c^5*d^3*x^2 + c^3*d^3)

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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(x^2*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^3,x, algorithm="fricas")

[Out]

integral(-(b*x^2*arcsin(c*x) + a*x^2)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {a x^{2}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac {b x^{2} \operatorname {asin}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx}{d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*asin(c*x))/(-c**2*d*x**2+d)**3,x)

[Out]

-(Integral(a*x**2/(c**6*x**6 - 3*c**4*x**4 + 3*c**2*x**2 - 1), x) + Integral(b*x**2*asin(c*x)/(c**6*x**6 - 3*c
**4*x**4 + 3*c**2*x**2 - 1), 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(x^2*(a+b*arcsin(c*x))/(-c^2*d*x^2+d)^3,x, algorithm="giac")

[Out]

integrate(-(b*arcsin(c*x) + a)*x^2/(c^2*d*x^2 - d)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^2\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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