3.2.64 \(\int \frac {(d-c^2 d x^2) (a+b \text {ArcSin}(c x))^2}{x^4} \, dx\) [164]

Optimal. Leaf size=176 \[ -\frac {b^2 c^2 d}{3 x}-\frac {b c d \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{3 x^2}+\frac {2 c^2 d (a+b \text {ArcSin}(c x))^2}{3 x}-\frac {d \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))^2}{3 x^3}+\frac {10}{3} b c^3 d (a+b \text {ArcSin}(c x)) \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right )-\frac {5}{3} i b^2 c^3 d \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )+\frac {5}{3} i b^2 c^3 d \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right ) \]

[Out]

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

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Rubi [A]
time = 0.27, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4785, 4723, 4803, 4268, 2317, 2438, 4781, 30} \begin {gather*} \frac {10}{3} b c^3 d \tanh ^{-1}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))-\frac {b c d \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))}{3 x^2}-\frac {d \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))^2}{3 x^3}+\frac {2 c^2 d (a+b \text {ArcSin}(c x))^2}{3 x}-\frac {5}{3} i b^2 c^3 d \text {Li}_2\left (-e^{i \text {ArcSin}(c x)}\right )+\frac {5}{3} i b^2 c^3 d \text {Li}_2\left (e^{i \text {ArcSin}(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*ArcSin[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*ArcSin[c*x]))/(3*x^2) + (2*c^2*d*(a + b*ArcSin[c*x])^2)/(
3*x) - (d*(1 - c^2*x^2)*(a + b*ArcSin[c*x])^2)/(3*x^3) + (10*b*c^3*d*(a + b*ArcSin[c*x])*ArcTanh[E^(I*ArcSin[c
*x])])/3 - ((5*I)/3)*b^2*c^3*d*PolyLog[2, -E^(I*ArcSin[c*x])] + ((5*I)/3)*b^2*c^3*d*PolyLog[2, E^(I*ArcSin[c*x
])]

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 4268

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

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSi
n[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSin[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 4781

Int[((a_.) + ArcSin[(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*ArcSin[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*ArcSin[c*x])^(n - 1), x], x] + Dist[(c^2/(f^2*(m + 1)))*S
imp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]], Int[(f*x)^(m + 2)*((a + b*ArcSin[c*x])^n/Sqrt[1 - c^2*x^2]), x], x]) /
; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1]

Rule 4785

Int[((a_.) + ArcSin[(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*ArcSin[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*ArcSin[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*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c,
d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1]

Rule 4803

Int[(((a_.) + ArcSin[(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*Sin[x]^m, x], x, ArcSin[c*x]], x] /; Free
Q[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]
time = 0.53, size = 266, normalized size = 1.51 \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 \text {ArcSin}(c x)+6 a b c^2 x^2 \text {ArcSin}(c x)-b^2 c x \sqrt {1-c^2 x^2} \text {ArcSin}(c x)-b^2 \text {ArcSin}(c x)^2+3 b^2 c^2 x^2 \text {ArcSin}(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 \text {ArcSin}(c x) \log \left (1-e^{i \text {ArcSin}(c x)}\right )+5 b^2 c^3 x^3 \text {ArcSin}(c x) \log \left (1+e^{i \text {ArcSin}(c x)}\right )-5 i b^2 c^3 x^3 \text {PolyLog}\left (2,-e^{i \text {ArcSin}(c x)}\right )+5 i b^2 c^3 x^3 \text {PolyLog}\left (2,e^{i \text {ArcSin}(c x)}\right )\right )}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(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*ArcSin[c*x] + 6*a*b*c^2*x^2*ArcSin[
c*x] - b^2*c*x*Sqrt[1 - c^2*x^2]*ArcSin[c*x] - b^2*ArcSin[c*x]^2 + 3*b^2*c^2*x^2*ArcSin[c*x]^2 + 5*a*b*c^3*x^3
*ArcTanh[Sqrt[1 - c^2*x^2]] - 5*b^2*c^3*x^3*ArcSin[c*x]*Log[1 - E^(I*ArcSin[c*x])] + 5*b^2*c^3*x^3*ArcSin[c*x]
*Log[1 + E^(I*ArcSin[c*x])] - (5*I)*b^2*c^3*x^3*PolyLog[2, -E^(I*ArcSin[c*x])] + (5*I)*b^2*c^3*x^3*PolyLog[2,
E^(I*ArcSin[c*x])]))/(3*x^3)

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Maple [A]
time = 0.57, size = 287, normalized size = 1.63

method	result	size
derivativedivides	\(c^{3} \left (-d \,a^{2} \left (\frac {1}{3 c^{3} x^{3}}-\frac {1}{c x}\right )+\frac {d \,b^{2} \arcsin \left (c x \right )^{2}}{c x}-\frac {d \,b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{3 c^{2} x^{2}}-\frac {d \,b^{2} \arcsin \left (c x \right )^{2}}{3 c^{3} x^{3}}-\frac {d \,b^{2}}{3 c x}-\frac {5 d \,b^{2} \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{3}+\frac {5 i d \,b^{2} \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3}+\frac {5 d \,b^{2} \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3}-\frac {5 i d \,b^{2} \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{3}-2 d a b \left (\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}-\frac {\arcsin \left (c x \right )}{c x}+\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {5 \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}\right )\right )\)	\(287\)
default	\(c^{3} \left (-d \,a^{2} \left (\frac {1}{3 c^{3} x^{3}}-\frac {1}{c x}\right )+\frac {d \,b^{2} \arcsin \left (c x \right )^{2}}{c x}-\frac {d \,b^{2} \arcsin \left (c x \right ) \sqrt {-c^{2} x^{2}+1}}{3 c^{2} x^{2}}-\frac {d \,b^{2} \arcsin \left (c x \right )^{2}}{3 c^{3} x^{3}}-\frac {d \,b^{2}}{3 c x}-\frac {5 d \,b^{2} \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{3}+\frac {5 i d \,b^{2} \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3}+\frac {5 d \,b^{2} \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{3}-\frac {5 i d \,b^{2} \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{3}-2 d a b \left (\frac {\arcsin \left (c x \right )}{3 c^{3} x^{3}}-\frac {\arcsin \left (c x \right )}{c x}+\frac {\sqrt {-c^{2} x^{2}+1}}{6 c^{2} x^{2}}-\frac {5 \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{6}\right )\right )\)	\(287\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*(-d*a^2*(1/3/c^3/x^3-1/c/x)+d*b^2/c/x*arcsin(c*x)^2-1/3*d*b^2/c^2/x^2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)-1/3*d
*b^2/c^3/x^3*arcsin(c*x)^2-1/3*d*b^2/c/x-5/3*d*b^2*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))+5/3*I*d*b^2*poly
log(2,I*c*x+(-c^2*x^2+1)^(1/2))+5/3*d*b^2*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-5/3*I*d*b^2*polylog(2,-I*
c*x-(-c^2*x^2+1)^(1/2))-2*d*a*b*(1/3/c^3/x^3*arcsin(c*x)-1/c/x*arcsin(c*x)+1/6/c^2/x^2*(-c^2*x^2+1)^(1/2)-5/6*
arctanh(1/(-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((-c^2*d*x^2+d)*(a+b*arcsin(c*x))^2/x^4,x, algorithm="maxima")

[Out]

2*(c*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + arcsin(c*x)/x)*a*b*c^2*d - 1/3*((c^2*log(2*sqrt(-c^2*x^2 +
1)/abs(x) + 2/abs(x)) + sqrt(-c^2*x^2 + 1)/x^2)*c + 2*arcsin(c*x)/x^3)*a*b*d + a^2*c^2*d/x - 1/3*a^2*d/x^3 + 1
/3*(3*x^3*integrate(2/3*(3*b^2*c^3*d*x^2 - b^2*c*d)*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan2(c*x, sqrt(c*x + 1)*sq
rt(-c*x + 1))/(c^2*x^5 - x^3), x) + (3*b^2*c^2*d*x^2 - b^2*d)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2)/x^
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((-c^2*d*x^2+d)*(a+b*arcsin(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)*arcsin(c*x)^2 + 2*(a*b*c^2*d*x^2 - a*b*d)*arcsin(c*
x))/x^4, x)

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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