3.1.43 \(\int \frac {a+b \text {ArcSin}(c x)}{x^2 (d-c^2 d x^2)^2} \, dx\) [43]

Optimal. Leaf size=186 \[ -\frac {b c}{2 d^2 \sqrt {1-c^2 x^2}}-\frac {a+b \text {ArcSin}(c x)}{d^2 x \left (1-c^2 x^2\right )}+\frac {3 c^2 x (a+b \text {ArcSin}(c x))}{2 d^2 \left (1-c^2 x^2\right )}-\frac {3 i c (a+b \text {ArcSin}(c x)) \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )}{d^2}-\frac {b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d^2}+\frac {3 i b c \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{2 d^2}-\frac {3 i b c \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{2 d^2} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {4789, 4747, 4749, 4266, 2317, 2438, 267, 272, 53, 65, 214} \begin {gather*} -\frac {3 i c \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{d^2}+\frac {3 c^2 x (a+b \text {ArcSin}(c x))}{2 d^2 \left (1-c^2 x^2\right )}-\frac {a+b \text {ArcSin}(c x)}{d^2 x \left (1-c^2 x^2\right )}+\frac {3 i b c \text {Li}_2\left (-i e^{i \text {ArcSin}(c x)}\right )}{2 d^2}-\frac {3 i b c \text {Li}_2\left (i e^{i \text {ArcSin}(c x)}\right )}{2 d^2}-\frac {b c}{2 d^2 \sqrt {1-c^2 x^2}}-\frac {b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*c)/(d^2*Sqrt[1 - c^2*x^2]) - (a + b*ArcSin[c*x])/(d^2*x*(1 - c^2*x^2)) + (3*c^2*x*(a + b*ArcSin[c*x]))
/(2*d^2*(1 - c^2*x^2)) - ((3*I)*c*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/d^2 - (b*c*ArcTanh[Sqrt[1 - c
^2*x^2]])/d^2 + (((3*I)/2)*b*c*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/d^2 - (((3*I)/2)*b*c*PolyLog[2, I*E^(I*ArcS
in[c*x])])/d^2

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

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 272

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

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 4789

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 + 1)*((a + b*ArcSin[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*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]) /; Free
Q[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && ILtQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.51, size = 348, normalized size = 1.87 \begin {gather*} -\frac {\frac {4 a}{x}+\frac {b c \sqrt {1-c^2 x^2}}{1-c x}+\frac {b c \sqrt {1-c^2 x^2}}{1+c x}+\frac {2 a c^2 x}{-1+c^2 x^2}+3 i b c \pi \text {ArcSin}(c x)+\frac {4 b \text {ArcSin}(c x)}{x}+\frac {b c \text {ArcSin}(c x)}{-1+c x}+\frac {b c \text {ArcSin}(c x)}{1+c x}+4 b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )-3 b c \pi \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-6 b c \text {ArcSin}(c x) \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-3 b c \pi \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+6 b c \text {ArcSin}(c x) \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+3 a c \log (1-c x)-3 a c \log (1+c x)+3 b c \pi \log \left (-\cos \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )+3 b c \pi \log \left (\sin \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )-6 i b c \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )+6 i b c \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{4 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.21, size = 325, normalized size = 1.75

method	result	size
derivativedivides	\(c \left (-\frac {a}{4 d^{2} \left (c x +1\right )}+\frac {3 a \ln \left (c x +1\right )}{4 d^{2}}-\frac {a}{4 d^{2} \left (c x -1\right )}-\frac {3 a \ln \left (c x -1\right )}{4 d^{2}}-\frac {a}{d^{2} c x}-\frac {3 b \arcsin \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-c^{2} x^{2}+1}}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \arcsin \left (c x \right )}{d^{2} c x \left (c^{2} x^{2}-1\right )}+\frac {b \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )}{d^{2}}-\frac {b \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}+\frac {3 b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}-\frac {3 b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}-\frac {3 i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}+\frac {3 i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}\right )\)	\(325\)
default	\(c \left (-\frac {a}{4 d^{2} \left (c x +1\right )}+\frac {3 a \ln \left (c x +1\right )}{4 d^{2}}-\frac {a}{4 d^{2} \left (c x -1\right )}-\frac {3 a \ln \left (c x -1\right )}{4 d^{2}}-\frac {a}{d^{2} c x}-\frac {3 b \arcsin \left (c x \right ) c x}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \sqrt {-c^{2} x^{2}+1}}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \arcsin \left (c x \right )}{d^{2} c x \left (c^{2} x^{2}-1\right )}+\frac {b \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )}{d^{2}}-\frac {b \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}+\frac {3 b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}-\frac {3 b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}-\frac {3 i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}+\frac {3 i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{2 d^{2}}\right )\)	\(325\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

-1/4*a*(2*(3*c^2*x^2 - 2)/(c^2*d^2*x^3 - d^2*x) - 3*c*log(c*x + 1)/d^2 + 3*c*log(c*x - 1)/d^2) + 1/4*(3*(c^3*x
^3 - c*x)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - 3*(c^3*x^3 - c*x)*arctan2(c*x, sqrt(c*x +
1)*sqrt(-c*x + 1))*log(-c*x + 1) - 2*(3*c^2*x^2 - 2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + 4*(c^2*d^2*x
^3 - d^2*x)*integrate(-1/4*(6*c^3*x^2 - 3*(c^4*x^3 - c^2*x)*log(c*x + 1) + 3*(c^4*x^3 - c^2*x)*log(-c*x + 1) -
4*c)*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^4*d^2*x^5 - 2*c^2*d^2*x^3 + d^2*x), x))*b/(c^2*d^2*x^3 - d^2*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*arcsin(c*x))/x^2/(-c^2*d*x^2+d)^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a/(c**4*x**6 - 2*c**2*x**4 + x**2), x) + Integral(b*asin(c*x)/(c**4*x**6 - 2*c**2*x**4 + x**2), x))/
d**2

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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((a+b*arcsin(c*x))/x^2/(-c^2*d*x^2+d)^2,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 {a+b\,\mathrm {asin}\left (c\,x\right )}{x^2\,{\left (d-c^2\,d\,x^2\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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