3.1.34 \(\int \frac {a+b \text {ArcSin}(c x)}{x^2 (d-c^2 d x^2)} \, dx\) [34]

Optimal. Leaf size=116 \[ -\frac {a+b \text {ArcSin}(c x)}{d x}-\frac {2 i c (a+b \text {ArcSin}(c x)) \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )}{d}-\frac {b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d}+\frac {i b c \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{d}-\frac {i b c \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{d} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4789, 4749, 4266, 2317, 2438, 272, 65, 214} \begin {gather*} -\frac {2 i c \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{d}-\frac {a+b \text {ArcSin}(c x)}{d x}+\frac {i b c \text {Li}_2\left (-i e^{i \text {ArcSin}(c x)}\right )}{d}-\frac {i b c \text {Li}_2\left (i e^{i \text {ArcSin}(c x)}\right )}{d}-\frac {b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 )} \, dx &=-\frac {a+b \sin ^{-1}(c x)}{d x}+c^2 \int \frac {a+b \sin ^{-1}(c x)}{d-c^2 d x^2} \, dx+\frac {(b c) \int \frac {1}{x \sqrt {1-c^2 x^2}} \, dx}{d}\\ &=-\frac {a+b \sin ^{-1}(c x)}{d x}+\frac {c \text {Subst}\left (\int (a+b x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac {(b c) \text {Subst}\left (\int \frac {1}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{2 d}\\ &=-\frac {a+b \sin ^{-1}(c x)}{d x}-\frac {2 i c \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d}-\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}-\frac {(b c) \text {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac {(b c) \text {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}\\ &=-\frac {a+b \sin ^{-1}(c x)}{d x}-\frac {2 i c \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d}+\frac {(i b c) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {(i b c) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{d}\\ &=-\frac {a+b \sin ^{-1}(c x)}{d x}-\frac {2 i c \left (a+b \sin ^{-1}(c x)\right ) \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {b c \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )}{d}+\frac {i b c \text {Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{d}-\frac {i b c \text {Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{d}\\ \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. \(259\) vs. \(2(116)=232\).
time = 0.24, size = 259, normalized size = 2.23 \begin {gather*} -\frac {2 a+2 b \text {ArcSin}(c x)+i b c \pi x \text {ArcSin}(c x)+2 b c x \tanh ^{-1}\left (\sqrt {1-c^2 x^2}\right )-b c \pi x \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-2 b c x \text {ArcSin}(c x) \log \left (1-i e^{i \text {ArcSin}(c x)}\right )-b c \pi x \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+2 b c x \text {ArcSin}(c x) \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+a c x \log (1-c x)-a c x \log (1+c x)+b c \pi x \log \left (-\cos \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )+b c \pi x \log \left (\sin \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )-2 i b c x \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )+2 i b c x \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{2 d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.18, size = 236, normalized size = 2.03

method	result	size
derivativedivides	\(c \left (-\frac {a}{d c x}+\frac {a \ln \left (c x +1\right )}{2 d}-\frac {a \ln \left (c x -1\right )}{2 d}-\frac {b \arcsin \left (c x \right )}{d c x}-\frac {b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {b \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )}{d}-\frac {b \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}\right )\)	\(236\)
default	\(c \left (-\frac {a}{d c x}+\frac {a \ln \left (c x +1\right )}{2 d}-\frac {a \ln \left (c x -1\right )}{2 d}-\frac {b \arcsin \left (c x \right )}{d c x}-\frac {b \arcsin \left (c x \right ) \ln \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {b \arcsin \left (c x \right ) \ln \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}+\frac {b \ln \left (i c x +\sqrt {-c^{2} x^{2}+1}-1\right )}{d}-\frac {b \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d}+\frac {i b \dilog \left (1+i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}-\frac {i b \dilog \left (1-i \left (i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d}\right )\)	\(236\)

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),x,method=_RETURNVERBOSE)

[Out]

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

[Out]

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

[Out]

integral(-(b*arcsin(c*x) + a)/(c^2*d*x^4 - d*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^{2} x^{4} - x^{2}}\, dx + \int \frac {b \operatorname {asin}{\left (c x \right )}}{c^{2} x^{4} - x^{2}}\, dx}{d} \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),x)

[Out]

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

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

[Out]

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

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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 )} \,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)),x)

[Out]

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

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