3.1.8 \(\int \frac {(d-c^2 d x^2) (a+b \text {ArcSin}(c x))}{x^3} \, dx\) [8]

Optimal. Leaf size=139 \[ -\frac {b c d \sqrt {1-c^2 x^2}}{2 x}-\frac {1}{2} b c^2 d \text {ArcSin}(c x)-\frac {d \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{2 x^2}+\frac {i c^2 d (a+b \text {ArcSin}(c x))^2}{2 b}-c^2 d (a+b \text {ArcSin}(c x)) \log \left (1-e^{2 i \text {ArcSin}(c x)}\right )+\frac {1}{2} i b c^2 d \text {PolyLog}\left (2,e^{2 i \text {ArcSin}(c x)}\right ) \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {4775, 283, 222, 4721, 3798, 2221, 2317, 2438} \begin {gather*} -\frac {d \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{2 x^2}+\frac {i c^2 d (a+b \text {ArcSin}(c x))^2}{2 b}-c^2 d \log \left (1-e^{2 i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))+\frac {1}{2} i b c^2 d \text {Li}_2\left (e^{2 i \text {ArcSin}(c x)}\right )-\frac {1}{2} b c^2 d \text {ArcSin}(c x)-\frac {b c d \sqrt {1-c^2 x^2}}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

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 3798

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(
m + 1))), x] - Dist[2*I, Int[(c + d*x)^m*E^(2*I*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x))))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4721

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n*Cot[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 4775

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((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])/(f*(m + 1))), x] + (-Dist[b*c*(d^p/(f*(m + 1))), Int[(f*x)^(m + 1)*
(1 - c^2*x^2)^(p - 1/2), x], x] - Dist[2*e*(p/(f^2*(m + 1))), Int[(f*x)^(m + 2)*(d + e*x^2)^(p - 1)*(a + b*Arc
Sin[c*x]), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0] && ILtQ[(m + 1)/2, 0]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 112, normalized size = 0.81 \begin {gather*} -\frac {a d}{2 x^2}-\frac {b c d \sqrt {1-c^2 x^2}}{2 x}-\frac {b d \text {ArcSin}(c x)}{2 x^2}-b c^2 d \text {ArcSin}(c x) \log \left (1-e^{2 i \text {ArcSin}(c x)}\right )-a c^2 d \log (x)+\frac {1}{2} i b c^2 d \left (\text {ArcSin}(c x)^2+\text {PolyLog}\left (2,e^{2 i \text {ArcSin}(c x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.29, size = 186, normalized size = 1.34

method	result	size
derivativedivides	\(c^{2} \left (-\frac {d a}{2 c^{2} x^{2}}-d a \ln \left (c x \right )+\frac {i b d \arcsin \left (c x \right )^{2}}{2}+\frac {i d b}{2}-\frac {d b \sqrt {-c^{2} x^{2}+1}}{2 c x}-\frac {d b \arcsin \left (c x \right )}{2 c^{2} x^{2}}-d b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-d b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )+i d b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+i d b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )\right )\)	\(186\)
default	\(c^{2} \left (-\frac {d a}{2 c^{2} x^{2}}-d a \ln \left (c x \right )+\frac {i b d \arcsin \left (c x \right )^{2}}{2}+\frac {i d b}{2}-\frac {d b \sqrt {-c^{2} x^{2}+1}}{2 c x}-\frac {d b \arcsin \left (c x \right )}{2 c^{2} x^{2}}-d b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-d b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )+i d b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+i d b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )\right )\)	\(186\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

-b*c^2*d*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/x, x) - a*c^2*d*log(x) - 1/2*b*d*(sqrt(-c^2*x^2
+ 1)*c/x + arcsin(c*x)/x^2) - 1/2*a*d/x^2

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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))/x^3,x, algorithm="fricas")

[Out]

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

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

[Out]

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

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

[Out]

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

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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 )\,\left (d-c^2\,d\,x^2\right )}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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