3.6 \(\int \frac {(d-c^2 d x^2) (a+b \sin ^{-1}(c x))}{x} \, dx\)

Optimal. Leaf size=121 \[ \frac {1}{2} d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {i d \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+d \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{4} b c d x \sqrt {1-c^2 x^2}-\frac {1}{2} i b d \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-\frac {1}{4} b d \sin ^{-1}(c x) \]

[Out]

-1/4*b*d*arcsin(c*x)+1/2*d*(-c^2*x^2+1)*(a+b*arcsin(c*x))-1/2*I*d*(a+b*arcsin(c*x))^2/b+d*(a+b*arcsin(c*x))*ln
(1-(I*c*x+(-c^2*x^2+1)^(1/2))^2)-1/2*I*b*d*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))^2)-1/4*b*c*d*x*(-c^2*x^2+1)^(1
/2)

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Rubi [A]  time = 0.12, antiderivative size = 121, 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 = {4683, 4625, 3717, 2190, 2279, 2391, 195, 216} \[ -\frac {1}{2} i b d \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )+\frac {1}{2} d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {i d \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+d \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{4} b c d x \sqrt {1-c^2 x^2}-\frac {1}{4} b d \sin ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 216

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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2279

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 2391

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 3717

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 4625

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

Rule 4683

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

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 99, normalized size = 0.82 \[ -\frac {1}{4} d \left (2 a c^2 x^2-4 a \log (x)+b c x \sqrt {1-c^2 x^2}+b \sin ^{-1}(c x) \left (2 c^2 x^2-4 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-1\right )+2 i b \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )+2 i b \sin ^{-1}(c x)^2\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {a c^{2} d x^{2} - a d + {\left (b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (c^{2} d x^{2} - d\right )} {\left (b \arcsin \left (c x\right ) + a\right )}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.24, size = 164, normalized size = 1.36 \[ -\frac {d a \,c^{2} x^{2}}{2}+d a \ln \left (c x \right )-\frac {i b d \arcsin \left (c x \right )^{2}}{2}+d b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )+d b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-i d b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-i d b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+\frac {d b \cos \left (2 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right )}{4}-\frac {d b \sin \left (2 \arcsin \left (c x \right )\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*d*a*c^2*x^2+d*a*ln(c*x)-1/2*I*b*d*arcsin(c*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))+1/4*d*b*cos(2*arcsin(c*x))*arcsin(c*x)-1/8*d*b*sin(2*arcsin(c*x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a c^{2} d x^{2} + a d \log \relax (x) - \int \frac {{\left (b c^{2} d x^{2} - b d\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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