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

Optimal. Leaf size=90 \[ -i b \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}+\log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2+\frac {1}{2} b^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c x)}\right ) \]

[Out]

-1/3*I*(a+b*arcsin(c*x))^3/b+(a+b*arcsin(c*x))^2*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))^2)-I*b*(a+b*arcsin(c*x))*poly
log(2,(I*c*x+(-c^2*x^2+1)^(1/2))^2)+1/2*b^2*polylog(3,(I*c*x+(-c^2*x^2+1)^(1/2))^2)

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Rubi [A]  time = 0.12, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4625, 3717, 2190, 2531, 2282, 6589} \[ -i b \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{2} b^2 \text {PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}+\log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])^2/x,x]

[Out]

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

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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

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 6589

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

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{x} \, dx &=\operatorname {Subst}\left (\int (a+b x)^2 \cot (x) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}-2 i \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)^2}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}+\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-(2 b) \operatorname {Subst}\left (\int (a+b x) \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}+\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )+\left (i b^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}+\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )+\frac {1}{2} b^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )\\ &=-\frac {i \left (a+b \sin ^{-1}(c x)\right )^3}{3 b}+\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-i b \left (a+b \sin ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )+\frac {1}{2} b^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 143, normalized size = 1.59 \[ a^2 \log (c x)+2 a b \left (\sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-\frac {1}{2} i \left (\sin ^{-1}(c x)^2+\text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )\right )\right )+b^2 \left (i \sin ^{-1}(c x) \text {Li}_2\left (e^{-2 i \sin ^{-1}(c x)}\right )+\frac {1}{2} \text {Li}_3\left (e^{-2 i \sin ^{-1}(c x)}\right )+\frac {1}{3} i \sin ^{-1}(c x)^3+\sin ^{-1}(c x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c x)}\right )-\frac {i \pi ^3}{24}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSin[c*x])^2/x,x]

[Out]

a^2*Log[c*x] + 2*a*b*(ArcSin[c*x]*Log[1 - E^((2*I)*ArcSin[c*x])] - (I/2)*(ArcSin[c*x]^2 + PolyLog[2, E^((2*I)*
ArcSin[c*x])])) + b^2*((-1/24*I)*Pi^3 + (I/3)*ArcSin[c*x]^3 + ArcSin[c*x]^2*Log[1 - E^((-2*I)*ArcSin[c*x])] +
I*ArcSin[c*x]*PolyLog[2, E^((-2*I)*ArcSin[c*x])] + PolyLog[3, E^((-2*I)*ArcSin[c*x])]/2)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsin(c*x) + a)^2/x, x)

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maple [B]  time = 0.05, size = 319, normalized size = 3.54 \[ a^{2} \ln \left (c x \right )-\frac {i b^{2} \arcsin \left (c x \right )^{3}}{3}+b^{2} \arcsin \left (c x \right )^{2} \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-2 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )+2 b^{2} \polylog \left (3, -i c x -\sqrt {-c^{2} x^{2}+1}\right )+b^{2} \arcsin \left (c x \right )^{2} \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-2 i b^{2} \arcsin \left (c x \right ) \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+2 b^{2} \polylog \left (3, i c x +\sqrt {-c^{2} x^{2}+1}\right )-i a b \arcsin \left (c x \right )^{2}-2 i a b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )-2 i a b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )+2 a b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+2 a b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*ln(c*x)-1/3*I*b^2*arcsin(c*x)^3+b^2*arcsin(c*x)^2*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-2*I*b^2*arcsin(c*x)*polyl
og(2,-I*c*x-(-c^2*x^2+1)^(1/2))+2*b^2*polylog(3,-I*c*x-(-c^2*x^2+1)^(1/2))+b^2*arcsin(c*x)^2*ln(1-I*c*x-(-c^2*
x^2+1)^(1/2))-2*I*b^2*arcsin(c*x)*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))+2*b^2*polylog(3,I*c*x+(-c^2*x^2+1)^(1/2)
)-I*a*b*arcsin(c*x)^2-2*I*a*b*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))-2*I*a*b*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))
+2*a*b*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))+2*a*b*arcsin(c*x)*ln(1+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 \[ a^{2} \log \relax (x) + \int \frac {b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 2 \, a b \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))^2/x,x)

[Out]

int((a + b*asin(c*x))^2/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(c*x))**2/x, x)

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