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

Optimal. Leaf size=132 \[ d \log (x) \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{2} e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {b e x \sqrt {1-c^2 x^2}}{4 c}-\frac {b e \sin ^{-1}(c x)}{4 c^2}-\frac {1}{2} i b d \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-\frac {1}{2} i b d \sin ^{-1}(c x)^2+b d \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-b d \log (x) \sin ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.24, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {14, 4731, 12, 6742, 321, 216, 2326, 4625, 3717, 2190, 2279, 2391} \[ -\frac {1}{2} i b d \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )+d \log (x) \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{2} e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {b e x \sqrt {1-c^2 x^2}}{4 c}-\frac {b e \sin ^{-1}(c x)}{4 c^2}-\frac {1}{2} i b d \sin ^{-1}(c x)^2+b d \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-b d \log (x) \sin ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 321

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

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 2326

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(ArcSin[(Rt[-e, 2]*x)/S
qrt[d]]*(a + b*Log[c*x^n]))/Rt[-e, 2], x] - Dist[(b*n)/Rt[-e, 2], Int[ArcSin[(Rt[-e, 2]*x)/Sqrt[d]]/x, x], x]
/; FreeQ[{a, b, c, d, e, n}, x] && GtQ[d, 0] && NegQ[e]

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 4731

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 -
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (GtQ[p, 0] ||
(IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{x} \, dx &=\frac {1}{2} e x^2 \left (a+b \sin ^{-1}(c x)\right )+d \left (a+b \sin ^{-1}(c x)\right ) \log (x)-(b c) \int \frac {e x^2+2 d \log (x)}{2 \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {1}{2} e x^2 \left (a+b \sin ^{-1}(c x)\right )+d \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\frac {1}{2} (b c) \int \frac {e x^2+2 d \log (x)}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {1}{2} e x^2 \left (a+b \sin ^{-1}(c x)\right )+d \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\frac {1}{2} (b c) \int \left (\frac {e x^2}{\sqrt {1-c^2 x^2}}+\frac {2 d \log (x)}{\sqrt {1-c^2 x^2}}\right ) \, dx\\ &=\frac {1}{2} e x^2 \left (a+b \sin ^{-1}(c x)\right )+d \left (a+b \sin ^{-1}(c x)\right ) \log (x)-(b c d) \int \frac {\log (x)}{\sqrt {1-c^2 x^2}} \, dx-\frac {1}{2} (b c e) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {b e x \sqrt {1-c^2 x^2}}{4 c}+\frac {1}{2} e x^2 \left (a+b \sin ^{-1}(c x)\right )-b d \sin ^{-1}(c x) \log (x)+d \left (a+b \sin ^{-1}(c x)\right ) \log (x)+(b d) \int \frac {\sin ^{-1}(c x)}{x} \, dx-\frac {(b e) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{4 c}\\ &=\frac {b e x \sqrt {1-c^2 x^2}}{4 c}-\frac {b e \sin ^{-1}(c x)}{4 c^2}+\frac {1}{2} e x^2 \left (a+b \sin ^{-1}(c x)\right )-b d \sin ^{-1}(c x) \log (x)+d \left (a+b \sin ^{-1}(c x)\right ) \log (x)+(b d) \operatorname {Subst}\left (\int x \cot (x) \, dx,x,\sin ^{-1}(c x)\right )\\ &=\frac {b e x \sqrt {1-c^2 x^2}}{4 c}-\frac {b e \sin ^{-1}(c x)}{4 c^2}-\frac {1}{2} i b d \sin ^{-1}(c x)^2+\frac {1}{2} e x^2 \left (a+b \sin ^{-1}(c x)\right )-b d \sin ^{-1}(c x) \log (x)+d \left (a+b \sin ^{-1}(c x)\right ) \log (x)-(2 i b d) \operatorname {Subst}\left (\int \frac {e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )\\ &=\frac {b e x \sqrt {1-c^2 x^2}}{4 c}-\frac {b e \sin ^{-1}(c x)}{4 c^2}-\frac {1}{2} i b d \sin ^{-1}(c x)^2+\frac {1}{2} e x^2 \left (a+b \sin ^{-1}(c x)\right )+b d \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-b d \sin ^{-1}(c x) \log (x)+d \left (a+b \sin ^{-1}(c x)\right ) \log (x)-(b d) \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=\frac {b e x \sqrt {1-c^2 x^2}}{4 c}-\frac {b e \sin ^{-1}(c x)}{4 c^2}-\frac {1}{2} i b d \sin ^{-1}(c x)^2+\frac {1}{2} e x^2 \left (a+b \sin ^{-1}(c x)\right )+b d \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-b d \sin ^{-1}(c x) \log (x)+d \left (a+b \sin ^{-1}(c x)\right ) \log (x)+\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 {b e x \sqrt {1-c^2 x^2}}{4 c}-\frac {b e \sin ^{-1}(c x)}{4 c^2}-\frac {1}{2} i b d \sin ^{-1}(c x)^2+\frac {1}{2} e x^2 \left (a+b \sin ^{-1}(c x)\right )+b d \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-b d \sin ^{-1}(c x) \log (x)+d \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\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 = 0.21, size = 108, normalized size = 0.82 \[ \frac {1}{2} \left (2 a d \log (x)+a e x^2+\frac {b e \left (c x \sqrt {1-c^2 x^2}-\sin ^{-1}(c x)\right )}{2 c^2}-i b d \left (\sin ^{-1}(c x)^2+\text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )\right )+2 b d \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+b e x^2 \sin ^{-1}(c x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a e x^{2} + a d + {\left (b e x^{2} + b d\right )} \arcsin \left (c x\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.26, size = 167, normalized size = 1.27 \[ \frac {a \,x^{2} e}{2}+d a \ln \left (c x \right )-\frac {i b d \arcsin \left (c x \right )^{2}}{2}+d b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+d b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i d b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-i d b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )-\frac {b \arcsin \left (c x \right ) e \cos \left (2 \arcsin \left (c x \right )\right )}{4 c^{2}}+\frac {b e \sin \left (2 \arcsin \left (c x \right )\right )}{8 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2*e+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*b/c^2*arcsin(c*x)*e*cos(2*arcsin(c*x))+1/8*b/c^2*e*sin(2*arcsin(c*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*e*x^2 + a*d*log(x) + integrate((b*e*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 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\left (e\,x^2+d\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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