3.22 \(\int \frac{(d+e x^2) (a+b \cos ^{-1}(c x))}{x^3} \, dx\)

Optimal. Leaf size=119 \[ \frac{1}{2} i b e \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )-\frac{d \left (a+b \cos ^{-1}(c x)\right )}{2 x^2}+e \log (x) \left (a+b \cos ^{-1}(c x)\right )+\frac{b c d \sqrt{1-c^2 x^2}}{2 x}+\frac{1}{2} i b e \sin ^{-1}(c x)^2-b e \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+b e \log (x) \sin ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.220723, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.526, Rules used = {14, 4732, 6742, 264, 2326, 4625, 3717, 2190, 2279, 2391} \[ \frac{1}{2} i b e \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )-\frac{d \left (a+b \cos ^{-1}(c x)\right )}{2 x^2}+e \log (x) \left (a+b \cos ^{-1}(c x)\right )+\frac{b c d \sqrt{1-c^2 x^2}}{2 x}+\frac{1}{2} i b e \sin ^{-1}(c x)^2-b e \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+b e \log (x) \sin ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*d*Sqrt[1 - c^2*x^2])/(2*x) - (d*(a + b*ArcCos[c*x]))/(2*x^2) + (I/2)*b*e*ArcSin[c*x]^2 - b*e*ArcSin[c*x]*
Log[1 - E^((2*I)*ArcSin[c*x])] + e*(a + b*ArcCos[c*x])*Log[x] + b*e*ArcSin[c*x]*Log[x] + (I/2)*b*e*PolyLog[2,
E^((2*I)*ArcSin[c*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 4732

Int[((a_.) + ArcCos[(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*ArcCos[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]]

Rule 264

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

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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.153997, size = 105, normalized size = 0.88 \[ -\frac{i b e x^2 \text{PolyLog}\left (2,-e^{2 i \cos ^{-1}(c x)}\right )+a d-2 a e x^2 \log (x)-b c d x \sqrt{1-c^2 x^2}+b \cos ^{-1}(c x) \left (d-2 e x^2 \log \left (1+e^{2 i \cos ^{-1}(c x)}\right )\right )+i b e x^2 \cos ^{-1}(c x)^2}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.313, size = 127, normalized size = 1.1 \begin{align*} -{\frac{ad}{2\,{x}^{2}}}+ae\ln \left ( cx \right ) -{\frac{i}{2}}b \left ( \arccos \left ( cx \right ) \right ) ^{2}e+{\frac{i}{2}}{c}^{2}bd+{\frac{bcd}{2\,x}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{b\arccos \left ( cx \right ) d}{2\,{x}^{2}}}+be\arccos \left ( cx \right ) \ln \left ( 1+ \left ( cx+i\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) -{\frac{i}{2}}be{\it polylog} \left ( 2,- \left ( cx+i\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)*(a+b*arccos(c*x))/x^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, b d{\left (\frac{\sqrt{-c^{2} x^{2} + 1} c}{x} - \frac{\arccos \left (c x\right )}{x^{2}}\right )} + b e \int \frac{\arctan \left (\sqrt{c x + 1} \sqrt{-c x + 1}, c x\right )}{x}\,{d x} + a e \log \left (x\right ) - \frac{a d}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arccos(c*x))/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a e x^{2} + a d +{\left (b e x^{2} + b d\right )} \arccos \left (c x\right )}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arccos(c*x))/x^3,x, algorithm="fricas")

[Out]

integral((a*e*x^2 + a*d + (b*e*x^2 + b*d)*arccos(c*x))/x^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{acos}{\left (c x \right )}\right ) \left (d + e x^{2}\right )}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(a+b*acos(c*x))/x**3,x)

[Out]

Integral((a + b*acos(c*x))*(d + e*x**2)/x**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}{\left (b \arccos \left (c x\right ) + a\right )}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arccos(c*x))/x^3,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(b*arccos(c*x) + a)/x^3, x)