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

Optimal. Leaf size=137 \[ -\frac{1}{2} i b e \text{PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right )-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-e \log \left (\frac{1}{x}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac{b c d \sqrt{1-\frac{1}{c^2 x^2}}}{4 x}-\frac{1}{4} b c^2 d \csc ^{-1}(c x)-\frac{1}{2} i b e \csc ^{-1}(c x)^2+b e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b e \log \left (\frac{1}{x}\right ) \csc ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.291375, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 13, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.684, Rules used = {5240, 14, 4732, 12, 6742, 321, 216, 2326, 4625, 3717, 2190, 2279, 2391} \[ -\frac{1}{2} i b e \text{PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right )-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-e \log \left (\frac{1}{x}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac{b c d \sqrt{1-\frac{1}{c^2 x^2}}}{4 x}-\frac{1}{4} b c^2 d \csc ^{-1}(c x)-\frac{1}{2} i b e \csc ^{-1}(c x)^2+b e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b e \log \left (\frac{1}{x}\right ) \csc ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5240

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> -Subst[Int[
((e + d*x^2)^p*(a + b*ArcCos[x/c])^n)/x^(m + 2*(p + 1)), x], x, 1/x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[n
, 0] && IntegerQ[m] && IntegerQ[p]

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 12

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

Rule 6742

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

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 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 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 \sec ^{-1}(c x)\right )}{x^3} \, dx &=-\operatorname{Subst}\left (\int \frac{\left (e+d x^2\right ) \left (a+b \cos ^{-1}\left (\frac{x}{c}\right )\right )}{x} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac{1}{x}\right )-\frac{b \operatorname{Subst}\left (\int \frac{d x^2+2 e \log (x)}{2 \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{c}\\ &=-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac{1}{x}\right )-\frac{b \operatorname{Subst}\left (\int \frac{d x^2+2 e \log (x)}{\sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{2 c}\\ &=-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac{1}{x}\right )-\frac{b \operatorname{Subst}\left (\int \left (\frac{d x^2}{\sqrt{1-\frac{x^2}{c^2}}}+\frac{2 e \log (x)}{\sqrt{1-\frac{x^2}{c^2}}}\right ) \, dx,x,\frac{1}{x}\right )}{2 c}\\ &=-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac{1}{x}\right )-\frac{(b d) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{2 c}-\frac{(b e) \operatorname{Subst}\left (\int \frac{\log (x)}{\sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{c}\\ &=\frac{b c d \sqrt{1-\frac{1}{c^2 x^2}}}{4 x}-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-b e \csc ^{-1}(c x) \log \left (\frac{1}{x}\right )-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac{1}{x}\right )-\frac{1}{4} (b c d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )+(b e) \operatorname{Subst}\left (\int \frac{\sin ^{-1}\left (\frac{x}{c}\right )}{x} \, dx,x,\frac{1}{x}\right )\\ &=\frac{b c d \sqrt{1-\frac{1}{c^2 x^2}}}{4 x}-\frac{1}{4} b c^2 d \csc ^{-1}(c x)-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-b e \csc ^{-1}(c x) \log \left (\frac{1}{x}\right )-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac{1}{x}\right )+(b e) \operatorname{Subst}\left (\int x \cot (x) \, dx,x,\csc ^{-1}(c x)\right )\\ &=\frac{b c d \sqrt{1-\frac{1}{c^2 x^2}}}{4 x}-\frac{1}{4} b c^2 d \csc ^{-1}(c x)-\frac{1}{2} i b e \csc ^{-1}(c x)^2-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-b e \csc ^{-1}(c x) \log \left (\frac{1}{x}\right )-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac{1}{x}\right )-(2 i b e) \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\csc ^{-1}(c x)\right )\\ &=\frac{b c d \sqrt{1-\frac{1}{c^2 x^2}}}{4 x}-\frac{1}{4} b c^2 d \csc ^{-1}(c x)-\frac{1}{2} i b e \csc ^{-1}(c x)^2-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}+b e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b e \csc ^{-1}(c x) \log \left (\frac{1}{x}\right )-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac{1}{x}\right )-(b e) \operatorname{Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\csc ^{-1}(c x)\right )\\ &=\frac{b c d \sqrt{1-\frac{1}{c^2 x^2}}}{4 x}-\frac{1}{4} b c^2 d \csc ^{-1}(c x)-\frac{1}{2} i b e \csc ^{-1}(c x)^2-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}+b e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b e \csc ^{-1}(c x) \log \left (\frac{1}{x}\right )-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac{1}{x}\right )+\frac{1}{2} (i b e) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i \csc ^{-1}(c x)}\right )\\ &=\frac{b c d \sqrt{1-\frac{1}{c^2 x^2}}}{4 x}-\frac{1}{4} b c^2 d \csc ^{-1}(c x)-\frac{1}{2} i b e \csc ^{-1}(c x)^2-\frac{d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}+b e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b e \csc ^{-1}(c x) \log \left (\frac{1}{x}\right )-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac{1}{x}\right )-\frac{1}{2} i b e \text{Li}_2\left (e^{2 i \csc ^{-1}(c x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.441772, size = 136, normalized size = 0.99 \[ \frac{1}{4} \left (2 i b e \left (\text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )+\sec ^{-1}(c x) \left (\sec ^{-1}(c x)+2 i \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )\right )\right )-\frac{2 a d}{x^2}+4 a e \log (x)+\frac{b c d \sqrt{1-\frac{1}{c^2 x^2}} \left (\frac{c^2 x^2 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{\sqrt{1-c^2 x^2}}+1\right )}{x}-\frac{2 b d \sec ^{-1}(c x)}{x^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.386, size = 145, normalized size = 1.1 \begin{align*} -{\frac{ad}{2\,{x}^{2}}}+ae\ln \left ( cx \right ) +{\frac{i}{2}}b \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}e+{\frac{bcd}{4\,x}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}+{\frac{{c}^{2}bd{\rm arcsec} \left (cx\right )}{4}}-{\frac{b{\rm arcsec} \left (cx\right )d}{2\,{x}^{2}}}-be{\rm arcsec} \left (cx\right )\ln \left ( 1+ \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) +{\frac{i}{2}}be{\it polylog} \left ( 2,- \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -{\left (c^{2} \int \frac{\sqrt{c x + 1} \sqrt{c x - 1} \log \left (x\right )}{c^{4} x^{3} - c^{2} x}\,{d x} - \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right ) \log \left (x\right )\right )} b e - \frac{1}{4} \, b d{\left (\frac{\frac{c^{4} x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c^{2} x^{2}{\left (\frac{1}{c^{2} x^{2}} - 1\right )} - 1} - c^{3} \arctan \left (c x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}\right )}{c} + \frac{2 \, \operatorname{arcsec}\left (c x\right )}{x^{2}}\right )} + 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*arcsec(c*x))/x^3,x, algorithm="maxima")

[Out]

-(c^2*integrate(sqrt(c*x + 1)*sqrt(c*x - 1)*log(x)/(c^4*x^3 - c^2*x), x) - arctan(sqrt(c*x + 1)*sqrt(c*x - 1))
*log(x))*b*e - 1/4*b*d*((c^4*x*sqrt(-1/(c^2*x^2) + 1)/(c^2*x^2*(1/(c^2*x^2) - 1) - 1) - c^3*arctan(c*x*sqrt(-1
/(c^2*x^2) + 1)))/c + 2*arcsec(c*x)/x^2) + 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 )} \operatorname{arcsec}\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*arcsec(c*x))/x^3,x, algorithm="fricas")

[Out]

integral((a*e*x^2 + a*d + (b*e*x^2 + b*d)*arcsec(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{asec}{\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*asec(c*x))/x**3,x)

[Out]

Integral((a + b*asec(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 \operatorname{arcsec}\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*arcsec(c*x))/x^3,x, algorithm="giac")

[Out]

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