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

Optimal. Leaf size=124 \[ -\frac{1}{2} i b d \text{PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right )-d \log \left (\frac{1}{x}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{2} e x^2 \left (a+b \sec ^{-1}(c x)\right )-\frac{b e x \sqrt{1-\frac{1}{c^2 x^2}}}{2 c}-\frac{1}{2} i b d \csc ^{-1}(c x)^2+b d \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b d \log \left (\frac{1}{x}\right ) \csc ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.283554, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 11, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.579, Rules used = {5240, 14, 4732, 6742, 264, 2326, 4625, 3717, 2190, 2279, 2391} \[ -\frac{1}{2} i b d \text{PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right )-d \log \left (\frac{1}{x}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{2} e x^2 \left (a+b \sec ^{-1}(c x)\right )-\frac{b e x \sqrt{1-\frac{1}{c^2 x^2}}}{2 c}-\frac{1}{2} i b d \csc ^{-1}(c x)^2+b d \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b d \log \left (\frac{1}{x}\right ) \csc ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.121672, size = 104, normalized size = 0.84 \[ \frac{i b c d \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )+2 a c d \log (x)+a c e x^2-b e x \sqrt{1-\frac{1}{c^2 x^2}}+b c \sec ^{-1}(c x) \left (e x^2-2 d \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )\right )+i b c d \sec ^{-1}(c x)^2}{2 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.531, size = 142, normalized size = 1.2 \begin{align*}{\frac{a{x}^{2}e}{2}}+ad\ln \left ( cx \right ) +{\frac{i}{2}}b \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}d+{\frac{b{\rm arcsec} \left (cx\right ){x}^{2}e}{2}}-{\frac{bxe}{2\,c}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}-{\frac{{\frac{i}{2}}be}{{c}^{2}}}-bd{\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}}bd{\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,x)

[Out]

1/2*a*x^2*e+a*d*ln(c*x)+1/2*I*b*arcsec(c*x)^2*d+1/2*b*arcsec(c*x)*x^2*e-1/2*b/c*((c^2*x^2-1)/c^2/x^2)^(1/2)*x*
e-1/2*I*b/c^2*e-b*d*arcsec(c*x)*ln(1+(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)+1/2*I*b*d*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*} \frac{1}{2} \, a e x^{2} + a d \log \left (x\right ) - \frac{-2 i \, b c^{2} e x^{2} \log \left (c\right ) - 2 i \, b c^{2} d \log \left (-c x + 1\right ) \log \left (x\right ) - 2 i \, b c^{2} d \log \left (x\right )^{2} - 2 i \, b c^{2} d{\rm Li}_2\left (c x\right ) - 2 i \, b c^{2} d{\rm Li}_2\left (-c x\right ) + i \,{\left (2 \,{\left ({\left (\log \left (c x + 1\right ) + \log \left (c x - 1\right ) - 2 \, \log \left (x\right )\right )} \log \left (x\right ) - \log \left (c x - 1\right ) \log \left (x\right ) + \log \left (-c x + 1\right ) \log \left (x\right ) + \log \left (x\right )^{2} +{\rm Li}_2\left (c x\right ) +{\rm Li}_2\left (-c x\right )\right )} b d + b e{\left (\frac{\log \left (c x + 1\right )}{c^{2}} + \frac{\log \left (c x - 1\right )}{c^{2}}\right )}\right )} c^{2} + 2 \,{\left (2 \, b d \int \frac{\sqrt{c x + 1} \sqrt{c x - 1} \log \left (x\right )}{c^{2} x^{3} - x}\,{d x} + \frac{\sqrt{c x + 1} \sqrt{c x - 1} b e}{c^{2}}\right )} c^{2} - i \, b e \log \left (c x - 1\right ) - 2 \,{\left (b c^{2} e x^{2} + 2 \, b c^{2} d \log \left (x\right )\right )} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right ) +{\left (i \, b c^{2} e x^{2} + 2 i \, b c^{2} d \log \left (x\right )\right )} \log \left (c^{2} x^{2}\right ) +{\left (-2 i \, b c^{2} d \log \left (x\right ) - i \, b e\right )} \log \left (c x + 1\right ) +{\left (-2 i \, b c^{2} e x^{2} - 4 i \, b c^{2} d \log \left (c\right )\right )} \log \left (x\right )}{4 \, c^{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,x, algorithm="maxima")

[Out]

1/2*a*e*x^2 + a*d*log(x) - 1/4*(-2*I*b*c^2*e*x^2*log(c) - 2*I*b*c^2*d*log(-c*x + 1)*log(x) - 2*I*b*c^2*d*log(x
)^2 - 2*I*b*c^2*d*dilog(c*x) - 2*I*b*c^2*d*dilog(-c*x) + I*(b*e*(log(c*x + 1)/c^2 + log(c*x - 1)/c^2) + 8*b*d*
integrate(1/2*log(x)/(c^2*x^3 - x), x))*c^2 + 4*c^2*integrate(1/2*(b*e*x^2 + 2*b*d*log(x))*sqrt(c*x + 1)*sqrt(
c*x - 1)/(c^2*x^3 - x), x) - I*b*e*log(c*x - 1) - 2*(b*c^2*e*x^2 + 2*b*c^2*d*log(x))*arctan(sqrt(c*x + 1)*sqrt
(c*x - 1)) + (I*b*c^2*e*x^2 + 2*I*b*c^2*d*log(x))*log(c^2*x^2) + (-2*I*b*c^2*d*log(x) - I*b*e)*log(c*x + 1) +
(-2*I*b*c^2*e*x^2 - 4*I*b*c^2*d*log(c))*log(x))/c^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}, 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,x, algorithm="fricas")

[Out]

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

[Out]

Integral((a + b*asec(c*x))*(d + e*x**2)/x, 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}\,{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,x, algorithm="giac")

[Out]

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