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

Optimal. Leaf size=247 \[ -\frac{i b \text{PolyLog}\left (2,-\frac{\left (e-\sqrt{e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac{i b \text{PolyLog}\left (2,-\frac{\left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac{i b \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{2 e}+\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{\left (e-\sqrt{e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{\left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac{\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{e} \]

[Out]

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

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Rubi [A]  time = 0.37276, antiderivative size = 247, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {5224, 2518} \[ -\frac{i b \text{PolyLog}\left (2,-\frac{\left (e-\sqrt{e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac{i b \text{PolyLog}\left (2,-\frac{\left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac{i b \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{2 e}+\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{\left (e-\sqrt{e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}+\frac{\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac{\left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )}{e}-\frac{\log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{e} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])/(d + e*x),x]

[Out]

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

Rule 5224

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[((a + b*ArcSec[c*x])*Log[1 + ((
e - Sqrt[-(c^2*d^2) + e^2])*E^(I*ArcSec[c*x]))/(c*d)])/e, x] + (-Dist[b/(c*e), Int[Log[1 + ((e - Sqrt[-(c^2*d^
2) + e^2])*E^(I*ArcSec[c*x]))/(c*d)]/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x], x] - Dist[b/(c*e), Int[Log[1 + ((e + Sqr
t[-(c^2*d^2) + e^2])*E^(I*ArcSec[c*x]))/(c*d)]/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x], x] + Dist[b/(c*e), Int[Log[1 +
E^(2*I*ArcSec[c*x])]/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x], x] + Simp[((a + b*ArcSec[c*x])*Log[1 + ((e + Sqrt[-(c^2
*d^2) + e^2])*E^(I*ArcSec[c*x]))/(c*d)])/e, x] - Simp[((a + b*ArcSec[c*x])*Log[1 + E^(2*I*ArcSec[c*x])])/e, x]
) /; FreeQ[{a, b, c, d, e}, x]

Rule 2518

Int[Log[v_]*(u_), x_Symbol] :> With[{w = DerivativeDivides[v, u*(1 - v), x]}, Simp[w*PolyLog[2, 1 - v], x] /;
!FalseQ[w]]

Rubi steps

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

Mathematica [A]  time = 0.587821, size = 333, normalized size = 1.35 \[ \frac{a \log (d+e x)}{e}+\frac{b \left (-i \left (\text{PolyLog}\left (2,\frac{\left (\sqrt{e^2-c^2 d^2}-e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )+\text{PolyLog}\left (2,-\frac{\left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )\right )+\frac{1}{2} i \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )+\log \left (1+\frac{\left (e-\sqrt{e^2-c^2 d^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right ) \left (2 \sin ^{-1}\left (\frac{\sqrt{\frac{e}{c d}+1}}{\sqrt{2}}\right )+\sec ^{-1}(c x)\right )+\log \left (1+\frac{\left (\sqrt{e^2-c^2 d^2}+e\right ) e^{i \sec ^{-1}(c x)}}{c d}\right ) \left (\sec ^{-1}(c x)-2 \sin ^{-1}\left (\frac{\sqrt{\frac{e}{c d}+1}}{\sqrt{2}}\right )\right )+4 i \sin ^{-1}\left (\frac{\sqrt{\frac{e}{c d}+1}}{\sqrt{2}}\right ) \tan ^{-1}\left (\frac{(e-c d) \tan \left (\frac{1}{2} \sec ^{-1}(c x)\right )}{\sqrt{e^2-c^2 d^2}}\right )-\sec ^{-1}(c x) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )\right )}{e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.374, size = 456, normalized size = 1.9 \begin{align*}{\frac{a\ln \left ( cex+dc \right ) }{e}}+{\frac{b{\rm arcsec} \left (cx\right )}{e}\ln \left ({ \left ( dc \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) +\sqrt{-{c}^{2}{d}^{2}+{e}^{2}}+e \right ) \left ( e+\sqrt{-{c}^{2}{d}^{2}+{e}^{2}} \right ) ^{-1}} \right ) }+{\frac{b{\rm arcsec} \left (cx\right )}{e}\ln \left ({ \left ( -dc \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) +\sqrt{-{c}^{2}{d}^{2}+{e}^{2}}-e \right ) \left ( -e+\sqrt{-{c}^{2}{d}^{2}+{e}^{2}} \right ) ^{-1}} \right ) }-{\frac{ib}{e}{\it dilog} \left ({ \left ( dc \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) +\sqrt{-{c}^{2}{d}^{2}+{e}^{2}}+e \right ) \left ( e+\sqrt{-{c}^{2}{d}^{2}+{e}^{2}} \right ) ^{-1}} \right ) }-{\frac{ib}{e}{\it dilog} \left ({ \left ( -dc \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) +\sqrt{-{c}^{2}{d}^{2}+{e}^{2}}-e \right ) \left ( -e+\sqrt{-{c}^{2}{d}^{2}+{e}^{2}} \right ) ^{-1}} \right ) }-{\frac{b{\rm arcsec} \left (cx\right )}{e}\ln \left ( 1+i \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) \right ) }-{\frac{b{\rm arcsec} \left (cx\right )}{e}\ln \left ( 1-i \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) \right ) }+{\frac{ib}{e}{\it dilog} \left ( 1+i \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) \right ) }+{\frac{ib}{e}{\it dilog} \left ( 1-i \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*integrate(arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(e*x + d), x) + a*log(e*x + d)/e

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arcsec(c*x) + a)/(e*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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