3.1.60 \(\int \frac {a+b \sec ^{-1}(c x)}{d+e x} \, dx\) [60]

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

[Out]

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

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Rubi [A]
time = 0.27, antiderivative size = 247, normalized size of antiderivative = 1.00, 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 = {5332, 2598} \begin {gather*} \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}-\frac {i b \text {Li}_2\left (-\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 {Li}_2\left (-\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 {Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )}{2 e} \end {gather*}

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 2598

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

Rule 5332

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 + 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^(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]

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*}

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Mathematica [A]
time = 0.44, size = 333, normalized size = 1.35 \begin {gather*} \frac {a \log (d+e x)}{e}+\frac {b \left (4 i \text {ArcSin}\left (\frac {\sqrt {1+\frac {e}{c d}}}{\sqrt {2}}\right ) \text {ArcTan}\left (\frac {(-c d+e) \tan \left (\frac {1}{2} \sec ^{-1}(c x)\right )}{\sqrt {-c^2 d^2+e^2}}\right )+\left (\sec ^{-1}(c x)+2 \text {ArcSin}\left (\frac {\sqrt {1+\frac {e}{c d}}}{\sqrt {2}}\right )\right ) \log \left (1+\frac {\left (e-\sqrt {-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )+\left (\sec ^{-1}(c x)-2 \text {ArcSin}\left (\frac {\sqrt {1+\frac {e}{c d}}}{\sqrt {2}}\right )\right ) \log \left (1+\frac {\left (e+\sqrt {-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )-\sec ^{-1}(c x) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )-i \left (\text {PolyLog}\left (2,\frac {\left (-e+\sqrt {-c^2 d^2+e^2}\right ) e^{i \sec ^{-1}(c x)}}{c d}\right )+\text {PolyLog}\left (2,-\frac {\left (e+\sqrt {-c^2 d^2+e^2}\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 )\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.64, size = 469, normalized size = 1.90

method	result	size
derivativedivides	\(\frac {\frac {a c \ln \left (c e x +c d \right )}{e}-\frac {b c \,\mathrm {arcsec}\left (c x \right ) \ln \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{e}-\frac {b c \,\mathrm {arcsec}\left (c x \right ) \ln \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{e}+\frac {i b c \dilog \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{e}+\frac {i b c \dilog \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{e}+\frac {b c \,\mathrm {arcsec}\left (c x \right ) \ln \left (\frac {-c d \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}-e}{-e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {b c \,\mathrm {arcsec}\left (c x \right ) \ln \left (\frac {c d \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}+e}{e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}-\frac {i b c \dilog \left (\frac {-c d \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}-e}{-e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}-\frac {i b c \dilog \left (\frac {c d \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}+e}{e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}}{c}\)	\(469\)
default	\(\frac {\frac {a c \ln \left (c e x +c d \right )}{e}-\frac {b c \,\mathrm {arcsec}\left (c x \right ) \ln \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{e}-\frac {b c \,\mathrm {arcsec}\left (c x \right ) \ln \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{e}+\frac {i b c \dilog \left (1+i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{e}+\frac {i b c \dilog \left (1-i \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )\right )}{e}+\frac {b c \,\mathrm {arcsec}\left (c x \right ) \ln \left (\frac {-c d \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}-e}{-e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}+\frac {b c \,\mathrm {arcsec}\left (c x \right ) \ln \left (\frac {c d \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}+e}{e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}-\frac {i b c \dilog \left (\frac {-c d \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}-e}{-e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}-\frac {i b c \dilog \left (\frac {c d \left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )+\sqrt {-c^{2} d^{2}+e^{2}}+e}{e +\sqrt {-c^{2} d^{2}+e^{2}}}\right )}{e}}{c}\)	\(469\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)/(x*e + d), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {asec}{\left (c x \right )}}{d + e x}\, dx \end {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(sageVARx)]s
ym2poly/r2sym(

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(1/(c*x)))/(d + e*x),x)

[Out]

int((a + b*acos(1/(c*x)))/(d + e*x), x)

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