3.1.93 \(\int \frac {a+b \sec ^{-1}(c x)}{d+e x^2} \, dx\) [93]

Optimal. Leaf size=509 \[ \frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {i b \text {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {i b \text {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {i b \text {PolyLog}\left (2,-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {i b \text {PolyLog}\left (2,\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {c^2 d+e}}\right )}{2 \sqrt {-d} \sqrt {e}} \]

[Out]

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

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Rubi [A]
time = 0.83, antiderivative size = 509, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {5338, 4758, 4826, 4616, 2221, 2317, 2438} \begin {gather*} \frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {c^2 d+e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {c^2 d+e}+\sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\left (a+b \sec ^{-1}(c x)\right ) \log \left (1+\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {c^2 d+e}+\sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {i b \text {Li}_2\left (-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {d c^2+e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {i b \text {Li}_2\left (\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}-\sqrt {d c^2+e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {i b \text {Li}_2\left (-\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {d c^2+e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {i b \text {Li}_2\left (\frac {c \sqrt {-d} e^{i \sec ^{-1}(c x)}}{\sqrt {e}+\sqrt {d c^2+e}}\right )}{2 \sqrt {-d} \sqrt {e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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]

Rule 4616

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)])/(Cos[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :>
Simp[I*((e + f*x)^(m + 1)/(b*f*(m + 1))), x] + (-Dist[I, Int[(e + f*x)^m*(E^(I*(c + d*x))/(a - Rt[a^2 - b^2, 2
] + b*E^(I*(c + d*x)))), x], x] - Dist[I, Int[(e + f*x)^m*(E^(I*(c + d*x))/(a + Rt[a^2 - b^2, 2] + b*E^(I*(c +
d*x)))), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

Rule 4758

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

Rule 4826

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Subst[Int[(a + b*x)^n*(Sin[x]/
(c*d + e*Cos[x])), x], x, ArcCos[c*x]] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[n, 0]

Rule 5338

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

Rubi steps

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

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Mathematica [A]
time = 0.32, size = 871, normalized size = 1.71 \begin {gather*} \frac {2 a \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )-4 b \text {ArcSin}\left (\frac {\sqrt {1-\frac {i \sqrt {e}}{c \sqrt {d}}}}{\sqrt {2}}\right ) \text {ArcTan}\left (\frac {\left (-i c \sqrt {d}+\sqrt {e}\right ) \tan \left (\frac {1}{2} \sec ^{-1}(c x)\right )}{\sqrt {c^2 d+e}}\right )+4 b \text {ArcSin}\left (\frac {\sqrt {1+\frac {i \sqrt {e}}{c \sqrt {d}}}}{\sqrt {2}}\right ) \text {ArcTan}\left (\frac {\left (i c \sqrt {d}+\sqrt {e}\right ) \tan \left (\frac {1}{2} \sec ^{-1}(c x)\right )}{\sqrt {c^2 d+e}}\right )-i b \sec ^{-1}(c x) \log \left (1+\frac {i \left (\sqrt {e}-\sqrt {c^2 d+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt {d}}\right )-2 i b \text {ArcSin}\left (\frac {\sqrt {1+\frac {i \sqrt {e}}{c \sqrt {d}}}}{\sqrt {2}}\right ) \log \left (1+\frac {i \left (\sqrt {e}-\sqrt {c^2 d+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt {d}}\right )+i b \sec ^{-1}(c x) \log \left (1+\frac {i \left (-\sqrt {e}+\sqrt {c^2 d+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt {d}}\right )+2 i b \text {ArcSin}\left (\frac {\sqrt {1-\frac {i \sqrt {e}}{c \sqrt {d}}}}{\sqrt {2}}\right ) \log \left (1+\frac {i \left (-\sqrt {e}+\sqrt {c^2 d+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt {d}}\right )+i b \sec ^{-1}(c x) \log \left (1-\frac {i \left (\sqrt {e}+\sqrt {c^2 d+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt {d}}\right )-2 i b \text {ArcSin}\left (\frac {\sqrt {1-\frac {i \sqrt {e}}{c \sqrt {d}}}}{\sqrt {2}}\right ) \log \left (1-\frac {i \left (\sqrt {e}+\sqrt {c^2 d+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt {d}}\right )-i b \sec ^{-1}(c x) \log \left (1+\frac {i \left (\sqrt {e}+\sqrt {c^2 d+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt {d}}\right )+2 i b \text {ArcSin}\left (\frac {\sqrt {1+\frac {i \sqrt {e}}{c \sqrt {d}}}}{\sqrt {2}}\right ) \log \left (1+\frac {i \left (\sqrt {e}+\sqrt {c^2 d+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt {d}}\right )+b \text {PolyLog}\left (2,-\frac {i \left (-\sqrt {e}+\sqrt {c^2 d+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt {d}}\right )-b \text {PolyLog}\left (2,\frac {i \left (-\sqrt {e}+\sqrt {c^2 d+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt {d}}\right )-b \text {PolyLog}\left (2,-\frac {i \left (\sqrt {e}+\sqrt {c^2 d+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt {d}}\right )+b \text {PolyLog}\left (2,\frac {i \left (\sqrt {e}+\sqrt {c^2 d+e}\right ) e^{i \sec ^{-1}(c x)}}{c \sqrt {d}}\right )}{2 \sqrt {d} \sqrt {e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 88.57, size = 281, normalized size = 0.55

method	result	size
derivativedivides	\(\frac {\frac {a c \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e}}+\frac {i b \,c^{2} \left (\munderset {\textit {\_R1} =\RootOf \left (c^{2} d \,\textit {\_Z}^{4}+\left (2 c^{2} d +4 e \right ) \textit {\_Z}^{2}+c^{2} d \right )}{\sum }\frac {\textit {\_R1} \left (i \mathrm {arcsec}\left (c x \right ) \ln \left (\frac {\textit {\_R1} -\frac {1}{c x}-i \sqrt {1-\frac {1}{c^{2} x^{2}}}}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -\frac {1}{c x}-i \sqrt {1-\frac {1}{c^{2} x^{2}}}}{\textit {\_R1}}\right )\right )}{\textit {\_R1}^{2} c^{2} d +c^{2} d +2 e}\right )}{2}-\frac {i b \,c^{2} \left (\munderset {\textit {\_R1} =\RootOf \left (c^{2} d \,\textit {\_Z}^{4}+\left (2 c^{2} d +4 e \right ) \textit {\_Z}^{2}+c^{2} d \right )}{\sum }\frac {i \mathrm {arcsec}\left (c x \right ) \ln \left (\frac {\textit {\_R1} -\frac {1}{c x}-i \sqrt {1-\frac {1}{c^{2} x^{2}}}}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -\frac {1}{c x}-i \sqrt {1-\frac {1}{c^{2} x^{2}}}}{\textit {\_R1}}\right )}{\textit {\_R1} \left (\textit {\_R1}^{2} c^{2} d +c^{2} d +2 e \right )}\right )}{2}}{c}\)	\(281\)
default	\(\frac {\frac {a c \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e}}+\frac {i b \,c^{2} \left (\munderset {\textit {\_R1} =\RootOf \left (c^{2} d \,\textit {\_Z}^{4}+\left (2 c^{2} d +4 e \right ) \textit {\_Z}^{2}+c^{2} d \right )}{\sum }\frac {\textit {\_R1} \left (i \mathrm {arcsec}\left (c x \right ) \ln \left (\frac {\textit {\_R1} -\frac {1}{c x}-i \sqrt {1-\frac {1}{c^{2} x^{2}}}}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -\frac {1}{c x}-i \sqrt {1-\frac {1}{c^{2} x^{2}}}}{\textit {\_R1}}\right )\right )}{\textit {\_R1}^{2} c^{2} d +c^{2} d +2 e}\right )}{2}-\frac {i b \,c^{2} \left (\munderset {\textit {\_R1} =\RootOf \left (c^{2} d \,\textit {\_Z}^{4}+\left (2 c^{2} d +4 e \right ) \textit {\_Z}^{2}+c^{2} d \right )}{\sum }\frac {i \mathrm {arcsec}\left (c x \right ) \ln \left (\frac {\textit {\_R1} -\frac {1}{c x}-i \sqrt {1-\frac {1}{c^{2} x^{2}}}}{\textit {\_R1}}\right )+\dilog \left (\frac {\textit {\_R1} -\frac {1}{c x}-i \sqrt {1-\frac {1}{c^{2} x^{2}}}}{\textit {\_R1}}\right )}{\textit {\_R1} \left (\textit {\_R1}^{2} c^{2} d +c^{2} d +2 e \right )}\right )}{2}}{c}\)	\(281\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(a*c/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))+1/2*I*b*c^2*sum(_R1/(_R1^2*c^2*d+c^2*d+2*e)*(I*arcsec(c*x)*ln((_R
1-1/c/x-I*(1-1/c^2/x^2)^(1/2))/_R1)+dilog((_R1-1/c/x-I*(1-1/c^2/x^2)^(1/2))/_R1)),_R1=RootOf(c^2*d*_Z^4+(2*c^2
*d+4*e)*_Z^2+c^2*d))-1/2*I*b*c^2*sum(1/_R1/(_R1^2*c^2*d+c^2*d+2*e)*(I*arcsec(c*x)*ln((_R1-1/c/x-I*(1-1/c^2/x^2
)^(1/2))/_R1)+dilog((_R1-1/c/x-I*(1-1/c^2/x^2)^(1/2))/_R1)),_R1=RootOf(c^2*d*_Z^4+(2*c^2*d+4*e)*_Z^2+c^2*d)))

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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^2+d),x, algorithm="maxima")

[Out]

a*arctan(x*e^(1/2)/sqrt(d))*e^(-1/2)/sqrt(d) + b*integrate(arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(x^2*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^2+d),x, algorithm="fricas")

[Out]

integral((b*arcsec(c*x) + a)/(x^2*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^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

int((a + b*acos(1/(c*x)))/(d + e*x^2), x)

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