\(\int \frac {(d+e x^2) (a+b \sec ^{-1}(c x))}{x} \, dx\) [79]

Optimal result

Integrand size = 19, antiderivative size = 124 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x} \, dx=-\frac {b e \sqrt {1-\frac {1}{c^2 x^2}} x}{2 c}-\frac {1}{2} i b d \csc ^{-1}(c x)^2+\frac {1}{2} e x^2 \left (a+b \sec ^{-1}(c x)\right )+b d \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b d \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-d \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\frac {1}{2} i b d \operatorname {PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {5348, 14, 4816, 6874, 270, 2363, 4721, 3798, 2221, 2317, 2438} \[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x} \, dx=-d \log \left (\frac {1}{x}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{2} e x^2 \left (a+b \sec ^{-1}(c x)\right )-\frac {b e x \sqrt {1-\frac {1}{c^2 x^2}}}{2 c}-\frac {1}{2} i b d \operatorname {PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right )-\frac {1}{2} i b d \csc ^{-1}(c x)^2+b d \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b d \log \left (\frac {1}{x}\right ) \csc ^{-1}(c x) \]

[In]

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

[Out]

-1/2*(b*e*Sqrt[1 - 1/(c^2*x^2)]*x)/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 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 270

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 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 2363

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-e, 2]*(x/Sqr
t[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 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 3798

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 4721

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n*Cot[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 4816

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 5348

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 6874

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

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\left (e+d x^2\right ) \left (a+b \arccos \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 \text {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 \text {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) \text {Subst}\left (\int \frac {\log (x)}{\sqrt {1-\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{c}+\frac {(b e) \text {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) \text {Subst}\left (\int \frac {\arcsin \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) \text {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) \text {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) \text {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) \text {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 \operatorname {PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.93 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x} \, dx=\frac {1}{2} a e x^2-\frac {b e x \sqrt {\frac {-1+c^2 x^2}{c^2 x^2}}}{2 c}+\frac {1}{2} b e x^2 \sec ^{-1}(c x)+\frac {1}{2} i b d \sec ^{-1}(c x)^2-b d \sec ^{-1}(c x) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )+a d \log (x)+\frac {1}{2} i b d \operatorname {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.86 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.10

method	result	size
parts	\(\frac {a e \,x^{2}}{2}+a d \ln \left (x \right )+b \left (\frac {i \operatorname {arcsec}\left (c x \right )^{2} d}{2}+\frac {e \left (c^{2} x^{2} \operatorname {arcsec}\left (c x \right )-x c \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}-i\right )}{2 c^{2}}-d \,\operatorname {arcsec}\left (c x \right ) \ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )+\frac {i d \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right )}{2}\right )\)	\(137\)
derivativedivides	\(\frac {a e \,x^{2}}{2}+a d \ln \left (c x \right )+\frac {b \left (\frac {i c^{2} d \operatorname {arcsec}\left (c x \right )^{2}}{2}+\frac {e \left (c^{2} x^{2} \operatorname {arcsec}\left (c x \right )-x c \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}-i\right )}{2}-\ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right ) c^{2} d \,\operatorname {arcsec}\left (c x \right )+\frac {i \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right ) c^{2} d}{2}\right )}{c^{2}}\)	\(148\)
default	\(\frac {a e \,x^{2}}{2}+a d \ln \left (c x \right )+\frac {b \left (\frac {i c^{2} d \operatorname {arcsec}\left (c x \right )^{2}}{2}+\frac {e \left (c^{2} x^{2} \operatorname {arcsec}\left (c x \right )-x c \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}-i\right )}{2}-\ln \left (1+{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right ) c^{2} d \,\operatorname {arcsec}\left (c x \right )+\frac {i \operatorname {polylog}\left (2, -{\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )}^{2}\right ) c^{2} d}{2}\right )}{c^{2}}\)	\(148\)

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}}{x} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x} \, dx=\int \frac {\left (a + b \operatorname {asec}{\left (c x \right )}\right ) \left (d + e x^{2}\right )}{x}\, dx \]

[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)

Maxima [F]

\[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x} \, dx=\int { \frac {{\left (e x^{2} + d\right )} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}}{x} \,d x } \]

[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 + 2*I*b*c^2*d*log(c))*log(x))/c^2

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> an error occurred running a Giac command:INPUT:sage2OUTPUT:Limit: Max order
reached or unable to make series expansion Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x} \, dx=\int \frac {\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{x} \,d x \]

[In]

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

[Out]

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