3.1.80 \(\int \frac {(d+e x^2) (a+b \sec ^{-1}(c x))}{x^3} \, dx\) [80]

Optimal. Leaf size=137 \[ \frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{4 x}-\frac {1}{4} b c^2 d \csc ^{-1}(c x)-\frac {1}{2} i b e \csc ^{-1}(c x)^2-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}+b e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b e \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\frac {1}{2} i b e \text {PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right ) \]

[Out]

-1/4*b*c^2*d*arccsc(c*x)-1/2*I*b*e*arccsc(c*x)^2-1/2*d*(a+b*arcsec(c*x))/x^2+b*e*arccsc(c*x)*ln(1-(I/c/x+(1-1/
c^2/x^2)^(1/2))^2)-b*e*arccsc(c*x)*ln(1/x)-e*(a+b*arcsec(c*x))*ln(1/x)-1/2*I*b*e*polylog(2,(I/c/x+(1-1/c^2/x^2
)^(1/2))^2)+1/4*b*c*d*(1-1/c^2/x^2)^(1/2)/x

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Rubi [A]
time = 0.22, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 13, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.684, Rules used = {5348, 14, 4816, 12, 6874, 327, 222, 2363, 4721, 3798, 2221, 2317, 2438} \begin {gather*} -\frac {d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-e \log \left (\frac {1}{x}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{4 x}-\frac {1}{4} b c^2 d \csc ^{-1}(c x)-\frac {1}{2} i b e \text {Li}_2\left (e^{2 i \csc ^{-1}(c x)}\right )-\frac {1}{2} i b e \csc ^{-1}(c x)^2+b e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b e \log \left (\frac {1}{x}\right ) \csc ^{-1}(c x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)*(a + b*ArcSec[c*x]))/x^3,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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*} \int \frac {\left (d+e x^2\right ) \left (a+b \sec ^{-1}(c x)\right )}{x^3} \, dx &=-\text {Subst}\left (\int \frac {\left (e+d x^2\right ) \left (a+b \cos ^{-1}\left (\frac {x}{c}\right )\right )}{x} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\frac {b \text {Subst}\left (\int \frac {d x^2+2 e \log (x)}{2 \sqrt {1-\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{c}\\ &=-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\frac {b \text {Subst}\left (\int \frac {d x^2+2 e \log (x)}{\sqrt {1-\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{2 c}\\ &=-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\frac {b \text {Subst}\left (\int \left (\frac {d x^2}{\sqrt {1-\frac {x^2}{c^2}}}+\frac {2 e \log (x)}{\sqrt {1-\frac {x^2}{c^2}}}\right ) \, dx,x,\frac {1}{x}\right )}{2 c}\\ &=-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\frac {(b d) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{2 c}-\frac {(b e) \text {Subst}\left (\int \frac {\log (x)}{\sqrt {1-\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{c}\\ &=\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{4 x}-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-b e \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\frac {1}{4} (b c d) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )+(b e) \text {Subst}\left (\int \frac {\sin ^{-1}\left (\frac {x}{c}\right )}{x} \, dx,x,\frac {1}{x}\right )\\ &=\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{4 x}-\frac {1}{4} b c^2 d \csc ^{-1}(c x)-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-b e \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )+(b e) \text {Subst}\left (\int x \cot (x) \, dx,x,\csc ^{-1}(c x)\right )\\ &=\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{4 x}-\frac {1}{4} b c^2 d \csc ^{-1}(c x)-\frac {1}{2} i b e \csc ^{-1}(c x)^2-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-b e \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-(2 i b e) \text {Subst}\left (\int \frac {e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\csc ^{-1}(c x)\right )\\ &=\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{4 x}-\frac {1}{4} b c^2 d \csc ^{-1}(c x)-\frac {1}{2} i b e \csc ^{-1}(c x)^2-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}+b e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b e \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-(b e) \text {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\csc ^{-1}(c x)\right )\\ &=\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{4 x}-\frac {1}{4} b c^2 d \csc ^{-1}(c x)-\frac {1}{2} i b e \csc ^{-1}(c x)^2-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}+b e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b e \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )+\frac {1}{2} (i b e) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \csc ^{-1}(c x)}\right )\\ &=\frac {b c d \sqrt {1-\frac {1}{c^2 x^2}}}{4 x}-\frac {1}{4} b c^2 d \csc ^{-1}(c x)-\frac {1}{2} i b e \csc ^{-1}(c x)^2-\frac {d \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}+b e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b e \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\frac {1}{2} i b e \text {Li}_2\left (e^{2 i \csc ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]
time = 0.39, size = 143, normalized size = 1.04 \begin {gather*} \frac {1}{4} \left (-\frac {2 a d}{x^2}-\frac {2 b d \sec ^{-1}(c x)}{x^2}+\frac {b d \left (-1+c^2 x^2+c^2 x^2 \sqrt {-1+c^2 x^2} \text {ArcTan}\left (\sqrt {-1+c^2 x^2}\right )\right )}{c \sqrt {1-\frac {1}{c^2 x^2}} x^3}+4 a e \log (x)+2 i b e \left (\sec ^{-1}(c x) \left (\sec ^{-1}(c x)+2 i \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )\right )+\text {PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)*(a + b*ArcSec[c*x]))/x^3,x]

[Out]

((-2*a*d)/x^2 - (2*b*d*ArcSec[c*x])/x^2 + (b*d*(-1 + c^2*x^2 + c^2*x^2*Sqrt[-1 + c^2*x^2]*ArcTan[Sqrt[-1 + c^2
*x^2]]))/(c*Sqrt[1 - 1/(c^2*x^2)]*x^3) + 4*a*e*Log[x] + (2*I)*b*e*(ArcSec[c*x]*(ArcSec[c*x] + (2*I)*Log[1 + E^
((2*I)*ArcSec[c*x])]) + PolyLog[2, -E^((2*I)*ArcSec[c*x])]))/4

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Maple [A]
time = 0.50, size = 166, normalized size = 1.21

method	result	size
derivativedivides	\(c^{2} \left (-\frac {a d}{2 c^{2} x^{2}}+\frac {a e \ln \left (c x \right )}{c^{2}}+\frac {i b \mathrm {arcsec}\left (c x \right )^{2} e}{2 c^{2}}+\frac {b d \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{4 c x}+\frac {b d \,\mathrm {arcsec}\left (c x \right )}{4}-\frac {b \,\mathrm {arcsec}\left (c x \right ) d}{2 c^{2} x^{2}}-\frac {b e \,\mathrm {arcsec}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{c^{2}}+\frac {i b e \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{2 c^{2}}\right )\)	\(166\)
default	\(c^{2} \left (-\frac {a d}{2 c^{2} x^{2}}+\frac {a e \ln \left (c x \right )}{c^{2}}+\frac {i b \mathrm {arcsec}\left (c x \right )^{2} e}{2 c^{2}}+\frac {b d \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{4 c x}+\frac {b d \,\mathrm {arcsec}\left (c x \right )}{4}-\frac {b \,\mathrm {arcsec}\left (c x \right ) d}{2 c^{2} x^{2}}-\frac {b e \,\mathrm {arcsec}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{c^{2}}+\frac {i b e \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{2 c^{2}}\right )\)	\(166\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)*(a+b*arcsec(c*x))/x^3,x,method=_RETURNVERBOSE)

[Out]

c^2*(-1/2*a*d/c^2/x^2+a/c^2*e*ln(c*x)+1/2*I*b/c^2*arcsec(c*x)^2*e+1/4*b*d/c/x*((c^2*x^2-1)/c^2/x^2)^(1/2)+1/4*
b*d*arcsec(c*x)-1/2*b*arcsec(c*x)*d/c^2/x^2-b/c^2*e*arcsec(c*x)*ln(1+(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)+1/2*I*b/
c^2*e*polylog(2,-(1/c/x+I*(1-1/c^2/x^2)^(1/2))^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((e*x^2+d)*(a+b*arcsec(c*x))/x^3,x, algorithm="maxima")

[Out]

-1/4*b*d*((c^4*x*sqrt(-1/(c^2*x^2) + 1)/(c^2*x^2*(1/(c^2*x^2) - 1) - 1) - c^3*arctan(c*x*sqrt(-1/(c^2*x^2) + 1
)))/c + 2*arcsec(c*x)/x^2) - (c^2*integrate(sqrt(c*x + 1)*sqrt(c*x - 1)*log(x)/(c^4*x^3 - c^2*x), x) - arctan(
sqrt(c*x + 1)*sqrt(c*x - 1))*log(x))*b*e + a*e*log(x) - 1/2*a*d/x^2

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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((e*x^2+d)*(a+b*arcsec(c*x))/x^3,x, algorithm="fricas")

[Out]

integral((a*x^2*e + a*d + (b*x^2*e + b*d)*arcsec(c*x))/x^3, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {asec}{\left (c x \right )}\right ) \left (d + e x^{2}\right )}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(a+b*asec(c*x))/x**3,x)

[Out]

Integral((a + b*asec(c*x))*(d + e*x**2)/x**3, x)

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Giac [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((e*x^2+d)*(a+b*arcsec(c*x))/x^3,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(b*arcsec(c*x) + a)/x^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (e\,x^2+d\right )\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*x^2)*(a + b*acos(1/(c*x))))/x^3,x)

[Out]

int(((d + e*x^2)*(a + b*acos(1/(c*x))))/x^3, x)

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