3.1.90 \(\int \frac {(d+e x^2)^2 (a+b \sec ^{-1}(c x))}{x^3} \, dx\) [90]

Optimal. Leaf size=189 \[ \frac {b c d^2 \sqrt {1-\frac {1}{c^2 x^2}}}{4 x}-\frac {b e^2 \sqrt {1-\frac {1}{c^2 x^2}} x}{2 c}-\frac {1}{4} b c^2 d^2 \csc ^{-1}(c x)-i b d e \csc ^{-1}(c x)^2-\frac {d^2 \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}+\frac {1}{2} e^2 x^2 \left (a+b \sec ^{-1}(c x)\right )+2 b d e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-2 b d e \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-2 d e \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-i b d e \text {PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right ) \]

[Out]

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

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Rubi [A]
time = 0.31, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 15, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5348, 272, 45, 4816, 12, 6874, 270, 327, 222, 2363, 4721, 3798, 2221, 2317, 2438} \begin {gather*} -\frac {d^2 \left (a+b \sec ^{-1}(c x)\right )}{2 x^2}-2 d e \log \left (\frac {1}{x}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{2} e^2 x^2 \left (a+b \sec ^{-1}(c x)\right )+\frac {b c d^2 \sqrt {1-\frac {1}{c^2 x^2}}}{4 x}-\frac {1}{4} b c^2 d^2 \csc ^{-1}(c x)-\frac {b e^2 x \sqrt {1-\frac {1}{c^2 x^2}}}{2 c}-i b d e \text {Li}_2\left (e^{2 i \csc ^{-1}(c x)}\right )-i b d e \csc ^{-1}(c x)^2+2 b d e \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-2 b d e \log \left (\frac {1}{x}\right ) \csc ^{-1}(c x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*d^2*Sqrt[1 - 1/(c^2*x^2)])/(4*x) - (b*e^2*Sqrt[1 - 1/(c^2*x^2)]*x)/(2*c) - (b*c^2*d^2*ArcCsc[c*x])/4 - I*
b*d*e*ArcCsc[c*x]^2 - (d^2*(a + b*ArcSec[c*x]))/(2*x^2) + (e^2*x^2*(a + b*ArcSec[c*x]))/2 + 2*b*d*e*ArcCsc[c*x
]*Log[1 - E^((2*I)*ArcCsc[c*x])] - 2*b*d*e*ArcCsc[c*x]*Log[x^(-1)] - 2*d*e*(a + b*ArcSec[c*x])*Log[x^(-1)] - I
*b*d*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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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

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Mathematica [A]
time = 0.50, size = 194, normalized size = 1.03 \begin {gather*} \frac {1}{4} \left (-\frac {2 a d^2}{x^2}+2 a e^2 x^2-\frac {2 b d^2 \sec ^{-1}(c x)}{x^2}+\frac {2 b e^2 x \left (-\sqrt {1-\frac {1}{c^2 x^2}}+c x \sec ^{-1}(c x)\right )}{c}+\frac {b d^2 \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}+8 a d e \log (x)+4 i b d 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)^2*(a + b*ArcSec[c*x]))/x^3,x]

[Out]

((-2*a*d^2)/x^2 + 2*a*e^2*x^2 - (2*b*d^2*ArcSec[c*x])/x^2 + (2*b*e^2*x*(-Sqrt[1 - 1/(c^2*x^2)] + c*x*ArcSec[c*
x]))/c + (b*d^2*(-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) + 8*a*d*e*Log[x] + (4*I)*b*d*e*(ArcSec[c*x]*(ArcSec[c*x] + (2*I)*Log[1 + E^((2*I)*ArcSec[c*x])]) + Pol
yLog[2, -E^((2*I)*ArcSec[c*x])]))/4

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Maple [A]
time = 1.04, size = 245, normalized size = 1.30

method	result	size
derivativedivides	\(c^{2} \left (\frac {a \,x^{2} e^{2}}{2 c^{2}}-\frac {a \,d^{2}}{2 c^{2} x^{2}}+\frac {2 a d e \ln \left (c x \right )}{c^{2}}+\frac {i b \mathrm {arcsec}\left (c x \right )^{2} d e}{c^{2}}+\frac {b \,d^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{4 c x}+\frac {b \,d^{2} \mathrm {arcsec}\left (c x \right )}{4}-\frac {b \,\mathrm {arcsec}\left (c x \right ) d^{2}}{2 c^{2} x^{2}}+\frac {b \,\mathrm {arcsec}\left (c x \right ) x^{2} e^{2}}{2 c^{2}}-\frac {b \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, e^{2} x}{2 c^{3}}-\frac {i b \,e^{2}}{2 c^{4}}-\frac {2 b d e \,\mathrm {arcsec}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{c^{2}}+\frac {i b d e \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{c^{2}}\right )\)	\(245\)
default	\(c^{2} \left (\frac {a \,x^{2} e^{2}}{2 c^{2}}-\frac {a \,d^{2}}{2 c^{2} x^{2}}+\frac {2 a d e \ln \left (c x \right )}{c^{2}}+\frac {i b \mathrm {arcsec}\left (c x \right )^{2} d e}{c^{2}}+\frac {b \,d^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{4 c x}+\frac {b \,d^{2} \mathrm {arcsec}\left (c x \right )}{4}-\frac {b \,\mathrm {arcsec}\left (c x \right ) d^{2}}{2 c^{2} x^{2}}+\frac {b \,\mathrm {arcsec}\left (c x \right ) x^{2} e^{2}}{2 c^{2}}-\frac {b \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, e^{2} x}{2 c^{3}}-\frac {i b \,e^{2}}{2 c^{4}}-\frac {2 b d e \,\mathrm {arcsec}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{c^{2}}+\frac {i b d e \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{c^{2}}\right )\)	\(245\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

-1/4*b*d^2*((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) + 1/2*a*x^2*e^2 + 2*a*d*e*log(x) - 1/2*a*d^2/x^2 - 1/4*(-2*I*b*c^2*x^2*e^2*log(c)
- 4*I*b*c^2*d*e*log(-c*x + 1)*log(x) - 4*I*b*c^2*d*e*log(x)^2 - 4*I*b*c^2*d*dilog(c*x)*e - 4*I*b*c^2*d*dilog(
-c*x)*e + I*(16*b*d*e*integrate(1/2*log(x)/(c^2*x^3 - x), x) + b*(log(c*x + 1)/c^2 + log(c*x - 1)/c^2)*e^2)*c^
2 + 4*c^2*integrate(1/2*(b*x^2*e^2 + 4*b*d*e*log(x))*sqrt(c*x + 1)*sqrt(c*x - 1)/(c^2*x^3 - x), x) - I*b*e^2*l
og(c*x - 1) - 2*(b*c^2*x^2*e^2 + 4*b*c^2*d*e*log(x))*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) + (I*b*c^2*x^2*e^2 +
4*I*b*c^2*d*e*log(x))*log(c^2*x^2) + (-4*I*b*c^2*d*e*log(x) - I*b*e^2)*log(c*x + 1) - 2*(I*b*c^2*x^2*e^2 + 4*I
*b*c^2*d*e*log(c))*log(x))/c^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)^2*(a+b*arcsec(c*x))/x^3,x, algorithm="fricas")

[Out]

integral((a*x^4*e^2 + 2*a*d*x^2*e + a*d^2 + (b*x^4*e^2 + 2*b*d*x^2*e + b*d^2)*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 )^{2}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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