3.1.89 \(\int \frac {(d+e x^2)^2 (a+b \sec ^{-1}(c x))}{x} \, dx\) [89]

Optimal. Leaf size=186 \[ -\frac {b e \left (6 c^2 d+e\right ) \sqrt {1-\frac {1}{c^2 x^2}} x}{6 c^3}-\frac {b e^2 \sqrt {1-\frac {1}{c^2 x^2}} x^3}{12 c}-\frac {1}{2} i b d^2 \csc ^{-1}(c x)^2+d e x^2 \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sec ^{-1}(c x)\right )+b d^2 \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b d^2 \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-d^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\frac {1}{2} i b d^2 \text {PolyLog}\left (2,e^{2 i \csc ^{-1}(c x)}\right ) \]

[Out]

-1/2*I*b*d^2*arccsc(c*x)^2+d*e*x^2*(a+b*arcsec(c*x))+1/4*e^2*x^4*(a+b*arcsec(c*x))+b*d^2*arccsc(c*x)*ln(1-(I/c
/x+(1-1/c^2/x^2)^(1/2))^2)-b*d^2*arccsc(c*x)*ln(1/x)-d^2*(a+b*arcsec(c*x))*ln(1/x)-1/2*I*b*d^2*polylog(2,(I/c/
x+(1-1/c^2/x^2)^(1/2))^2)-1/6*b*e*(6*c^2*d+e)*x*(1-1/c^2/x^2)^(1/2)/c^3-1/12*b*e^2*x^3*(1-1/c^2/x^2)^(1/2)/c

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Rubi [A]
time = 0.31, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 13, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {5348, 272, 45, 4816, 6874, 464, 270, 2363, 4721, 3798, 2221, 2317, 2438} \begin {gather*} -d^2 \log \left (\frac {1}{x}\right ) \left (a+b \sec ^{-1}(c x)\right )+d e x^2 \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sec ^{-1}(c x)\right )-\frac {b e^2 x^3 \sqrt {1-\frac {1}{c^2 x^2}}}{12 c}-\frac {b e x \sqrt {1-\frac {1}{c^2 x^2}} \left (6 c^2 d+e\right )}{6 c^3}-\frac {1}{2} i b d^2 \text {Li}_2\left (e^{2 i \csc ^{-1}(c x)}\right )-\frac {1}{2} i b d^2 \csc ^{-1}(c x)^2+b d^2 \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b d^2 \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,x]

[Out]

-1/6*(b*e*(6*c^2*d + e)*Sqrt[1 - 1/(c^2*x^2)]*x)/c^3 - (b*e^2*Sqrt[1 - 1/(c^2*x^2)]*x^3)/(12*c) - (I/2)*b*d^2*
ArcCsc[c*x]^2 + d*e*x^2*(a + b*ArcSec[c*x]) + (e^2*x^4*(a + b*ArcSec[c*x]))/4 + b*d^2*ArcCsc[c*x]*Log[1 - E^((
2*I)*ArcCsc[c*x])] - b*d^2*ArcCsc[c*x]*Log[x^(-1)] - d^2*(a + b*ArcSec[c*x])*Log[x^(-1)] - (I/2)*b*d^2*PolyLog
[2, E^((2*I)*ArcCsc[c*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 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 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -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*} \int \frac {\left (d+e x^2\right )^2 \left (a+b \sec ^{-1}(c x)\right )}{x} \, 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^5} \, dx,x,\frac {1}{x}\right )\\ &=d e x^2 \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sec ^{-1}(c x)\right )-d^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\frac {b \text {Subst}\left (\int \frac {-\frac {e \left (e+4 d x^2\right )}{4 x^4}+d^2 \log (x)}{\sqrt {1-\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{c}\\ &=d e x^2 \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sec ^{-1}(c x)\right )-d^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\frac {b \text {Subst}\left (\int \left (-\frac {e \left (e+4 d x^2\right )}{4 x^4 \sqrt {1-\frac {x^2}{c^2}}}+\frac {d^2 \log (x)}{\sqrt {1-\frac {x^2}{c^2}}}\right ) \, dx,x,\frac {1}{x}\right )}{c}\\ &=d e x^2 \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sec ^{-1}(c x)\right )-d^2 \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 {\log (x)}{\sqrt {1-\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{c}+\frac {(b e) \text {Subst}\left (\int \frac {e+4 d x^2}{x^4 \sqrt {1-\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{4 c}\\ &=-\frac {b e^2 \sqrt {1-\frac {1}{c^2 x^2}} x^3}{12 c}+d e x^2 \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sec ^{-1}(c x)\right )-b d^2 \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-d^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )+\left (b d^2\right ) \text {Subst}\left (\int \frac {\sin ^{-1}\left (\frac {x}{c}\right )}{x} \, dx,x,\frac {1}{x}\right )+\frac {\left (b e \left (6 c^2 d+e\right )\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1-\frac {x^2}{c^2}}} \, dx,x,\frac {1}{x}\right )}{6 c^3}\\ &=-\frac {b e \left (6 c^2 d+e\right ) \sqrt {1-\frac {1}{c^2 x^2}} x}{6 c^3}-\frac {b e^2 \sqrt {1-\frac {1}{c^2 x^2}} x^3}{12 c}+d e x^2 \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sec ^{-1}(c x)\right )-b d^2 \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-d^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )+\left (b d^2\right ) \text {Subst}\left (\int x \cot (x) \, dx,x,\csc ^{-1}(c x)\right )\\ &=-\frac {b e \left (6 c^2 d+e\right ) \sqrt {1-\frac {1}{c^2 x^2}} x}{6 c^3}-\frac {b e^2 \sqrt {1-\frac {1}{c^2 x^2}} x^3}{12 c}-\frac {1}{2} i b d^2 \csc ^{-1}(c x)^2+d e x^2 \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sec ^{-1}(c x)\right )-b d^2 \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-d^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\left (2 i b d^2\right ) \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 \left (6 c^2 d+e\right ) \sqrt {1-\frac {1}{c^2 x^2}} x}{6 c^3}-\frac {b e^2 \sqrt {1-\frac {1}{c^2 x^2}} x^3}{12 c}-\frac {1}{2} i b d^2 \csc ^{-1}(c x)^2+d e x^2 \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sec ^{-1}(c x)\right )+b d^2 \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b d^2 \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-d^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\left (b d^2\right ) \text {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\csc ^{-1}(c x)\right )\\ &=-\frac {b e \left (6 c^2 d+e\right ) \sqrt {1-\frac {1}{c^2 x^2}} x}{6 c^3}-\frac {b e^2 \sqrt {1-\frac {1}{c^2 x^2}} x^3}{12 c}-\frac {1}{2} i b d^2 \csc ^{-1}(c x)^2+d e x^2 \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sec ^{-1}(c x)\right )+b d^2 \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b d^2 \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-d^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )+\frac {1}{2} \left (i b d^2\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \csc ^{-1}(c x)}\right )\\ &=-\frac {b e \left (6 c^2 d+e\right ) \sqrt {1-\frac {1}{c^2 x^2}} x}{6 c^3}-\frac {b e^2 \sqrt {1-\frac {1}{c^2 x^2}} x^3}{12 c}-\frac {1}{2} i b d^2 \csc ^{-1}(c x)^2+d e x^2 \left (a+b \sec ^{-1}(c x)\right )+\frac {1}{4} e^2 x^4 \left (a+b \sec ^{-1}(c x)\right )+b d^2 \csc ^{-1}(c x) \log \left (1-e^{2 i \csc ^{-1}(c x)}\right )-b d^2 \csc ^{-1}(c x) \log \left (\frac {1}{x}\right )-d^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (\frac {1}{x}\right )-\frac {1}{2} i b d^2 \text {Li}_2\left (e^{2 i \csc ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 160, normalized size = 0.86 \begin {gather*} a d e x^2+\frac {1}{4} a e^2 x^4-\frac {b e^2 \sqrt {1-\frac {1}{c^2 x^2}} x \left (2+c^2 x^2\right )}{12 c^3}+\frac {1}{4} b e^2 x^4 \sec ^{-1}(c x)+\frac {b d e x \left (-\sqrt {1-\frac {1}{c^2 x^2}}+c x \sec ^{-1}(c x)\right )}{c}+a d^2 \log (x)+\frac {1}{2} i b d^2 \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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a*d*e*x^2 + (a*e^2*x^4)/4 - (b*e^2*Sqrt[1 - 1/(c^2*x^2)]*x*(2 + c^2*x^2))/(12*c^3) + (b*e^2*x^4*ArcSec[c*x])/4
+ (b*d*e*x*(-Sqrt[1 - 1/(c^2*x^2)] + c*x*ArcSec[c*x]))/c + a*d^2*Log[x] + (I/2)*b*d^2*(ArcSec[c*x]*(ArcSec[c*
x] + (2*I)*Log[1 + E^((2*I)*ArcSec[c*x])]) + PolyLog[2, -E^((2*I)*ArcSec[c*x])])

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Maple [A]
time = 1.70, size = 242, normalized size = 1.30

method	result	size
derivativedivides	\(a d e \,x^{2}+\frac {a \,e^{2} x^{4}}{4}+a \,d^{2} \ln \left (c x \right )+\frac {i b \mathrm {arcsec}\left (c x \right )^{2} d^{2}}{2}+b \,\mathrm {arcsec}\left (c x \right ) d e \,x^{2}+\frac {b \,\mathrm {arcsec}\left (c x \right ) e^{2} x^{4}}{4}-\frac {b \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, d e x}{c}-\frac {b \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, e^{2} x^{3}}{12 c}-\frac {i b d e}{c^{2}}-\frac {b \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, e^{2} x}{6 c^{3}}-\frac {i b \,e^{2}}{6 c^{4}}-b \,d^{2} \mathrm {arcsec}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )+\frac {i b \,d^{2} \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{2}\)	\(242\)
default	\(a d e \,x^{2}+\frac {a \,e^{2} x^{4}}{4}+a \,d^{2} \ln \left (c x \right )+\frac {i b \mathrm {arcsec}\left (c x \right )^{2} d^{2}}{2}+b \,\mathrm {arcsec}\left (c x \right ) d e \,x^{2}+\frac {b \,\mathrm {arcsec}\left (c x \right ) e^{2} x^{4}}{4}-\frac {b \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, d e x}{c}-\frac {b \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, e^{2} x^{3}}{12 c}-\frac {i b d e}{c^{2}}-\frac {b \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, e^{2} x}{6 c^{3}}-\frac {i b \,e^{2}}{6 c^{4}}-b \,d^{2} \mathrm {arcsec}\left (c x \right ) \ln \left (1+\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )+\frac {i b \,d^{2} \polylog \left (2, -\left (\frac {1}{c x}+i \sqrt {1-\frac {1}{c^{2} x^{2}}}\right )^{2}\right )}{2}\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d*e*x^2+1/4*a*e^2*x^4+a*d^2*ln(c*x)+1/2*I*b*arcsec(c*x)^2*d^2+b*arcsec(c*x)*d*e*x^2+1/4*b*arcsec(c*x)*e^2*x^
4-b/c*((c^2*x^2-1)/c^2/x^2)^(1/2)*d*e*x-1/12*b/c*((c^2*x^2-1)/c^2/x^2)^(1/2)*e^2*x^3-I*b/c^2*d*e-1/6*b/c^3*((c
^2*x^2-1)/c^2/x^2)^(1/2)*e^2*x-1/6*I*b/c^4*e^2-b*d^2*arcsec(c*x)*ln(1+(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)+1/2*I*b
*d^2*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,x, algorithm="maxima")

[Out]

1/4*a*x^4*e^2 + a*d*x^2*e + a*d^2*log(x) - 1/8*(-2*I*b*c^4*x^4*e^2*log(c) - 4*I*b*c^4*d^2*log(-c*x + 1)*log(x)
- 4*I*b*c^4*d^2*log(x)^2 - 4*I*b*c^4*d^2*dilog(c*x) - 4*I*b*c^4*d^2*dilog(-c*x) + I*(4*b*d*(log(c*x + 1)/c^2
+ log(c*x - 1)/c^2)*e + 32*b*d^2*integrate(1/4*log(x)/(c^2*x^3 - x), x) + b*(x^2/c^2 + log(c*x + 1)/c^4 + log(
c*x - 1)/c^4)*e^2)*c^4 + 8*c^4*integrate(1/4*(b*x^4*e^2 + 4*b*d*x^2*e + 4*b*d^2*log(x))*sqrt(c*x + 1)*sqrt(c*x
- 1)/(c^2*x^3 - x), x) + (-8*I*b*c^4*d*e*log(c) - I*b*c^2*e^2)*x^2 - 2*(b*c^4*x^4*e^2 + 4*b*c^4*d*x^2*e + 4*b
*c^4*d^2*log(x))*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) + (I*b*c^4*x^4*e^2 + 4*I*b*c^4*d*x^2*e + 4*I*b*c^4*d^2*lo
g(x))*log(c^2*x^2) + (-4*I*b*c^4*d^2*log(x) - 4*I*b*c^2*d*e - I*b*e^2)*log(c*x + 1) + (-4*I*b*c^2*d*e - I*b*e^
2)*log(c*x - 1) - 2*(I*b*c^4*x^4*e^2 + 4*I*b*c^4*d*x^2*e + 4*I*b*c^4*d^2*log(c))*log(x))/c^4

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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,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, 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}\, 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,x)

[Out]

Integral((a + b*asec(c*x))*(d + e*x**2)**2/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((e*x^2+d)^2*(a+b*arcsec(c*x))/x,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)]U
ndef/Unsigned

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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} \,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,x)

[Out]

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

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