3.149 \(\int \frac {a+b \sec ^{-1}(c x)}{x^2 (d+e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=274 \[ -\frac {2 e x \left (a+b \sec ^{-1}(c x)\right )}{d^2 \sqrt {d+e x^2}}-\frac {a+b \sec ^{-1}(c x)}{d x \sqrt {d+e x^2}}+\frac {b c \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{d^2 \sqrt {c^2 x^2}}+\frac {b x \sqrt {1-c^2 x^2} \left (c^2 d+2 e\right ) \sqrt {\frac {e x^2}{d}+1} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{d^2 \sqrt {c^2 x^2} \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}-\frac {b c^2 x \sqrt {1-c^2 x^2} \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{d^2 \sqrt {c^2 x^2} \sqrt {c^2 x^2-1} \sqrt {\frac {e x^2}{d}+1}} \]

[Out]

(-a-b*arcsec(c*x))/d/x/(e*x^2+d)^(1/2)-2*e*x*(a+b*arcsec(c*x))/d^2/(e*x^2+d)^(1/2)+b*c*(c^2*x^2-1)^(1/2)*(e*x^
2+d)^(1/2)/d^2/(c^2*x^2)^(1/2)-b*c^2*x*EllipticE(c*x,(-e/c^2/d)^(1/2))*(-c^2*x^2+1)^(1/2)*(e*x^2+d)^(1/2)/d^2/
(c^2*x^2)^(1/2)/(c^2*x^2-1)^(1/2)/(1+e*x^2/d)^(1/2)+b*(c^2*d+2*e)*x*EllipticF(c*x,(-e/c^2/d)^(1/2))*(-c^2*x^2+
1)^(1/2)*(1+e*x^2/d)^(1/2)/d^2/(c^2*x^2)^(1/2)/(c^2*x^2-1)^(1/2)/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {271, 191, 5238, 12, 583, 524, 427, 426, 424, 421, 419} \[ -\frac {2 e x \left (a+b \sec ^{-1}(c x)\right )}{d^2 \sqrt {d+e x^2}}-\frac {a+b \sec ^{-1}(c x)}{d x \sqrt {d+e x^2}}+\frac {b c \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{d^2 \sqrt {c^2 x^2}}+\frac {b x \sqrt {1-c^2 x^2} \left (c^2 d+2 e\right ) \sqrt {\frac {e x^2}{d}+1} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{d^2 \sqrt {c^2 x^2} \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}-\frac {b c^2 x \sqrt {1-c^2 x^2} \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{d^2 \sqrt {c^2 x^2} \sqrt {c^2 x^2-1} \sqrt {\frac {e x^2}{d}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*Sqrt[-1 + c^2*x^2]*Sqrt[d + e*x^2])/(d^2*Sqrt[c^2*x^2]) - (a + b*ArcSec[c*x])/(d*x*Sqrt[d + e*x^2]) - (2*
e*x*(a + b*ArcSec[c*x]))/(d^2*Sqrt[d + e*x^2]) - (b*c^2*x*Sqrt[1 - c^2*x^2]*Sqrt[d + e*x^2]*EllipticE[ArcSin[c
*x], -(e/(c^2*d))])/(d^2*Sqrt[c^2*x^2]*Sqrt[-1 + c^2*x^2]*Sqrt[1 + (e*x^2)/d]) + (b*(c^2*d + 2*e)*x*Sqrt[1 - c
^2*x^2]*Sqrt[1 + (e*x^2)/d]*EllipticF[ArcSin[c*x], -(e/(c^2*d))])/(d^2*Sqrt[c^2*x^2]*Sqrt[-1 + c^2*x^2]*Sqrt[d
+ e*x^2])

Rule 12

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

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 271

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

Rule 419

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

Rule 421

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d*x^2)/c]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d*x^2)/c]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 424

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

Rule 426

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b*x^2)/a]
, Int[Sqrt[1 + (b*x^2)/a]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 427

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + (d*x^2)/c]/Sqrt[c + d*x^2]
, Int[Sqrt[a + b*x^2]/Sqrt[1 + (d*x^2)/c], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] &&  !GtQ[c, 0]

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 5238

Int[((a_.) + ArcSec[(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*ArcSec[c*x], u, x] - Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&  !(ILtQ
[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rubi steps

\begin {align*} \int \frac {a+b \sec ^{-1}(c x)}{x^2 \left (d+e x^2\right )^{3/2}} \, dx &=-\frac {a+b \sec ^{-1}(c x)}{d x \sqrt {d+e x^2}}-\frac {2 e x \left (a+b \sec ^{-1}(c x)\right )}{d^2 \sqrt {d+e x^2}}-\frac {(b c x) \int \frac {-d-2 e x^2}{d^2 x^2 \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}} \, dx}{\sqrt {c^2 x^2}}\\ &=-\frac {a+b \sec ^{-1}(c x)}{d x \sqrt {d+e x^2}}-\frac {2 e x \left (a+b \sec ^{-1}(c x)\right )}{d^2 \sqrt {d+e x^2}}-\frac {(b c x) \int \frac {-d-2 e x^2}{x^2 \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}} \, dx}{d^2 \sqrt {c^2 x^2}}\\ &=\frac {b c \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{d^2 \sqrt {c^2 x^2}}-\frac {a+b \sec ^{-1}(c x)}{d x \sqrt {d+e x^2}}-\frac {2 e x \left (a+b \sec ^{-1}(c x)\right )}{d^2 \sqrt {d+e x^2}}-\frac {(b c x) \int \frac {-2 d e+c^2 d e x^2}{\sqrt {-1+c^2 x^2} \sqrt {d+e x^2}} \, dx}{d^3 \sqrt {c^2 x^2}}\\ &=\frac {b c \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{d^2 \sqrt {c^2 x^2}}-\frac {a+b \sec ^{-1}(c x)}{d x \sqrt {d+e x^2}}-\frac {2 e x \left (a+b \sec ^{-1}(c x)\right )}{d^2 \sqrt {d+e x^2}}-\frac {\left (b c^3 x\right ) \int \frac {\sqrt {d+e x^2}}{\sqrt {-1+c^2 x^2}} \, dx}{d^2 \sqrt {c^2 x^2}}+\frac {\left (b c \left (c^2 d+2 e\right ) x\right ) \int \frac {1}{\sqrt {-1+c^2 x^2} \sqrt {d+e x^2}} \, dx}{d^2 \sqrt {c^2 x^2}}\\ &=\frac {b c \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{d^2 \sqrt {c^2 x^2}}-\frac {a+b \sec ^{-1}(c x)}{d x \sqrt {d+e x^2}}-\frac {2 e x \left (a+b \sec ^{-1}(c x)\right )}{d^2 \sqrt {d+e x^2}}-\frac {\left (b c^3 x \sqrt {1-c^2 x^2}\right ) \int \frac {\sqrt {d+e x^2}}{\sqrt {1-c^2 x^2}} \, dx}{d^2 \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2}}+\frac {\left (b c \left (c^2 d+2 e\right ) x \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {1}{\sqrt {-1+c^2 x^2} \sqrt {1+\frac {e x^2}{d}}} \, dx}{d^2 \sqrt {c^2 x^2} \sqrt {d+e x^2}}\\ &=\frac {b c \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{d^2 \sqrt {c^2 x^2}}-\frac {a+b \sec ^{-1}(c x)}{d x \sqrt {d+e x^2}}-\frac {2 e x \left (a+b \sec ^{-1}(c x)\right )}{d^2 \sqrt {d+e x^2}}-\frac {\left (b c^3 x \sqrt {1-c^2 x^2} \sqrt {d+e x^2}\right ) \int \frac {\sqrt {1+\frac {e x^2}{d}}}{\sqrt {1-c^2 x^2}} \, dx}{d^2 \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2} \sqrt {1+\frac {e x^2}{d}}}+\frac {\left (b c \left (c^2 d+2 e\right ) x \sqrt {1-c^2 x^2} \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {1}{\sqrt {1-c^2 x^2} \sqrt {1+\frac {e x^2}{d}}} \, dx}{d^2 \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}\\ &=\frac {b c \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{d^2 \sqrt {c^2 x^2}}-\frac {a+b \sec ^{-1}(c x)}{d x \sqrt {d+e x^2}}-\frac {2 e x \left (a+b \sec ^{-1}(c x)\right )}{d^2 \sqrt {d+e x^2}}-\frac {b c^2 x \sqrt {1-c^2 x^2} \sqrt {d+e x^2} E\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{d^2 \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2} \sqrt {1+\frac {e x^2}{d}}}+\frac {b \left (c^2 d+2 e\right ) x \sqrt {1-c^2 x^2} \sqrt {1+\frac {e x^2}{d}} F\left (\sin ^{-1}(c x)|-\frac {e}{c^2 d}\right )}{d^2 \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.48, size = 212, normalized size = 0.77 \[ \frac {-a \left (d+2 e x^2\right )+b c x \sqrt {1-\frac {1}{c^2 x^2}} \left (d+e x^2\right )-b \sec ^{-1}(c x) \left (d+2 e x^2\right )}{d^2 x \sqrt {d+e x^2}}-\frac {i b c x \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {\frac {e x^2}{d}+1} \left (c^2 d E\left (i \sinh ^{-1}\left (\sqrt {-c^2} x\right )|-\frac {e}{c^2 d}\right )-\left (c^2 d+2 e\right ) F\left (i \sinh ^{-1}\left (\sqrt {-c^2} x\right )|-\frac {e}{c^2 d}\right )\right )}{\sqrt {-c^2} d^2 \sqrt {1-c^2 x^2} \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*Sqrt[1 - 1/(c^2*x^2)]*x*(d + e*x^2) - a*(d + 2*e*x^2) - b*(d + 2*e*x^2)*ArcSec[c*x])/(d^2*x*Sqrt[d + e*x^
2]) - (I*b*c*Sqrt[1 - 1/(c^2*x^2)]*x*Sqrt[1 + (e*x^2)/d]*(c^2*d*EllipticE[I*ArcSinh[Sqrt[-c^2]*x], -(e/(c^2*d)
)] - (c^2*d + 2*e)*EllipticF[I*ArcSinh[Sqrt[-c^2]*x], -(e/(c^2*d))]))/(Sqrt[-c^2]*d^2*Sqrt[1 - c^2*x^2]*Sqrt[d
+ e*x^2])

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {e x^{2} + d} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )}}{e^{2} x^{6} + 2 \, d e x^{4} + d^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(e*x^2 + d)*(b*arcsec(c*x) + a)/(e^2*x^6 + 2*d*e*x^4 + d^2*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arcsec}\left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.12, size = 0, normalized size = 0.00 \[ \int \frac {a +b \,\mathrm {arcsec}\left (c x \right )}{x^{2} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -a {\left (\frac {2 \, e x}{\sqrt {e x^{2} + d} d^{2}} + \frac {1}{\sqrt {e x^{2} + d} d x}\right )} - \frac {{\left ({\left (2 \, e x^{2} + d\right )} \sqrt {e x^{2} + d} \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right ) - {\left (2 \, e x^{2} \log \relax (c) + d \log \relax (c) + {\left (2 \, e x^{2} + d\right )} \log \relax (x)\right )} \sqrt {e x^{2} + d}\right )} b}{d^{2} e x^{3} + d^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(2*e*x/(sqrt(e*x^2 + d)*d^2) + 1/(sqrt(e*x^2 + d)*d*x)) - ((2*e*x^2 + d)*sqrt(e*x^2 + d)*arctan(sqrt(c*x +
1)*sqrt(c*x - 1)) + (d^2*e*x^3 + d^3*x)*integrate((c^2*d^2*x^2*log(c) - d^2*log(c) - (2*c^2*e^2*x^6 + 3*c^2*d*
e*x^4 - (c^2*log(c) - c^2)*d^2*x^2 + d^2*log(c))*e^(log(c*x + 1) + log(c*x - 1)) + (c^2*d^2*x^2 - d^2 + (c^2*d
^2*x^2 - d^2)*e^(log(c*x + 1) + log(c*x - 1)))*log(x))/((c^2*d^2*e*x^6 - d^3*x^2 + (c^2*d^3 - d^2*e)*x^4 + (c^
2*d^2*e*x^6 - d^3*x^2 + (c^2*d^3 - d^2*e)*x^4)*e^(log(c*x + 1) + log(c*x - 1)))*sqrt(e*x^2 + d)), x))*b/(d^2*e
*x^3 + d^3*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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