3.2.48 \(\int \frac {a+b \sec ^{-1}(c x)}{(d+e x^2)^{3/2}} \, dx\) [148]

Optimal. Leaf size=109 \[ \frac {x \left (a+b \sec ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {b x \sqrt {1-c^2 x^2} \sqrt {1+\frac {e x^2}{d}} F\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{d \sqrt {c^2 x^2} \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}} \]

[Out]

x*(a+b*arcsec(c*x))/d/(e*x^2+d)^(1/2)-b*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/(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 = 0.06, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {197, 5336, 12, 432, 430} \begin {gather*} \frac {x \left (a+b \sec ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {b x \sqrt {1-c^2 x^2} \sqrt {\frac {e x^2}{d}+1} F\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{d \sqrt {c^2 x^2} \sqrt {c^2 x^2-1} \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*ArcSec[c*x]))/(d*Sqrt[d + e*x^2]) - (b*x*Sqrt[1 - c^2*x^2]*Sqrt[1 + (e*x^2)/d]*EllipticF[ArcSin[c*x]
, -(e/(c^2*d))])/(d*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 197

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 430

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

Rule 432

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

Rule 5336

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^p, x]}, Dist[a + b*ArcSec[c*x], u, x] - Dist[b*c*(x/Sqrt[c^2*x^2]), Int[SimplifyIntegrand[u/(x*Sqrt[c^2*x^2
- 1]), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rubi steps

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

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Mathematica [A]
time = 0.69, size = 113, normalized size = 1.04 \begin {gather*} \frac {x \left (a+b \sec ^{-1}(c x)\right )}{d \sqrt {d+e x^2}}-\frac {b c \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {1-c^2 x^2} \sqrt {1+\frac {e x^2}{d}} F\left (\text {ArcSin}(c x)\left |-\frac {e}{c^2 d}\right .\right )}{d \left (-c+c^3 x^2\right ) \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*ArcSec[c*x]))/(d*Sqrt[d + e*x^2]) - (b*c*Sqrt[1 - 1/(c^2*x^2)]*x*Sqrt[1 - c^2*x^2]*Sqrt[1 + (e*x^2)/
d]*EllipticF[ArcSin[c*x], -(e/(c^2*d))])/(d*(-c + c^3*x^2)*Sqrt[d + e*x^2])

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Maple [F]
time = 1.22, size = 0, normalized size = 0.00 \[\int \frac {a +b \,\mathrm {arcsec}\left (c x \right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]
time = 0.32, size = 77, normalized size = 0.71 \begin {gather*} \frac {{\left (b x^{2} e + b d\right )} \sqrt {-d} {\rm ellipticF}\left (c x, -\frac {e}{c^{2} d}\right ) + {\left (b c d x \operatorname {arcsec}\left (c x\right ) + a c d x\right )} \sqrt {x^{2} e + d}}{c d^{2} x^{2} e + c d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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