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

Optimal. Leaf size=80 \[ -\frac {a+b \sec ^{-1}(c x)}{e \sqrt {d+e x^2}}-\frac {b c x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{\sqrt {d} e \sqrt {c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5236, 446, 93, 204} \[ -\frac {a+b \sec ^{-1}(c x)}{e \sqrt {d+e x^2}}-\frac {b c x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{\sqrt {d} e \sqrt {c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSec[c*x])/(e*Sqrt[d + e*x^2])) - (b*c*x*ArcTan[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[-1 + c^2*x^2])])/(Sqr
t[d]*e*Sqrt[c^2*x^2])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5236

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^(p +
1)*(a + b*ArcSec[c*x]))/(2*e*(p + 1)), x] - Dist[(b*c*x)/(2*e*(p + 1)*Sqrt[c^2*x^2]), Int[(d + e*x^2)^(p + 1)/
(x*Sqrt[c^2*x^2 - 1]), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 95, normalized size = 1.19 \[ \frac {b c x \sqrt {1-\frac {1}{c^2 x^2}} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {c^2 x^2-1}}{\sqrt {d+e x^2}}\right )}{\sqrt {d} e \sqrt {c^2 x^2-1}}-\frac {a+b \sec ^{-1}(c x)}{e \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSec[c*x])/(e*Sqrt[d + e*x^2])) + (b*c*Sqrt[1 - 1/(c^2*x^2)]*x*ArcTan[(Sqrt[d]*Sqrt[-1 + c^2*x^2])/
Sqrt[d + e*x^2]])/(Sqrt[d]*e*Sqrt[-1 + c^2*x^2])

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fricas [A]  time = 0.64, size = 283, normalized size = 3.54 \[ \left [-\frac {{\left (b e x^{2} + b d\right )} \sqrt {-d} \log \left (\frac {{\left (c^{4} d^{2} - 6 \, c^{2} d e + e^{2}\right )} x^{4} - 8 \, {\left (c^{2} d^{2} - d e\right )} x^{2} - 4 \, \sqrt {c^{2} x^{2} - 1} {\left ({\left (c^{2} d - e\right )} x^{2} - 2 \, d\right )} \sqrt {e x^{2} + d} \sqrt {-d} + 8 \, d^{2}}{x^{4}}\right ) + 4 \, \sqrt {e x^{2} + d} {\left (b d \operatorname {arcsec}\left (c x\right ) + a d\right )}}{4 \, {\left (d e^{2} x^{2} + d^{2} e\right )}}, -\frac {{\left (b e x^{2} + b d\right )} \sqrt {d} \arctan \left (-\frac {\sqrt {c^{2} x^{2} - 1} {\left ({\left (c^{2} d - e\right )} x^{2} - 2 \, d\right )} \sqrt {e x^{2} + d} \sqrt {d}}{2 \, {\left (c^{2} d e x^{4} + {\left (c^{2} d^{2} - d e\right )} x^{2} - d^{2}\right )}}\right ) + 2 \, \sqrt {e x^{2} + d} {\left (b d \operatorname {arcsec}\left (c x\right ) + a d\right )}}{2 \, {\left (d e^{2} x^{2} + d^{2} e\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((b*e*x^2 + b*d)*sqrt(-d)*log(((c^4*d^2 - 6*c^2*d*e + e^2)*x^4 - 8*(c^2*d^2 - d*e)*x^2 - 4*sqrt(c^2*x^2
- 1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(-d) + 8*d^2)/x^4) + 4*sqrt(e*x^2 + d)*(b*d*arcsec(c*x) + a*d
))/(d*e^2*x^2 + d^2*e), -1/2*((b*e*x^2 + b*d)*sqrt(d)*arctan(-1/2*sqrt(c^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)*sq
rt(e*x^2 + d)*sqrt(d)/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) + 2*sqrt(e*x^2 + d)*(b*d*arcsec(c*x) + a*d))/
(d*e^2*x^2 + d^2*e)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 3.36, size = 0, normalized size = 0.00 \[ \int \frac {x \left (a +b \,\mathrm {arcsec}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*(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 \[ -\frac {-\frac {1}{2} \, {\left ({\left (\frac {c^{4} d {\left (\frac {2 \, e}{{\left (c^{4} d + c^{2} e\right )} \sqrt {e x^{2} + d}} + \frac {e \log \left (\frac {\sqrt {e x^{2} + d} c^{2} - \sqrt {c^{2} d + e} c}{\sqrt {e x^{2} + d} c^{2} + \sqrt {c^{2} d + e} c}\right )}{{\left (c^{2} d + e\right )}^{\frac {3}{2}} c}\right )}}{e^{2}} + \frac {c^{4} {\left (\frac {2 \, d e}{{\left (c^{4} d + c^{2} e\right )} \sqrt {e x^{2} + d}} - \frac {e^{2} \log \left (\frac {\sqrt {e x^{2} + d} c^{2} - \sqrt {c^{2} d + e} c}{\sqrt {e x^{2} + d} c^{2} + \sqrt {c^{2} d + e} c}\right )}{{\left (c^{4} d + c^{2} e\right )} \sqrt {c^{2} d + e} c}\right )} \log \relax (c)}{e^{2}} - \frac {c^{4} {\left (\frac {2 \, d e}{{\left (c^{4} d + c^{2} e\right )} \sqrt {e x^{2} + d}} - \frac {e^{2} \log \left (\frac {\sqrt {e x^{2} + d} c^{2} - \sqrt {c^{2} d + e} c}{\sqrt {e x^{2} + d} c^{2} + \sqrt {c^{2} d + e} c}\right )}{{\left (c^{4} d + c^{2} e\right )} \sqrt {c^{2} d + e} c}\right )}}{e^{2}} - \frac {c^{2} d {\left (\frac {c \log \left (\frac {\sqrt {e x^{2} + d} c^{2} - \sqrt {c^{2} d + e} c}{\sqrt {e x^{2} + d} c^{2} + \sqrt {c^{2} d + e} c}\right )}{{\left (c^{2} d + e\right )}^{\frac {3}{2}}} - \frac {2 \, e}{{\left (c^{4} d^{2} + c^{2} d e\right )} \sqrt {e x^{2} + d}} - \frac {\log \left (\frac {\sqrt {e x^{2} + d} - \sqrt {d}}{\sqrt {e x^{2} + d} + \sqrt {d}}\right )}{c^{2} d^{\frac {3}{2}}}\right )}}{e} + \frac {c^{2} {\left (\frac {2 \, e}{{\left (c^{4} d + c^{2} e\right )} \sqrt {e x^{2} + d}} + \frac {e \log \left (\frac {\sqrt {e x^{2} + d} c^{2} - \sqrt {c^{2} d + e} c}{\sqrt {e x^{2} + d} c^{2} + \sqrt {c^{2} d + e} c}\right )}{{\left (c^{2} d + e\right )}^{\frac {3}{2}} c}\right )} \log \relax (c)}{e} - \frac {c^{2} {\left (\frac {2 \, e}{{\left (c^{4} d + c^{2} e\right )} \sqrt {e x^{2} + d}} + \frac {e \log \left (\frac {\sqrt {e x^{2} + d} c^{2} - \sqrt {c^{2} d + e} c}{\sqrt {e x^{2} + d} c^{2} + \sqrt {c^{2} d + e} c}\right )}{{\left (c^{2} d + e\right )}^{\frac {3}{2}} c}\right )}}{e} - 2 \, \int \frac {\sqrt {e x^{2} + d} x \log \relax (x)}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}\,{d x}\right )} \sqrt {e x^{2} + d} e - 2 \, \arctan \left (\sqrt {c x + 1} \sqrt {c x - 1}\right )\right )} b}{\sqrt {e x^{2} + d} e} - \frac {a}{\sqrt {e x^{2} + d} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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