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

Optimal. Leaf size=132 \[ \frac{\sqrt{d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e}+\frac{b c \sqrt{d} x \tan ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{c^2 x^2-1}}\right )}{e \sqrt{c^2 x^2}}-\frac{b x \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{c^2 x^2-1}}{c \sqrt{d+e x^2}}\right )}{\sqrt{e} \sqrt{c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.145577, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {5236, 446, 105, 63, 217, 206, 93, 204} \[ \frac{\sqrt{d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e}+\frac{b c \sqrt{d} x \tan ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{c^2 x^2-1}}\right )}{e \sqrt{c^2 x^2}}-\frac{b x \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{c^2 x^2-1}}{c \sqrt{d+e x^2}}\right )}{\sqrt{e} \sqrt{c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

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

Rule 206

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

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])

Rubi steps

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

Mathematica [A]  time = 0.219815, size = 211, normalized size = 1.6 \[ \frac{\sqrt{d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e}-\frac{b x \sqrt{1-\frac{1}{c^2 x^2}} \left (c^3 \sqrt{d} \sqrt{d+e x^2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{c^2 x^2-1}}{\sqrt{d+e x^2}}\right )+\sqrt{c^2} \sqrt{e} \sqrt{c^2 d+e} \sqrt{\frac{c^2 \left (d+e x^2\right )}{c^2 d+e}} \sinh ^{-1}\left (\frac{c \sqrt{e} \sqrt{c^2 x^2-1}}{\sqrt{c^2} \sqrt{c^2 d+e}}\right )\right )}{c^2 e \sqrt{c^2 x^2-1} \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.591, size = 0, normalized size = 0. \begin{align*} \int{x \left ( a+b{\rm arcsec} \left (cx\right ) \right ){\frac{1}{\sqrt{e{x}^{2}+d}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsec(c*x))/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.59961, size = 1925, normalized size = 14.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsec(c*x))/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(b*c*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) + b*sqrt(e)*log(8*c^4*e^2*x^4 + c^4*d^2 - 6*c^2*d*e + 8
*(c^4*d*e - c^2*e^2)*x^2 - 4*(2*c^3*e*x^2 + c^3*d - c*e)*sqrt(c^2*x^2 - 1)*sqrt(e*x^2 + d)*sqrt(e) + e^2) + 4*
sqrt(e*x^2 + d)*(b*c*arcsec(c*x) + a*c))/(c*e), 1/4*(2*b*c*sqrt(d)*arctan(-1/2*sqrt(c^2*x^2 - 1)*((c^2*d - e)*
x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(d)/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) + b*sqrt(e)*log(8*c^4*e^2*x^4 +
c^4*d^2 - 6*c^2*d*e + 8*(c^4*d*e - c^2*e^2)*x^2 - 4*(2*c^3*e*x^2 + c^3*d - c*e)*sqrt(c^2*x^2 - 1)*sqrt(e*x^2 +
d)*sqrt(e) + e^2) + 4*sqrt(e*x^2 + d)*(b*c*arcsec(c*x) + a*c))/(c*e), 1/4*(b*c*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) + 2*b*sqrt(-e)*arctan(1/2*(2*c^2*e*x^2 + c^2*d - e)*sqrt(c^2*x^2 - 1)*sqrt(e*x^2 + d)*sqrt(-e)/
(c^3*e^2*x^4 - c*d*e + (c^3*d*e - c*e^2)*x^2)) + 4*sqrt(e*x^2 + d)*(b*c*arcsec(c*x) + a*c))/(c*e), 1/2*(b*c*sq
rt(d)*arctan(-1/2*sqrt(c^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(d)/(c^2*d*e*x^4 + (c^2*d^2 -
d*e)*x^2 - d^2)) + b*sqrt(-e)*arctan(1/2*(2*c^2*e*x^2 + c^2*d - e)*sqrt(c^2*x^2 - 1)*sqrt(e*x^2 + d)*sqrt(-e)/
(c^3*e^2*x^4 - c*d*e + (c^3*d*e - c*e^2)*x^2)) + 2*sqrt(e*x^2 + d)*(b*c*arcsec(c*x) + a*c))/(c*e)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*asec(c*x))/(e*x**2+d)**(1/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )} x}{\sqrt{e x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arcsec(c*x))/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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