3.30 \(\int \frac{(a+b \sec ^{-1}(c x))^3}{x^3} \, dx\)

Optimal. Leaf size=137 \[ -\frac{3}{4} b^2 \left (c^2-\frac{1}{x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac{3 b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{4 x}+\frac{1}{2} \left (c^2-\frac{1}{x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )^3-\frac{1}{4} c^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{3 b^3 c \sqrt{1-\frac{1}{c^2 x^2}}}{8 x}+\frac{3}{8} b^3 c^2 \sec ^{-1}(c x) \]

[Out]

(-3*b^3*c*Sqrt[1 - 1/(c^2*x^2)])/(8*x) + (3*b^3*c^2*ArcSec[c*x])/8 - (3*b^2*(c^2 - x^(-2))*(a + b*ArcSec[c*x])
)/4 + (3*b*c*Sqrt[1 - 1/(c^2*x^2)]*(a + b*ArcSec[c*x])^2)/(4*x) - (c^2*(a + b*ArcSec[c*x])^3)/4 + ((c^2 - x^(-
2))*(a + b*ArcSec[c*x])^3)/2

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Rubi [A]  time = 0.10455, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {5222, 4404, 3311, 32, 2635, 8} \[ -\frac{3}{4} b^2 \left (c^2-\frac{1}{x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac{3 b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{4 x}+\frac{1}{2} \left (c^2-\frac{1}{x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )^3-\frac{1}{4} c^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{3 b^3 c \sqrt{1-\frac{1}{c^2 x^2}}}{8 x}+\frac{3}{8} b^3 c^2 \sec ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])^3/x^3,x]

[Out]

(-3*b^3*c*Sqrt[1 - 1/(c^2*x^2)])/(8*x) + (3*b^3*c^2*ArcSec[c*x])/8 - (3*b^2*(c^2 - x^(-2))*(a + b*ArcSec[c*x])
)/4 + (3*b*c*Sqrt[1 - 1/(c^2*x^2)]*(a + b*ArcSec[c*x])^2)/(4*x) - (c^2*(a + b*ArcSec[c*x])^3)/4 + ((c^2 - x^(-
2))*(a + b*ArcSec[c*x])^3)/2

Rule 5222

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*S
ec[x]^(m + 1)*Tan[x], x], x, ArcSec[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (GtQ[n,
0] || LtQ[m, -1])

Rule 4404

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[((c +
d*x)^m*Sin[a + b*x]^(n + 1))/(b*(n + 1)), x] - Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n +
1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sec ^{-1}(c x)\right )^3}{x^3} \, dx &=c^2 \operatorname{Subst}\left (\int (a+b x)^3 \cos (x) \sin (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{1}{2} \left (c^2-\frac{1}{x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )^3-\frac{1}{2} \left (3 b c^2\right ) \operatorname{Subst}\left (\int (a+b x)^2 \sin ^2(x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=-\frac{3}{4} b^2 \left (c^2-\frac{1}{x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac{3 b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{4 x}+\frac{1}{2} \left (c^2-\frac{1}{x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )^3-\frac{1}{4} \left (3 b c^2\right ) \operatorname{Subst}\left (\int (a+b x)^2 \, dx,x,\sec ^{-1}(c x)\right )+\frac{1}{4} \left (3 b^3 c^2\right ) \operatorname{Subst}\left (\int \sin ^2(x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=-\frac{3 b^3 c \sqrt{1-\frac{1}{c^2 x^2}}}{8 x}-\frac{3}{4} b^2 \left (c^2-\frac{1}{x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac{3 b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{4 x}-\frac{1}{4} c^2 \left (a+b \sec ^{-1}(c x)\right )^3+\frac{1}{2} \left (c^2-\frac{1}{x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )^3+\frac{1}{8} \left (3 b^3 c^2\right ) \operatorname{Subst}\left (\int 1 \, dx,x,\sec ^{-1}(c x)\right )\\ &=-\frac{3 b^3 c \sqrt{1-\frac{1}{c^2 x^2}}}{8 x}+\frac{3}{8} b^3 c^2 \sec ^{-1}(c x)-\frac{3}{4} b^2 \left (c^2-\frac{1}{x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )+\frac{3 b c \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{4 x}-\frac{1}{4} c^2 \left (a+b \sec ^{-1}(c x)\right )^3+\frac{1}{2} \left (c^2-\frac{1}{x^2}\right ) \left (a+b \sec ^{-1}(c x)\right )^3\\ \end{align*}

Mathematica [A]  time = 0.202228, size = 185, normalized size = 1.35 \[ \frac{3 b c^2 x^2 \left (b^2-2 a^2\right ) \sin ^{-1}\left (\frac{1}{c x}\right )+6 b \sec ^{-1}(c x) \left (-2 a^2+2 a b c x \sqrt{1-\frac{1}{c^2 x^2}}+b^2\right )+6 a^2 b c x \sqrt{1-\frac{1}{c^2 x^2}}-4 a^3+6 b^2 \sec ^{-1}(c x)^2 \left (a \left (c^2 x^2-2\right )+b c x \sqrt{1-\frac{1}{c^2 x^2}}\right )+6 a b^2-3 b^3 c x \sqrt{1-\frac{1}{c^2 x^2}}+2 b^3 \left (c^2 x^2-2\right ) \sec ^{-1}(c x)^3}{8 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*a^3 + 6*a*b^2 + 6*a^2*b*c*Sqrt[1 - 1/(c^2*x^2)]*x - 3*b^3*c*Sqrt[1 - 1/(c^2*x^2)]*x + 6*b*(-2*a^2 + b^2 +
2*a*b*c*Sqrt[1 - 1/(c^2*x^2)]*x)*ArcSec[c*x] + 6*b^2*(b*c*Sqrt[1 - 1/(c^2*x^2)]*x + a*(-2 + c^2*x^2))*ArcSec[c
*x]^2 + 2*b^3*(-2 + c^2*x^2)*ArcSec[c*x]^3 + 3*b*(-2*a^2 + b^2)*c^2*x^2*ArcSin[1/(c*x)])/(8*x^2)

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Maple [B]  time = 0.323, size = 324, normalized size = 2.4 \begin{align*} -{\frac{{a}^{3}}{2\,{x}^{2}}}-{\frac{{b}^{3} \left ({\rm arcsec} \left (cx\right ) \right ) ^{3}}{2\,{x}^{2}}}+{\frac{{c}^{2}{b}^{3} \left ({\rm arcsec} \left (cx\right ) \right ) ^{3}}{4}}+{\frac{3\,c{b}^{3} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}}{4\,x}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}+{\frac{3\,{b}^{3}{\rm arcsec} \left (cx\right )}{4\,{x}^{2}}}-{\frac{3\,c{b}^{3}}{8\,x}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}-{\frac{3\,{b}^{3}{c}^{2}{\rm arcsec} \left (cx\right )}{8}}-{\frac{3\,a{b}^{2} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}}{2\,{x}^{2}}}+{\frac{3\,{c}^{2}a{b}^{2} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}}{4}}+{\frac{3\,a{b}^{2}c{\rm arcsec} \left (cx\right )}{2\,x}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}-{\frac{3\,{c}^{2}a{b}^{2}}{4}}+{\frac{3\,a{b}^{2}}{4\,{x}^{2}}}-{\frac{3\,{a}^{2}b{\rm arcsec} \left (cx\right )}{2\,{x}^{2}}}-{\frac{3\,{a}^{2}cb}{4\,x}\sqrt{{c}^{2}{x}^{2}-1}\arctan \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{3\,{a}^{2}cb}{4\,x}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{3\,{a}^{2}b}{4\,c{x}^{3}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a^3/x^2-1/2*b^3/x^2*arcsec(c*x)^3+1/4*c^2*b^3*arcsec(c*x)^3+3/4*c*b^3*arcsec(c*x)^2/x*((c^2*x^2-1)/c^2/x^
2)^(1/2)+3/4*b^3/x^2*arcsec(c*x)-3/8*c*b^3*((c^2*x^2-1)/c^2/x^2)^(1/2)/x-3/8*b^3*c^2*arcsec(c*x)-3/2*a*b^2/x^2
*arcsec(c*x)^2+3/4*c^2*a*b^2*arcsec(c*x)^2+3/2*c*a*b^2*arcsec(c*x)/x*((c^2*x^2-1)/c^2/x^2)^(1/2)-3/4*c^2*a*b^2
+3/4*a*b^2/x^2-3/2*a^2*b/x^2*arcsec(c*x)-3/4*c*a^2*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*arctan(1/
(c^2*x^2-1)^(1/2))+3/4*c*a^2*b/((c^2*x^2-1)/c^2/x^2)^(1/2)/x-3/4/c*a^2*b/((c^2*x^2-1)/c^2/x^2)^(1/2)/x^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*a^2*b*((c^4*x*sqrt(-1/(c^2*x^2) + 1)/(c^2*x^2*(1/(c^2*x^2) - 1) - 1) - c^3*arctan(c*x*sqrt(-1/(c^2*x^2) +
1)))/c + 2*arcsec(c*x)/x^2) - 1/2*a^3/x^2 - 1/8*(4*b^3*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^3 - 3*b^3*arctan(s
qrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)^2 + 12*(a*b^2*c^2*(log(c*x + 1) + log(c*x - 1) - 2*log(x))*log(c)^2 +
16*b^3*c^2*integrate(1/8*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(c^2*x^5 - x^3), x)*log(c)^2 - 16*b^3*c^2*in
tegrate(1/8*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)/(c^2*x^5 - x^3), x)*log(c) + 32*b^3*c^2*integ
rate(1/8*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(x)/(c^2*x^5 - x^3), x)*log(c) - 16*a*b^2*c^2*integrate(1/
8*x^2*log(c^2*x^2)/(c^2*x^5 - x^3), x)*log(c) + 32*a*b^2*c^2*integrate(1/8*x^2*log(x)/(c^2*x^5 - x^3), x)*log(
c) - 16*b^3*c^2*integrate(1/8*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)*log(x)/(c^2*x^5 - x^3), x)
+ 16*b^3*c^2*integrate(1/8*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(x)^2/(c^2*x^5 - x^3), x) - 16*a*b^2*c^2
*integrate(1/8*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2/(c^2*x^5 - x^3), x) + 8*b^3*c^2*integrate(1/8*x^2*arc
tan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)/(c^2*x^5 - x^3), x) + 4*a*b^2*c^2*integrate(1/8*x^2*log(c^2*x^2)
^2/(c^2*x^5 - x^3), x) - 16*a*b^2*c^2*integrate(1/8*x^2*log(c^2*x^2)*log(x)/(c^2*x^5 - x^3), x) + 16*a*b^2*c^2
*integrate(1/8*x^2*log(x)^2/(c^2*x^5 - x^3), x) - (c^2*log(c*x + 1) + c^2*log(c*x - 1) - 2*c^2*log(x) + 1/x^2)
*a*b^2*log(c)^2 - 16*b^3*integrate(1/8*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))/(c^2*x^5 - x^3), x)*log(c)^2 + 16*b
^3*integrate(1/8*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)/(c^2*x^5 - x^3), x)*log(c) - 32*b^3*integrat
e(1/8*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(x)/(c^2*x^5 - x^3), x)*log(c) + 16*a*b^2*integrate(1/8*log(c^2*x
^2)/(c^2*x^5 - x^3), x)*log(c) - 32*a*b^2*integrate(1/8*log(x)/(c^2*x^5 - x^3), x)*log(c) - 8*b^3*integrate(1/
8*sqrt(c*x + 1)*sqrt(c*x - 1)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2/(c^2*x^5 - x^3), x) + 2*b^3*integrate(1/8*
sqrt(c*x + 1)*sqrt(c*x - 1)*log(c^2*x^2)^2/(c^2*x^5 - x^3), x) + 16*b^3*integrate(1/8*arctan(sqrt(c*x + 1)*sqr
t(c*x - 1))*log(c^2*x^2)*log(x)/(c^2*x^5 - x^3), x) - 16*b^3*integrate(1/8*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))
*log(x)^2/(c^2*x^5 - x^3), x) + 16*a*b^2*integrate(1/8*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2/(c^2*x^5 - x^3),
x) - 8*b^3*integrate(1/8*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)/(c^2*x^5 - x^3), x) - 4*a*b^2*integr
ate(1/8*log(c^2*x^2)^2/(c^2*x^5 - x^3), x) + 16*a*b^2*integrate(1/8*log(c^2*x^2)*log(x)/(c^2*x^5 - x^3), x) -
16*a*b^2*integrate(1/8*log(x)^2/(c^2*x^5 - x^3), x))*x^2)/x^2

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Fricas [A]  time = 2.21898, size = 342, normalized size = 2.5 \begin{align*} \frac{2 \,{\left (b^{3} c^{2} x^{2} - 2 \, b^{3}\right )} \operatorname{arcsec}\left (c x\right )^{3} - 4 \, a^{3} + 6 \, a b^{2} + 6 \,{\left (a b^{2} c^{2} x^{2} - 2 \, a b^{2}\right )} \operatorname{arcsec}\left (c x\right )^{2} + 3 \,{\left ({\left (2 \, a^{2} b - b^{3}\right )} c^{2} x^{2} - 4 \, a^{2} b + 2 \, b^{3}\right )} \operatorname{arcsec}\left (c x\right ) + 3 \,{\left (2 \, b^{3} \operatorname{arcsec}\left (c x\right )^{2} + 4 \, a b^{2} \operatorname{arcsec}\left (c x\right ) + 2 \, a^{2} b - b^{3}\right )} \sqrt{c^{2} x^{2} - 1}}{8 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(b^3*c^2*x^2 - 2*b^3)*arcsec(c*x)^3 - 4*a^3 + 6*a*b^2 + 6*(a*b^2*c^2*x^2 - 2*a*b^2)*arcsec(c*x)^2 + 3*(
(2*a^2*b - b^3)*c^2*x^2 - 4*a^2*b + 2*b^3)*arcsec(c*x) + 3*(2*b^3*arcsec(c*x)^2 + 4*a*b^2*arcsec(c*x) + 2*a^2*
b - b^3)*sqrt(c^2*x^2 - 1))/x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asec(c*x))**3/x**3, 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 )}^{3}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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