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

Optimal. Leaf size=80 \[ \frac {6 b^2 \left (a+b \sec ^{-1}(c x)\right )}{x}+3 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2-\frac {\left (a+b \sec ^{-1}(c x)\right )^3}{x}-6 b^3 c \sqrt {1-\frac {1}{c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {5222, 3296, 2637} \[ \frac {6 b^2 \left (a+b \sec ^{-1}(c x)\right )}{x}+3 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2-\frac {\left (a+b \sec ^{-1}(c x)\right )^3}{x}-6 b^3 c \sqrt {1-\frac {1}{c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])^3/x^2,x]

[Out]

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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])

Rubi steps

\begin {align*} \int \frac {\left (a+b \sec ^{-1}(c x)\right )^3}{x^2} \, dx &=c \operatorname {Subst}\left (\int (a+b x)^3 \sin (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=-\frac {\left (a+b \sec ^{-1}(c x)\right )^3}{x}+(3 b c) \operatorname {Subst}\left (\int (a+b x)^2 \cos (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=3 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2-\frac {\left (a+b \sec ^{-1}(c x)\right )^3}{x}-\left (6 b^2 c\right ) \operatorname {Subst}\left (\int (a+b x) \sin (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac {6 b^2 \left (a+b \sec ^{-1}(c x)\right )}{x}+3 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2-\frac {\left (a+b \sec ^{-1}(c x)\right )^3}{x}-\left (6 b^3 c\right ) \operatorname {Subst}\left (\int \cos (x) \, dx,x,\sec ^{-1}(c x)\right )\\ &=-6 b^3 c \sqrt {1-\frac {1}{c^2 x^2}}+\frac {6 b^2 \left (a+b \sec ^{-1}(c x)\right )}{x}+3 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2-\frac {\left (a+b \sec ^{-1}(c x)\right )^3}{x}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 141, normalized size = 1.76 \[ \frac {-a^3+3 b \sec ^{-1}(c x) \left (-a^2+2 a b c x \sqrt {1-\frac {1}{c^2 x^2}}+2 b^2\right )+3 a^2 b c x \sqrt {1-\frac {1}{c^2 x^2}}+3 b^2 \sec ^{-1}(c x)^2 \left (b c x \sqrt {1-\frac {1}{c^2 x^2}}-a\right )+6 a b^2-6 b^3 c x \sqrt {1-\frac {1}{c^2 x^2}}-b^3 \sec ^{-1}(c x)^3}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-a^3 + 6*a*b^2 + 3*a^2*b*c*Sqrt[1 - 1/(c^2*x^2)]*x - 6*b^3*c*Sqrt[1 - 1/(c^2*x^2)]*x + 3*b*(-a^2 + 2*b^2 + 2*
a*b*c*Sqrt[1 - 1/(c^2*x^2)]*x)*ArcSec[c*x] + 3*b^2*(-a + b*c*Sqrt[1 - 1/(c^2*x^2)]*x)*ArcSec[c*x]^2 - b^3*ArcS
ec[c*x]^3)/x

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fricas [A]  time = 0.69, size = 98, normalized size = 1.22 \[ -\frac {b^{3} \operatorname {arcsec}\left (c x\right )^{3} + 3 \, a b^{2} \operatorname {arcsec}\left (c x\right )^{2} + a^{3} - 6 \, a b^{2} + 3 \, {\left (a^{2} b - 2 \, b^{3}\right )} \operatorname {arcsec}\left (c x\right ) - 3 \, {\left (b^{3} \operatorname {arcsec}\left (c x\right )^{2} + 2 \, a b^{2} \operatorname {arcsec}\left (c x\right ) + a^{2} b - 2 \, b^{3}\right )} \sqrt {c^{2} x^{2} - 1}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.19, size = 196, normalized size = 2.45 \[ {\left (3 \, b^{3} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} \arccos \left (\frac {1}{c x}\right )^{2} + 6 \, a b^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} \arccos \left (\frac {1}{c x}\right ) - \frac {b^{3} \arccos \left (\frac {1}{c x}\right )^{3}}{c x} + 3 \, a^{2} b \sqrt {-\frac {1}{c^{2} x^{2}} + 1} - 6 \, b^{3} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} - \frac {3 \, a b^{2} \arccos \left (\frac {1}{c x}\right )^{2}}{c x} - \frac {3 \, a^{2} b \arccos \left (\frac {1}{c x}\right )}{c x} + \frac {6 \, b^{3} \arccos \left (\frac {1}{c x}\right )}{c x} - \frac {a^{3}}{c x} + \frac {6 \, a b^{2}}{c x}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.18, size = 198, normalized size = 2.48 \[ c \left (-\frac {a^{3}}{c x}+b^{3} \left (-\frac {\mathrm {arcsec}\left (c x \right )^{3}}{c x}+3 \mathrm {arcsec}\left (c x \right )^{2} \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}-6 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}+\frac {6 \,\mathrm {arcsec}\left (c x \right )}{c x}\right )+3 a \,b^{2} \left (-\frac {\mathrm {arcsec}\left (c x \right )^{2}}{c x}+\frac {2}{c x}+2 \,\mathrm {arcsec}\left (c x \right ) \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\right )+3 a^{2} b \left (-\frac {\mathrm {arcsec}\left (c x \right )}{c x}+\frac {c^{2} x^{2}-1}{\sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{2} x^{2}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.33, size = 146, normalized size = 1.82 \[ -\frac {b^{3} \operatorname {arcsec}\left (c x\right )^{3}}{x} + 3 \, {\left (c \sqrt {-\frac {1}{c^{2} x^{2}} + 1} - \frac {\operatorname {arcsec}\left (c x\right )}{x}\right )} a^{2} b + 6 \, {\left (c \sqrt {-\frac {1}{c^{2} x^{2}} + 1} \operatorname {arcsec}\left (c x\right ) + \frac {1}{x}\right )} a b^{2} + 3 \, {\left (c \sqrt {-\frac {1}{c^{2} x^{2}} + 1} \operatorname {arcsec}\left (c x\right )^{2} - 2 \, c \sqrt {-\frac {1}{c^{2} x^{2}} + 1} + \frac {2 \, \operatorname {arcsec}\left (c x\right )}{x}\right )} b^{3} - \frac {3 \, a b^{2} \operatorname {arcsec}\left (c x\right )^{2}}{x} - \frac {a^{3}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.80, size = 156, normalized size = 1.95 \[ \frac {b^3\,\left (6\,\mathrm {acos}\left (\frac {1}{c\,x}\right )-{\mathrm {acos}\left (\frac {1}{c\,x}\right )}^3\right )}{x}-\frac {a^3}{x}+3\,a^2\,b\,c\,\left (\sqrt {1-\frac {1}{c^2\,x^2}}-\frac {\mathrm {acos}\left (\frac {1}{c\,x}\right )}{c\,x}\right )+b^3\,c\,\sqrt {1-\frac {1}{c^2\,x^2}}\,\left (3\,{\mathrm {acos}\left (\frac {1}{c\,x}\right )}^2-6\right )+3\,a\,b^2\,c\,\left (2\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\,\sqrt {1-\frac {1}{c^2\,x^2}}-\frac {{\mathrm {acos}\left (\frac {1}{c\,x}\right )}^2-2}{c\,x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(1/(c*x)))^3/x^2,x)

[Out]

(b^3*(6*acos(1/(c*x)) - acos(1/(c*x))^3))/x - a^3/x + 3*a^2*b*c*((1 - 1/(c^2*x^2))^(1/2) - acos(1/(c*x))/(c*x)
) + b^3*c*(1 - 1/(c^2*x^2))^(1/2)*(3*acos(1/(c*x))^2 - 6) + 3*a*b^2*c*(2*acos(1/(c*x))*(1 - 1/(c^2*x^2))^(1/2)
- (acos(1/(c*x))^2 - 2)/(c*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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