3.12 \(\int \frac{a+b \sec ^{-1}(c x)}{x^5} \, dx\)

Optimal. Leaf size=76 \[ -\frac{a+b \sec ^{-1}(c x)}{4 x^4}+\frac{3 b c^3 \sqrt{1-\frac{1}{c^2 x^2}}}{32 x}+\frac{b c \sqrt{1-\frac{1}{c^2 x^2}}}{16 x^3}-\frac{3}{32} b c^4 \csc ^{-1}(c x) \]

[Out]

(b*c*Sqrt[1 - 1/(c^2*x^2)])/(16*x^3) + (3*b*c^3*Sqrt[1 - 1/(c^2*x^2)])/(32*x) - (3*b*c^4*ArcCsc[c*x])/32 - (a
+ b*ArcSec[c*x])/(4*x^4)

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Rubi [A]  time = 0.0453043, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5220, 335, 321, 216} \[ -\frac{a+b \sec ^{-1}(c x)}{4 x^4}+\frac{3 b c^3 \sqrt{1-\frac{1}{c^2 x^2}}}{32 x}+\frac{b c \sqrt{1-\frac{1}{c^2 x^2}}}{16 x^3}-\frac{3}{32} b c^4 \csc ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSec[c*x])/x^5,x]

[Out]

(b*c*Sqrt[1 - 1/(c^2*x^2)])/(16*x^3) + (3*b*c^3*Sqrt[1 - 1/(c^2*x^2)])/(32*x) - (3*b*c^4*ArcCsc[c*x])/32 - (a
+ b*ArcSec[c*x])/(4*x^4)

Rule 5220

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

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 321

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

Rule 216

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

Rubi steps

\begin{align*} \int \frac{a+b \sec ^{-1}(c x)}{x^5} \, dx &=-\frac{a+b \sec ^{-1}(c x)}{4 x^4}+\frac{b \int \frac{1}{\sqrt{1-\frac{1}{c^2 x^2}} x^6} \, dx}{4 c}\\ &=-\frac{a+b \sec ^{-1}(c x)}{4 x^4}-\frac{b \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{4 c}\\ &=\frac{b c \sqrt{1-\frac{1}{c^2 x^2}}}{16 x^3}-\frac{a+b \sec ^{-1}(c x)}{4 x^4}-\frac{1}{16} (3 b c) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{b c \sqrt{1-\frac{1}{c^2 x^2}}}{16 x^3}+\frac{3 b c^3 \sqrt{1-\frac{1}{c^2 x^2}}}{32 x}-\frac{a+b \sec ^{-1}(c x)}{4 x^4}-\frac{1}{32} \left (3 b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{b c \sqrt{1-\frac{1}{c^2 x^2}}}{16 x^3}+\frac{3 b c^3 \sqrt{1-\frac{1}{c^2 x^2}}}{32 x}-\frac{3}{32} b c^4 \csc ^{-1}(c x)-\frac{a+b \sec ^{-1}(c x)}{4 x^4}\\ \end{align*}

Mathematica [A]  time = 0.0587985, size = 78, normalized size = 1.03 \[ -\frac{a}{4 x^4}+b \left (\frac{3 c^3}{32 x}+\frac{c}{16 x^3}\right ) \sqrt{\frac{c^2 x^2-1}{c^2 x^2}}-\frac{3}{32} b c^4 \sin ^{-1}\left (\frac{1}{c x}\right )-\frac{b \sec ^{-1}(c x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSec[c*x])/x^5,x]

[Out]

-a/(4*x^4) + b*(c/(16*x^3) + (3*c^3)/(32*x))*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)] - (b*ArcSec[c*x])/(4*x^4) - (3*b*c
^4*ArcSin[1/(c*x)])/32

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Maple [B]  time = 0.16, size = 147, normalized size = 1.9 \begin{align*} -{\frac{a}{4\,{x}^{4}}}-{\frac{b{\rm arcsec} \left (cx\right )}{4\,{x}^{4}}}-{\frac{3\,b{c}^{3}}{32\,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\,b{c}^{3}}{32\,x}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{cb}{32\,{x}^{3}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{b}{16\,c{x}^{5}}{\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))/x^5,x)

[Out]

-1/4*a/x^4-1/4*b/x^4*arcsec(c*x)-3/32*c^3*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))+3/32*c^3*b/((c^2*x^2-1)/c^2/x^2)^(1/2)/x-1/32*c*b/((c^2*x^2-1)/c^2/x^2)^(1/2)/x^3-1/16/c*b/((c^2*x^2
-1)/c^2/x^2)^(1/2)/x^5

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Maxima [A]  time = 1.49674, size = 169, normalized size = 2.22 \begin{align*} \frac{1}{32} \, b{\left (\frac{3 \, c^{5} \arctan \left (c x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}\right ) + \frac{3 \, c^{8} x^{3}{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} + 5 \, c^{6} x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c^{4} x^{4}{\left (\frac{1}{c^{2} x^{2}} - 1\right )}^{2} - 2 \, c^{2} x^{2}{\left (\frac{1}{c^{2} x^{2}} - 1\right )} + 1}}{c} - \frac{8 \, \operatorname{arcsec}\left (c x\right )}{x^{4}}\right )} - \frac{a}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x^5,x, algorithm="maxima")

[Out]

1/32*b*((3*c^5*arctan(c*x*sqrt(-1/(c^2*x^2) + 1)) + (3*c^8*x^3*(-1/(c^2*x^2) + 1)^(3/2) + 5*c^6*x*sqrt(-1/(c^2
*x^2) + 1))/(c^4*x^4*(1/(c^2*x^2) - 1)^2 - 2*c^2*x^2*(1/(c^2*x^2) - 1) + 1))/c - 8*arcsec(c*x)/x^4) - 1/4*a/x^
4

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Fricas [A]  time = 2.68544, size = 122, normalized size = 1.61 \begin{align*} \frac{{\left (3 \, b c^{4} x^{4} - 8 \, b\right )} \operatorname{arcsec}\left (c x\right ) +{\left (3 \, b c^{2} x^{2} + 2 \, b\right )} \sqrt{c^{2} x^{2} - 1} - 8 \, a}{32 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x^5,x, algorithm="fricas")

[Out]

1/32*((3*b*c^4*x^4 - 8*b)*arcsec(c*x) + (3*b*c^2*x^2 + 2*b)*sqrt(c^2*x^2 - 1) - 8*a)/x^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{asec}{\left (c x \right )}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asec(c*x))/x**5,x)

[Out]

Integral((a + b*asec(c*x))/x**5, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arcsec}\left (c x\right ) + a}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsec(c*x))/x^5,x, algorithm="giac")

[Out]

integrate((b*arcsec(c*x) + a)/x^5, x)