3.1.43 \(\int \frac {1}{x^3 (a+b \sec ^{-1}(c x))^2} \, dx\) [43]

Optimal. Leaf size=84 \[ \frac {c^2 \cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^2}-\frac {c^2 \sin \left (2 \sec ^{-1}(c x)\right )}{2 b \left (a+b \sec ^{-1}(c x)\right )}+\frac {c^2 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^2} \]

[Out]

c^2*Ci(2*a/b+2*arcsec(c*x))*cos(2*a/b)/b^2+c^2*Si(2*a/b+2*arcsec(c*x))*sin(2*a/b)/b^2-1/2*c^2*sin(2*arcsec(c*x
))/b/(a+b*arcsec(c*x))

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Rubi [A]
time = 0.12, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5330, 4491, 12, 3378, 3384, 3380, 3383} \begin {gather*} \frac {c^2 \cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^2}+\frac {c^2 \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \sec ^{-1}(c x)\right )}{b^2}-\frac {c^2 \sin \left (2 \sec ^{-1}(c x)\right )}{2 b \left (a+b \sec ^{-1}(c x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*ArcSec[c*x])^2),x]

[Out]

(c^2*Cos[(2*a)/b]*CosIntegral[(2*a)/b + 2*ArcSec[c*x]])/b^2 - (c^2*Sin[2*ArcSec[c*x]])/(2*b*(a + b*ArcSec[c*x]
)) + (c^2*Sin[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSec[c*x]])/b^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3378

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

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3383

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3384

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

Rule 4491

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

Rule 5330

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

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Mathematica [A]
time = 0.26, size = 80, normalized size = 0.95 \begin {gather*} \frac {c \left (-\frac {b \sqrt {1-\frac {1}{c^2 x^2}}}{a x+b x \sec ^{-1}(c x)}+c \cos \left (\frac {2 a}{b}\right ) \text {CosIntegral}\left (2 \left (\frac {a}{b}+\sec ^{-1}(c x)\right )\right )+c \sin \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\sec ^{-1}(c x)\right )\right )\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*ArcSec[c*x])^2),x]

[Out]

(c*(-((b*Sqrt[1 - 1/(c^2*x^2)])/(a*x + b*x*ArcSec[c*x])) + c*Cos[(2*a)/b]*CosIntegral[2*(a/b + ArcSec[c*x])] +
c*Sin[(2*a)/b]*SinIntegral[2*(a/b + ArcSec[c*x])]))/b^2

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Maple [A]
time = 0.14, size = 77, normalized size = 0.92

method	result	size
derivativedivides	\(c^{2} \left (-\frac {\sin \left (2 \,\mathrm {arcsec}\left (c x \right )\right )}{2 \left (a +b \,\mathrm {arcsec}\left (c x \right )\right ) b}+\frac {\cosineIntegral \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) \cos \left (\frac {2 a}{b}\right )+\sinIntegral \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) \sin \left (\frac {2 a}{b}\right )}{b^{2}}\right )\)	\(77\)
default	\(c^{2} \left (-\frac {\sin \left (2 \,\mathrm {arcsec}\left (c x \right )\right )}{2 \left (a +b \,\mathrm {arcsec}\left (c x \right )\right ) b}+\frac {\cosineIntegral \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) \cos \left (\frac {2 a}{b}\right )+\sinIntegral \left (\frac {2 a}{b}+2 \,\mathrm {arcsec}\left (c x \right )\right ) \sin \left (\frac {2 a}{b}\right )}{b^{2}}\right )\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(a+b*arcsec(c*x))^2,x,method=_RETURNVERBOSE)

[Out]

c^2*(-1/2*sin(2*arcsec(c*x))/(a+b*arcsec(c*x))/b+(Ci(2*a/b+2*arcsec(c*x))*cos(2*a/b)+Si(2*a/b+2*arcsec(c*x))*s
in(2*a/b))/b^2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*arcsec(c*x))^2,x, algorithm="maxima")

[Out]

-(4*sqrt(c*x + 1)*sqrt(c*x - 1)*(b*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) + a) + (4*b^3*x^2*arctan(sqrt(c*x + 1)*
sqrt(c*x - 1))^2 + b^3*x^2*log(c^2*x^2)^2 + 8*b^3*x^2*log(c)*log(x) + 4*b^3*x^2*log(x)^2 + 8*a*b^2*x^2*arctan(
sqrt(c*x + 1)*sqrt(c*x - 1)) + 4*(b^3*log(c)^2 + a^2*b)*x^2 - 4*(b^3*x^2*log(c) + b^3*x^2*log(x))*log(c^2*x^2)
)*integrate(4*(a*c^2*x^2 + (b*c^2*x^2 - 2*b)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) - 2*a)*sqrt(c*x + 1)*sqrt(c*x
- 1)/(4*(b^3*c^2*log(c)^2 + a^2*b*c^2)*x^5 - 4*(b^3*log(c)^2 + a^2*b)*x^3 + 4*(b^3*c^2*x^5 - b^3*x^3)*arctan(
sqrt(c*x + 1)*sqrt(c*x - 1))^2 + (b^3*c^2*x^5 - b^3*x^3)*log(c^2*x^2)^2 + 4*(b^3*c^2*x^5 - b^3*x^3)*log(x)^2 +
8*(a*b^2*c^2*x^5 - a*b^2*x^3)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) - 4*(b^3*c^2*x^5*log(c) - b^3*x^3*log(c) +
(b^3*c^2*x^5 - b^3*x^3)*log(x))*log(c^2*x^2) + 8*(b^3*c^2*x^5*log(c) - b^3*x^3*log(c))*log(x)), x))/(4*b^3*x^2
*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2 + b^3*x^2*log(c^2*x^2)^2 + 8*b^3*x^2*log(c)*log(x) + 4*b^3*x^2*log(x)^2
+ 8*a*b^2*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) + 4*(b^3*log(c)^2 + a^2*b)*x^2 - 4*(b^3*x^2*log(c) + b^3*x^
2*log(x))*log(c^2*x^2))

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*arcsec(c*x))^2,x, algorithm="fricas")

[Out]

integral(1/(b^2*x^3*arcsec(c*x)^2 + 2*a*b*x^3*arcsec(c*x) + a^2*x^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \left (a + b \operatorname {asec}{\left (c x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(a+b*asec(c*x))**2,x)

[Out]

Integral(1/(x**3*(a + b*asec(c*x))**2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 357 vs. \(2 (82) = 164\).
time = 0.41, size = 357, normalized size = 4.25 \begin {gather*} {\left (\frac {2 \, b c \arccos \left (\frac {1}{c x}\right ) \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} + \frac {2 \, b c \arccos \left (\frac {1}{c x}\right ) \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} + \frac {2 \, a c \cos \left (\frac {a}{b}\right )^{2} \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} + \frac {2 \, a c \cos \left (\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} - \frac {b c \arccos \left (\frac {1}{c x}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} - \frac {a c \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (\frac {1}{c x}\right )\right )}{b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}} - \frac {b \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{{\left (b^{3} \arccos \left (\frac {1}{c x}\right ) + a b^{2}\right )} x}\right )} c \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*arcsec(c*x))^2,x, algorithm="giac")

[Out]

(2*b*c*arccos(1/(c*x))*cos(a/b)^2*cos_integral(2*a/b + 2*arccos(1/(c*x)))/(b^3*arccos(1/(c*x)) + a*b^2) + 2*b*
c*arccos(1/(c*x))*cos(a/b)*sin(a/b)*sin_integral(2*a/b + 2*arccos(1/(c*x)))/(b^3*arccos(1/(c*x)) + a*b^2) + 2*
a*c*cos(a/b)^2*cos_integral(2*a/b + 2*arccos(1/(c*x)))/(b^3*arccos(1/(c*x)) + a*b^2) + 2*a*c*cos(a/b)*sin(a/b)
*sin_integral(2*a/b + 2*arccos(1/(c*x)))/(b^3*arccos(1/(c*x)) + a*b^2) - b*c*arccos(1/(c*x))*cos_integral(2*a/
b + 2*arccos(1/(c*x)))/(b^3*arccos(1/(c*x)) + a*b^2) - a*c*cos_integral(2*a/b + 2*arccos(1/(c*x)))/(b^3*arccos
(1/(c*x)) + a*b^2) - b*sqrt(-1/(c^2*x^2) + 1)/((b^3*arccos(1/(c*x)) + a*b^2)*x))*c

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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