3.50 \(\int \frac{1}{x^4 (a+b \sec ^{-1}(c x))^3} \, dx\)

Optimal. Leaf size=228 \[ \frac{c^3 \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{8 b^3}+\frac{9 c^3 \sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 a}{b}+3 \sec ^{-1}(c x)\right )}{8 b^3}-\frac{c^3 \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{8 b^3}-\frac{9 c^3 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sec ^{-1}(c x)\right )}{8 b^3}-\frac{c^2}{8 b^2 x \left (a+b \sec ^{-1}(c x)\right )}-\frac{3 c^3 \cos \left (3 \sec ^{-1}(c x)\right )}{8 b^2 \left (a+b \sec ^{-1}(c x)\right )}-\frac{c^3 \sqrt{1-\frac{1}{c^2 x^2}}}{8 b \left (a+b \sec ^{-1}(c x)\right )^2}-\frac{c^3 \sin \left (3 \sec ^{-1}(c x)\right )}{8 b \left (a+b \sec ^{-1}(c x)\right )^2} \]

[Out]

-(c^3*Sqrt[1 - 1/(c^2*x^2)])/(8*b*(a + b*ArcSec[c*x])^2) - c^2/(8*b^2*x*(a + b*ArcSec[c*x])) - (3*c^3*Cos[3*Ar
cSec[c*x]])/(8*b^2*(a + b*ArcSec[c*x])) + (c^3*CosIntegral[a/b + ArcSec[c*x]]*Sin[a/b])/(8*b^3) + (9*c^3*CosIn
tegral[(3*a)/b + 3*ArcSec[c*x]]*Sin[(3*a)/b])/(8*b^3) - (c^3*Sin[3*ArcSec[c*x]])/(8*b*(a + b*ArcSec[c*x])^2) -
(c^3*Cos[a/b]*SinIntegral[a/b + ArcSec[c*x]])/(8*b^3) - (9*c^3*Cos[(3*a)/b]*SinIntegral[(3*a)/b + 3*ArcSec[c*
x]])/(8*b^3)

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Rubi [A]  time = 0.312718, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {5222, 4406, 3297, 3303, 3299, 3302} \[ \frac{c^3 \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{8 b^3}+\frac{9 c^3 \sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (\frac{3 a}{b}+3 \sec ^{-1}(c x)\right )}{8 b^3}-\frac{c^3 \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{8 b^3}-\frac{9 c^3 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sec ^{-1}(c x)\right )}{8 b^3}-\frac{c^2}{8 b^2 x \left (a+b \sec ^{-1}(c x)\right )}-\frac{3 c^3 \cos \left (3 \sec ^{-1}(c x)\right )}{8 b^2 \left (a+b \sec ^{-1}(c x)\right )}-\frac{c^3 \sqrt{1-\frac{1}{c^2 x^2}}}{8 b \left (a+b \sec ^{-1}(c x)\right )^2}-\frac{c^3 \sin \left (3 \sec ^{-1}(c x)\right )}{8 b \left (a+b \sec ^{-1}(c x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*ArcSec[c*x])^3),x]

[Out]

-(c^3*Sqrt[1 - 1/(c^2*x^2)])/(8*b*(a + b*ArcSec[c*x])^2) - c^2/(8*b^2*x*(a + b*ArcSec[c*x])) - (3*c^3*Cos[3*Ar
cSec[c*x]])/(8*b^2*(a + b*ArcSec[c*x])) + (c^3*CosIntegral[a/b + ArcSec[c*x]]*Sin[a/b])/(8*b^3) + (9*c^3*CosIn
tegral[(3*a)/b + 3*ArcSec[c*x]]*Sin[(3*a)/b])/(8*b^3) - (c^3*Sin[3*ArcSec[c*x]])/(8*b*(a + b*ArcSec[c*x])^2) -
(c^3*Cos[a/b]*SinIntegral[a/b + ArcSec[c*x]])/(8*b^3) - (9*c^3*Cos[(3*a)/b]*SinIntegral[(3*a)/b + 3*ArcSec[c*
x]])/(8*b^3)

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 4406

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 3297

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 3303

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 3299

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 3302

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]

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (a+b \sec ^{-1}(c x)\right )^3} \, dx &=c^3 \operatorname{Subst}\left (\int \frac{\cos ^2(x) \sin (x)}{(a+b x)^3} \, dx,x,\sec ^{-1}(c x)\right )\\ &=c^3 \operatorname{Subst}\left (\int \left (\frac{\sin (x)}{4 (a+b x)^3}+\frac{\sin (3 x)}{4 (a+b x)^3}\right ) \, dx,x,\sec ^{-1}(c x)\right )\\ &=\frac{1}{4} c^3 \operatorname{Subst}\left (\int \frac{\sin (x)}{(a+b x)^3} \, dx,x,\sec ^{-1}(c x)\right )+\frac{1}{4} c^3 \operatorname{Subst}\left (\int \frac{\sin (3 x)}{(a+b x)^3} \, dx,x,\sec ^{-1}(c x)\right )\\ &=-\frac{c^3 \sqrt{1-\frac{1}{c^2 x^2}}}{8 b \left (a+b \sec ^{-1}(c x)\right )^2}-\frac{c^3 \sin \left (3 \sec ^{-1}(c x)\right )}{8 b \left (a+b \sec ^{-1}(c x)\right )^2}+\frac{c^3 \operatorname{Subst}\left (\int \frac{\cos (x)}{(a+b x)^2} \, dx,x,\sec ^{-1}(c x)\right )}{8 b}+\frac{\left (3 c^3\right ) \operatorname{Subst}\left (\int \frac{\cos (3 x)}{(a+b x)^2} \, dx,x,\sec ^{-1}(c x)\right )}{8 b}\\ &=-\frac{c^3 \sqrt{1-\frac{1}{c^2 x^2}}}{8 b \left (a+b \sec ^{-1}(c x)\right )^2}-\frac{c^2}{8 b^2 x \left (a+b \sec ^{-1}(c x)\right )}-\frac{3 c^3 \cos \left (3 \sec ^{-1}(c x)\right )}{8 b^2 \left (a+b \sec ^{-1}(c x)\right )}-\frac{c^3 \sin \left (3 \sec ^{-1}(c x)\right )}{8 b \left (a+b \sec ^{-1}(c x)\right )^2}-\frac{c^3 \operatorname{Subst}\left (\int \frac{\sin (x)}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )}{8 b^2}-\frac{\left (9 c^3\right ) \operatorname{Subst}\left (\int \frac{\sin (3 x)}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )}{8 b^2}\\ &=-\frac{c^3 \sqrt{1-\frac{1}{c^2 x^2}}}{8 b \left (a+b \sec ^{-1}(c x)\right )^2}-\frac{c^2}{8 b^2 x \left (a+b \sec ^{-1}(c x)\right )}-\frac{3 c^3 \cos \left (3 \sec ^{-1}(c x)\right )}{8 b^2 \left (a+b \sec ^{-1}(c x)\right )}-\frac{c^3 \sin \left (3 \sec ^{-1}(c x)\right )}{8 b \left (a+b \sec ^{-1}(c x)\right )^2}-\frac{\left (c^3 \cos \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )}{8 b^2}-\frac{\left (9 c^3 \cos \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sin \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )}{8 b^2}+\frac{\left (c^3 \sin \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )}{8 b^2}+\frac{\left (9 c^3 \sin \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cos \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sec ^{-1}(c x)\right )}{8 b^2}\\ &=-\frac{c^3 \sqrt{1-\frac{1}{c^2 x^2}}}{8 b \left (a+b \sec ^{-1}(c x)\right )^2}-\frac{c^2}{8 b^2 x \left (a+b \sec ^{-1}(c x)\right )}-\frac{3 c^3 \cos \left (3 \sec ^{-1}(c x)\right )}{8 b^2 \left (a+b \sec ^{-1}(c x)\right )}+\frac{c^3 \text{Ci}\left (\frac{a}{b}+\sec ^{-1}(c x)\right ) \sin \left (\frac{a}{b}\right )}{8 b^3}+\frac{9 c^3 \text{Ci}\left (\frac{3 a}{b}+3 \sec ^{-1}(c x)\right ) \sin \left (\frac{3 a}{b}\right )}{8 b^3}-\frac{c^3 \sin \left (3 \sec ^{-1}(c x)\right )}{8 b \left (a+b \sec ^{-1}(c x)\right )^2}-\frac{c^3 \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )}{8 b^3}-\frac{9 c^3 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (\frac{3 a}{b}+3 \sec ^{-1}(c x)\right )}{8 b^3}\\ \end{align*}

Mathematica [A]  time = 0.444805, size = 169, normalized size = 0.74 \[ \frac{-\frac{4 b^2 c \sqrt{1-\frac{1}{c^2 x^2}}}{x^2 \left (a+b \sec ^{-1}(c x)\right )^2}+c^3 \sin \left (\frac{a}{b}\right ) \text{CosIntegral}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )+9 c^3 \sin \left (\frac{3 a}{b}\right ) \text{CosIntegral}\left (3 \left (\frac{a}{b}+\sec ^{-1}(c x)\right )\right )-c^3 \cos \left (\frac{a}{b}\right ) \text{Si}\left (\frac{a}{b}+\sec ^{-1}(c x)\right )-9 c^3 \cos \left (\frac{3 a}{b}\right ) \text{Si}\left (3 \left (\frac{a}{b}+\sec ^{-1}(c x)\right )\right )+\frac{8 b c^2}{a x+b x \sec ^{-1}(c x)}-\frac{12 b}{x^3 \left (a+b \sec ^{-1}(c x)\right )}}{8 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*ArcSec[c*x])^3),x]

[Out]

((-4*b^2*c*Sqrt[1 - 1/(c^2*x^2)])/(x^2*(a + b*ArcSec[c*x])^2) - (12*b)/(x^3*(a + b*ArcSec[c*x])) + (8*b*c^2)/(
a*x + b*x*ArcSec[c*x]) + c^3*CosIntegral[a/b + ArcSec[c*x]]*Sin[a/b] + 9*c^3*CosIntegral[3*(a/b + ArcSec[c*x])
]*Sin[(3*a)/b] - c^3*Cos[a/b]*SinIntegral[a/b + ArcSec[c*x]] - 9*c^3*Cos[(3*a)/b]*SinIntegral[3*(a/b + ArcSec[
c*x])])/(8*b^3)

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Maple [A]  time = 0.25, size = 307, normalized size = 1.4 \begin{align*}{c}^{3} \left ( -{\frac{\sin \left ( 3\,{\rm arcsec} \left (cx\right ) \right ) }{8\, \left ( a+b{\rm arcsec} \left (cx\right ) \right ) ^{2}b}}-{\frac{3}{ \left ( 8\,a+8\,b{\rm arcsec} \left (cx\right ) \right ){b}^{3}} \left ( 3\,{\it Si} \left ( 3\,{\frac{a}{b}}+3\,{\rm arcsec} \left (cx\right ) \right ) \cos \left ( 3\,{\frac{a}{b}} \right ){\rm arcsec} \left (cx\right )b-3\,{\it Ci} \left ( 3\,{\frac{a}{b}}+3\,{\rm arcsec} \left (cx\right ) \right ) \sin \left ( 3\,{\frac{a}{b}} \right ){\rm arcsec} \left (cx\right )b+3\,{\it Si} \left ( 3\,{\frac{a}{b}}+3\,{\rm arcsec} \left (cx\right ) \right ) \cos \left ( 3\,{\frac{a}{b}} \right ) a-3\,{\it Ci} \left ( 3\,{\frac{a}{b}}+3\,{\rm arcsec} \left (cx\right ) \right ) \sin \left ( 3\,{\frac{a}{b}} \right ) a+\cos \left ( 3\,{\rm arcsec} \left (cx\right ) \right ) b \right ) }-{\frac{1}{8\, \left ( a+b{\rm arcsec} \left (cx\right ) \right ) ^{2}b}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}-{\frac{1}{8\,cx \left ( a+b{\rm arcsec} \left (cx\right ) \right ){b}^{3}} \left ({\it Si} \left ({\frac{a}{b}}+{\rm arcsec} \left (cx\right ) \right ) \cos \left ({\frac{a}{b}} \right ){\rm arcsec} \left (cx\right )cxb-{\it Ci} \left ({\frac{a}{b}}+{\rm arcsec} \left (cx\right ) \right ) \sin \left ({\frac{a}{b}} \right ){\rm arcsec} \left (cx\right )cxb+{\it Si} \left ({\frac{a}{b}}+{\rm arcsec} \left (cx\right ) \right ) \cos \left ({\frac{a}{b}} \right ) cxa-{\it Ci} \left ({\frac{a}{b}}+{\rm arcsec} \left (cx\right ) \right ) \sin \left ({\frac{a}{b}} \right ) cxa+b \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(a+b*arcsec(c*x))^3,x)

[Out]

c^3*(-1/8*sin(3*arcsec(c*x))/(a+b*arcsec(c*x))^2/b-3/8*(3*Si(3*a/b+3*arcsec(c*x))*cos(3*a/b)*arcsec(c*x)*b-3*C
i(3*a/b+3*arcsec(c*x))*sin(3*a/b)*arcsec(c*x)*b+3*Si(3*a/b+3*arcsec(c*x))*cos(3*a/b)*a-3*Ci(3*a/b+3*arcsec(c*x
))*sin(3*a/b)*a+cos(3*arcsec(c*x))*b)/(a+b*arcsec(c*x))/b^3-1/8*((c^2*x^2-1)/c^2/x^2)^(1/2)/(a+b*arcsec(c*x))^
2/b-1/8*(Si(a/b+arcsec(c*x))*cos(a/b)*arcsec(c*x)*c*x*b-Ci(a/b+arcsec(c*x))*sin(a/b)*arcsec(c*x)*c*x*b+Si(a/b+
arcsec(c*x))*cos(a/b)*c*x*a-Ci(a/b+arcsec(c*x))*sin(a/b)*c*x*a+b)/c/x/(a+b*arcsec(c*x))/b^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(1/x^4/(a+b*arcsec(c*x))^3,x, algorithm="maxima")

[Out]

-(24*a*b^2*log(c)^2 - 8*(2*b^3*c^2*x^2 - 3*b^3)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^3 + 24*a^3 - 16*(a*b^2*c^2
*log(c)^2 + a^3*c^2)*x^2 - 24*(2*a*b^2*c^2*x^2 - 3*a*b^2)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2 - 2*(2*a*b^2*c
^2*x^2 - 3*a*b^2)*log(c^2*x^2)^2 - 8*(2*a*b^2*c^2*x^2 - 3*a*b^2)*log(x)^2 + 2*(4*b^3*arctan(sqrt(c*x + 1)*sqrt
(c*x - 1))^2 - b^3*log(c^2*x^2)^2 - 4*b^3*log(c)^2 - 8*b^3*log(c)*log(x) - 4*b^3*log(x)^2 + 8*a*b^2*arctan(sqr
t(c*x + 1)*sqrt(c*x - 1)) + 4*a^2*b + 4*(b^3*log(c) + b^3*log(x))*log(c^2*x^2))*sqrt(c*x + 1)*sqrt(c*x - 1) +
2*(12*b^3*log(c)^2 + 36*a^2*b - 8*(b^3*c^2*log(c)^2 + 3*a^2*b*c^2)*x^2 - (2*b^3*c^2*x^2 - 3*b^3)*log(c^2*x^2)^
2 - 4*(2*b^3*c^2*x^2 - 3*b^3)*log(x)^2 + 4*(2*b^3*c^2*x^2*log(c) - 3*b^3*log(c) + (2*b^3*c^2*x^2 - 3*b^3)*log(
x))*log(c^2*x^2) - 8*(2*b^3*c^2*x^2*log(c) - 3*b^3*log(c))*log(x))*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) - (16*b
^6*x^3*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^4 + b^6*x^3*log(c^2*x^2)^4 + 64*b^6*x^3*log(c)*log(x)^3 + 16*b^6*x^
3*log(x)^4 + 64*a*b^5*x^3*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^3 + 32*(3*b^6*log(c)^2 + a^2*b^4)*x^3*log(x)^2 +
64*(b^6*log(c)^3 + a^2*b^4*log(c))*x^3*log(x) + 16*(b^6*log(c)^4 + 2*a^2*b^4*log(c)^2 + a^4*b^2)*x^3 - 8*(b^6
*x^3*log(c) + b^6*x^3*log(x))*log(c^2*x^2)^3 + 8*(b^6*x^3*log(c^2*x^2)^2 + 8*b^6*x^3*log(c)*log(x) + 4*b^6*x^3
*log(x)^2 + 4*(b^6*log(c)^2 + 3*a^2*b^4)*x^3 - 4*(b^6*x^3*log(c) + b^6*x^3*log(x))*log(c^2*x^2))*arctan(sqrt(c
*x + 1)*sqrt(c*x - 1))^2 + 8*(6*b^6*x^3*log(c)*log(x) + 3*b^6*x^3*log(x)^2 + (3*b^6*log(c)^2 + a^2*b^4)*x^3)*l
og(c^2*x^2)^2 + 16*(a*b^5*x^3*log(c^2*x^2)^2 + 8*a*b^5*x^3*log(c)*log(x) + 4*a*b^5*x^3*log(x)^2 + 4*(a*b^5*log
(c)^2 + a^3*b^3)*x^3 - 4*(a*b^5*x^3*log(c) + a*b^5*x^3*log(x))*log(c^2*x^2))*arctan(sqrt(c*x + 1)*sqrt(c*x - 1
)) - 32*(3*b^6*x^3*log(c)*log(x)^2 + b^6*x^3*log(x)^3 + (3*b^6*log(c)^2 + a^2*b^4)*x^3*log(x) + (b^6*log(c)^3
+ a^2*b^4*log(c))*x^3)*log(c^2*x^2))*integrate(2*(2*a*c^2*x^2 + (2*b*c^2*x^2 - 9*b)*arctan(sqrt(c*x + 1)*sqrt(
c*x - 1)) - 9*a)/(4*b^4*x^4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^2 + b^4*x^4*log(c^2*x^2)^2 + 8*b^4*x^4*log(c)*
log(x) + 4*b^4*x^4*log(x)^2 + 8*a*b^3*x^4*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) + 4*(b^4*log(c)^2 + a^2*b^2)*x^4
- 4*(b^4*x^4*log(c) + b^4*x^4*log(x))*log(c^2*x^2)), x) + 8*(2*a*b^2*c^2*x^2*log(c) - 3*a*b^2*log(c) + (2*a*b
^2*c^2*x^2 - 3*a*b^2)*log(x))*log(c^2*x^2) - 16*(2*a*b^2*c^2*x^2*log(c) - 3*a*b^2*log(c))*log(x))/(16*b^6*x^3*
arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^4 + b^6*x^3*log(c^2*x^2)^4 + 64*b^6*x^3*log(c)*log(x)^3 + 16*b^6*x^3*log(x
)^4 + 64*a*b^5*x^3*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^3 + 32*(3*b^6*log(c)^2 + a^2*b^4)*x^3*log(x)^2 + 64*(b^
6*log(c)^3 + a^2*b^4*log(c))*x^3*log(x) + 16*(b^6*log(c)^4 + 2*a^2*b^4*log(c)^2 + a^4*b^2)*x^3 - 8*(b^6*x^3*lo
g(c) + b^6*x^3*log(x))*log(c^2*x^2)^3 + 8*(b^6*x^3*log(c^2*x^2)^2 + 8*b^6*x^3*log(c)*log(x) + 4*b^6*x^3*log(x)
^2 + 4*(b^6*log(c)^2 + 3*a^2*b^4)*x^3 - 4*(b^6*x^3*log(c) + b^6*x^3*log(x))*log(c^2*x^2))*arctan(sqrt(c*x + 1)
*sqrt(c*x - 1))^2 + 8*(6*b^6*x^3*log(c)*log(x) + 3*b^6*x^3*log(x)^2 + (3*b^6*log(c)^2 + a^2*b^4)*x^3)*log(c^2*
x^2)^2 + 16*(a*b^5*x^3*log(c^2*x^2)^2 + 8*a*b^5*x^3*log(c)*log(x) + 4*a*b^5*x^3*log(x)^2 + 4*(a*b^5*log(c)^2 +
a^3*b^3)*x^3 - 4*(a*b^5*x^3*log(c) + a*b^5*x^3*log(x))*log(c^2*x^2))*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) - 32
*(3*b^6*x^3*log(c)*log(x)^2 + b^6*x^3*log(x)^3 + (3*b^6*log(c)^2 + a^2*b^4)*x^3*log(x) + (b^6*log(c)^3 + a^2*b
^4*log(c))*x^3)*log(c^2*x^2))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{b^{3} x^{4} \operatorname{arcsec}\left (c x\right )^{3} + 3 \, a b^{2} x^{4} \operatorname{arcsec}\left (c x\right )^{2} + 3 \, a^{2} b x^{4} \operatorname{arcsec}\left (c x\right ) + a^{3} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arcsec(c*x))^3,x, algorithm="fricas")

[Out]

integral(1/(b^3*x^4*arcsec(c*x)^3 + 3*a*b^2*x^4*arcsec(c*x)^2 + 3*a^2*b*x^4*arcsec(c*x) + a^3*x^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \left (a + b \operatorname{asec}{\left (c x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(a+b*asec(c*x))**3,x)

[Out]

Integral(1/(x**4*(a + b*asec(c*x))**3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )}^{3} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arcsec(c*x))^3,x, algorithm="giac")

[Out]

integrate(1/((b*arcsec(c*x) + a)^3*x^4), x)