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

Optimal. Leaf size=181 \[ -\frac{\left (2-5 a^2\right ) b^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a^2 \left (1-a^2\right )^2 x}-\frac{b^3 \sec ^{-1}(a+b x)}{3 a^3}+\frac{\left (6 a^4-5 a^2+2\right ) b^3 \tan ^{-1}\left (\frac{\sqrt{a+1} \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{1-a}}\right )}{3 a^3 \left (1-a^2\right )^{5/2}}+\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a \left (1-a^2\right ) x^2}-\frac{\sec ^{-1}(a+b x)}{3 x^3} \]

[Out]

(b*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)])/(6*a*(1 - a^2)*x^2) - ((2 - 5*a^2)*b^2*(a + b*x)*Sqrt[1 - (a + b*x)^(-2
)])/(6*a^2*(1 - a^2)^2*x) - (b^3*ArcSec[a + b*x])/(3*a^3) - ArcSec[a + b*x]/(3*x^3) + ((2 - 5*a^2 + 6*a^4)*b^3
*ArcTan[(Sqrt[1 + a]*Tan[ArcSec[a + b*x]/2])/Sqrt[1 - a]])/(3*a^3*(1 - a^2)^(5/2))

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Rubi [A]  time = 0.287762, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {5258, 4426, 3785, 4060, 3919, 3831, 2659, 205} \[ -\frac{\left (2-5 a^2\right ) b^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a^2 \left (1-a^2\right )^2 x}-\frac{b^3 \sec ^{-1}(a+b x)}{3 a^3}+\frac{\left (6 a^4-5 a^2+2\right ) b^3 \tan ^{-1}\left (\frac{\sqrt{a+1} \tan \left (\frac{1}{2} \sec ^{-1}(a+b x)\right )}{\sqrt{1-a}}\right )}{3 a^3 \left (1-a^2\right )^{5/2}}+\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a \left (1-a^2\right ) x^2}-\frac{\sec ^{-1}(a+b x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[a + b*x]/x^4,x]

[Out]

(b*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)])/(6*a*(1 - a^2)*x^2) - ((2 - 5*a^2)*b^2*(a + b*x)*Sqrt[1 - (a + b*x)^(-2
)])/(6*a^2*(1 - a^2)^2*x) - (b^3*ArcSec[a + b*x])/(3*a^3) - ArcSec[a + b*x]/(3*x^3) + ((2 - 5*a^2 + 6*a^4)*b^3
*ArcTan[(Sqrt[1 + a]*Tan[ArcSec[a + b*x]/2])/Sqrt[1 - a]])/(3*a^3*(1 - a^2)^(5/2))

Rule 5258

Int[((a_.) + ArcSec[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d^(m + 1),
Subst[Int[(a + b*x)^p*Sec[x]*Tan[x]*(d*e - c*f + f*Sec[x])^m, x], x, ArcSec[c + d*x]], x] /; FreeQ[{a, b, c,
d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rule 4426

Int[((e_.) + (f_.)*(x_))^(m_.)*Sec[(c_.) + (d_.)*(x_)]*((a_) + (b_.)*Sec[(c_.) + (d_.)*(x_)])^(n_.)*Tan[(c_.)
+ (d_.)*(x_)], x_Symbol] :> Simp[((e + f*x)^m*(a + b*Sec[c + d*x])^(n + 1))/(b*d*(n + 1)), x] - Dist[(f*m)/(b*
d*(n + 1)), Int[(e + f*x)^(m - 1)*(a + b*Sec[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && I
GtQ[m, 0] && NeQ[n, -1]

Rule 3785

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n +
1))/(a*d*(n + 1)*(a^2 - b^2)), x] + Dist[1/(a*(n + 1)*(a^2 - b^2)), Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^
2 - b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 4060

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(a*f*(m + 1
)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m +
1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

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

Rubi steps

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

Mathematica [C]  time = 0.405984, size = 241, normalized size = 1.33 \[ \frac{1}{6} \left (-\frac{b \sqrt{\frac{a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} \left (-a^2 \left (5 b^2 x^2+1\right )-4 a^3 b x+a^4+a b x+2 b^2 x^2\right )}{a^2 \left (a^2-1\right )^2 x^2}-\frac{i \left (6 a^4-5 a^2+2\right ) b^3 \log \left (\frac{12 a^3 \left (a^2-1\right )^2 \left (\sqrt{\frac{a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} (a+b x)+\frac{i \left (a^2+a b x-1\right )}{\sqrt{1-a^2}}\right )}{\left (6 a^4-5 a^2+2\right ) b^3 x}\right )}{a^3 \left (1-a^2\right )^{5/2}}+\frac{2 b^3 \sin ^{-1}\left (\frac{1}{a+b x}\right )}{a^3}-\frac{2 \sec ^{-1}(a+b x)}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSec[a + b*x]/x^4,x]

[Out]

(-((b*Sqrt[(-1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2]*(a^4 + a*b*x - 4*a^3*b*x + 2*b^2*x^2 - a^2*(1 + 5*b^2*x
^2)))/(a^2*(-1 + a^2)^2*x^2)) - (2*ArcSec[a + b*x])/x^3 + (2*b^3*ArcSin[(a + b*x)^(-1)])/a^3 - (I*(2 - 5*a^2 +
6*a^4)*b^3*Log[(12*a^3*(-1 + a^2)^2*((I*(-1 + a^2 + a*b*x))/Sqrt[1 - a^2] + (a + b*x)*Sqrt[(-1 + a^2 + 2*a*b*
x + b^2*x^2)/(a + b*x)^2]))/((2 - 5*a^2 + 6*a^4)*b^3*x)])/(a^3*(1 - a^2)^(5/2)))/6

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Maple [B]  time = 0.226, size = 760, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*arcsec(b*x+a)/x^3+1/3*b^3*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*a/(a^2-1)^2*arcta
n(1/(-1+(b*x+a)^2)^(1/2))-b^3*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*a^3/(a^2-1)^(7/2)*
ln(2*((a^2-1)^(1/2)*(-1+(b*x+a)^2)^(1/2)+a*(b*x+a)-1)/b/x)-2/3*b^3*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a
)^2)^(1/2)/(b*x+a)/a/(a^2-1)^2*arctan(1/(-1+(b*x+a)^2)^(1/2))+5/6*b^2*(-1+(b*x+a)^2)/((-1+(b*x+a)^2)/(b*x+a)^2
)^(1/2)/(b*x+a)/(a^2-1)^2/x+11/6*b^3*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*a/(a^2-1)^(
7/2)*ln(2*((a^2-1)^(1/2)*(-1+(b*x+a)^2)^(1/2)+a*(b*x+a)-1)/b/x)-1/6*b*(-1+(b*x+a)^2)/((-1+(b*x+a)^2)/(b*x+a)^2
)^(1/2)/(b*x+a)*a/(a^2-1)^2/x^2+1/3*b^3*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)/a^3/(a^2
-1)^2*arctan(1/(-1+(b*x+a)^2)^(1/2))-1/3*b^2*(-1+(b*x+a)^2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)/a^2/(a^2-
1)^2/x-7/6*b^3*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)/a/(a^2-1)^(7/2)*ln(2*((a^2-1)^(1/
2)*(-1+(b*x+a)^2)^(1/2)+a*(b*x+a)-1)/b/x)+1/6*b*(-1+(b*x+a)^2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)/a/(a^2
-1)^2/x^2+1/3*b^3*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)/a^3/(a^2-1)^(7/2)*ln(2*((a^2-1
)^(1/2)*(-1+(b*x+a)^2)^(1/2)+a*(b*x+a)-1)/b/x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{x^{3} \int \frac{{\left (b^{2} x + a b\right )} e^{\left (\frac{1}{2} \, \log \left (b x + a + 1\right ) + \frac{1}{2} \, \log \left (b x + a - 1\right )\right )}}{b^{2} x^{5} + 2 \, a b x^{4} +{\left (a^{2} - 1\right )} x^{3} +{\left (b^{2} x^{5} + 2 \, a b x^{4} +{\left (a^{2} - 1\right )} x^{3}\right )}{\left (b x + a + 1\right )}{\left (b x + a - 1\right )}}\,{d x} - \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(b*x+a)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 3.04645, size = 1241, normalized size = 6.86 \begin{align*} \left [\frac{{\left (6 \, a^{4} - 5 \, a^{2} + 2\right )} \sqrt{a^{2} - 1} b^{3} x^{3} \log \left (\frac{a^{2} b x + a^{3} + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{\left (a^{2} - \sqrt{a^{2} - 1} a - 1\right )} -{\left (a b x + a^{2} - 1\right )} \sqrt{a^{2} - 1} - a}{x}\right ) - 4 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} b^{3} x^{3} \arctan \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) +{\left (5 \, a^{5} - 7 \, a^{3} + 2 \, a\right )} b^{3} x^{3} - 2 \,{\left (a^{9} - 3 \, a^{7} + 3 \, a^{5} - a^{3}\right )} \operatorname{arcsec}\left (b x + a\right ) +{\left ({\left (5 \, a^{5} - 7 \, a^{3} + 2 \, a\right )} b^{2} x^{2} -{\left (a^{6} - 2 \, a^{4} + a^{2}\right )} b x\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{6 \,{\left (a^{9} - 3 \, a^{7} + 3 \, a^{5} - a^{3}\right )} x^{3}}, \frac{2 \,{\left (6 \, a^{4} - 5 \, a^{2} + 2\right )} \sqrt{-a^{2} + 1} b^{3} x^{3} \arctan \left (-\frac{\sqrt{-a^{2} + 1} b x - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1} \sqrt{-a^{2} + 1}}{a^{2} - 1}\right ) - 4 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} b^{3} x^{3} \arctan \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) +{\left (5 \, a^{5} - 7 \, a^{3} + 2 \, a\right )} b^{3} x^{3} - 2 \,{\left (a^{9} - 3 \, a^{7} + 3 \, a^{5} - a^{3}\right )} \operatorname{arcsec}\left (b x + a\right ) +{\left ({\left (5 \, a^{5} - 7 \, a^{3} + 2 \, a\right )} b^{2} x^{2} -{\left (a^{6} - 2 \, a^{4} + a^{2}\right )} b x\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{6 \,{\left (a^{9} - 3 \, a^{7} + 3 \, a^{5} - a^{3}\right )} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*((6*a^4 - 5*a^2 + 2)*sqrt(a^2 - 1)*b^3*x^3*log((a^2*b*x + a^3 + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*(a^2 -
sqrt(a^2 - 1)*a - 1) - (a*b*x + a^2 - 1)*sqrt(a^2 - 1) - a)/x) - 4*(a^6 - 3*a^4 + 3*a^2 - 1)*b^3*x^3*arctan(-b
*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)) + (5*a^5 - 7*a^3 + 2*a)*b^3*x^3 - 2*(a^9 - 3*a^7 + 3*a^5 - a^3)*ar
csec(b*x + a) + ((5*a^5 - 7*a^3 + 2*a)*b^2*x^2 - (a^6 - 2*a^4 + a^2)*b*x)*sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))/(
(a^9 - 3*a^7 + 3*a^5 - a^3)*x^3), 1/6*(2*(6*a^4 - 5*a^2 + 2)*sqrt(-a^2 + 1)*b^3*x^3*arctan(-(sqrt(-a^2 + 1)*b*
x - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*sqrt(-a^2 + 1))/(a^2 - 1)) - 4*(a^6 - 3*a^4 + 3*a^2 - 1)*b^3*x^3*arctan(
-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)) + (5*a^5 - 7*a^3 + 2*a)*b^3*x^3 - 2*(a^9 - 3*a^7 + 3*a^5 - a^3)*
arcsec(b*x + a) + ((5*a^5 - 7*a^3 + 2*a)*b^2*x^2 - (a^6 - 2*a^4 + a^2)*b*x)*sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))
/((a^9 - 3*a^7 + 3*a^5 - a^3)*x^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asec(a + b*x)/x**4, x)

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Giac [B]  time = 1.84108, size = 792, normalized size = 4.38 \begin{align*} \frac{1}{3} \, b{\left (\frac{{\left (6 \, a^{4} b^{8} - 5 \, a^{2} b^{8} + 2 \, b^{8}\right )} \arctan \left (-\frac{x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{\sqrt{-a^{2} + 1}}\right )}{{\left (a^{7} b^{6} \mathrm{sgn}\left (b x + a\right ) - 2 \, a^{5} b^{6} \mathrm{sgn}\left (b x + a\right ) + a^{3} b^{6} \mathrm{sgn}\left (b x + a\right )\right )} \sqrt{-a^{2} + 1}} - \frac{2 \, b^{2} \arctan \left (-\frac{{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} b + a{\left | b \right |}}{b}\right )}{a^{3} \mathrm{sgn}\left (b x + a\right )} - \frac{4 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{3} a^{3} b^{8} - 8 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} a^{5} b^{8} + 2 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{2} a^{4} b^{7}{\left | b \right |} - 6 \, a^{6} b^{7}{\left | b \right |} -{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{3} a b^{8} + 11 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} a^{3} b^{8} - 4 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{2} a^{2} b^{7}{\left | b \right |} + 14 \, a^{4} b^{7}{\left | b \right |} - 3 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} a b^{8} + 2 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{2} b^{7}{\left | b \right |} - 10 \, a^{2} b^{7}{\left | b \right |} + 2 \, b^{7}{\left | b \right |}}{{\left (a^{6} b^{6} \mathrm{sgn}\left (b x + a\right ) - 2 \, a^{4} b^{6} \mathrm{sgn}\left (b x + a\right ) + a^{2} b^{6} \mathrm{sgn}\left (b x + a\right )\right )}{\left ({\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{2} - a^{2} + 1\right )}^{2}}\right )} - \frac{\arccos \left (\frac{1}{b x + a}\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*((6*a^4*b^8 - 5*a^2*b^8 + 2*b^8)*arctan(-(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))/sqrt(-a^2 + 1))/
((a^7*b^6*sgn(b*x + a) - 2*a^5*b^6*sgn(b*x + a) + a^3*b^6*sgn(b*x + a))*sqrt(-a^2 + 1)) - 2*b^2*arctan(-((x*ab
s(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))*b + a*abs(b))/b)/(a^3*sgn(b*x + a)) - (4*(x*abs(b) - sqrt(b^2*x^2 +
2*a*b*x + a^2 - 1))^3*a^3*b^8 - 8*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))*a^5*b^8 + 2*(x*abs(b) - sqrt(
b^2*x^2 + 2*a*b*x + a^2 - 1))^2*a^4*b^7*abs(b) - 6*a^6*b^7*abs(b) - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 -
1))^3*a*b^8 + 11*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))*a^3*b^8 - 4*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*
x + a^2 - 1))^2*a^2*b^7*abs(b) + 14*a^4*b^7*abs(b) - 3*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))*a*b^8 +
2*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))^2*b^7*abs(b) - 10*a^2*b^7*abs(b) + 2*b^7*abs(b))/((a^6*b^6*sg
n(b*x + a) - 2*a^4*b^6*sgn(b*x + a) + a^2*b^6*sgn(b*x + a))*((x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))^2
- a^2 + 1)^2)) - 1/3*arccos(1/(b*x + a))/x^3