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

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

[Out]

-1/3*b^3*arcsec(b*x+a)/a^3-1/3*arcsec(b*x+a)/x^3+1/3*(6*a^4-5*a^2+2)*b^3*arctan((1+a)^(1/2)*tan(1/2*arcsec(b*x
+a))/(1-a)^(1/2))/a^3/(-a^2+1)^(5/2)+1/6*b*(b*x+a)*(1-1/(b*x+a)^2)^(1/2)/a/(-a^2+1)/x^2-1/6*(-5*a^2+2)*b^2*(b*
x+a)*(1-1/(b*x+a)^2)^(1/2)/a^2/(-a^2+1)^2/x

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Rubi [A]  time = 0.29, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, 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 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]

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 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 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 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 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 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 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]

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*}

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Mathematica [C]  time = 0.43, size = 241, normalized size = 1.33 \[ \frac {1}{6} \left (\frac {2 b^3 \sin ^{-1}\left (\frac {1}{a+b x}\right )}{a^3}-\frac {b \sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} \left (a^4-4 a^3 b x-a^2 \left (5 b^2 x^2+1\right )+a b x+2 b^2 x^2\right )}{a^2 \left (a^2-1\right )^2 x^2}-\frac {i \left (6 a^4-5 a^2+2\right ) b^3 \log \left (\frac {12 a^3 \left (a^2-1\right )^2 \left (\sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} (a+b x)+\frac {i \left (a^2+a b x-1\right )}{\sqrt {1-a^2}}\right )}{\left (6 a^4-5 a^2+2\right ) b^3 x}\right )}{a^3 \left (1-a^2\right )^{5/2}}-\frac {2 \sec ^{-1}(a+b x)}{x^3}\right ) \]

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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fricas [A]  time = 1.93, size = 548, normalized size = 3.03 \[ \left [\frac {{\left (6 \, a^{4} - 5 \, a^{2} + 2\right )} \sqrt {a^{2} - 1} b^{3} x^{3} \log \left (\frac {a^{2} b x + a^{3} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} {\left (a^{2} - \sqrt {a^{2} - 1} a - 1\right )} - {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1} - a}{x}\right ) - 4 \, {\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} b^{3} x^{3} \arctan \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + {\left (5 \, a^{5} - 7 \, a^{3} + 2 \, a\right )} b^{3} x^{3} - 2 \, {\left (a^{9} - 3 \, a^{7} + 3 \, a^{5} - a^{3}\right )} \operatorname {arcsec}\left (b x + a\right ) + {\left ({\left (5 \, a^{5} - 7 \, a^{3} + 2 \, a\right )} b^{2} x^{2} - {\left (a^{6} - 2 \, a^{4} + a^{2}\right )} b x\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{6 \, {\left (a^{9} - 3 \, a^{7} + 3 \, a^{5} - a^{3}\right )} x^{3}}, \frac {2 \, {\left (6 \, a^{4} - 5 \, a^{2} + 2\right )} \sqrt {-a^{2} + 1} b^{3} x^{3} \arctan \left (-\frac {\sqrt {-a^{2} + 1} b x - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} \sqrt {-a^{2} + 1}}{a^{2} - 1}\right ) - 4 \, {\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} b^{3} x^{3} \arctan \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + {\left (5 \, a^{5} - 7 \, a^{3} + 2 \, a\right )} b^{3} x^{3} - 2 \, {\left (a^{9} - 3 \, a^{7} + 3 \, a^{5} - a^{3}\right )} \operatorname {arcsec}\left (b x + a\right ) + {\left ({\left (5 \, a^{5} - 7 \, a^{3} + 2 \, a\right )} b^{2} x^{2} - {\left (a^{6} - 2 \, a^{4} + a^{2}\right )} b x\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{6 \, {\left (a^{9} - 3 \, a^{7} + 3 \, a^{5} - a^{3}\right )} x^{3}}\right ] \]

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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giac [B]  time = 0.22, size = 451, normalized size = 2.49 \[ \frac {1}{3} \, b {\left (\frac {{\left (6 \, a^{4} b^{2} - 5 \, a^{2} b^{2} + 2 \, b^{2}\right )} \arctan \left (\frac {{\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + a}{\sqrt {-a^{2} + 1}}\right )}{{\left (a^{7} - 2 \, a^{5} + a^{3}\right )} \sqrt {-a^{2} + 1}} + \frac {4 \, {\left (b x + a\right )}^{3} a^{3} b^{2} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}^{3} + 10 \, {\left (b x + a\right )}^{2} a^{4} b^{2} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}^{2} - {\left (b x + a\right )}^{3} a b^{2} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}^{3} + {\left (b x + a\right )}^{2} a^{2} b^{2} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}^{2} + 16 \, {\left (b x + a\right )} a^{3} b^{2} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} - 2 \, {\left (b x + a\right )}^{2} b^{2} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}^{2} - 7 \, {\left (b x + a\right )} a b^{2} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + 5 \, a^{2} b^{2} - 2 \, b^{2}}{{\left (a^{6} - 2 \, a^{4} + a^{2}\right )} {\left ({\left (b x + a\right )}^{2} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}^{2} + 2 \, {\left (b x + a\right )} a {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + 1\right )}^{2}} - \frac {{\left (\frac {3 \, a b^{2}}{b x + a} - \frac {3 \, a^{2} b^{2}}{{\left (b x + a\right )}^{2}} - b^{2}\right )} \arccos \left (-\frac {1}{{\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a}\right )}{a^{3} {\left (\frac {a}{b x + a} - 1\right )}^{3}}\right )} \]

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^2 - 5*a^2*b^2 + 2*b^2)*arctan(((b*x + a)*(sqrt(-1/(b*x + a)^2 + 1) - 1) + a)/sqrt(-a^2 + 1))/(
(a^7 - 2*a^5 + a^3)*sqrt(-a^2 + 1)) + (4*(b*x + a)^3*a^3*b^2*(sqrt(-1/(b*x + a)^2 + 1) - 1)^3 + 10*(b*x + a)^2
*a^4*b^2*(sqrt(-1/(b*x + a)^2 + 1) - 1)^2 - (b*x + a)^3*a*b^2*(sqrt(-1/(b*x + a)^2 + 1) - 1)^3 + (b*x + a)^2*a
^2*b^2*(sqrt(-1/(b*x + a)^2 + 1) - 1)^2 + 16*(b*x + a)*a^3*b^2*(sqrt(-1/(b*x + a)^2 + 1) - 1) - 2*(b*x + a)^2*
b^2*(sqrt(-1/(b*x + a)^2 + 1) - 1)^2 - 7*(b*x + a)*a*b^2*(sqrt(-1/(b*x + a)^2 + 1) - 1) + 5*a^2*b^2 - 2*b^2)/(
(a^6 - 2*a^4 + a^2)*((b*x + a)^2*(sqrt(-1/(b*x + a)^2 + 1) - 1)^2 + 2*(b*x + a)*a*(sqrt(-1/(b*x + a)^2 + 1) -
1) + 1)^2) - (3*a*b^2/(b*x + a) - 3*a^2*b^2/(b*x + a)^2 - b^2)*arccos(-1/((b*x + a)*(a/(b*x + a) - 1) - a))/(a
^3*(a/(b*x + a) - 1)^3))

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maple [B]  time = 0.07, size = 760, normalized size = 4.20 \[ -\frac {\mathrm {arcsec}\left (b x +a \right )}{3 x^{3}}+\frac {b^{3} \sqrt {-1+\left (b x +a \right )^{2}}\, a \arctan \left (\frac {1}{\sqrt {-1+\left (b x +a \right )^{2}}}\right )}{3 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) \left (a^{2}-1\right )^{2}}-\frac {b^{3} \sqrt {-1+\left (b x +a \right )^{2}}\, a^{3} \ln \left (\frac {2 \sqrt {a^{2}-1}\, \sqrt {-1+\left (b x +a \right )^{2}}+2 a \left (b x +a \right )-2}{b x}\right )}{\sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) \left (a^{2}-1\right )^{\frac {7}{2}}}-\frac {2 b^{3} \sqrt {-1+\left (b x +a \right )^{2}}\, \arctan \left (\frac {1}{\sqrt {-1+\left (b x +a \right )^{2}}}\right )}{3 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a \left (a^{2}-1\right )^{2}}+\frac {5 b^{2} \left (-1+\left (b x +a \right )^{2}\right )}{6 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) \left (a^{2}-1\right )^{2} x}+\frac {11 b^{3} \sqrt {-1+\left (b x +a \right )^{2}}\, a \ln \left (\frac {2 \sqrt {a^{2}-1}\, \sqrt {-1+\left (b x +a \right )^{2}}+2 a \left (b x +a \right )-2}{b x}\right )}{6 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) \left (a^{2}-1\right )^{\frac {7}{2}}}-\frac {b \left (-1+\left (b x +a \right )^{2}\right ) a}{6 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) \left (a^{2}-1\right )^{2} x^{2}}+\frac {b^{3} \sqrt {-1+\left (b x +a \right )^{2}}\, \arctan \left (\frac {1}{\sqrt {-1+\left (b x +a \right )^{2}}}\right )}{3 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a^{3} \left (a^{2}-1\right )^{2}}-\frac {b^{2} \left (-1+\left (b x +a \right )^{2}\right )}{3 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a^{2} \left (a^{2}-1\right )^{2} x}-\frac {7 b^{3} \sqrt {-1+\left (b x +a \right )^{2}}\, \ln \left (\frac {2 \sqrt {a^{2}-1}\, \sqrt {-1+\left (b x +a \right )^{2}}+2 a \left (b x +a \right )-2}{b x}\right )}{6 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a \left (a^{2}-1\right )^{\frac {7}{2}}}+\frac {b \left (-1+\left (b x +a \right )^{2}\right )}{6 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a \left (a^{2}-1\right )^{2} x^{2}}+\frac {b^{3} \sqrt {-1+\left (b x +a \right )^{2}}\, \ln \left (\frac {2 \sqrt {a^{2}-1}\, \sqrt {-1+\left (b x +a \right )^{2}}+2 a \left (b x +a \right )-2}{b x}\right )}{3 \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a^{3} \left (a^{2}-1\right )^{\frac {7}{2}}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \frac {x^{3} \int \frac {{\left (b^{2} x + a b\right )} e^{\left (\frac {1}{2} \, \log \left (b x + a + 1\right ) + \frac {1}{2} \, \log \left (b x + a - 1\right )\right )}}{b^{2} x^{5} + 2 \, a b x^{4} + {\left (a^{2} - 1\right )} x^{3} + {\left (b^{2} x^{5} + 2 \, a b x^{4} + {\left (a^{2} - 1\right )} x^{3}\right )} {\left (b x + a + 1\right )} {\left (b x + a - 1\right )}}\,{d x} - \arctan \left (\sqrt {b x + a + 1} \sqrt {b x + a - 1}\right )}{3 \, x^{3}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acos}\left (\frac {1}{a+b\,x}\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(1/(a + b*x))/x^4,x)

[Out]

int(acos(1/(a + b*x))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asec}{\left (a + b x \right )}}{x^{4}}\, dx \]

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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