3.19 \(\int x^3 \sec ^{-1}(a+b x) \, dx\)

Optimal. Leaf size=155 \[ -\frac {a^4 \sec ^{-1}(a+b x)}{4 b^4}-\frac {\left (17 a^2+2\right ) (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{12 b^4}+\frac {\left (2 a^2+1\right ) a \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{2 b^4}+\frac {a (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}}}{3 b^4}-\frac {x^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{12 b^2}+\frac {1}{4} x^4 \sec ^{-1}(a+b x) \]

[Out]

-1/4*a^4*arcsec(b*x+a)/b^4+1/4*x^4*arcsec(b*x+a)+1/2*a*(2*a^2+1)*arctanh((1-1/(b*x+a)^2)^(1/2))/b^4-1/12*(17*a
^2+2)*(b*x+a)*(1-1/(b*x+a)^2)^(1/2)/b^4-1/12*x^2*(b*x+a)*(1-1/(b*x+a)^2)^(1/2)/b^2+1/3*a*(b*x+a)^2*(1-1/(b*x+a
)^2)^(1/2)/b^4

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Rubi [A]  time = 0.14, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5258, 4426, 3782, 4048, 3770, 3767, 8} \[ -\frac {\left (17 a^2+2\right ) (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{12 b^4}-\frac {a^4 \sec ^{-1}(a+b x)}{4 b^4}+\frac {\left (2 a^2+1\right ) a \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{2 b^4}-\frac {x^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{12 b^2}+\frac {a (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}}}{3 b^4}+\frac {1}{4} x^4 \sec ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSec[a + b*x],x]

[Out]

-((2 + 17*a^2)*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)])/(12*b^4) - (x^2*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)])/(12*b^2
) + (a*(a + b*x)^2*Sqrt[1 - (a + b*x)^(-2)])/(3*b^4) - (a^4*ArcSec[a + b*x])/(4*b^4) + (x^4*ArcSec[a + b*x])/4
+ (a*(1 + 2*a^2)*ArcTanh[Sqrt[1 - (a + b*x)^(-2)]])/(2*b^4)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3782

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[1/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2) +
3*a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]

Rule 4048

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

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 x^3 \sec ^{-1}(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int x \sec (x) (-a+\sec (x))^3 \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^4}\\ &=\frac {1}{4} x^4 \sec ^{-1}(a+b x)-\frac {\operatorname {Subst}\left (\int (-a+\sec (x))^4 \, dx,x,\sec ^{-1}(a+b x)\right )}{4 b^4}\\ &=-\frac {x^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{12 b^2}+\frac {1}{4} x^4 \sec ^{-1}(a+b x)-\frac {\operatorname {Subst}\left (\int (-a+\sec (x)) \left (-3 a^3+\left (2+9 a^2\right ) \sec (x)-8 a \sec ^2(x)\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{12 b^4}\\ &=-\frac {x^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{12 b^2}+\frac {a (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}}}{3 b^4}+\frac {1}{4} x^4 \sec ^{-1}(a+b x)-\frac {\operatorname {Subst}\left (\int \left (6 a^4-12 a \left (1+2 a^2\right ) \sec (x)+2 \left (2+17 a^2\right ) \sec ^2(x)\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{24 b^4}\\ &=-\frac {x^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{12 b^2}+\frac {a (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}}}{3 b^4}-\frac {a^4 \sec ^{-1}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \sec ^{-1}(a+b x)+\frac {\left (a \left (1+2 a^2\right )\right ) \operatorname {Subst}\left (\int \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{2 b^4}-\frac {\left (2+17 a^2\right ) \operatorname {Subst}\left (\int \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{12 b^4}\\ &=-\frac {x^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{12 b^2}+\frac {a (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}}}{3 b^4}-\frac {a^4 \sec ^{-1}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \sec ^{-1}(a+b x)+\frac {a \left (1+2 a^2\right ) \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{2 b^4}+\frac {\left (2+17 a^2\right ) \operatorname {Subst}\left (\int 1 \, dx,x,-(a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}\right )}{12 b^4}\\ &=-\frac {\left (2+17 a^2\right ) (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{12 b^4}-\frac {x^2 (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{12 b^2}+\frac {a (a+b x)^2 \sqrt {1-\frac {1}{(a+b x)^2}}}{3 b^4}-\frac {a^4 \sec ^{-1}(a+b x)}{4 b^4}+\frac {1}{4} x^4 \sec ^{-1}(a+b x)+\frac {a \left (1+2 a^2\right ) \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{2 b^4}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 150, normalized size = 0.97 \[ \frac {3 a^4 \sin ^{-1}\left (\frac {1}{a+b x}\right )+6 \left (2 a^2+1\right ) a \log \left ((a+b x) \left (\sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}}+1\right )\right )-\sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} \left (13 a^3+9 a^2 b x-3 a b^2 x^2+2 a+b^3 x^3+2 b x\right )+3 b^4 x^4 \sec ^{-1}(a+b x)}{12 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSec[a + b*x],x]

[Out]

(-(Sqrt[(-1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2]*(2*a + 13*a^3 + 2*b*x + 9*a^2*b*x - 3*a*b^2*x^2 + b^3*x^3)
) + 3*b^4*x^4*ArcSec[a + b*x] + 3*a^4*ArcSin[(a + b*x)^(-1)] + 6*a*(1 + 2*a^2)*Log[(a + b*x)*(1 + Sqrt[(-1 + a
^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2])])/(12*b^4)

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fricas [A]  time = 2.07, size = 130, normalized size = 0.84 \[ \frac {3 \, b^{4} x^{4} \operatorname {arcsec}\left (b x + a\right ) - 6 \, a^{4} \arctan \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - 6 \, {\left (2 \, a^{3} + a\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} {\left (b^{2} x^{2} - 4 \, a b x + 13 \, a^{2} + 2\right )}}{12 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b^4*x^4*arcsec(b*x + a) - 6*a^4*arctan(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)) - 6*(2*a^3 + a)*l
og(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)) - sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)*(b^2*x^2 - 4*a*b*x + 13*a
^2 + 2))/b^4

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giac [B]  time = 0.19, size = 299, normalized size = 1.93 \[ -\frac {1}{96} \, b {\left (\frac {24 \, {\left (b x + a\right )}^{4} {\left (\frac {4 \, a}{b x + a} - \frac {6 \, a^{2}}{{\left (b x + a\right )}^{2}} + \frac {4 \, a^{3}}{{\left (b x + a\right )}^{3}} - 1\right )} \arccos \left (-\frac {1}{{\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a}\right )}{b^{5}} + \frac {{\left (b x + a\right )}^{3} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}^{3} + 12 \, {\left (b x + a\right )}^{2} a {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}^{2} + 72 \, {\left (b x + a\right )} a^{2} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + 9 \, {\left (b x + a\right )} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + 48 \, {\left (2 \, a^{3} + a\right )} \log \left (-{\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} {\left | b x + a \right |}\right ) - \frac {9 \, {\left (8 \, a^{2} + 1\right )} {\left (b x + a\right )}^{2} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}^{2} + 12 \, {\left (b x + a\right )} a {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + 1}{{\left (b x + a\right )}^{3} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}^{3}}}{b^{5}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*b*(24*(b*x + a)^4*(4*a/(b*x + a) - 6*a^2/(b*x + a)^2 + 4*a^3/(b*x + a)^3 - 1)*arccos(-1/((b*x + a)*(a/(b
*x + a) - 1) - a))/b^5 + ((b*x + a)^3*(sqrt(-1/(b*x + a)^2 + 1) - 1)^3 + 12*(b*x + a)^2*a*(sqrt(-1/(b*x + a)^2
+ 1) - 1)^2 + 72*(b*x + a)*a^2*(sqrt(-1/(b*x + a)^2 + 1) - 1) + 9*(b*x + a)*(sqrt(-1/(b*x + a)^2 + 1) - 1) +
48*(2*a^3 + a)*log(-(sqrt(-1/(b*x + a)^2 + 1) - 1)*abs(b*x + a)) - (9*(8*a^2 + 1)*(b*x + a)^2*(sqrt(-1/(b*x +
a)^2 + 1) - 1)^2 + 12*(b*x + a)*a*(sqrt(-1/(b*x + a)^2 + 1) - 1) + 1)/((b*x + a)^3*(sqrt(-1/(b*x + a)^2 + 1) -
1)^3))/b^5)

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maple [B]  time = 0.06, size = 359, normalized size = 2.32 \[ \frac {x^{4} \mathrm {arcsec}\left (b x +a \right )}{4}-\frac {\left (-1+\left (b x +a \right )^{2}\right ) x^{2}}{12 b^{2} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}+\frac {\sqrt {-1+\left (b x +a \right )^{2}}\, a^{4} \arctan \left (\frac {1}{\sqrt {-1+\left (b x +a \right )^{2}}}\right )}{4 b^{4} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}+\frac {\left (-1+\left (b x +a \right )^{2}\right ) x a}{3 b^{3} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}+\frac {\sqrt {-1+\left (b x +a \right )^{2}}\, a^{3} \ln \left (b x +a +\sqrt {-1+\left (b x +a \right )^{2}}\right )}{b^{4} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}-\frac {13 \left (-1+\left (b x +a \right )^{2}\right ) a^{2}}{12 b^{4} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}+\frac {\sqrt {-1+\left (b x +a \right )^{2}}\, a \ln \left (b x +a +\sqrt {-1+\left (b x +a \right )^{2}}\right )}{2 b^{4} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}-\frac {-1+\left (b x +a \right )^{2}}{6 b^{4} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsec(b*x+a),x)

[Out]

1/4*x^4*arcsec(b*x+a)-1/12/b^2*(-1+(b*x+a)^2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*x^2+1/4/b^4*(-1+(b*x+a)
^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*a^4*arctan(1/(-1+(b*x+a)^2)^(1/2))+1/3/b^3*(-1+(b*x+a)^2)/(
(-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*x*a+1/b^4*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a
)*a^3*ln(b*x+a+(-1+(b*x+a)^2)^(1/2))-13/12/b^4*(-1+(b*x+a)^2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*a^2+1/2
/b^4*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*a*ln(b*x+a+(-1+(b*x+a)^2)^(1/2))-1/6/b^4*(-
1+(b*x+a)^2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\mathrm {acos}\left (\frac {1}{a+b\,x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acos(1/(a + b*x)),x)

[Out]

int(x^3*acos(1/(a + b*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {asec}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asec(b*x+a),x)

[Out]

Integral(x**3*asec(a + b*x), x)

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