3.20 \(\int x^2 \sec ^{-1}(a+b x) \, dx\)

Optimal. Leaf size=116 \[ \frac {a^3 \sec ^{-1}(a+b x)}{3 b^3}-\frac {\left (6 a^2+1\right ) \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{6 b^3}+\frac {5 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{6 b^3}-\frac {x (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{6 b^2}+\frac {1}{3} x^3 \sec ^{-1}(a+b x) \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5258, 4426, 3782, 3770, 3767, 8} \[ \frac {a^3 \sec ^{-1}(a+b x)}{3 b^3}-\frac {\left (6 a^2+1\right ) \tanh ^{-1}\left (\sqrt {1-\frac {1}{(a+b x)^2}}\right )}{6 b^3}+\frac {5 a (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{6 b^3}-\frac {x (a+b x) \sqrt {1-\frac {1}{(a+b x)^2}}}{6 b^2}+\frac {1}{3} x^3 \sec ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcSec[a + b*x],x]

[Out]

(5*a*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)])/(6*b^3) - (x*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)])/(6*b^2) + (a^3*ArcSe
c[a + b*x])/(3*b^3) + (x^3*ArcSec[a + b*x])/3 - ((1 + 6*a^2)*ArcTanh[Sqrt[1 - (a + b*x)^(-2)]])/(6*b^3)

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

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Mathematica [A]  time = 0.18, size = 131, normalized size = 1.13 \[ \frac {-2 a^3 \sin ^{-1}\left (\frac {1}{a+b x}\right )+\left (5 a^2+4 a b x-b^2 x^2\right ) \sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}}-\left (6 a^2+1\right ) \log \left ((a+b x) \left (\sqrt {\frac {a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}}+1\right )\right )+2 b^3 x^3 \sec ^{-1}(a+b x)}{6 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcSec[a + b*x],x]

[Out]

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

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fricas [A]  time = 2.18, size = 117, normalized size = 1.01 \[ \frac {2 \, b^{3} x^{3} \operatorname {arcsec}\left (b x + a\right ) + 4 \, a^{3} \arctan \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + {\left (6 \, a^{2} + 1\right )} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} - 1} {\left (b x - 5 \, a\right )}}{6 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.17, size = 204, normalized size = 1.76 \[ -\frac {1}{24} \, b {\left (\frac {8 \, {\left (b x + a\right )}^{3} {\left (\frac {3 \, a}{b x + a} - \frac {3 \, a^{2}}{{\left (b x + a\right )}^{2}} - 1\right )} \arccos \left (-\frac {1}{{\left (b x + a\right )} {\left (\frac {a}{b x + a} - 1\right )} - a}\right )}{b^{4}} - \frac {{\left (b x + a\right )}^{2} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}^{2} + 12 \, {\left (b x + a\right )} a {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + 4 \, {\left (6 \, a^{2} + 1\right )} \log \left (-{\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} {\left | b x + a \right |}\right ) - \frac {12 \, {\left (b x + a\right )} a {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )} + 1}{{\left (b x + a\right )}^{2} {\left (\sqrt {-\frac {1}{{\left (b x + a\right )}^{2}} + 1} - 1\right )}^{2}}}{b^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 273, normalized size = 2.35 \[ \frac {x^{3} \mathrm {arcsec}\left (b x +a \right )}{3}-\frac {\sqrt {-1+\left (b x +a \right )^{2}}\, a^{3} \arctan \left (\frac {1}{\sqrt {-1+\left (b x +a \right )^{2}}}\right )}{3 b^{3} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}-\frac {\left (-1+\left (b x +a \right )^{2}\right ) x}{6 b^{2} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}-\frac {\sqrt {-1+\left (b x +a \right )^{2}}\, a^{2} \ln \left (b x +a +\sqrt {-1+\left (b x +a \right )^{2}}\right )}{b^{3} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}+\frac {5 \left (-1+\left (b x +a \right )^{2}\right ) a}{6 b^{3} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )}-\frac {\sqrt {-1+\left (b x +a \right )^{2}}\, \ln \left (b x +a +\sqrt {-1+\left (b x +a \right )^{2}}\right )}{6 b^{3} \sqrt {\frac {-1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arcsec(b*x+a),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) - integrate(1/3*(b^2*x^4 + a*b*x^3)*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 \[ \int x^2\,\mathrm {acos}\left (\frac {1}{a+b\,x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acos(1/(a + b*x)),x)

[Out]

int(x^2*acos(1/(a + b*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \operatorname {asec}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*asec(b*x+a),x)

[Out]

Integral(x**2*asec(a + b*x), x)

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