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

Optimal. Leaf size=197 \[ \frac{\left (53 a^2+20\right ) a (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{30 b^5}-\frac{\left (58 a^2+9\right ) (a+b x)^2 \sqrt{1-\frac{1}{(a+b x)^2}}}{120 b^5}+\frac{a^5 \sec ^{-1}(a+b x)}{5 b^5}-\frac{\left (40 a^4+40 a^2+3\right ) \tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{40 b^5}+\frac{11 a x^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{60 b^3}-\frac{x^3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{20 b^2}+\frac{1}{5} x^5 \sec ^{-1}(a+b x) \]

[Out]

(a*(20 + 53*a^2)*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)])/(30*b^5) + (11*a*x^2*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)])/
(60*b^3) - (x^3*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)])/(20*b^2) - ((9 + 58*a^2)*(a + b*x)^2*Sqrt[1 - (a + b*x)^(-
2)])/(120*b^5) + (a^5*ArcSec[a + b*x])/(5*b^5) + (x^5*ArcSec[a + b*x])/5 - ((3 + 40*a^2 + 40*a^4)*ArcTanh[Sqrt
[1 - (a + b*x)^(-2)]])/(40*b^5)

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Rubi [A]  time = 0.231645, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {5258, 4426, 3782, 4056, 4048, 3770, 3767, 8} \[ \frac{\left (53 a^2+20\right ) a (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{30 b^5}-\frac{\left (58 a^2+9\right ) (a+b x)^2 \sqrt{1-\frac{1}{(a+b x)^2}}}{120 b^5}+\frac{a^5 \sec ^{-1}(a+b x)}{5 b^5}-\frac{\left (40 a^4+40 a^2+3\right ) \tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{40 b^5}+\frac{11 a x^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{60 b^3}-\frac{x^3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{20 b^2}+\frac{1}{5} x^5 \sec ^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^4*ArcSec[a + b*x],x]

[Out]

(a*(20 + 53*a^2)*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)])/(30*b^5) + (11*a*x^2*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)])/
(60*b^3) - (x^3*(a + b*x)*Sqrt[1 - (a + b*x)^(-2)])/(20*b^2) - ((9 + 58*a^2)*(a + b*x)^2*Sqrt[1 - (a + b*x)^(-
2)])/(120*b^5) + (a^5*ArcSec[a + b*x])/(5*b^5) + (x^5*ArcSec[a + b*x])/5 - ((3 + 40*a^2 + 40*a^4)*ArcTanh[Sqrt
[1 - (a + b*x)^(-2)]])/(40*b^5)

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 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 4056

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

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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 8

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

Rubi steps

\begin{align*} \int x^4 \sec ^{-1}(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x \sec (x) (-a+\sec (x))^4 \tan (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{b^5}\\ &=\frac{1}{5} x^5 \sec ^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int (-a+\sec (x))^5 \, dx,x,\sec ^{-1}(a+b x)\right )}{5 b^5}\\ &=-\frac{x^3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{20 b^2}+\frac{1}{5} x^5 \sec ^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int (-a+\sec (x))^2 \left (-4 a^3+3 \left (1+4 a^2\right ) \sec (x)-11 a \sec ^2(x)\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{20 b^5}\\ &=\frac{11 a x^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{60 b^3}-\frac{x^3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{20 b^2}+\frac{1}{5} x^5 \sec ^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int (-a+\sec (x)) \left (12 a^4-a \left (31+48 a^2\right ) \sec (x)+\left (9+58 a^2\right ) \sec ^2(x)\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{60 b^5}\\ &=\frac{11 a x^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{60 b^3}-\frac{x^3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{20 b^2}-\frac{\left (9+58 a^2\right ) (a+b x)^2 \sqrt{1-\frac{1}{(a+b x)^2}}}{120 b^5}+\frac{1}{5} x^5 \sec ^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int \left (-24 a^5+3 \left (3+40 a^2+40 a^4\right ) \sec (x)-4 a \left (20+53 a^2\right ) \sec ^2(x)\right ) \, dx,x,\sec ^{-1}(a+b x)\right )}{120 b^5}\\ &=\frac{11 a x^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{60 b^3}-\frac{x^3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{20 b^2}-\frac{\left (9+58 a^2\right ) (a+b x)^2 \sqrt{1-\frac{1}{(a+b x)^2}}}{120 b^5}+\frac{a^5 \sec ^{-1}(a+b x)}{5 b^5}+\frac{1}{5} x^5 \sec ^{-1}(a+b x)+\frac{\left (a \left (20+53 a^2\right )\right ) \operatorname{Subst}\left (\int \sec ^2(x) \, dx,x,\sec ^{-1}(a+b x)\right )}{30 b^5}-\frac{\left (3+40 a^2+40 a^4\right ) \operatorname{Subst}\left (\int \sec (x) \, dx,x,\sec ^{-1}(a+b x)\right )}{40 b^5}\\ &=\frac{11 a x^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{60 b^3}-\frac{x^3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{20 b^2}-\frac{\left (9+58 a^2\right ) (a+b x)^2 \sqrt{1-\frac{1}{(a+b x)^2}}}{120 b^5}+\frac{a^5 \sec ^{-1}(a+b x)}{5 b^5}+\frac{1}{5} x^5 \sec ^{-1}(a+b x)-\frac{\left (3+40 a^2+40 a^4\right ) \tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{40 b^5}-\frac{\left (a \left (20+53 a^2\right )\right ) \operatorname{Subst}\left (\int 1 \, dx,x,-(a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}\right )}{30 b^5}\\ &=\frac{a \left (20+53 a^2\right ) (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{30 b^5}+\frac{11 a x^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{60 b^3}-\frac{x^3 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{20 b^2}-\frac{\left (9+58 a^2\right ) (a+b x)^2 \sqrt{1-\frac{1}{(a+b x)^2}}}{120 b^5}+\frac{a^5 \sec ^{-1}(a+b x)}{5 b^5}+\frac{1}{5} x^5 \sec ^{-1}(a+b x)-\frac{\left (3+40 a^2+40 a^4\right ) \tanh ^{-1}\left (\sqrt{1-\frac{1}{(a+b x)^2}}\right )}{40 b^5}\\ \end{align*}

Mathematica [A]  time = 0.182892, size = 173, normalized size = 0.88 \[ \frac{\sqrt{\frac{a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} \left (-9 \left (4 a^2+1\right ) b^2 x^2+2 a \left (48 a^2+31\right ) b x+a^2 \left (154 a^2+71\right )+16 a b^3 x^3-6 b^4 x^4\right )-3 \left (40 a^4+40 a^2+3\right ) \log \left ((a+b x) \left (\sqrt{\frac{a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}}+1\right )\right )-24 a^5 \sin ^{-1}\left (\frac{1}{a+b x}\right )+24 b^5 x^5 \sec ^{-1}(a+b x)}{120 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*ArcSec[a + b*x],x]

[Out]

(Sqrt[(-1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2]*(a^2*(71 + 154*a^2) + 2*a*(31 + 48*a^2)*b*x - 9*(1 + 4*a^2)*
b^2*x^2 + 16*a*b^3*x^3 - 6*b^4*x^4) + 24*b^5*x^5*ArcSec[a + b*x] - 24*a^5*ArcSin[(a + b*x)^(-1)] - 3*(3 + 40*a
^2 + 40*a^4)*Log[(a + b*x)*(1 + Sqrt[(-1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2])])/(120*b^5)

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Maple [B]  time = 0.237, size = 509, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/b^5*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*a^5*arctan(1/(-1+(b*x+a)^2)^(1/2))-1/b^
5*(-1+(b*x+a)^2)^(1/2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*a^4*ln(b*x+a+(-1+(b*x+a)^2)^(1/2))-3/40/b^4*(-
1+(b*x+a)^2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*x-1/b^5*(-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))-1/20/b^2*(-1+(b*x+a)^2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x
+a)*x^3+11/60/b^3*(-1+(b*x+a)^2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*x^2*a-29/60/b^4*(-1+(b*x+a)^2)/((-1+
(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*x*a^2+1/5*x^5*arcsec(b*x+a)+77/60/b^5*(-1+(b*x+a)^2)/((-1+(b*x+a)^2)/(b*x+
a)^2)^(1/2)/(b*x+a)*a^3+71/120/b^5*(-1+(b*x+a)^2)/((-1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*a-3/40/b^5*(-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., size = 0, normalized size = 0. \begin{align*} \frac{1}{5} \, x^{5} \arctan \left (\sqrt{b x + a + 1} \sqrt{b x + a - 1}\right ) - \int \frac{{\left (b^{2} x^{6} + a b x^{5}\right )} e^{\left (\frac{1}{2} \, \log \left (b x + a + 1\right ) + \frac{1}{2} \, \log \left (b x + a - 1\right )\right )}}{5 \,{\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5*arctan(sqrt(b*x + a + 1)*sqrt(b*x + a - 1)) - integrate(1/5*(b^2*x^6 + a*b*x^5)*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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Fricas [A]  time = 2.4978, size = 374, normalized size = 1.9 \begin{align*} \frac{24 \, b^{5} x^{5} \operatorname{arcsec}\left (b x + a\right ) + 48 \, a^{5} \arctan \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + 3 \,{\left (40 \, a^{4} + 40 \, a^{2} + 3\right )} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) -{\left (6 \, b^{3} x^{3} - 22 \, a b^{2} x^{2} - 154 \, a^{3} +{\left (58 \, a^{2} + 9\right )} b x - 71 \, a\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{120 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(24*b^5*x^5*arcsec(b*x + a) + 48*a^5*arctan(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)) + 3*(40*a^4 +
40*a^2 + 3)*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1)) - (6*b^3*x^3 - 22*a*b^2*x^2 - 154*a^3 + (58*a^2
+ 9)*b*x - 71*a)*sqrt(b^2*x^2 + 2*a*b*x + a^2 - 1))/b^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4*asec(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4*arcsec(b*x + a), x)