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

Optimal. Leaf size=126 \[ \frac{3 i b^3 \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{2 c^2}-\frac{3 b^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac{3 b x \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac{3 i b \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^2}+\frac{1}{2} x^2 \left (a+b \sec ^{-1}(c x)\right )^3 \]

[Out]

(((3*I)/2)*b*(a + b*ArcSec[c*x])^2)/c^2 - (3*b*Sqrt[1 - 1/(c^2*x^2)]*x*(a + b*ArcSec[c*x])^2)/(2*c) + (x^2*(a
+ b*ArcSec[c*x])^3)/2 - (3*b^2*(a + b*ArcSec[c*x])*Log[1 + E^((2*I)*ArcSec[c*x])])/c^2 + (((3*I)/2)*b^3*PolyLo
g[2, -E^((2*I)*ArcSec[c*x])])/c^2

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Rubi [A]  time = 0.142431, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {5222, 4409, 4184, 3719, 2190, 2279, 2391} \[ \frac{3 i b^3 \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )}{2 c^2}-\frac{3 b^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right ) \left (a+b \sec ^{-1}(c x)\right )}{c^2}-\frac{3 b x \sqrt{1-\frac{1}{c^2 x^2}} \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac{3 i b \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^2}+\frac{1}{2} x^2 \left (a+b \sec ^{-1}(c x)\right )^3 \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcSec[c*x])^3,x]

[Out]

(((3*I)/2)*b*(a + b*ArcSec[c*x])^2)/c^2 - (3*b*Sqrt[1 - 1/(c^2*x^2)]*x*(a + b*ArcSec[c*x])^2)/(2*c) + (x^2*(a
+ b*ArcSec[c*x])^3)/2 - (3*b^2*(a + b*ArcSec[c*x])*Log[1 + E^((2*I)*ArcSec[c*x])])/c^2 + (((3*I)/2)*b^3*PolyLo
g[2, -E^((2*I)*ArcSec[c*x])])/c^2

Rule 5222

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*S
ec[x]^(m + 1)*Tan[x], x], x, ArcSec[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (GtQ[n,
0] || LtQ[m, -1])

Rule 4409

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
((c + d*x)^m*Sec[a + b*x]^n)/(b*n), x] - Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int x \left (a+b \sec ^{-1}(c x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int (a+b x)^3 \sec ^2(x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )}{c^2}\\ &=\frac{1}{2} x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{(3 b) \operatorname{Subst}\left (\int (a+b x)^2 \sec ^2(x) \, dx,x,\sec ^{-1}(c x)\right )}{2 c^2}\\ &=-\frac{3 b \sqrt{1-\frac{1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac{1}{2} x^2 \left (a+b \sec ^{-1}(c x)\right )^3+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int (a+b x) \tan (x) \, dx,x,\sec ^{-1}(c x)\right )}{c^2}\\ &=\frac{3 i b \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^2}-\frac{3 b \sqrt{1-\frac{1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac{1}{2} x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{\left (6 i b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)}{1+e^{2 i x}} \, dx,x,\sec ^{-1}(c x)\right )}{c^2}\\ &=\frac{3 i b \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^2}-\frac{3 b \sqrt{1-\frac{1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac{1}{2} x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{3 b^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{c^2}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sec ^{-1}(c x)\right )}{c^2}\\ &=\frac{3 i b \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^2}-\frac{3 b \sqrt{1-\frac{1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac{1}{2} x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{3 b^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{c^2}-\frac{\left (3 i b^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \sec ^{-1}(c x)}\right )}{2 c^2}\\ &=\frac{3 i b \left (a+b \sec ^{-1}(c x)\right )^2}{2 c^2}-\frac{3 b \sqrt{1-\frac{1}{c^2 x^2}} x \left (a+b \sec ^{-1}(c x)\right )^2}{2 c}+\frac{1}{2} x^2 \left (a+b \sec ^{-1}(c x)\right )^3-\frac{3 b^2 \left (a+b \sec ^{-1}(c x)\right ) \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )}{c^2}+\frac{3 i b^3 \text{Li}_2\left (-e^{2 i \sec ^{-1}(c x)}\right )}{2 c^2}\\ \end{align*}

Mathematica [A]  time = 0.459848, size = 184, normalized size = 1.46 \[ \frac{3 i b^3 \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}(c x)}\right )+a \left (a c x \left (a c x-3 b \sqrt{1-\frac{1}{c^2 x^2}}\right )-6 b^2 \log \left (\frac{1}{c x}\right )\right )-3 b^2 \sec ^{-1}(c x)^2 \left (-a c^2 x^2+b \left (c x \sqrt{1-\frac{1}{c^2 x^2}}-i\right )\right )-3 b \sec ^{-1}(c x) \left (a c x \left (2 b \sqrt{1-\frac{1}{c^2 x^2}}-a c x\right )+2 b^2 \log \left (1+e^{2 i \sec ^{-1}(c x)}\right )\right )+b^3 c^2 x^2 \sec ^{-1}(c x)^3}{2 c^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x*(a + b*ArcSec[c*x])^3,x]

[Out]

(-3*b^2*(-(a*c^2*x^2) + b*(-I + c*Sqrt[1 - 1/(c^2*x^2)]*x))*ArcSec[c*x]^2 + b^3*c^2*x^2*ArcSec[c*x]^3 - 3*b*Ar
cSec[c*x]*(a*c*x*(2*b*Sqrt[1 - 1/(c^2*x^2)] - a*c*x) + 2*b^2*Log[1 + E^((2*I)*ArcSec[c*x])]) + a*(a*c*x*(-3*b*
Sqrt[1 - 1/(c^2*x^2)] + a*c*x) - 6*b^2*Log[1/(c*x)]) + (3*I)*b^3*PolyLog[2, -E^((2*I)*ArcSec[c*x])])/(2*c^2)

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Maple [A]  time = 0.401, size = 285, normalized size = 2.3 \begin{align*}{\frac{{x}^{2}{a}^{3}}{2}}+{\frac{{x}^{2}{b}^{3} \left ({\rm arcsec} \left (cx\right ) \right ) ^{3}}{2}}-{\frac{3\,{b}^{3} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}x}{2\,c}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}+{\frac{{\frac{3\,i}{2}}{b}^{3} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}}{{c}^{2}}}-3\,{\frac{{b}^{3}{\rm arcsec} \left (cx\right )}{{c}^{2}}\ln \left ( 1+ \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) }+{\frac{{\frac{3\,i}{2}}{b}^{3}}{{c}^{2}}{\it polylog} \left ( 2,- \left ({\frac{1}{cx}}+i\sqrt{1-{\frac{1}{{c}^{2}{x}^{2}}}} \right ) ^{2} \right ) }+{\frac{3\,{x}^{2}a{b}^{2} \left ({\rm arcsec} \left (cx\right ) \right ) ^{2}}{2}}-3\,{\frac{a{b}^{2}{\rm arcsec} \left (cx\right )x}{c}\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}-3\,{\frac{a{b}^{2}}{{c}^{2}}\ln \left ({\frac{1}{cx}} \right ) }+{\frac{3\,{a}^{2}b{x}^{2}{\rm arcsec} \left (cx\right )}{2}}-{\frac{3\,{a}^{2}bx}{2\,c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{3\,{a}^{2}b}{2\,{c}^{3}x}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*a^3+1/2*x^2*b^3*arcsec(c*x)^3-3/2/c*b^3*arcsec(c*x)^2*((c^2*x^2-1)/c^2/x^2)^(1/2)*x+3/2*I/c^2*b^3*arcs
ec(c*x)^2-3/c^2*b^3*arcsec(c*x)*ln(1+(1/c/x+I*(1-1/c^2/x^2)^(1/2))^2)+3/2*I*b^3*polylog(2,-(1/c/x+I*(1-1/c^2/x
^2)^(1/2))^2)/c^2+3/2*x^2*a*b^2*arcsec(c*x)^2-3/c*a*b^2*((c^2*x^2-1)/c^2/x^2)^(1/2)*arcsec(c*x)*x-3/c^2*a*b^2*
ln(1/c/x)+3/2*a^2*b*x^2*arcsec(c*x)-3/2/c*a^2*b/((c^2*x^2-1)/c^2/x^2)^(1/2)*x+3/2/c^3*a^2*b/((c^2*x^2-1)/c^2/x
^2)^(1/2)/x

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3}{2} \, a b^{2} x^{2} \operatorname{arcsec}\left (c x\right )^{2} + \frac{1}{2} \, a^{3} x^{2} + \frac{3}{2} \,{\left (x^{2} \operatorname{arcsec}\left (c x\right ) - \frac{x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c}\right )} a^{2} b - 3 \,{\left (\frac{x \sqrt{-\frac{1}{c^{2} x^{2}} + 1} \operatorname{arcsec}\left (c x\right )}{c} - \frac{\log \left (x\right )}{c^{2}}\right )} a b^{2} + \frac{1}{8} \,{\left (4 \, x^{2} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )^{3} - 3 \, x^{2} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right ) \log \left (c^{2} x^{2}\right )^{2} - 8 \, \int \frac{3 \,{\left ({\left (4 \, x \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )^{2} - x \log \left (c^{2} x^{2}\right )^{2}\right )} \sqrt{c x + 1} \sqrt{c x - 1} + 4 \,{\left (2 \, c^{2} x^{3} \log \left (c\right )^{2} - 2 \, x \log \left (c\right )^{2} + 2 \,{\left (c^{2} x^{3} - x\right )} \log \left (x\right )^{2} -{\left ({\left (2 \, c^{2} \log \left (c\right ) + c^{2}\right )} x^{3} - x{\left (2 \, \log \left (c\right ) + 1\right )} + 2 \,{\left (c^{2} x^{3} - x\right )} \log \left (x\right )\right )} \log \left (c^{2} x^{2}\right ) + 4 \,{\left (c^{2} x^{3} \log \left (c\right ) - x \log \left (c\right )\right )} \log \left (x\right )\right )} \arctan \left (\sqrt{c x + 1} \sqrt{c x - 1}\right )\right )}}{8 \,{\left (c^{2} x^{2} - 1\right )}}\,{d x}\right )} b^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*a*b^2*x^2*arcsec(c*x)^2 + 1/2*a^3*x^2 + 3/2*(x^2*arcsec(c*x) - x*sqrt(-1/(c^2*x^2) + 1)/c)*a^2*b - 3*(x*sq
rt(-1/(c^2*x^2) + 1)*arcsec(c*x)/c - log(x)/c^2)*a*b^2 + 1/8*(4*x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))^3 - 3*
x^2*arctan(sqrt(c*x + 1)*sqrt(c*x - 1))*log(c^2*x^2)^2 - 8*integrate(3/8*((4*x*arctan(sqrt(c*x + 1)*sqrt(c*x -
1))^2 - x*log(c^2*x^2)^2)*sqrt(c*x + 1)*sqrt(c*x - 1) + 4*(2*c^2*x^3*log(c)^2 - 2*x*log(c)^2 + 2*(c^2*x^3 - x
)*log(x)^2 - ((2*c^2*log(c) + c^2)*x^3 - x*(2*log(c) + 1) + 2*(c^2*x^3 - x)*log(x))*log(c^2*x^2) + 4*(c^2*x^3*
log(c) - x*log(c))*log(x))*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)))/(c^2*x^2 - 1), x))*b^3

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{3} x \operatorname{arcsec}\left (c x\right )^{3} + 3 \, a b^{2} x \operatorname{arcsec}\left (c x\right )^{2} + 3 \, a^{2} b x \operatorname{arcsec}\left (c x\right ) + a^{3} x, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b^3*x*arcsec(c*x)^3 + 3*a*b^2*x*arcsec(c*x)^2 + 3*a^2*b*x*arcsec(c*x) + a^3*x, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (a + b \operatorname{asec}{\left (c x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*(a + b*asec(c*x))**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )}^{3} x\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsec(c*x) + a)^3*x, x)