3.71 \(\int (d+e x^2) (a+b \sec ^{-1}(c x)) \, dx\)

Optimal. Leaf size=109 \[ d x \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac{b x \left (6 c^2 d+e\right ) \tanh ^{-1}\left (\frac{c x}{\sqrt{c^2 x^2-1}}\right )}{6 c^2 \sqrt{c^2 x^2}}-\frac{b e x^2 \sqrt{c^2 x^2-1}}{6 c \sqrt{c^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0513458, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {5228, 12, 388, 217, 206} \[ d x \left (a+b \sec ^{-1}(c x)\right )+\frac{1}{3} e x^3 \left (a+b \sec ^{-1}(c x)\right )-\frac{b x \left (6 c^2 d+e\right ) \tanh ^{-1}\left (\frac{c x}{\sqrt{c^2 x^2-1}}\right )}{6 c^2 \sqrt{c^2 x^2}}-\frac{b e x^2 \sqrt{c^2 x^2-1}}{6 c \sqrt{c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)*(a + b*ArcSec[c*x]),x]

[Out]

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

Rule 5228

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^p, x]}, Dist[a + b*ArcSec[c*x], u, x] - Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyIntegrand[u/(x*Sqrt[c^2*x^2
- 1]), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.26434, size = 150, normalized size = 1.38 \[ a d x+\frac{1}{3} a e x^3-\frac{b d x \sqrt{1-\frac{1}{c^2 x^2}} \tanh ^{-1}\left (\frac{c x}{\sqrt{c^2 x^2-1}}\right )}{\sqrt{c^2 x^2-1}}-\frac{b e x^2 \sqrt{\frac{c^2 x^2-1}{c^2 x^2}}}{6 c}-\frac{b e \log \left (x \left (\sqrt{\frac{c^2 x^2-1}{c^2 x^2}}+1\right )\right )}{6 c^3}+b d x \sec ^{-1}(c x)+\frac{1}{3} b e x^3 \sec ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^2)*(a + b*ArcSec[c*x]),x]

[Out]

a*d*x + (a*e*x^3)/3 - (b*e*x^2*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)])/(6*c) + b*d*x*ArcSec[c*x] + (b*e*x^3*ArcSec[c*x
])/3 - (b*d*Sqrt[1 - 1/(c^2*x^2)]*x*ArcTanh[(c*x)/Sqrt[-1 + c^2*x^2]])/Sqrt[-1 + c^2*x^2] - (b*e*Log[x*(1 + Sq
rt[(-1 + c^2*x^2)/(c^2*x^2)])])/(6*c^3)

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Maple [B]  time = 0.171, size = 195, normalized size = 1.8 \begin{align*}{\frac{a{x}^{3}e}{3}}+adx+{\frac{b{\rm arcsec} \left (cx\right ){x}^{3}e}{3}}+b{\rm arcsec} \left (cx\right )xd-{\frac{bd}{{c}^{2}x}\sqrt{{c}^{2}{x}^{2}-1}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}-1} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{be{x}^{2}}{6\,c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{be}{6\,{c}^{3}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{be}{6\,{c}^{4}x}\sqrt{{c}^{2}{x}^{2}-1}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}-1} \right ){\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((e*x^2+d)*(a+b*arcsec(c*x)),x)

[Out]

1/3*a*x^3*e+a*d*x+1/3*b*arcsec(c*x)*x^3*e+b*arcsec(c*x)*x*d-1/c^2*b/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*(c^2*x^2-1)^
(1/2)*d*ln(c*x+(c^2*x^2-1)^(1/2))-1/6/c*b/((c^2*x^2-1)/c^2/x^2)^(1/2)*e*x^2+1/6/c^3*b/((c^2*x^2-1)/c^2/x^2)^(1
/2)*e-1/6/c^4*b*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*e*ln(c*x+(c^2*x^2-1)^(1/2))

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Maxima [A]  time = 0.96545, size = 208, normalized size = 1.91 \begin{align*} \frac{1}{3} \, a e x^{3} + \frac{1}{12} \,{\left (4 \, x^{3} \operatorname{arcsec}\left (c x\right ) - \frac{\frac{2 \, \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c^{2}{\left (\frac{1}{c^{2} x^{2}} - 1\right )} + c^{2}} + \frac{\log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right )}{c^{2}} - \frac{\log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} - 1\right )}{c^{2}}}{c}\right )} b e + a d x + \frac{{\left (2 \, c x \operatorname{arcsec}\left (c x\right ) - \log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right ) + \log \left (-\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right )\right )} b d}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*e*x^3 + 1/12*(4*x^3*arcsec(c*x) - (2*sqrt(-1/(c^2*x^2) + 1)/(c^2*(1/(c^2*x^2) - 1) + c^2) + log(sqrt(-1/
(c^2*x^2) + 1) + 1)/c^2 - log(sqrt(-1/(c^2*x^2) + 1) - 1)/c^2)/c)*b*e + a*d*x + 1/2*(2*c*x*arcsec(c*x) - log(s
qrt(-1/(c^2*x^2) + 1) + 1) + log(-sqrt(-1/(c^2*x^2) + 1) + 1))*b*d/c

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Fricas [A]  time = 2.54092, size = 327, normalized size = 3. \begin{align*} \frac{2 \, a c^{3} e x^{3} + 6 \, a c^{3} d x - \sqrt{c^{2} x^{2} - 1} b c e x + 2 \,{\left (b c^{3} e x^{3} + 3 \, b c^{3} d x - 3 \, b c^{3} d - b c^{3} e\right )} \operatorname{arcsec}\left (c x\right ) + 4 \,{\left (3 \, b c^{3} d + b c^{3} e\right )} \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) +{\left (6 \, b c^{2} d + b e\right )} \log \left (-c x + \sqrt{c^{2} x^{2} - 1}\right )}{6 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a*c^3*e*x^3 + 6*a*c^3*d*x - sqrt(c^2*x^2 - 1)*b*c*e*x + 2*(b*c^3*e*x^3 + 3*b*c^3*d*x - 3*b*c^3*d - b*c^
3*e)*arcsec(c*x) + 4*(3*b*c^3*d + b*c^3*e)*arctan(-c*x + sqrt(c^2*x^2 - 1)) + (6*b*c^2*d + b*e)*log(-c*x + sqr
t(c^2*x^2 - 1)))/c^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{asec}{\left (c x \right )}\right ) \left (d + e x^{2}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*(a+b*asec(c*x)),x)

[Out]

Integral((a + b*asec(c*x))*(d + e*x**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)*(b*arcsec(c*x) + a), x)