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

Optimal. Leaf size=124 \[ \frac{(d+e x)^3 \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac{b \left (6 c^2 d^2+e^2\right ) \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{6 c^3}-\frac{b d e x \sqrt{1-\frac{1}{c^2 x^2}}}{c}-\frac{b e^2 x^2 \sqrt{1-\frac{1}{c^2 x^2}}}{6 c}+\frac{b d^3 \csc ^{-1}(c x)}{3 e} \]

[Out]

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

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Rubi [A]  time = 0.266256, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.562, Rules used = {5226, 1568, 1475, 1807, 844, 216, 266, 63, 208} \[ \frac{(d+e x)^3 \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac{b \left (6 c^2 d^2+e^2\right ) \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{6 c^3}-\frac{b d e x \sqrt{1-\frac{1}{c^2 x^2}}}{c}-\frac{b e^2 x^2 \sqrt{1-\frac{1}{c^2 x^2}}}{6 c}+\frac{b d^3 \csc ^{-1}(c x)}{3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5226

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + b
*ArcSec[c*x]))/(e*(m + 1)), x] - Dist[b/(c*e*(m + 1)), Int[(d + e*x)^(m + 1)/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x],
x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[m, -1]

Rule 1568

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Int[x^(m + mn*q
)*(e + d/x^mn)^q*(a + c*x^n2)^p, x] /; FreeQ[{a, c, d, e, m, mn, p}, x] && EqQ[n2, -2*mn] && IntegerQ[q] && (P
osQ[n2] ||  !IntegerQ[p])

Rule 1475

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x
] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 216

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int (d+e x)^2 \left (a+b \sec ^{-1}(c x)\right ) \, dx &=\frac{(d+e x)^3 \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac{b \int \frac{(d+e x)^3}{\sqrt{1-\frac{1}{c^2 x^2}} x^2} \, dx}{3 c e}\\ &=\frac{(d+e x)^3 \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac{b \int \frac{\left (e+\frac{d}{x}\right )^3 x}{\sqrt{1-\frac{1}{c^2 x^2}}} \, dx}{3 c e}\\ &=\frac{(d+e x)^3 \left (a+b \sec ^{-1}(c x)\right )}{3 e}+\frac{b \operatorname{Subst}\left (\int \frac{(e+d x)^3}{x^3 \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{3 c e}\\ &=-\frac{b e^2 \sqrt{1-\frac{1}{c^2 x^2}} x^2}{6 c}+\frac{(d+e x)^3 \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac{b \operatorname{Subst}\left (\int \frac{-6 d e^2-e \left (6 d^2+\frac{e^2}{c^2}\right ) x-2 d^3 x^2}{x^2 \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{6 c e}\\ &=-\frac{b d e \sqrt{1-\frac{1}{c^2 x^2}} x}{c}-\frac{b e^2 \sqrt{1-\frac{1}{c^2 x^2}} x^2}{6 c}+\frac{(d+e x)^3 \left (a+b \sec ^{-1}(c x)\right )}{3 e}+\frac{b \operatorname{Subst}\left (\int \frac{e \left (6 d^2+\frac{e^2}{c^2}\right )+2 d^3 x}{x \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{6 c e}\\ &=-\frac{b d e \sqrt{1-\frac{1}{c^2 x^2}} x}{c}-\frac{b e^2 \sqrt{1-\frac{1}{c^2 x^2}} x^2}{6 c}+\frac{(d+e x)^3 \left (a+b \sec ^{-1}(c x)\right )}{3 e}+\frac{\left (b d^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{3 c e}+\frac{\left (b \left (6 c^2 d^2+e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{6 c^3}\\ &=-\frac{b d e \sqrt{1-\frac{1}{c^2 x^2}} x}{c}-\frac{b e^2 \sqrt{1-\frac{1}{c^2 x^2}} x^2}{6 c}+\frac{b d^3 \csc ^{-1}(c x)}{3 e}+\frac{(d+e x)^3 \left (a+b \sec ^{-1}(c x)\right )}{3 e}+\frac{\left (b \left (6 c^2 d^2+e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x}{c^2}}} \, dx,x,\frac{1}{x^2}\right )}{12 c^3}\\ &=-\frac{b d e \sqrt{1-\frac{1}{c^2 x^2}} x}{c}-\frac{b e^2 \sqrt{1-\frac{1}{c^2 x^2}} x^2}{6 c}+\frac{b d^3 \csc ^{-1}(c x)}{3 e}+\frac{(d+e x)^3 \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac{\left (b \left (6 c^2 d^2+e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{c^2-c^2 x^2} \, dx,x,\sqrt{1-\frac{1}{c^2 x^2}}\right )}{6 c}\\ &=-\frac{b d e \sqrt{1-\frac{1}{c^2 x^2}} x}{c}-\frac{b e^2 \sqrt{1-\frac{1}{c^2 x^2}} x^2}{6 c}+\frac{b d^3 \csc ^{-1}(c x)}{3 e}+\frac{(d+e x)^3 \left (a+b \sec ^{-1}(c x)\right )}{3 e}-\frac{b \left (6 c^2 d^2+e^2\right ) \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{6 c^3}\\ \end{align*}

Mathematica [A]  time = 0.176971, size = 124, normalized size = 1. \[ \frac{c^2 x \left (2 a c \left (3 d^2+3 d e x+e^2 x^2\right )-b e \sqrt{1-\frac{1}{c^2 x^2}} (6 d+e x)\right )-b \left (6 c^2 d^2+e^2\right ) \log \left (x \left (\sqrt{1-\frac{1}{c^2 x^2}}+1\right )\right )+2 b c^3 x \sec ^{-1}(c x) \left (3 d^2+3 d e x+e^2 x^2\right )}{6 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.129, size = 362, normalized size = 2.9 \begin{align*}{\frac{a{e}^{2}{x}^{3}}{3}}+ae{x}^{2}d+ax{d}^{2}+{\frac{a{d}^{3}}{3\,e}}+{\frac{b{e}^{2}{\rm arcsec} \left (cx\right ){x}^{3}}{3}}+be{\rm arcsec} \left (cx\right ){x}^{2}d+b{\rm arcsec} \left (cx\right )x{d}^{2}+{\frac{b{\rm arcsec} \left (cx\right ){d}^{3}}{3\,e}}+{\frac{b{d}^{3}}{3\,cex}\sqrt{{c}^{2}{x}^{2}-1}\arctan \left ({\frac{1}{\sqrt{{c}^{2}{x}^{2}-1}}} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{b{d}^{2}}{{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{b{e}^{2}{x}^{2}}{6\,c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{b{e}^{2}}{6\,{c}^{3}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{bexd}{c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{bed}{{c}^{3}x}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{b{e}^{2}}{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+d)^2*(a+b*arcsec(c*x)),x)

[Out]

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

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Maxima [A]  time = 1.01039, size = 270, normalized size = 2.18 \begin{align*} \frac{1}{3} \, a e^{2} x^{3} + a d e x^{2} +{\left (x^{2} \operatorname{arcsec}\left (c x\right ) - \frac{x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c}\right )} b d e + \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^{2} + a d^{2} 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}}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 3.90167, size = 462, normalized size = 3.73 \begin{align*} \frac{2 \, a c^{3} e^{2} x^{3} + 6 \, a c^{3} d e x^{2} + 6 \, a c^{3} d^{2} x + 2 \,{\left (b c^{3} e^{2} x^{3} + 3 \, b c^{3} d e x^{2} + 3 \, b c^{3} d^{2} x - 3 \, b c^{3} d^{2} - 3 \, b c^{3} d e - b c^{3} e^{2}\right )} \operatorname{arcsec}\left (c x\right ) + 4 \,{\left (3 \, b c^{3} d^{2} + 3 \, b c^{3} d e + b c^{3} e^{2}\right )} \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) +{\left (6 \, b c^{2} d^{2} + b e^{2}\right )} \log \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) -{\left (b c e^{2} x + 6 \, b c d e\right )} \sqrt{c^{2} x^{2} - 1}}{6 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a*c^3*e^2*x^3 + 6*a*c^3*d*e*x^2 + 6*a*c^3*d^2*x + 2*(b*c^3*e^2*x^3 + 3*b*c^3*d*e*x^2 + 3*b*c^3*d^2*x -
3*b*c^3*d^2 - 3*b*c^3*d*e - b*c^3*e^2)*arcsec(c*x) + 4*(3*b*c^3*d^2 + 3*b*c^3*d*e + b*c^3*e^2)*arctan(-c*x + s
qrt(c^2*x^2 - 1)) + (6*b*c^2*d^2 + b*e^2)*log(-c*x + sqrt(c^2*x^2 - 1)) - (b*c*e^2*x + 6*b*c*d*e)*sqrt(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\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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