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

Optimal. Leaf size=167 \[ \frac{(d+e x)^4 \left (a+b \sec ^{-1}(c x)\right )}{4 e}-\frac{b e x \sqrt{1-\frac{1}{c^2 x^2}} \left (9 c^2 d^2+e^2\right )}{6 c^3}-\frac{b d \left (2 c^2 d^2+e^2\right ) \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{2 c^3}-\frac{b d e^2 x^2 \sqrt{1-\frac{1}{c^2 x^2}}}{2 c}-\frac{b e^3 x^3 \sqrt{1-\frac{1}{c^2 x^2}}}{12 c}+\frac{b d^4 \csc ^{-1}(c x)}{4 e} \]

[Out]

-(b*e*(9*c^2*d^2 + e^2)*Sqrt[1 - 1/(c^2*x^2)]*x)/(6*c^3) - (b*d*e^2*Sqrt[1 - 1/(c^2*x^2)]*x^2)/(2*c) - (b*e^3*
Sqrt[1 - 1/(c^2*x^2)]*x^3)/(12*c) + (b*d^4*ArcCsc[c*x])/(4*e) + ((d + e*x)^4*(a + b*ArcSec[c*x]))/(4*e) - (b*d
*(2*c^2*d^2 + e^2)*ArcTanh[Sqrt[1 - 1/(c^2*x^2)]])/(2*c^3)

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Rubi [A]  time = 0.401404, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 11, 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} \[ \frac{(d+e x)^4 \left (a+b \sec ^{-1}(c x)\right )}{4 e}-\frac{b e x \sqrt{1-\frac{1}{c^2 x^2}} \left (9 c^2 d^2+e^2\right )}{6 c^3}-\frac{b d \left (2 c^2 d^2+e^2\right ) \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{2 c^3}-\frac{b d e^2 x^2 \sqrt{1-\frac{1}{c^2 x^2}}}{2 c}-\frac{b e^3 x^3 \sqrt{1-\frac{1}{c^2 x^2}}}{12 c}+\frac{b d^4 \csc ^{-1}(c x)}{4 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*(a + b*ArcSec[c*x]),x]

[Out]

-(b*e*(9*c^2*d^2 + e^2)*Sqrt[1 - 1/(c^2*x^2)]*x)/(6*c^3) - (b*d*e^2*Sqrt[1 - 1/(c^2*x^2)]*x^2)/(2*c) - (b*e^3*
Sqrt[1 - 1/(c^2*x^2)]*x^3)/(12*c) + (b*d^4*ArcCsc[c*x])/(4*e) + ((d + e*x)^4*(a + b*ArcSec[c*x]))/(4*e) - (b*d
*(2*c^2*d^2 + e^2)*ArcTanh[Sqrt[1 - 1/(c^2*x^2)]])/(2*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)^3 \left (a+b \sec ^{-1}(c x)\right ) \, dx &=\frac{(d+e x)^4 \left (a+b \sec ^{-1}(c x)\right )}{4 e}-\frac{b \int \frac{(d+e x)^4}{\sqrt{1-\frac{1}{c^2 x^2}} x^2} \, dx}{4 c e}\\ &=\frac{(d+e x)^4 \left (a+b \sec ^{-1}(c x)\right )}{4 e}-\frac{b \int \frac{\left (e+\frac{d}{x}\right )^4 x^2}{\sqrt{1-\frac{1}{c^2 x^2}}} \, dx}{4 c e}\\ &=\frac{(d+e x)^4 \left (a+b \sec ^{-1}(c x)\right )}{4 e}+\frac{b \operatorname{Subst}\left (\int \frac{(e+d x)^4}{x^4 \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{4 c e}\\ &=-\frac{b e^3 \sqrt{1-\frac{1}{c^2 x^2}} x^3}{12 c}+\frac{(d+e x)^4 \left (a+b \sec ^{-1}(c x)\right )}{4 e}-\frac{b \operatorname{Subst}\left (\int \frac{-12 d e^3-2 e^2 \left (9 d^2+\frac{e^2}{c^2}\right ) x-12 d^3 e x^2-3 d^4 x^3}{x^3 \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{12 c e}\\ &=-\frac{b d e^2 \sqrt{1-\frac{1}{c^2 x^2}} x^2}{2 c}-\frac{b e^3 \sqrt{1-\frac{1}{c^2 x^2}} x^3}{12 c}+\frac{(d+e x)^4 \left (a+b \sec ^{-1}(c x)\right )}{4 e}+\frac{b \operatorname{Subst}\left (\int \frac{4 e^2 \left (9 d^2+\frac{e^2}{c^2}\right )+12 d e \left (2 d^2+\frac{e^2}{c^2}\right ) x+6 d^4 x^2}{x^2 \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{24 c e}\\ &=-\frac{b e \left (9 c^2 d^2+e^2\right ) \sqrt{1-\frac{1}{c^2 x^2}} x}{6 c^3}-\frac{b d e^2 \sqrt{1-\frac{1}{c^2 x^2}} x^2}{2 c}-\frac{b e^3 \sqrt{1-\frac{1}{c^2 x^2}} x^3}{12 c}+\frac{(d+e x)^4 \left (a+b \sec ^{-1}(c x)\right )}{4 e}-\frac{b \operatorname{Subst}\left (\int \frac{-12 d e \left (2 d^2+\frac{e^2}{c^2}\right )-6 d^4 x}{x \sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{24 c e}\\ &=-\frac{b e \left (9 c^2 d^2+e^2\right ) \sqrt{1-\frac{1}{c^2 x^2}} x}{6 c^3}-\frac{b d e^2 \sqrt{1-\frac{1}{c^2 x^2}} x^2}{2 c}-\frac{b e^3 \sqrt{1-\frac{1}{c^2 x^2}} x^3}{12 c}+\frac{(d+e x)^4 \left (a+b \sec ^{-1}(c x)\right )}{4 e}+\frac{\left (b d^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{c^2}}} \, dx,x,\frac{1}{x}\right )}{4 c e}+\frac{\left (b d \left (2 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 )}{2 c^3}\\ &=-\frac{b e \left (9 c^2 d^2+e^2\right ) \sqrt{1-\frac{1}{c^2 x^2}} x}{6 c^3}-\frac{b d e^2 \sqrt{1-\frac{1}{c^2 x^2}} x^2}{2 c}-\frac{b e^3 \sqrt{1-\frac{1}{c^2 x^2}} x^3}{12 c}+\frac{b d^4 \csc ^{-1}(c x)}{4 e}+\frac{(d+e x)^4 \left (a+b \sec ^{-1}(c x)\right )}{4 e}+\frac{\left (b d \left (2 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 )}{4 c^3}\\ &=-\frac{b e \left (9 c^2 d^2+e^2\right ) \sqrt{1-\frac{1}{c^2 x^2}} x}{6 c^3}-\frac{b d e^2 \sqrt{1-\frac{1}{c^2 x^2}} x^2}{2 c}-\frac{b e^3 \sqrt{1-\frac{1}{c^2 x^2}} x^3}{12 c}+\frac{b d^4 \csc ^{-1}(c x)}{4 e}+\frac{(d+e x)^4 \left (a+b \sec ^{-1}(c x)\right )}{4 e}-\frac{\left (b d \left (2 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 )}{2 c}\\ &=-\frac{b e \left (9 c^2 d^2+e^2\right ) \sqrt{1-\frac{1}{c^2 x^2}} x}{6 c^3}-\frac{b d e^2 \sqrt{1-\frac{1}{c^2 x^2}} x^2}{2 c}-\frac{b e^3 \sqrt{1-\frac{1}{c^2 x^2}} x^3}{12 c}+\frac{b d^4 \csc ^{-1}(c x)}{4 e}+\frac{(d+e x)^4 \left (a+b \sec ^{-1}(c x)\right )}{4 e}-\frac{b d \left (2 c^2 d^2+e^2\right ) \tanh ^{-1}\left (\sqrt{1-\frac{1}{c^2 x^2}}\right )}{2 c^3}\\ \end{align*}

Mathematica [A]  time = 0.278559, size = 166, normalized size = 0.99 \[ \frac{3 a c^3 x \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )-b e x \sqrt{1-\frac{1}{c^2 x^2}} \left (c^2 \left (18 d^2+6 d e x+e^2 x^2\right )+2 e^2\right )-6 b d \left (2 c^2 d^2+e^2\right ) \log \left (x \left (\sqrt{1-\frac{1}{c^2 x^2}}+1\right )\right )+3 b c^3 x \sec ^{-1}(c x) \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )}{12 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3*(a + b*ArcSec[c*x]),x]

[Out]

(3*a*c^3*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) - b*e*Sqrt[1 - 1/(c^2*x^2)]*x*(2*e^2 + c^2*(18*d^2 + 6*
d*e*x + e^2*x^2)) + 3*b*c^3*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3)*ArcSec[c*x] - 6*b*d*(2*c^2*d^2 + e^2
)*Log[(1 + Sqrt[1 - 1/(c^2*x^2)])*x])/(12*c^3)

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Maple [B]  time = 0.135, size = 486, normalized size = 2.9 \begin{align*}{\frac{a{e}^{3}{x}^{4}}{4}}+a{e}^{2}{x}^{3}d+{\frac{3\,ae{x}^{2}{d}^{2}}{2}}+ax{d}^{3}+{\frac{a{d}^{4}}{4\,e}}+{\frac{b{e}^{3}{\rm arcsec} \left (cx\right ){x}^{4}}{4}}+b{e}^{2}{\rm arcsec} \left (cx\right ){x}^{3}d+{\frac{3\,be{\rm arcsec} \left (cx\right ){x}^{2}{d}^{2}}{2}}+b{\rm arcsec} \left (cx\right )x{d}^{3}+{\frac{b{\rm arcsec} \left (cx\right ){d}^{4}}{4\,e}}+{\frac{b{d}^{4}}{4\,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}^{3}}{{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}^{3}{x}^{3}}{12\,c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{b{e}^{3}x}{12\,{c}^{3}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{b{e}^{2}d{x}^{2}}{2\,c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{b{e}^{2}d}{2\,{c}^{3}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{3\,bex{d}^{2}}{2\,c}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{3\,be{d}^{2}}{2\,{c}^{3}x}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}-{\frac{b{e}^{2}d}{2\,{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}}}}}}}+{\frac{b{e}^{3}}{6\,{c}^{5}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((e*x+d)^3*(a+b*arcsec(c*x)),x)

[Out]

1/4*a*e^3*x^4+a*e^2*x^3*d+3/2*a*e*x^2*d^2+a*x*d^3+1/4*a/e*d^4+1/4*b*e^3*arcsec(c*x)*x^4+b*e^2*arcsec(c*x)*x^3*
d+3/2*b*e*arcsec(c*x)*x^2*d^2+b*arcsec(c*x)*x*d^3+1/4*b/e*arcsec(c*x)*d^4+1/4/c*b/e*(c^2*x^2-1)^(1/2)/((c^2*x^
2-1)/c^2/x^2)^(1/2)/x*d^4*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^3*ln(c*x+(c^2*x^2-1)^(1/2))-1/12/c*b*e^3/((c^2*x^2-1)/c^2/x^2)^(1/2)*x^3-1/12/c^3*b*e^3/((c^2*x^2-1)/c^2/x^2
)^(1/2)*x-1/2/c*b*e^2/((c^2*x^2-1)/c^2/x^2)^(1/2)*d*x^2+1/2/c^3*b*e^2/((c^2*x^2-1)/c^2/x^2)^(1/2)*d-3/2/c*b*e/
((c^2*x^2-1)/c^2/x^2)^(1/2)*x*d^2+3/2/c^3*b*e/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*d^2-1/2/c^4*b*e^2*(c^2*x^2-1)^(1/2
)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*d*ln(c*x+(c^2*x^2-1)^(1/2))+1/6/c^5*b*e^3/((c^2*x^2-1)/c^2/x^2)^(1/2)/x

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Maxima [A]  time = 0.97603, size = 367, normalized size = 2.2 \begin{align*} \frac{1}{4} \, a e^{3} x^{4} + a d e^{2} x^{3} + \frac{3}{2} \, a d^{2} e 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 )} b d^{2} e + \frac{1}{4} \,{\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 d e^{2} + \frac{1}{12} \,{\left (3 \, x^{4} \operatorname{arcsec}\left (c x\right ) - \frac{c^{2} x^{3}{\left (-\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} + 3 \, x \sqrt{-\frac{1}{c^{2} x^{2}} + 1}}{c^{3}}\right )} b e^{3} + a d^{3} 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^{3}}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 4.56251, size = 633, normalized size = 3.79 \begin{align*} \frac{3 \, a c^{4} e^{3} x^{4} + 12 \, a c^{4} d e^{2} x^{3} + 18 \, a c^{4} d^{2} e x^{2} + 12 \, a c^{4} d^{3} x + 3 \,{\left (b c^{4} e^{3} x^{4} + 4 \, b c^{4} d e^{2} x^{3} + 6 \, b c^{4} d^{2} e x^{2} + 4 \, b c^{4} d^{3} x - 4 \, b c^{4} d^{3} - 6 \, b c^{4} d^{2} e - 4 \, b c^{4} d e^{2} - b c^{4} e^{3}\right )} \operatorname{arcsec}\left (c x\right ) + 6 \,{\left (4 \, b c^{4} d^{3} + 6 \, b c^{4} d^{2} e + 4 \, b c^{4} d e^{2} + b c^{4} e^{3}\right )} \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + 6 \,{\left (2 \, b c^{3} d^{3} + b c d e^{2}\right )} \log \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) -{\left (b c^{2} e^{3} x^{2} + 6 \, b c^{2} d e^{2} x + 18 \, b c^{2} d^{2} e + 2 \, b e^{3}\right )} \sqrt{c^{2} x^{2} - 1}}{12 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*c^4*e^3*x^4 + 12*a*c^4*d*e^2*x^3 + 18*a*c^4*d^2*e*x^2 + 12*a*c^4*d^3*x + 3*(b*c^4*e^3*x^4 + 4*b*c^4*
d*e^2*x^3 + 6*b*c^4*d^2*e*x^2 + 4*b*c^4*d^3*x - 4*b*c^4*d^3 - 6*b*c^4*d^2*e - 4*b*c^4*d*e^2 - b*c^4*e^3)*arcse
c(c*x) + 6*(4*b*c^4*d^3 + 6*b*c^4*d^2*e + 4*b*c^4*d*e^2 + b*c^4*e^3)*arctan(-c*x + sqrt(c^2*x^2 - 1)) + 6*(2*b
*c^3*d^3 + b*c*d*e^2)*log(-c*x + sqrt(c^2*x^2 - 1)) - (b*c^2*e^3*x^2 + 6*b*c^2*d*e^2*x + 18*b*c^2*d^2*e + 2*b*
e^3)*sqrt(c^2*x^2 - 1))/c^4

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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 )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asec(c*x))*(d + e*x)**3, x)

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

[Out]

integrate((e*x + d)^3*(b*arcsec(c*x) + a), x)