3.1 \(\int (d+e x)^2 (A+B x+C x^2) \sqrt{d^2-e^2 x^2} \, dx\)

Optimal. Leaf size=236 \[ \frac{d^2 x \sqrt{d^2-e^2 x^2} \left (10 A e^2+4 B d e+3 C d^2\right )}{16 e^2}-\frac{d \left (d^2-e^2 x^2\right )^{3/2} \left (e (10 A e+7 B d)+4 C d^2\right )}{15 e^3}-\frac{x \left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+3 C d^2\right )}{8 e^2}+\frac{d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right ) \left (10 A e^2+4 B d e+3 C d^2\right )}{16 e^3}-\frac{x^2 \left (d^2-e^2 x^2\right )^{3/2} (B e+2 C d)}{5 e}-\frac{1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2} \]

[Out]

(d^2*(3*C*d^2 + 4*B*d*e + 10*A*e^2)*x*Sqrt[d^2 - e^2*x^2])/(16*e^2) - (d*(4*C*d^2 + e*(7*B*d + 10*A*e))*(d^2 -
e^2*x^2)^(3/2))/(15*e^3) - ((3*C*d^2 + 2*e*(2*B*d + A*e))*x*(d^2 - e^2*x^2)^(3/2))/(8*e^2) - ((2*C*d + B*e)*x
^2*(d^2 - e^2*x^2)^(3/2))/(5*e) - (C*x^3*(d^2 - e^2*x^2)^(3/2))/6 + (d^4*(3*C*d^2 + 4*B*d*e + 10*A*e^2)*ArcTan
[(e*x)/Sqrt[d^2 - e^2*x^2]])/(16*e^3)

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Rubi [A]  time = 0.469868, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {1815, 641, 195, 217, 203} \[ \frac{d^2 x \sqrt{d^2-e^2 x^2} \left (10 A e^2+4 B d e+3 C d^2\right )}{16 e^2}-\frac{d \left (d^2-e^2 x^2\right )^{3/2} \left (e (10 A e+7 B d)+4 C d^2\right )}{15 e^3}-\frac{x \left (d^2-e^2 x^2\right )^{3/2} \left (2 e (A e+2 B d)+3 C d^2\right )}{8 e^2}+\frac{d^4 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right ) \left (10 A e^2+4 B d e+3 C d^2\right )}{16 e^3}-\frac{x^2 \left (d^2-e^2 x^2\right )^{3/2} (B e+2 C d)}{5 e}-\frac{1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*(A + B*x + C*x^2)*Sqrt[d^2 - e^2*x^2],x]

[Out]

(d^2*(3*C*d^2 + 4*B*d*e + 10*A*e^2)*x*Sqrt[d^2 - e^2*x^2])/(16*e^2) - (d*(4*C*d^2 + e*(7*B*d + 10*A*e))*(d^2 -
e^2*x^2)^(3/2))/(15*e^3) - ((3*C*d^2 + 2*e*(2*B*d + A*e))*x*(d^2 - e^2*x^2)^(3/2))/(8*e^2) - ((2*C*d + B*e)*x
^2*(d^2 - e^2*x^2)^(3/2))/(5*e) - (C*x^3*(d^2 - e^2*x^2)^(3/2))/6 + (d^4*(3*C*d^2 + 4*B*d*e + 10*A*e^2)*ArcTan
[(e*x)/Sqrt[d^2 - e^2*x^2]])/(16*e^3)

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rule 641

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 203

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

Rubi steps

\begin{align*} \int (d+e x)^2 \left (A+B x+C x^2\right ) \sqrt{d^2-e^2 x^2} \, dx &=-\frac{1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}-\frac{\int \sqrt{d^2-e^2 x^2} \left (-6 A d^2 e^2-6 d e^2 (B d+2 A e) x-3 e^2 \left (3 C d^2+2 e (2 B d+A e)\right ) x^2-6 e^3 (2 C d+B e) x^3\right ) \, dx}{6 e^2}\\ &=-\frac{(2 C d+B e) x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac{1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac{\int \sqrt{d^2-e^2 x^2} \left (30 A d^2 e^4+6 d e^3 \left (4 C d^2+e (7 B d+10 A e)\right ) x+15 e^4 \left (3 C d^2+2 e (2 B d+A e)\right ) x^2\right ) \, dx}{30 e^4}\\ &=-\frac{\left (3 C d^2+2 e (2 B d+A e)\right ) x \left (d^2-e^2 x^2\right )^{3/2}}{8 e^2}-\frac{(2 C d+B e) x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac{1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}-\frac{\int \left (-15 d^2 e^4 \left (3 C d^2+4 B d e+10 A e^2\right )-24 d e^5 \left (4 C d^2+e (7 B d+10 A e)\right ) x\right ) \sqrt{d^2-e^2 x^2} \, dx}{120 e^6}\\ &=-\frac{d \left (4 C d^2+e (7 B d+10 A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{15 e^3}-\frac{\left (3 C d^2+2 e (2 B d+A e)\right ) x \left (d^2-e^2 x^2\right )^{3/2}}{8 e^2}-\frac{(2 C d+B e) x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac{1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac{\left (d^2 \left (3 C d^2+4 B d e+10 A e^2\right )\right ) \int \sqrt{d^2-e^2 x^2} \, dx}{8 e^2}\\ &=\frac{d^2 \left (3 C d^2+4 B d e+10 A e^2\right ) x \sqrt{d^2-e^2 x^2}}{16 e^2}-\frac{d \left (4 C d^2+e (7 B d+10 A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{15 e^3}-\frac{\left (3 C d^2+2 e (2 B d+A e)\right ) x \left (d^2-e^2 x^2\right )^{3/2}}{8 e^2}-\frac{(2 C d+B e) x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac{1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac{\left (d^4 \left (3 C d^2+4 B d e+10 A e^2\right )\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{16 e^2}\\ &=\frac{d^2 \left (3 C d^2+4 B d e+10 A e^2\right ) x \sqrt{d^2-e^2 x^2}}{16 e^2}-\frac{d \left (4 C d^2+e (7 B d+10 A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{15 e^3}-\frac{\left (3 C d^2+2 e (2 B d+A e)\right ) x \left (d^2-e^2 x^2\right )^{3/2}}{8 e^2}-\frac{(2 C d+B e) x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac{1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac{\left (d^4 \left (3 C d^2+4 B d e+10 A e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )}{16 e^2}\\ &=\frac{d^2 \left (3 C d^2+4 B d e+10 A e^2\right ) x \sqrt{d^2-e^2 x^2}}{16 e^2}-\frac{d \left (4 C d^2+e (7 B d+10 A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{15 e^3}-\frac{\left (3 C d^2+2 e (2 B d+A e)\right ) x \left (d^2-e^2 x^2\right )^{3/2}}{8 e^2}-\frac{(2 C d+B e) x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac{1}{6} C x^3 \left (d^2-e^2 x^2\right )^{3/2}+\frac{d^4 \left (3 C d^2+4 B d e+10 A e^2\right ) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{16 e^3}\\ \end{align*}

Mathematica [A]  time = 0.51806, size = 226, normalized size = 0.96 \[ \frac{\sqrt{d^2-e^2 x^2} \left (\sqrt{1-\frac{e^2 x^2}{d^2}} \left (2 e \left (5 A e \left (9 d^2 e x-16 d^3+16 d e^2 x^2+6 e^3 x^3\right )+B \left (32 d^2 e^2 x^2-30 d^3 e x-56 d^4+60 d e^3 x^3+24 e^4 x^4\right )\right )+C \left (-32 d^3 e^2 x^2+50 d^2 e^3 x^3-45 d^4 e x-64 d^5+96 d e^4 x^4+40 e^5 x^5\right )\right )+15 \sin ^{-1}\left (\frac{e x}{d}\right ) \left (2 d^3 e (5 A e+2 B d)+3 C d^5\right )\right )}{240 e^3 \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2*(A + B*x + C*x^2)*Sqrt[d^2 - e^2*x^2],x]

[Out]

(Sqrt[d^2 - e^2*x^2]*(Sqrt[1 - (e^2*x^2)/d^2]*(C*(-64*d^5 - 45*d^4*e*x - 32*d^3*e^2*x^2 + 50*d^2*e^3*x^3 + 96*
d*e^4*x^4 + 40*e^5*x^5) + 2*e*(5*A*e*(-16*d^3 + 9*d^2*e*x + 16*d*e^2*x^2 + 6*e^3*x^3) + B*(-56*d^4 - 30*d^3*e*
x + 32*d^2*e^2*x^2 + 60*d*e^3*x^3 + 24*e^4*x^4))) + 15*(3*C*d^5 + 2*d^3*e*(2*B*d + 5*A*e))*ArcSin[(e*x)/d]))/(
240*e^3*Sqrt[1 - (e^2*x^2)/d^2])

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Maple [A]  time = 0.087, size = 371, normalized size = 1.6 \begin{align*} -{\frac{C{x}^{3}}{6} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{3\,C{d}^{2}x}{8\,{e}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{3\,C{d}^{4}x}{16\,{e}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{3\,C{d}^{6}}{16\,{e}^{2}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{{x}^{2}B}{5} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{2\,Cd{x}^{2}}{5\,e} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{7\,B{d}^{2}}{15\,{e}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{4\,{d}^{3}C}{15\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{xA}{4} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{Bdx}{2\,e} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{d}^{2}xA}{8}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{{d}^{3}xB}{4\,e}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{5\,{d}^{4}A}{8}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+{\frac{{d}^{5}B}{4\,e}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{2\,Ad}{3\,e} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2*(C*x^2+B*x+A)*(-e^2*x^2+d^2)^(1/2),x)

[Out]

-1/6*C*x^3*(-e^2*x^2+d^2)^(3/2)-3/8/e^2*C*d^2*x*(-e^2*x^2+d^2)^(3/2)+3/16/e^2*C*d^4*x*(-e^2*x^2+d^2)^(1/2)+3/1
6/e^2*C*d^6/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))-1/5*x^2*(-e^2*x^2+d^2)^(3/2)*B-2/5*x^2*(-e^
2*x^2+d^2)^(3/2)/e*d*C-7/15*d^2/e^2*(-e^2*x^2+d^2)^(3/2)*B-4/15*d^3/e^3*(-e^2*x^2+d^2)^(3/2)*C-1/4*x*(-e^2*x^2
+d^2)^(3/2)*A-1/2*x*(-e^2*x^2+d^2)^(3/2)/e*B*d+5/8*d^2*x*(-e^2*x^2+d^2)^(1/2)*A+1/4*d^3/e*x*(-e^2*x^2+d^2)^(1/
2)*B+5/8*d^4/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))*A+1/4*d^5/e/(e^2)^(1/2)*arctan((e^2)^(1/2)
*x/(-e^2*x^2+d^2)^(1/2))*B-2/3*(-e^2*x^2+d^2)^(3/2)/e*A*d

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Maxima [A]  time = 1.51519, size = 505, normalized size = 2.14 \begin{align*} -\frac{1}{6} \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} C x^{3} + \frac{A d^{4} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{2 \, \sqrt{e^{2}}} + \frac{C d^{6} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{16 \, \sqrt{e^{2}} e^{2}} + \frac{1}{2} \, \sqrt{-e^{2} x^{2} + d^{2}} A d^{2} x + \frac{\sqrt{-e^{2} x^{2} + d^{2}} C d^{4} x}{16 \, e^{2}} - \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} C d^{2} x}{8 \, e^{2}} + \frac{{\left (C d^{2} + 2 \, B d e + A e^{2}\right )} d^{4} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{8 \, \sqrt{e^{2}} e^{2}} - \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} B d^{2}}{3 \, e^{2}} - \frac{2 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} A d}{3 \, e} + \frac{\sqrt{-e^{2} x^{2} + d^{2}}{\left (C d^{2} + 2 \, B d e + A e^{2}\right )} d^{2} x}{8 \, e^{2}} - \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}}{\left (2 \, C d e + B e^{2}\right )} x^{2}}{5 \, e^{2}} - \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}}{\left (C d^{2} + 2 \, B d e + A e^{2}\right )} x}{4 \, e^{2}} - \frac{2 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}}{\left (2 \, C d e + B e^{2}\right )} d^{2}}{15 \, e^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(C*x^2+B*x+A)*(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

-1/6*(-e^2*x^2 + d^2)^(3/2)*C*x^3 + 1/2*A*d^4*arcsin(e^2*x/sqrt(d^2*e^2))/sqrt(e^2) + 1/16*C*d^6*arcsin(e^2*x/
sqrt(d^2*e^2))/(sqrt(e^2)*e^2) + 1/2*sqrt(-e^2*x^2 + d^2)*A*d^2*x + 1/16*sqrt(-e^2*x^2 + d^2)*C*d^4*x/e^2 - 1/
8*(-e^2*x^2 + d^2)^(3/2)*C*d^2*x/e^2 + 1/8*(C*d^2 + 2*B*d*e + A*e^2)*d^4*arcsin(e^2*x/sqrt(d^2*e^2))/(sqrt(e^2
)*e^2) - 1/3*(-e^2*x^2 + d^2)^(3/2)*B*d^2/e^2 - 2/3*(-e^2*x^2 + d^2)^(3/2)*A*d/e + 1/8*sqrt(-e^2*x^2 + d^2)*(C
*d^2 + 2*B*d*e + A*e^2)*d^2*x/e^2 - 1/5*(-e^2*x^2 + d^2)^(3/2)*(2*C*d*e + B*e^2)*x^2/e^2 - 1/4*(-e^2*x^2 + d^2
)^(3/2)*(C*d^2 + 2*B*d*e + A*e^2)*x/e^2 - 2/15*(-e^2*x^2 + d^2)^(3/2)*(2*C*d*e + B*e^2)*d^2/e^4

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Fricas [A]  time = 2.39827, size = 464, normalized size = 1.97 \begin{align*} -\frac{30 \,{\left (3 \, C d^{6} + 4 \, B d^{5} e + 10 \, A d^{4} e^{2}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) -{\left (40 \, C e^{5} x^{5} - 64 \, C d^{5} - 112 \, B d^{4} e - 160 \, A d^{3} e^{2} + 48 \,{\left (2 \, C d e^{4} + B e^{5}\right )} x^{4} + 10 \,{\left (5 \, C d^{2} e^{3} + 12 \, B d e^{4} + 6 \, A e^{5}\right )} x^{3} - 32 \,{\left (C d^{3} e^{2} - 2 \, B d^{2} e^{3} - 5 \, A d e^{4}\right )} x^{2} - 15 \,{\left (3 \, C d^{4} e + 4 \, B d^{3} e^{2} - 6 \, A d^{2} e^{3}\right )} x\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{240 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(C*x^2+B*x+A)*(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/240*(30*(3*C*d^6 + 4*B*d^5*e + 10*A*d^4*e^2)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (40*C*e^5*x^5 - 64
*C*d^5 - 112*B*d^4*e - 160*A*d^3*e^2 + 48*(2*C*d*e^4 + B*e^5)*x^4 + 10*(5*C*d^2*e^3 + 12*B*d*e^4 + 6*A*e^5)*x^
3 - 32*(C*d^3*e^2 - 2*B*d^2*e^3 - 5*A*d*e^4)*x^2 - 15*(3*C*d^4*e + 4*B*d^3*e^2 - 6*A*d^2*e^3)*x)*sqrt(-e^2*x^2
+ d^2))/e^3

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Sympy [C]  time = 21.467, size = 1239, normalized size = 5.25 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2*(C*x**2+B*x+A)*(-e**2*x**2+d**2)**(1/2),x)

[Out]

A*d**2*Piecewise((-I*d**2*acosh(e*x/d)/(2*e) - I*d*x/(2*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**3/(2*d*sqrt(-1
+ e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (d**2*asin(e*x/d)/(2*e) + d*x*sqrt(1 - e**2*x**2/d**2)/2, T
rue)) + 2*A*d*e*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) + A*e
**2*Piecewise((-I*d**4*acosh(e*x/d)/(8*e**3) + I*d**3*x/(8*e**2*sqrt(-1 + e**2*x**2/d**2)) - 3*I*d*x**3/(8*sqr
t(-1 + e**2*x**2/d**2)) + I*e**2*x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (d**4*as
in(e*x/d)/(8*e**3) - d**3*x/(8*e**2*sqrt(1 - e**2*x**2/d**2)) + 3*d*x**3/(8*sqrt(1 - e**2*x**2/d**2)) - e**2*x
**5/(4*d*sqrt(1 - e**2*x**2/d**2)), True)) + B*d**2*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2
*x**2)**(3/2)/(3*e**2), True)) + 2*B*d*e*Piecewise((-I*d**4*acosh(e*x/d)/(8*e**3) + I*d**3*x/(8*e**2*sqrt(-1 +
e**2*x**2/d**2)) - 3*I*d*x**3/(8*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)), Ab
s(e**2*x**2)/Abs(d**2) > 1), (d**4*asin(e*x/d)/(8*e**3) - d**3*x/(8*e**2*sqrt(1 - e**2*x**2/d**2)) + 3*d*x**3/
(8*sqrt(1 - e**2*x**2/d**2)) - e**2*x**5/(4*d*sqrt(1 - e**2*x**2/d**2)), True)) + B*e**2*Piecewise((-2*d**4*sq
rt(d**2 - e**2*x**2)/(15*e**4) - d**2*x**2*sqrt(d**2 - e**2*x**2)/(15*e**2) + x**4*sqrt(d**2 - e**2*x**2)/5, N
e(e, 0)), (x**4*sqrt(d**2)/4, True)) + C*d**2*Piecewise((-I*d**4*acosh(e*x/d)/(8*e**3) + I*d**3*x/(8*e**2*sqrt
(-1 + e**2*x**2/d**2)) - 3*I*d*x**3/(8*sqrt(-1 + e**2*x**2/d**2)) + I*e**2*x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)
), Abs(e**2*x**2)/Abs(d**2) > 1), (d**4*asin(e*x/d)/(8*e**3) - d**3*x/(8*e**2*sqrt(1 - e**2*x**2/d**2)) + 3*d*
x**3/(8*sqrt(1 - e**2*x**2/d**2)) - e**2*x**5/(4*d*sqrt(1 - e**2*x**2/d**2)), True)) + 2*C*d*e*Piecewise((-2*d
**4*sqrt(d**2 - e**2*x**2)/(15*e**4) - d**2*x**2*sqrt(d**2 - e**2*x**2)/(15*e**2) + x**4*sqrt(d**2 - e**2*x**2
)/5, Ne(e, 0)), (x**4*sqrt(d**2)/4, True)) + C*e**2*Piecewise((-I*d**6*acosh(e*x/d)/(16*e**5) + I*d**5*x/(16*e
**4*sqrt(-1 + e**2*x**2/d**2)) - I*d**3*x**3/(48*e**2*sqrt(-1 + e**2*x**2/d**2)) - 5*I*d*x**5/(24*sqrt(-1 + e*
*2*x**2/d**2)) + I*e**2*x**7/(6*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) > 1), (d**6*asin(e*x/d)
/(16*e**5) - d**5*x/(16*e**4*sqrt(1 - e**2*x**2/d**2)) + d**3*x**3/(48*e**2*sqrt(1 - e**2*x**2/d**2)) + 5*d*x*
*5/(24*sqrt(1 - e**2*x**2/d**2)) - e**2*x**7/(6*d*sqrt(1 - e**2*x**2/d**2)), True))

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Giac [A]  time = 1.15015, size = 266, normalized size = 1.13 \begin{align*} \frac{1}{16} \,{\left (3 \, C d^{6} + 4 \, B d^{5} e + 10 \, A d^{4} e^{2}\right )} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-3\right )} \mathrm{sgn}\left (d\right ) + \frac{1}{240} \, \sqrt{-x^{2} e^{2} + d^{2}}{\left ({\left (2 \,{\left ({\left (4 \,{\left (5 \, C x e^{2} + 6 \,{\left (2 \, C d e^{9} + B e^{10}\right )} e^{\left (-8\right )}\right )} x + 5 \,{\left (5 \, C d^{2} e^{8} + 12 \, B d e^{9} + 6 \, A e^{10}\right )} e^{\left (-8\right )}\right )} x - 16 \,{\left (C d^{3} e^{7} - 2 \, B d^{2} e^{8} - 5 \, A d e^{9}\right )} e^{\left (-8\right )}\right )} x - 15 \,{\left (3 \, C d^{4} e^{6} + 4 \, B d^{3} e^{7} - 6 \, A d^{2} e^{8}\right )} e^{\left (-8\right )}\right )} x - 16 \,{\left (4 \, C d^{5} e^{5} + 7 \, B d^{4} e^{6} + 10 \, A d^{3} e^{7}\right )} e^{\left (-8\right )}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(C*x^2+B*x+A)*(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

1/16*(3*C*d^6 + 4*B*d^5*e + 10*A*d^4*e^2)*arcsin(x*e/d)*e^(-3)*sgn(d) + 1/240*sqrt(-x^2*e^2 + d^2)*((2*((4*(5*
C*x*e^2 + 6*(2*C*d*e^9 + B*e^10)*e^(-8))*x + 5*(5*C*d^2*e^8 + 12*B*d*e^9 + 6*A*e^10)*e^(-8))*x - 16*(C*d^3*e^7
- 2*B*d^2*e^8 - 5*A*d*e^9)*e^(-8))*x - 15*(3*C*d^4*e^6 + 4*B*d^3*e^7 - 6*A*d^2*e^8)*e^(-8))*x - 16*(4*C*d^5*e
^5 + 7*B*d^4*e^6 + 10*A*d^3*e^7)*e^(-8))