### 3.10 $$\int \frac{(d+e x)^3 (A+B x+C x^2)}{\sqrt{d^2-e^2 x^2}} \, dx$$

Optimal. Leaf size=236 $-\frac{x^2 \sqrt{d^2-e^2 x^2} \left (5 e (A e+3 B d)+19 C d^2\right )}{15 e}-\frac{d x \sqrt{d^2-e^2 x^2} \left (12 A e^2+15 B d e+13 C d^2\right )}{8 e^2}-\frac{d^2 \sqrt{d^2-e^2 x^2} \left (55 A e^2+45 B d e+38 C d^2\right )}{15 e^3}+\frac{d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right ) \left (20 A e^2+15 B d e+13 C d^2\right )}{8 e^3}-\frac{1}{4} x^3 \sqrt{d^2-e^2 x^2} (B e+3 C d)-\frac{1}{5} C e x^4 \sqrt{d^2-e^2 x^2}$

[Out]

-(d^2*(38*C*d^2 + 45*B*d*e + 55*A*e^2)*Sqrt[d^2 - e^2*x^2])/(15*e^3) - (d*(13*C*d^2 + 15*B*d*e + 12*A*e^2)*x*S
qrt[d^2 - e^2*x^2])/(8*e^2) - ((19*C*d^2 + 5*e*(3*B*d + A*e))*x^2*Sqrt[d^2 - e^2*x^2])/(15*e) - ((3*C*d + B*e)
*x^3*Sqrt[d^2 - e^2*x^2])/4 - (C*e*x^4*Sqrt[d^2 - e^2*x^2])/5 + (d^3*(13*C*d^2 + 15*B*d*e + 20*A*e^2)*ArcTan[(
e*x)/Sqrt[d^2 - e^2*x^2]])/(8*e^3)

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Rubi [A]  time = 0.656543, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 34, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.118, Rules used = {1815, 641, 217, 203} $-\frac{x^2 \sqrt{d^2-e^2 x^2} \left (5 e (A e+3 B d)+19 C d^2\right )}{15 e}-\frac{d x \sqrt{d^2-e^2 x^2} \left (12 A e^2+15 B d e+13 C d^2\right )}{8 e^2}-\frac{d^2 \sqrt{d^2-e^2 x^2} \left (55 A e^2+45 B d e+38 C d^2\right )}{15 e^3}+\frac{d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right ) \left (20 A e^2+15 B d e+13 C d^2\right )}{8 e^3}-\frac{1}{4} x^3 \sqrt{d^2-e^2 x^2} (B e+3 C d)-\frac{1}{5} C e x^4 \sqrt{d^2-e^2 x^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[((d + e*x)^3*(A + B*x + C*x^2))/Sqrt[d^2 - e^2*x^2],x]

[Out]

-(d^2*(38*C*d^2 + 45*B*d*e + 55*A*e^2)*Sqrt[d^2 - e^2*x^2])/(15*e^3) - (d*(13*C*d^2 + 15*B*d*e + 12*A*e^2)*x*S
qrt[d^2 - e^2*x^2])/(8*e^2) - ((19*C*d^2 + 5*e*(3*B*d + A*e))*x^2*Sqrt[d^2 - e^2*x^2])/(15*e) - ((3*C*d + B*e)
*x^3*Sqrt[d^2 - e^2*x^2])/4 - (C*e*x^4*Sqrt[d^2 - e^2*x^2])/5 + (d^3*(13*C*d^2 + 15*B*d*e + 20*A*e^2)*ArcTan[(
e*x)/Sqrt[d^2 - e^2*x^2]])/(8*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 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 \frac{(d+e x)^3 \left (A+B x+C x^2\right )}{\sqrt{d^2-e^2 x^2}} \, dx &=-\frac{1}{5} C e x^4 \sqrt{d^2-e^2 x^2}-\frac{\int \frac{-5 A d^3 e^2-5 d^2 e^2 (B d+3 A e) x-5 d e^2 \left (C d^2+3 e (B d+A e)\right ) x^2-e^3 \left (19 C d^2+5 e (3 B d+A e)\right ) x^3-5 e^4 (3 C d+B e) x^4}{\sqrt{d^2-e^2 x^2}} \, dx}{5 e^2}\\ &=-\frac{1}{4} (3 C d+B e) x^3 \sqrt{d^2-e^2 x^2}-\frac{1}{5} C e x^4 \sqrt{d^2-e^2 x^2}+\frac{\int \frac{20 A d^3 e^4+20 d^2 e^4 (B d+3 A e) x+5 d e^4 \left (13 C d^2+15 B d e+12 A e^2\right ) x^2+4 e^5 \left (19 C d^2+5 e (3 B d+A e)\right ) x^3}{\sqrt{d^2-e^2 x^2}} \, dx}{20 e^4}\\ &=-\frac{\left (19 C d^2+5 e (3 B d+A e)\right ) x^2 \sqrt{d^2-e^2 x^2}}{15 e}-\frac{1}{4} (3 C d+B e) x^3 \sqrt{d^2-e^2 x^2}-\frac{1}{5} C e x^4 \sqrt{d^2-e^2 x^2}-\frac{\int \frac{-60 A d^3 e^6-4 d^2 e^5 \left (38 C d^2+45 B d e+55 A e^2\right ) x-15 d e^6 \left (13 C d^2+15 B d e+12 A e^2\right ) x^2}{\sqrt{d^2-e^2 x^2}} \, dx}{60 e^6}\\ &=-\frac{d \left (13 C d^2+15 B d e+12 A e^2\right ) x \sqrt{d^2-e^2 x^2}}{8 e^2}-\frac{\left (19 C d^2+5 e (3 B d+A e)\right ) x^2 \sqrt{d^2-e^2 x^2}}{15 e}-\frac{1}{4} (3 C d+B e) x^3 \sqrt{d^2-e^2 x^2}-\frac{1}{5} C e x^4 \sqrt{d^2-e^2 x^2}+\frac{\int \frac{15 d^3 e^6 \left (13 C d^2+15 B d e+20 A e^2\right )+8 d^2 e^7 \left (38 C d^2+45 B d e+55 A e^2\right ) x}{\sqrt{d^2-e^2 x^2}} \, dx}{120 e^8}\\ &=-\frac{d^2 \left (38 C d^2+45 B d e+55 A e^2\right ) \sqrt{d^2-e^2 x^2}}{15 e^3}-\frac{d \left (13 C d^2+15 B d e+12 A e^2\right ) x \sqrt{d^2-e^2 x^2}}{8 e^2}-\frac{\left (19 C d^2+5 e (3 B d+A e)\right ) x^2 \sqrt{d^2-e^2 x^2}}{15 e}-\frac{1}{4} (3 C d+B e) x^3 \sqrt{d^2-e^2 x^2}-\frac{1}{5} C e x^4 \sqrt{d^2-e^2 x^2}+\frac{\left (d^3 \left (13 C d^2+15 B d e+20 A e^2\right )\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{8 e^2}\\ &=-\frac{d^2 \left (38 C d^2+45 B d e+55 A e^2\right ) \sqrt{d^2-e^2 x^2}}{15 e^3}-\frac{d \left (13 C d^2+15 B d e+12 A e^2\right ) x \sqrt{d^2-e^2 x^2}}{8 e^2}-\frac{\left (19 C d^2+5 e (3 B d+A e)\right ) x^2 \sqrt{d^2-e^2 x^2}}{15 e}-\frac{1}{4} (3 C d+B e) x^3 \sqrt{d^2-e^2 x^2}-\frac{1}{5} C e x^4 \sqrt{d^2-e^2 x^2}+\frac{\left (d^3 \left (13 C d^2+15 B d e+20 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 )}{8 e^2}\\ &=-\frac{d^2 \left (38 C d^2+45 B d e+55 A e^2\right ) \sqrt{d^2-e^2 x^2}}{15 e^3}-\frac{d \left (13 C d^2+15 B d e+12 A e^2\right ) x \sqrt{d^2-e^2 x^2}}{8 e^2}-\frac{\left (19 C d^2+5 e (3 B d+A e)\right ) x^2 \sqrt{d^2-e^2 x^2}}{15 e}-\frac{1}{4} (3 C d+B e) x^3 \sqrt{d^2-e^2 x^2}-\frac{1}{5} C e x^4 \sqrt{d^2-e^2 x^2}+\frac{d^3 \left (13 C d^2+15 B d e+20 A e^2\right ) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^3}\\ \end{align*}

Mathematica [A]  time = 0.453851, size = 174, normalized size = 0.74 $\frac{15 d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right ) \left (5 e (4 A e+3 B d)+13 C d^2\right )-\sqrt{d^2-e^2 x^2} \left (5 e \left (4 A e \left (22 d^2+9 d e x+2 e^2 x^2\right )+3 B \left (15 d^2 e x+24 d^3+8 d e^2 x^2+2 e^3 x^3\right )\right )+C \left (152 d^2 e^2 x^2+195 d^3 e x+304 d^4+90 d e^3 x^3+24 e^4 x^4\right )\right )}{120 e^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[((d + e*x)^3*(A + B*x + C*x^2))/Sqrt[d^2 - e^2*x^2],x]

[Out]

(-(Sqrt[d^2 - e^2*x^2]*(C*(304*d^4 + 195*d^3*e*x + 152*d^2*e^2*x^2 + 90*d*e^3*x^3 + 24*e^4*x^4) + 5*e*(4*A*e*(
22*d^2 + 9*d*e*x + 2*e^2*x^2) + 3*B*(24*d^3 + 15*d^2*e*x + 8*d*e^2*x^2 + 2*e^3*x^3)))) + 15*d^3*(13*C*d^2 + 5*
e*(3*B*d + 4*A*e))*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(120*e^3)

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Maple [A]  time = 0.066, size = 374, normalized size = 1.6 \begin{align*} -{\frac{Ce{x}^{4}}{5}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{19\,C{d}^{2}{x}^{2}}{15\,e}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{38\,C{d}^{4}}{15\,{e}^{3}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{{x}^{3}eB}{4}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{3\,d{x}^{3}C}{4}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{15\,{d}^{2}xB}{8\,e}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{\frac{13\,C{d}^{3}x}{8\,{e}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{15\,{d}^{4}B}{8\,e}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}+{\frac{13\,{d}^{5}C}{8\,{e}^{2}}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}}-{\frac{e{x}^{2}A}{3}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-{x}^{2}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}dB-{\frac{11\,{d}^{2}A}{3\,e}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}-3\,{\frac{{d}^{3}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}B}{{e}^{2}}}-{\frac{3\,Adx}{2}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{5\,{d}^{3}A}{2}\arctan \left ({x\sqrt{{e}^{2}}{\frac{1}{\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}}} \right ){\frac{1}{\sqrt{{e}^{2}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(C*x^2+B*x+A)/(-e^2*x^2+d^2)^(1/2),x)

[Out]

-1/5*C*e*x^4*(-e^2*x^2+d^2)^(1/2)-19/15/e*C*d^2*x^2*(-e^2*x^2+d^2)^(1/2)-38/15/e^3*C*d^4*(-e^2*x^2+d^2)^(1/2)-
1/4*x^3*e*(-e^2*x^2+d^2)^(1/2)*B-3/4*x^3*(-e^2*x^2+d^2)^(1/2)*d*C-15/8*d^2/e*x*(-e^2*x^2+d^2)^(1/2)*B-13/8*d^3
/e^2*x*(-e^2*x^2+d^2)^(1/2)*C+15/8*d^4/e/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))*B+13/8*d^5/e^2
/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))*C-1/3*x^2*e*(-e^2*x^2+d^2)^(1/2)*A-x^2*(-e^2*x^2+d^2)^
(1/2)*d*B-11/3*d^2/e*(-e^2*x^2+d^2)^(1/2)*A-3*d^3/e^2*(-e^2*x^2+d^2)^(1/2)*B-3/2*A*d*x*(-e^2*x^2+d^2)^(1/2)+5/
2*A*d^3/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))

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Maxima [B]  time = 1.5055, size = 575, normalized size = 2.44 \begin{align*} -\frac{1}{5} \, \sqrt{-e^{2} x^{2} + d^{2}} C e x^{4} - \frac{4 \, \sqrt{-e^{2} x^{2} + d^{2}} C d^{2} x^{2}}{15 \, e} + \frac{A d^{3} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{\sqrt{e^{2}}} - \frac{8 \, \sqrt{-e^{2} x^{2} + d^{2}} C d^{4}}{15 \, e^{3}} - \frac{\sqrt{-e^{2} x^{2} + d^{2}} B d^{3}}{e^{2}} - \frac{3 \, \sqrt{-e^{2} x^{2} + d^{2}} A d^{2}}{e} - \frac{{\left (3 \, C d e^{2} + B e^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}} x^{3}}{4 \, e^{2}} - \frac{{\left (3 \, C d^{2} e + 3 \, B d e^{2} + A e^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}} x^{2}}{3 \, e^{2}} + \frac{3 \,{\left (3 \, C d e^{2} + B e^{3}\right )} d^{4} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{8 \, \sqrt{e^{2}} e^{4}} + \frac{{\left (C d^{3} + 3 \, B d^{2} e + 3 \, A d e^{2}\right )} d^{2} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{2 \, \sqrt{e^{2}} e^{2}} - \frac{3 \,{\left (3 \, C d e^{2} + B e^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}} d^{2} x}{8 \, e^{4}} - \frac{{\left (C d^{3} + 3 \, B d^{2} e + 3 \, A d e^{2}\right )} \sqrt{-e^{2} x^{2} + d^{2}} x}{2 \, e^{2}} - \frac{2 \,{\left (3 \, C d^{2} e + 3 \, B d e^{2} + A e^{3}\right )} \sqrt{-e^{2} x^{2} + d^{2}} d^{2}}{3 \, e^{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(C*x^2+B*x+A)/(-e^2*x^2+d^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.35253, size = 406, normalized size = 1.72 \begin{align*} -\frac{30 \,{\left (13 \, C d^{5} + 15 \, B d^{4} e + 20 \, A d^{3} e^{2}\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) +{\left (24 \, C e^{4} x^{4} + 304 \, C d^{4} + 360 \, B d^{3} e + 440 \, A d^{2} e^{2} + 30 \,{\left (3 \, C d e^{3} + B e^{4}\right )} x^{3} + 8 \,{\left (19 \, C d^{2} e^{2} + 15 \, B d e^{3} + 5 \, A e^{4}\right )} x^{2} + 15 \,{\left (13 \, C d^{3} e + 15 \, B d^{2} e^{2} + 12 \, A d e^{3}\right )} x\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{120 \, e^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(C*x^2+B*x+A)/(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/120*(30*(13*C*d^5 + 15*B*d^4*e + 20*A*d^3*e^2)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (24*C*e^4*x^4 +
304*C*d^4 + 360*B*d^3*e + 440*A*d^2*e^2 + 30*(3*C*d*e^3 + B*e^4)*x^3 + 8*(19*C*d^2*e^2 + 15*B*d*e^3 + 5*A*e^4)
*x^2 + 15*(13*C*d^3*e + 15*B*d^2*e^2 + 12*A*d*e^3)*x)*sqrt(-e^2*x^2 + d^2))/e^3

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Sympy [A]  time = 20.234, size = 1277, normalized size = 5.41 \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)**3*(C*x**2+B*x+A)/(-e**2*x**2+d**2)**(1/2),x)

[Out]

A*d**3*Piecewise((sqrt(d**2/e**2)*asin(x*sqrt(e**2/d**2))/sqrt(d**2), (d**2 > 0) & (e**2 > 0)), (sqrt(-d**2/e*
*2)*asinh(x*sqrt(-e**2/d**2))/sqrt(d**2), (d**2 > 0) & (e**2 < 0)), (sqrt(d**2/e**2)*acosh(x*sqrt(e**2/d**2))/
sqrt(-d**2), (d**2 < 0) & (e**2 < 0))) + 3*A*d**2*e*Piecewise((x**2/(2*sqrt(d**2)), Eq(e**2, 0)), (-sqrt(d**2
- e**2*x**2)/e**2, True)) + 3*A*d*e**2*Piecewise((-I*d**2*acosh(e*x/d)/(2*e**3) - I*d*x*sqrt(-1 + e**2*x**2/d*
*2)/(2*e**2), Abs(e**2*x**2)/Abs(d**2) > 1), (d**2*asin(e*x/d)/(2*e**3) - d*x/(2*e**2*sqrt(1 - e**2*x**2/d**2)
) + x**3/(2*d*sqrt(1 - e**2*x**2/d**2)), True)) + A*e**3*Piecewise((-2*d**2*sqrt(d**2 - e**2*x**2)/(3*e**4) -
x**2*sqrt(d**2 - e**2*x**2)/(3*e**2), Ne(e, 0)), (x**4/(4*sqrt(d**2)), True)) + B*d**3*Piecewise((x**2/(2*sqrt
(d**2)), Eq(e**2, 0)), (-sqrt(d**2 - e**2*x**2)/e**2, True)) + 3*B*d**2*e*Piecewise((-I*d**2*acosh(e*x/d)/(2*e
**3) - I*d*x*sqrt(-1 + e**2*x**2/d**2)/(2*e**2), Abs(e**2*x**2)/Abs(d**2) > 1), (d**2*asin(e*x/d)/(2*e**3) - d
*x/(2*e**2*sqrt(1 - e**2*x**2/d**2)) + x**3/(2*d*sqrt(1 - e**2*x**2/d**2)), True)) + 3*B*d*e**2*Piecewise((-2*
d**2*sqrt(d**2 - e**2*x**2)/(3*e**4) - x**2*sqrt(d**2 - e**2*x**2)/(3*e**2), Ne(e, 0)), (x**4/(4*sqrt(d**2)),
True)) + B*e**3*Piecewise((-3*I*d**4*acosh(e*x/d)/(8*e**5) + 3*I*d**3*x/(8*e**4*sqrt(-1 + e**2*x**2/d**2)) - I
*d*x**3/(8*e**2*sqrt(-1 + e**2*x**2/d**2)) - I*x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2)
> 1), (3*d**4*asin(e*x/d)/(8*e**5) - 3*d**3*x/(8*e**4*sqrt(1 - e**2*x**2/d**2)) + d*x**3/(8*e**2*sqrt(1 - e**2
*x**2/d**2)) + x**5/(4*d*sqrt(1 - e**2*x**2/d**2)), True)) + C*d**3*Piecewise((-I*d**2*acosh(e*x/d)/(2*e**3) -
I*d*x*sqrt(-1 + e**2*x**2/d**2)/(2*e**2), Abs(e**2*x**2)/Abs(d**2) > 1), (d**2*asin(e*x/d)/(2*e**3) - d*x/(2*
e**2*sqrt(1 - e**2*x**2/d**2)) + x**3/(2*d*sqrt(1 - e**2*x**2/d**2)), True)) + 3*C*d**2*e*Piecewise((-2*d**2*s
qrt(d**2 - e**2*x**2)/(3*e**4) - x**2*sqrt(d**2 - e**2*x**2)/(3*e**2), Ne(e, 0)), (x**4/(4*sqrt(d**2)), True))
+ 3*C*d*e**2*Piecewise((-3*I*d**4*acosh(e*x/d)/(8*e**5) + 3*I*d**3*x/(8*e**4*sqrt(-1 + e**2*x**2/d**2)) - I*d
*x**3/(8*e**2*sqrt(-1 + e**2*x**2/d**2)) - I*x**5/(4*d*sqrt(-1 + e**2*x**2/d**2)), Abs(e**2*x**2)/Abs(d**2) >
1), (3*d**4*asin(e*x/d)/(8*e**5) - 3*d**3*x/(8*e**4*sqrt(1 - e**2*x**2/d**2)) + d*x**3/(8*e**2*sqrt(1 - e**2*x
**2/d**2)) + x**5/(4*d*sqrt(1 - e**2*x**2/d**2)), True)) + C*e**3*Piecewise((-8*d**4*sqrt(d**2 - e**2*x**2)/(1
5*e**6) - 4*d**2*x**2*sqrt(d**2 - e**2*x**2)/(15*e**4) - x**4*sqrt(d**2 - e**2*x**2)/(5*e**2), Ne(e, 0)), (x**
6/(6*sqrt(d**2)), True))

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Giac [A]  time = 1.23252, size = 224, normalized size = 0.95 \begin{align*} \frac{1}{8} \,{\left (13 \, C d^{5} + 15 \, B d^{4} e + 20 \, A d^{3} e^{2}\right )} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-3\right )} \mathrm{sgn}\left (d\right ) - \frac{1}{120} \, \sqrt{-x^{2} e^{2} + d^{2}}{\left ({\left (2 \,{\left (3 \,{\left (4 \, C x e + 5 \,{\left (3 \, C d e^{6} + B e^{7}\right )} e^{\left (-6\right )}\right )} x + 4 \,{\left (19 \, C d^{2} e^{5} + 15 \, B d e^{6} + 5 \, A e^{7}\right )} e^{\left (-6\right )}\right )} x + 15 \,{\left (13 \, C d^{3} e^{4} + 15 \, B d^{2} e^{5} + 12 \, A d e^{6}\right )} e^{\left (-6\right )}\right )} x + 8 \,{\left (38 \, C d^{4} e^{3} + 45 \, B d^{3} e^{4} + 55 \, A d^{2} e^{5}\right )} e^{\left (-6\right )}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(C*x^2+B*x+A)/(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

1/8*(13*C*d^5 + 15*B*d^4*e + 20*A*d^3*e^2)*arcsin(x*e/d)*e^(-3)*sgn(d) - 1/120*sqrt(-x^2*e^2 + d^2)*((2*(3*(4*
C*x*e + 5*(3*C*d*e^6 + B*e^7)*e^(-6))*x + 4*(19*C*d^2*e^5 + 15*B*d*e^6 + 5*A*e^7)*e^(-6))*x + 15*(13*C*d^3*e^4
+ 15*B*d^2*e^5 + 12*A*d*e^6)*e^(-6))*x + 8*(38*C*d^4*e^3 + 45*B*d^3*e^4 + 55*A*d^2*e^5)*e^(-6))