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

Optimal. Leaf size=186 \[ \frac{d x \sqrt{d^2-e^2 x^2} \left (e (4 A e+B d)+C d^2\right )}{8 e^2}-\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (5 e (A e+B d)+2 C d^2\right )}{15 e^3}+\frac{d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right ) \left (e (4 A e+B d)+C d^2\right )}{8 e^3}-\frac{x \left (d^2-e^2 x^2\right )^{3/2} (B e+C d)}{4 e^2}-\frac{C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.227334, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {1815, 641, 195, 217, 203} \[ \frac{d x \sqrt{d^2-e^2 x^2} \left (e (4 A e+B d)+C d^2\right )}{8 e^2}-\frac{\left (d^2-e^2 x^2\right )^{3/2} \left (5 e (A e+B d)+2 C d^2\right )}{15 e^3}+\frac{d^3 \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right ) \left (e (4 A e+B d)+C d^2\right )}{8 e^3}-\frac{x \left (d^2-e^2 x^2\right )^{3/2} (B e+C d)}{4 e^2}-\frac{C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(C*d^2 + e*(B*d + 4*A*e))*x*Sqrt[d^2 - e^2*x^2])/(8*e^2) - ((2*C*d^2 + 5*e*(B*d + A*e))*(d^2 - e^2*x^2)^(3/
2))/(15*e^3) - ((C*d + B*e)*x*(d^2 - e^2*x^2)^(3/2))/(4*e^2) - (C*x^2*(d^2 - e^2*x^2)^(3/2))/(5*e) + (d^3*(C*d
^2 + e*(B*d + 4*A*e))*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 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) \left (A+B x+C x^2\right ) \sqrt{d^2-e^2 x^2} \, dx &=-\frac{C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}-\frac{\int \sqrt{d^2-e^2 x^2} \left (-5 A d e^2-e \left (2 C d^2+5 e (B d+A e)\right ) x-5 e^2 (C d+B e) x^2\right ) \, dx}{5 e^2}\\ &=-\frac{(C d+B e) x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac{C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}+\frac{\int \left (5 d e^2 \left (C d^2+e (B d+4 A e)\right )+4 e^3 \left (2 C d^2+5 e (B d+A e)\right ) x\right ) \sqrt{d^2-e^2 x^2} \, dx}{20 e^4}\\ &=-\frac{\left (2 C d^2+5 e (B d+A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{15 e^3}-\frac{(C d+B e) x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac{C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}+\frac{\left (d \left (C d^2+e (B d+4 A e)\right )\right ) \int \sqrt{d^2-e^2 x^2} \, dx}{4 e^2}\\ &=\frac{d \left (C d^2+e (B d+4 A e)\right ) x \sqrt{d^2-e^2 x^2}}{8 e^2}-\frac{\left (2 C d^2+5 e (B d+A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{15 e^3}-\frac{(C d+B e) x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac{C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}+\frac{\left (d^3 \left (C d^2+e (B d+4 A e)\right )\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{8 e^2}\\ &=\frac{d \left (C d^2+e (B d+4 A e)\right ) x \sqrt{d^2-e^2 x^2}}{8 e^2}-\frac{\left (2 C d^2+5 e (B d+A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{15 e^3}-\frac{(C d+B e) x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac{C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}+\frac{\left (d^3 \left (C d^2+e (B d+4 A e)\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 \left (C d^2+e (B d+4 A e)\right ) x \sqrt{d^2-e^2 x^2}}{8 e^2}-\frac{\left (2 C d^2+5 e (B d+A e)\right ) \left (d^2-e^2 x^2\right )^{3/2}}{15 e^3}-\frac{(C d+B e) x \left (d^2-e^2 x^2\right )^{3/2}}{4 e^2}-\frac{C x^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e}+\frac{d^3 \left (C d^2+e (B d+4 A e)\right ) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{8 e^3}\\ \end{align*}

Mathematica [A]  time = 0.308719, size = 190, normalized size = 1.02 \[ \frac{\sqrt{d^2-e^2 x^2} \left (\sqrt{1-\frac{e^2 x^2}{d^2}} \left (5 e \left (4 A e \left (-2 d^2+3 d e x+2 e^2 x^2\right )+B \left (-3 d^2 e x-8 d^3+8 d e^2 x^2+6 e^3 x^3\right )\right )+C \left (-8 d^2 e^2 x^2-15 d^3 e x-16 d^4+30 d e^3 x^3+24 e^4 x^4\right )\right )+15 \sin ^{-1}\left (\frac{e x}{d}\right ) \left (d^2 e (4 A e+B d)+C d^4\right )\right )}{120 e^3 \sqrt{1-\frac{e^2 x^2}{d^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)*(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*(-16*d^4 - 15*d^3*e*x - 8*d^2*e^2*x^2 + 30*d*e^3*x^3 + 24*e^4
*x^4) + 5*e*(4*A*e*(-2*d^2 + 3*d*e*x + 2*e^2*x^2) + B*(-8*d^3 - 3*d^2*e*x + 8*d*e^2*x^2 + 6*e^3*x^3))) + 15*(C
*d^4 + d^2*e*(B*d + 4*A*e))*ArcSin[(e*x)/d]))/(120*e^3*Sqrt[1 - (e^2*x^2)/d^2])

________________________________________________________________________________________

Maple [A]  time = 0.059, size = 304, normalized size = 1.6 \begin{align*} -{\frac{C{x}^{2}}{5\,e} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{2\,C{d}^{2}}{15\,{e}^{3}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{Bx}{4\,e} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{Cdx}{4\,{e}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{{d}^{2}xB}{8\,e}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{C{d}^{3}x}{8\,{e}^{2}}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{{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{{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{A}{3\,e} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{Bd}{3\,{e}^{2}} \left ( -{x}^{2}{e}^{2}+{d}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{Adx}{2}\sqrt{-{x}^{2}{e}^{2}+{d}^{2}}}+{\frac{A{d}^{3}}{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)*(C*x^2+B*x+A)*(-e^2*x^2+d^2)^(1/2),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.78755, size = 304, normalized size = 1.63 \begin{align*} \frac{A d^{3} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{2 \, \sqrt{e^{2}}} + \frac{1}{2} \, \sqrt{-e^{2} x^{2} + d^{2}} A d x - \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} C x^{2}}{5 \, e} + \frac{{\left (C d + B e\right )} d^{4} \arcsin \left (\frac{e^{2} x}{\sqrt{d^{2} e^{2}}}\right )}{8 \, \sqrt{e^{2}} e^{2}} + \frac{\sqrt{-e^{2} x^{2} + d^{2}}{\left (C d + B e\right )} d^{2} x}{8 \, e^{2}} - \frac{2 \,{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} C d^{2}}{15 \, e^{3}} - \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} B d}{3 \, e^{2}} - \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}} A}{3 \, e} - \frac{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac{3}{2}}{\left (C d + B e\right )} x}{4 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.51104, size = 375, normalized size = 2.02 \begin{align*} -\frac{30 \,{\left (C d^{5} + B d^{4} e + 4 \, 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} - 16 \, C d^{4} - 40 \, B d^{3} e - 40 \, A d^{2} e^{2} + 30 \,{\left (C d e^{3} + B e^{4}\right )} x^{3} - 8 \,{\left (C d^{2} e^{2} - 5 \, B d e^{3} - 5 \, A e^{4}\right )} x^{2} - 15 \,{\left (C d^{3} e + B d^{2} e^{2} - 4 \, 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)*(C*x^2+B*x+A)*(-e^2*x^2+d^2)^(1/2),x, algorithm="fricas")

[Out]

-1/120*(30*(C*d^5 + B*d^4*e + 4*A*d^3*e^2)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) - (24*C*e^4*x^4 - 16*C*d^
4 - 40*B*d^3*e - 40*A*d^2*e^2 + 30*(C*d*e^3 + B*e^4)*x^3 - 8*(C*d^2*e^2 - 5*B*d*e^3 - 5*A*e^4)*x^2 - 15*(C*d^3
*e + B*d^2*e^2 - 4*A*d*e^3)*x)*sqrt(-e^2*x^2 + d^2))/e^3

________________________________________________________________________________________

Sympy [C]  time = 11.4995, size = 675, normalized size = 3.63 \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)*(C*x**2+B*x+A)*(-e**2*x**2+d**2)**(1/2),x)

[Out]

A*d*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, True
)) + A*e*Piecewise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) + B*d*Piecew
ise((x**2*sqrt(d**2)/2, Eq(e**2, 0)), (-(d**2 - e**2*x**2)**(3/2)/(3*e**2), True)) + B*e*Piecewise((-I*d**4*ac
osh(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)) + C*d*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)) + C*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))

________________________________________________________________________________________

Giac [A]  time = 1.18175, size = 216, normalized size = 1.16 \begin{align*} \frac{1}{8} \,{\left (C d^{5} + B d^{4} e + 4 \, 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 (C d e^{6} + B e^{7}\right )} e^{\left (-6\right )}\right )} x - 4 \,{\left (C d^{2} e^{5} - 5 \, B d e^{6} - 5 \, A e^{7}\right )} e^{\left (-6\right )}\right )} x - 15 \,{\left (C d^{3} e^{4} + B d^{2} e^{5} - 4 \, A d e^{6}\right )} e^{\left (-6\right )}\right )} x - 8 \,{\left (2 \, C d^{4} e^{3} + 5 \, B d^{3} e^{4} + 5 \, 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)*(C*x^2+B*x+A)*(-e^2*x^2+d^2)^(1/2),x, algorithm="giac")

[Out]

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