3.1.6 \(\int \frac {(A+B x+C x^2) \sqrt {d^2-e^2 x^2}}{(d+e x)^3} \, dx\) [6]

Optimal. Leaf size=149 \[ \frac {2 (3 C d-B e) \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{3 d e^3 (d+e x)^3}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^2}+\frac {(3 C d-B e) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {1653, 807, 677, 223, 209} \begin {gather*} -\frac {\left (d^2-e^2 x^2\right )^{3/2} \left (A e^2-B d e+C d^2\right )}{3 d e^3 (d+e x)^3}+\frac {\text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right ) (3 C d-B e)}{e^3}+\frac {2 \sqrt {d^2-e^2 x^2} (3 C d-B e)}{e^3 (d+e x)}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

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

Rule 223

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 677

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + c*x^2)^p/(e
*(m + p + 1))), x] - Dist[c*(p/(e^2*(m + p + 1))), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1
, 0] && IntegerQ[2*p]

Rule 807

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

Rule 1653

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x)^(m + q - 1)*((a + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q + 2*p + 1))), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {\left (A+B x+C x^2\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^3} \, dx &=-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^2}-\frac {\int \frac {\left (e^2 \left (2 C d^2-A e^2\right )+e^3 (3 C d-B e) x\right ) \sqrt {d^2-e^2 x^2}}{(d+e x)^3} \, dx}{e^4}\\ &=-\frac {\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{3 d e^3 (d+e x)^3}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^2}-\frac {(3 C d-B e) \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^2} \, dx}{e^2}\\ &=\frac {2 (3 C d-B e) \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{3 d e^3 (d+e x)^3}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^2}+\frac {(3 C d-B e) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^2}\\ &=\frac {2 (3 C d-B e) \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{3 d e^3 (d+e x)^3}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^2}+\frac {(3 C d-B e) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^2}\\ &=\frac {2 (3 C d-B e) \sqrt {d^2-e^2 x^2}}{e^3 (d+e x)}-\frac {\left (C d^2-B d e+A e^2\right ) \left (d^2-e^2 x^2\right )^{3/2}}{3 d e^3 (d+e x)^3}-\frac {C \left (d^2-e^2 x^2\right )^{3/2}}{e^3 (d+e x)^2}+\frac {(3 C d-B e) \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3}\\ \end {align*}

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Mathematica [A]
time = 0.63, size = 133, normalized size = 0.89 \begin {gather*} \frac {\frac {e \sqrt {d^2-e^2 x^2} \left (C d \left (14 d^2+19 d e x+3 e^2 x^2\right )+e (A e (-d+e x)-B d (5 d+7 e x))\right )}{d (d+e x)^2}-3 \sqrt {-e^2} (-3 C d+B e) \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{3 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*Sqrt[d^2 - e^2*x^2]*(C*d*(14*d^2 + 19*d*e*x + 3*e^2*x^2) + e*(A*e*(-d + e*x) - B*d*(5*d + 7*e*x))))/(d*(d
+ e*x)^2) - 3*Sqrt[-e^2]*(-3*C*d + B*e)*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/(3*e^4)

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Maple [A]
time = 0.10, size = 278, normalized size = 1.87

method	result	size
default	\(\frac {C \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{e^{3}}-\frac {\left (A \,e^{2}-B d e +C \,d^{2}\right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{3 e^{6} d \left (x +\frac {d}{e}\right )^{3}}+\frac {\left (B e -2 C d \right ) \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{d e \left (x +\frac {d}{e}\right )^{2}}-\frac {e \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d}\right )}{e^{4}}\)	\(278\)
risch	\(\frac {C \sqrt {-e^{2} x^{2}+d^{2}}}{e^{3}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right ) B}{e \sqrt {e^{2}}}+\frac {3 \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right ) C d}{e^{2} \sqrt {e^{2}}}-\frac {2 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}\, A}{3 e^{3} \left (x +\frac {d}{e}\right )^{2}}+\frac {2 d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}\, B}{3 e^{4} \left (x +\frac {d}{e}\right )^{2}}-\frac {2 d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}\, C}{3 e^{5} \left (x +\frac {d}{e}\right )^{2}}+\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}\, A}{3 e^{2} d \left (x +\frac {d}{e}\right )}-\frac {7 \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}\, B}{3 e^{3} \left (x +\frac {d}{e}\right )}+\frac {13 d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}\, C}{3 e^{4} \left (x +\frac {d}{e}\right )}\)	\(354\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

C/e^3*((-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+d*e/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-(x+d/e)^2*e^2+2*d*e*(x+d/e)
)^(1/2)))-1/3*(A*e^2-B*d*e+C*d^2)/e^6/d/(x+d/e)^3*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(3/2)+(B*e-2*C*d)/e^4*(-1/d/e
/(x+d/e)^2*(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(3/2)-e/d*((-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2)+d*e/(e^2)^(1/2)*arct
an((e^2)^(1/2)*x/(-(x+d/e)^2*e^2+2*d*e*(x+d/e))^(1/2))))

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [A]
time = 0.36, size = 244, normalized size = 1.64 \begin {gather*} \frac {14 \, C d^{4} - A x^{2} e^{4} - 6 \, {\left (3 \, C d^{4} - B d x^{2} e^{3} + {\left (3 \, C d^{2} x^{2} - 2 \, B d^{2} x\right )} e^{2} + {\left (6 \, C d^{3} x - B d^{3}\right )} e\right )} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) - {\left (5 \, B d x^{2} + 2 \, A d x\right )} e^{3} + {\left (14 \, C d^{2} x^{2} - 10 \, B d^{2} x - A d^{2}\right )} e^{2} + {\left (28 \, C d^{3} x - 5 \, B d^{3}\right )} e + {\left (14 \, C d^{3} + A x e^{3} + {\left (3 \, C d x^{2} - 7 \, B d x - A d\right )} e^{2} + {\left (19 \, C d^{2} x - 5 \, B d^{2}\right )} e\right )} \sqrt {-x^{2} e^{2} + d^{2}}}{3 \, {\left (d x^{2} e^{5} + 2 \, d^{2} x e^{4} + d^{3} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(14*C*d^4 - A*x^2*e^4 - 6*(3*C*d^4 - B*d*x^2*e^3 + (3*C*d^2*x^2 - 2*B*d^2*x)*e^2 + (6*C*d^3*x - B*d^3)*e)*
arctan(-(d - sqrt(-x^2*e^2 + d^2))*e^(-1)/x) - (5*B*d*x^2 + 2*A*d*x)*e^3 + (14*C*d^2*x^2 - 10*B*d^2*x - A*d^2)
*e^2 + (28*C*d^3*x - 5*B*d^3)*e + (14*C*d^3 + A*x*e^3 + (3*C*d*x^2 - 7*B*d*x - A*d)*e^2 + (19*C*d^2*x - 5*B*d^
2)*e)*sqrt(-x^2*e^2 + d^2))/(d*x^2*e^5 + 2*d^2*x*e^4 + d^3*e^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- \left (- d + e x\right ) \left (d + e x\right )} \left (A + B x + C x^{2}\right )}{\left (d + e x\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-(-d + e*x)*(d + e*x))*(A + B*x + C*x**2)/(d + e*x)**3, x)

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Giac [A]
time = 4.48, size = 256, normalized size = 1.72 \begin {gather*} {\left (3 \, C d - B e\right )} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-3\right )} \mathrm {sgn}\left (d\right ) + \sqrt {-x^{2} e^{2} + d^{2}} C e^{\left (-3\right )} - \frac {2 \, {\left (\frac {24 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} C d^{2} e^{\left (-2\right )}}{x} + \frac {9 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} C d^{2} e^{\left (-4\right )}}{x^{2}} + 11 \, C d^{2} - 5 \, B d e - \frac {12 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} B d e^{\left (-1\right )}}{x} - \frac {3 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} B d e^{\left (-3\right )}}{x^{2}} - A e^{2} - \frac {3 \, {\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )}^{2} A e^{\left (-2\right )}}{x^{2}}\right )} e^{\left (-3\right )}}{3 \, d {\left (\frac {{\left (d e + \sqrt {-x^{2} e^{2} + d^{2}} e\right )} e^{\left (-2\right )}}{x} + 1\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*C*d - B*e)*arcsin(x*e/d)*e^(-3)*sgn(d) + sqrt(-x^2*e^2 + d^2)*C*e^(-3) - 2/3*(24*(d*e + sqrt(-x^2*e^2 + d^2
)*e)*C*d^2*e^(-2)/x + 9*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*C*d^2*e^(-4)/x^2 + 11*C*d^2 - 5*B*d*e - 12*(d*e + sqr
t(-x^2*e^2 + d^2)*e)*B*d*e^(-1)/x - 3*(d*e + sqrt(-x^2*e^2 + d^2)*e)^2*B*d*e^(-3)/x^2 - A*e^2 - 3*(d*e + sqrt(
-x^2*e^2 + d^2)*e)^2*A*e^(-2)/x^2)*e^(-3)/(d*((d*e + sqrt(-x^2*e^2 + d^2)*e)*e^(-2)/x + 1)^3)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {d^2-e^2\,x^2}\,\left (C\,x^2+B\,x+A\right )}{{\left (d+e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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