3.2176 \(\int \frac{(f+g x) \sqrt{c d^2-b d e-b e^2 x-c e^2 x^2}}{(d+e x)^2} \, dx\)

Optimal. Leaf size=200 \[ -\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{e^2 (d+e x)^2 (2 c d-b e)}-\frac{\sqrt{d (c d-b e)-b e^2 x-c e^2 x^2} (b e g-4 c d g+2 c e f)}{e^2 (2 c d-b e)}-\frac{(b e g-4 c d g+2 c e f) \tan ^{-1}\left (\frac{e (b+2 c x)}{2 \sqrt{c} \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{2 \sqrt{c} e^2} \]

[Out]

-(((2*c*e*f - 4*c*d*g + b*e*g)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(e^2*(2*c*d - b*e))) - (2*(e*f - d*g
)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(e^2*(2*c*d - b*e)*(d + e*x)^2) - ((2*c*e*f - 4*c*d*g + b*e*g)*
ArcTan[(e*(b + 2*c*x))/(2*Sqrt[c]*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])])/(2*Sqrt[c]*e^2)

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Rubi [A]  time = 0.338962, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 44, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {792, 664, 621, 204} \[ -\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{e^2 (d+e x)^2 (2 c d-b e)}-\frac{\sqrt{d (c d-b e)-b e^2 x-c e^2 x^2} (b e g-4 c d g+2 c e f)}{e^2 (2 c d-b e)}-\frac{(b e g-4 c d g+2 c e f) \tan ^{-1}\left (\frac{e (b+2 c x)}{2 \sqrt{c} \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{2 \sqrt{c} e^2} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x)*Sqrt[c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2])/(d + e*x)^2,x]

[Out]

-(((2*c*e*f - 4*c*d*g + b*e*g)*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(e^2*(2*c*d - b*e))) - (2*(e*f - d*g
)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(e^2*(2*c*d - b*e)*(d + e*x)^2) - ((2*c*e*f - 4*c*d*g + b*e*g)*
ArcTan[(e*(b + 2*c*x))/(2*Sqrt[c]*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])])/(2*Sqrt[c]*e^2)

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[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 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{(f+g x) \sqrt{c d^2-b d e-b e^2 x-c e^2 x^2}}{(d+e x)^2} \, dx &=-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{e^2 (2 c d-b e) (d+e x)^2}-\frac{(2 c e f-4 c d g+b e g) \int \frac{\sqrt{c d^2-b d e-b e^2 x-c e^2 x^2}}{d+e x} \, dx}{e (2 c d-b e)}\\ &=-\frac{(2 c e f-4 c d g+b e g) \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 (2 c d-b e)}-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{e^2 (2 c d-b e) (d+e x)^2}+\frac{((-2 c d+b e) (2 c e f-4 c d g+b e g)) \int \frac{1}{\sqrt{c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{2 e (2 c d-b e)}\\ &=-\frac{(2 c e f-4 c d g+b e g) \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 (2 c d-b e)}-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{e^2 (2 c d-b e) (d+e x)^2}+\frac{((-2 c d+b e) (2 c e f-4 c d g+b e g)) \operatorname{Subst}\left (\int \frac{1}{-4 c e^2-x^2} \, dx,x,\frac{-b e^2-2 c e^2 x}{\sqrt{c d^2-b d e-b e^2 x-c e^2 x^2}}\right )}{e (2 c d-b e)}\\ &=-\frac{(2 c e f-4 c d g+b e g) \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 (2 c d-b e)}-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{e^2 (2 c d-b e) (d+e x)^2}-\frac{(2 c e f-4 c d g+b e g) \tan ^{-1}\left (\frac{e (b+2 c x)}{2 \sqrt{c} \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{2 \sqrt{c} e^2}\\ \end{align*}

Mathematica [A]  time = 0.790027, size = 185, normalized size = 0.92 \[ \frac{\sqrt{(d+e x) (c (d-e x)-b e)} \left (-\frac{\sqrt{e} (b e-2 c d) (3 d g-2 e f+e g x)}{d+e x}-\frac{\sqrt{e (2 c d-b e)} (b e g-4 c d g+2 c e f) \sin ^{-1}\left (\frac{\sqrt{c} \sqrt{e} \sqrt{d+e x}}{\sqrt{e (2 c d-b e)}}\right )}{\sqrt{c} \sqrt{d+e x} \sqrt{\frac{b e-c d+c e x}{b e-2 c d}}}\right )}{e^{5/2} (2 c d-b e)} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x)*Sqrt[c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2])/(d + e*x)^2,x]

[Out]

(Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))]*(-((Sqrt[e]*(-2*c*d + b*e)*(-2*e*f + 3*d*g + e*g*x))/(d + e*x)) - (Sqr
t[e*(2*c*d - b*e)]*(2*c*e*f - 4*c*d*g + b*e*g)*ArcSin[(Sqrt[c]*Sqrt[e]*Sqrt[d + e*x])/Sqrt[e*(2*c*d - b*e)]])/
(Sqrt[c]*Sqrt[d + e*x]*Sqrt[(-(c*d) + b*e + c*e*x)/(-2*c*d + b*e)])))/(e^(5/2)*(2*c*d - b*e))

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Maple [B]  time = 0.014, size = 880, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)/(e*x+d)^2,x)

[Out]

g/e^2*(-(x+d/e)^2*c*e^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2)-1/2*g/(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+d/e-1/2*(-
b*e^2+2*c*d*e)/c/e^2)/(-(x+d/e)^2*c*e^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2))*b+g/e/(c*e^2)^(1/2)*arctan((c*e^2)^(1
/2)*(x+d/e-1/2*(-b*e^2+2*c*d*e)/c/e^2)/(-(x+d/e)^2*c*e^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2))*c*d+2/e^3/(-b*e^2+2*
c*d*e)/(x+d/e)^2*(-(x+d/e)^2*c*e^2+(-b*e^2+2*c*d*e)*(x+d/e))^(3/2)*d*g-2/e^2/(-b*e^2+2*c*d*e)/(x+d/e)^2*(-(x+d
/e)^2*c*e^2+(-b*e^2+2*c*d*e)*(x+d/e))^(3/2)*f+2/e*c/(-b*e^2+2*c*d*e)*(-(x+d/e)^2*c*e^2+(-b*e^2+2*c*d*e)*(x+d/e
))^(1/2)*d*g-2*c/(-b*e^2+2*c*d*e)*(-(x+d/e)^2*c*e^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2)*f-e*c/(-b*e^2+2*c*d*e)/(c*
e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+d/e-1/2*(-b*e^2+2*c*d*e)/c/e^2)/(-(x+d/e)^2*c*e^2+(-b*e^2+2*c*d*e)*(x+d/e))
^(1/2))*b*d*g+e^2*c/(-b*e^2+2*c*d*e)/(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+d/e-1/2*(-b*e^2+2*c*d*e)/c/e^2)/(-(
x+d/e)^2*c*e^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2))*b*f+2*c^2/(-b*e^2+2*c*d*e)/(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*
(x+d/e-1/2*(-b*e^2+2*c*d*e)/c/e^2)/(-(x+d/e)^2*c*e^2+(-b*e^2+2*c*d*e)*(x+d/e))^(1/2))*d^2*g-2*e*c^2/(-b*e^2+2*
c*d*e)/(c*e^2)^(1/2)*arctan((c*e^2)^(1/2)*(x+d/e-1/2*(-b*e^2+2*c*d*e)/c/e^2)/(-(x+d/e)^2*c*e^2+(-b*e^2+2*c*d*e
)*(x+d/e))^(1/2))*d*f

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)/(e*x+d)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 5.87423, size = 861, normalized size = 4.3 \begin{align*} \left [-\frac{{\left (2 \, c d e f -{\left (4 \, c d^{2} - b d e\right )} g +{\left (2 \, c e^{2} f -{\left (4 \, c d e - b e^{2}\right )} g\right )} x\right )} \sqrt{-c} \log \left (8 \, c^{2} e^{2} x^{2} + 8 \, b c e^{2} x - 4 \, c^{2} d^{2} + 4 \, b c d e + b^{2} e^{2} + 4 \, \sqrt{-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}{\left (2 \, c e x + b e\right )} \sqrt{-c}\right ) - 4 \, \sqrt{-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}{\left (c e g x - 2 \, c e f + 3 \, c d g\right )}}{4 \,{\left (c e^{3} x + c d e^{2}\right )}}, \frac{{\left (2 \, c d e f -{\left (4 \, c d^{2} - b d e\right )} g +{\left (2 \, c e^{2} f -{\left (4 \, c d e - b e^{2}\right )} g\right )} x\right )} \sqrt{c} \arctan \left (\frac{\sqrt{-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}{\left (2 \, c e x + b e\right )} \sqrt{c}}{2 \,{\left (c^{2} e^{2} x^{2} + b c e^{2} x - c^{2} d^{2} + b c d e\right )}}\right ) + 2 \, \sqrt{-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}{\left (c e g x - 2 \, c e f + 3 \, c d g\right )}}{2 \,{\left (c e^{3} x + c d e^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)/(e*x+d)^2,x, algorithm="fricas")

[Out]

[-1/4*((2*c*d*e*f - (4*c*d^2 - b*d*e)*g + (2*c*e^2*f - (4*c*d*e - b*e^2)*g)*x)*sqrt(-c)*log(8*c^2*e^2*x^2 + 8*
b*c*e^2*x - 4*c^2*d^2 + 4*b*c*d*e + b^2*e^2 + 4*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2*c*e*x + b*e)*sqr
t(-c)) - 4*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(c*e*g*x - 2*c*e*f + 3*c*d*g))/(c*e^3*x + c*d*e^2), 1/2*
((2*c*d*e*f - (4*c*d^2 - b*d*e)*g + (2*c*e^2*f - (4*c*d*e - b*e^2)*g)*x)*sqrt(c)*arctan(1/2*sqrt(-c*e^2*x^2 -
b*e^2*x + c*d^2 - b*d*e)*(2*c*e*x + b*e)*sqrt(c)/(c^2*e^2*x^2 + b*c*e^2*x - c^2*d^2 + b*c*d*e)) + 2*sqrt(-c*e^
2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(c*e*g*x - 2*c*e*f + 3*c*d*g))/(c*e^3*x + c*d*e^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{- \left (d + e x\right ) \left (b e - c d + c e x\right )} \left (f + g x\right )}{\left (d + e x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(1/2)/(e*x+d)**2,x)

[Out]

Integral(sqrt(-(d + e*x)*(b*e - c*d + c*e*x))*(f + g*x)/(d + e*x)**2, x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)/(e*x+d)^2,x, algorithm="giac")

[Out]

Timed out