3.2179 \(\int \frac{(f+g x) \sqrt{c d^2-b d e-b e^2 x-c e^2 x^2}}{(d+e x)^5} \, dx\)

Optimal. Leaf size=210 \[ -\frac{4 c \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2} (-7 b e g+10 c d g+4 c e f)}{105 e^2 (d+e x)^3 (2 c d-b e)^3}-\frac{2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2} (-7 b e g+10 c d g+4 c e f)}{35 e^2 (d+e x)^4 (2 c d-b e)^2}-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{7 e^2 (d+e x)^5 (2 c d-b e)} \]

[Out]

(-2*(e*f - d*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(7*e^2*(2*c*d - b*e)*(d + e*x)^5) - (2*(4*c*e*f +
10*c*d*g - 7*b*e*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(35*e^2*(2*c*d - b*e)^2*(d + e*x)^4) - (4*c*
(4*c*e*f + 10*c*d*g - 7*b*e*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(105*e^2*(2*c*d - b*e)^3*(d + e*x)
^3)

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Rubi [A]  time = 0.339418, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 44, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.068, Rules used = {792, 658, 650} \[ -\frac{4 c \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2} (-7 b e g+10 c d g+4 c e f)}{105 e^2 (d+e x)^3 (2 c d-b e)^3}-\frac{2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2} (-7 b e g+10 c d g+4 c e f)}{35 e^2 (d+e x)^4 (2 c d-b e)^2}-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{7 e^2 (d+e x)^5 (2 c d-b e)} \]

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)^5,x]

[Out]

(-2*(e*f - d*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(7*e^2*(2*c*d - b*e)*(d + e*x)^5) - (2*(4*c*e*f +
10*c*d*g - 7*b*e*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(35*e^2*(2*c*d - b*e)^2*(d + e*x)^4) - (4*c*
(4*c*e*f + 10*c*d*g - 7*b*e*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(105*e^2*(2*c*d - b*e)^3*(d + e*x)
^3)

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 658

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

Rule 650

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] &&
EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 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)^5} \, 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}}{7 e^2 (2 c d-b e) (d+e x)^5}+\frac{(4 c e f+10 c d g-7 b e g) \int \frac{\sqrt{c d^2-b d e-b e^2 x-c e^2 x^2}}{(d+e x)^4} \, dx}{7 e (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}}{7 e^2 (2 c d-b e) (d+e x)^5}-\frac{2 (4 c e f+10 c d g-7 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{35 e^2 (2 c d-b e)^2 (d+e x)^4}+\frac{(2 c (4 c e f+10 c d g-7 b e g)) \int \frac{\sqrt{c d^2-b d e-b e^2 x-c e^2 x^2}}{(d+e x)^3} \, dx}{35 e (2 c d-b e)^2}\\ &=-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{7 e^2 (2 c d-b e) (d+e x)^5}-\frac{2 (4 c e f+10 c d g-7 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{35 e^2 (2 c d-b e)^2 (d+e x)^4}-\frac{4 c (4 c e f+10 c d g-7 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{105 e^2 (2 c d-b e)^3 (d+e x)^3}\\ \end{align*}

Mathematica [A]  time = 0.139721, size = 154, normalized size = 0.73 \[ \frac{2 ((d+e x) (c (d-e x)-b e))^{3/2} \left (3 b^2 e^2 (2 d g+5 e f+7 e g x)-2 b c e \left (13 d^2 g+d e (36 f+50 g x)+e^2 x (6 f+7 g x)\right )+4 c^2 \left (d^2 e (23 f+25 g x)+5 d^3 g+5 d e^2 x (2 f+g x)+2 e^3 f x^2\right )\right )}{105 e^2 (d+e x)^5 (b e-2 c d)^3} \]

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)^5,x]

[Out]

(2*((d + e*x)*(-(b*e) + c*(d - e*x)))^(3/2)*(3*b^2*e^2*(5*e*f + 2*d*g + 7*e*g*x) + 4*c^2*(5*d^3*g + 2*e^3*f*x^
2 + 5*d*e^2*x*(2*f + g*x) + d^2*e*(23*f + 25*g*x)) - 2*b*c*e*(13*d^2*g + e^2*x*(6*f + 7*g*x) + d*e*(36*f + 50*
g*x))))/(105*e^2*(-2*c*d + b*e)^3*(d + e*x)^5)

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Maple [A]  time = 0.009, size = 236, normalized size = 1.1 \begin{align*} -{\frac{ \left ( 2\,cex+2\,be-2\,cd \right ) \left ( -14\,bc{e}^{3}g{x}^{2}+20\,{c}^{2}d{e}^{2}g{x}^{2}+8\,{c}^{2}{e}^{3}f{x}^{2}+21\,{b}^{2}{e}^{3}gx-100\,bcd{e}^{2}gx-12\,bc{e}^{3}fx+100\,{c}^{2}{d}^{2}egx+40\,{c}^{2}d{e}^{2}fx+6\,{b}^{2}d{e}^{2}g+15\,{b}^{2}{e}^{3}f-26\,bc{d}^{2}eg-72\,bcd{e}^{2}f+20\,{c}^{2}{d}^{3}g+92\,{c}^{2}{d}^{2}ef \right ) }{105\, \left ( ex+d \right ) ^{4} \left ({b}^{3}{e}^{3}-6\,{b}^{2}cd{e}^{2}+12\,b{c}^{2}{d}^{2}e-8\,{c}^{3}{d}^{3} \right ){e}^{2}}\sqrt{-c{e}^{2}{x}^{2}-b{e}^{2}x-bde+c{d}^{2}}} \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)^5,x)

[Out]

-2/105*(c*e*x+b*e-c*d)*(-14*b*c*e^3*g*x^2+20*c^2*d*e^2*g*x^2+8*c^2*e^3*f*x^2+21*b^2*e^3*g*x-100*b*c*d*e^2*g*x-
12*b*c*e^3*f*x+100*c^2*d^2*e*g*x+40*c^2*d*e^2*f*x+6*b^2*d*e^2*g+15*b^2*e^3*f-26*b*c*d^2*e*g-72*b*c*d*e^2*f+20*
c^2*d^3*g+92*c^2*d^2*e*f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(1/2)/(e*x+d)^4/(b^3*e^3-6*b^2*c*d*e^2+12*b*c^2*d^2
*e-8*c^3*d^3)/e^2

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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)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [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)^5,x, algorithm="fricas")

[Out]

Timed out

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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 )^{5}}\, 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)**5,x)

[Out]

Integral(sqrt(-(d + e*x)*(b*e - c*d + c*e*x))*(f + g*x)/(d + e*x)**5, x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \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)^5,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError