3.2226 \(\int \frac {(d+e x)^2 (f+g x)}{(c d^2-b d e-b e^2 x-c e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=146 \[ \frac {2 (d+e x)^2 (-b e g+c d g+c e f)}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}+\frac {2 (d+e x) (b e g-4 c d g+2 c e f)}{3 c e^2 (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}} \]

[Out]

2/3*(-b*e*g+c*d*g+c*e*f)*(e*x+d)^2/c/e^2/(-b*e+2*c*d)/(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(3/2)+2/3*(b*e*g-4*c*d*
g+2*c*e*f)*(e*x+d)/c/e^2/(-b*e+2*c*d)^2/(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {788, 636} \[ \frac {2 (d+e x)^2 (-b e g+c d g+c e f)}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}+\frac {2 (d+e x) (b e g-4 c d g+2 c e f)}{3 c e^2 (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^2*(f + g*x))/(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(5/2),x]

[Out]

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

Rule 636

Int[((d_.) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-2*(b*d - 2*a*e + (2*c*
d - b*e)*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] &&
NeQ[b^2 - 4*a*c, 0]

Rule 788

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^2 (f+g x)}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}} \, dx &=\frac {2 (c e f+c d g-b e g) (d+e x)^2}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}+\frac {(2 c e f-4 c d g+b e g) \int \frac {d+e x}{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}} \, dx}{3 c e (2 c d-b e)}\\ &=\frac {2 (c e f+c d g-b e g) (d+e x)^2}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}+\frac {2 (2 c e f-4 c d g+b e g) (d+e x)}{3 c e^2 (2 c d-b e)^2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 100, normalized size = 0.68 \[ \frac {2 (d+e x) \left (b e (-2 d g+3 e f+e g x)+2 c \left (d^2 g-2 d e (f+g x)+e^2 f x\right )\right )}{3 e^2 (b e-2 c d)^2 (b e-c d+c e x) \sqrt {(d+e x) (c (d-e x)-b e)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^2*(f + g*x))/(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(5/2),x]

[Out]

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

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fricas [A]  time = 6.03, size = 227, normalized size = 1.55 \[ \frac {2 \, \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} {\left ({\left (4 \, c d e - 3 \, b e^{2}\right )} f - 2 \, {\left (c d^{2} - b d e\right )} g - {\left (2 \, c e^{2} f - {\left (4 \, c d e - b e^{2}\right )} g\right )} x\right )}}{3 \, {\left (4 \, c^{4} d^{4} e^{2} - 12 \, b c^{3} d^{3} e^{3} + 13 \, b^{2} c^{2} d^{2} e^{4} - 6 \, b^{3} c d e^{5} + b^{4} e^{6} + {\left (4 \, c^{4} d^{2} e^{4} - 4 \, b c^{3} d e^{5} + b^{2} c^{2} e^{6}\right )} x^{2} - 2 \, {\left (4 \, c^{4} d^{3} e^{3} - 8 \, b c^{3} d^{2} e^{4} + 5 \, b^{2} c^{2} d e^{5} - b^{3} c e^{6}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x, algorithm="fricas")

[Out]

2/3*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*((4*c*d*e - 3*b*e^2)*f - 2*(c*d^2 - b*d*e)*g - (2*c*e^2*f - (4*
c*d*e - b*e^2)*g)*x)/(4*c^4*d^4*e^2 - 12*b*c^3*d^3*e^3 + 13*b^2*c^2*d^2*e^4 - 6*b^3*c*d*e^5 + b^4*e^6 + (4*c^4
*d^2*e^4 - 4*b*c^3*d*e^5 + b^2*c^2*e^6)*x^2 - 2*(4*c^4*d^3*e^3 - 8*b*c^3*d^2*e^4 + 5*b^2*c^2*d*e^5 - b^3*c*e^6
)*x)

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giac [B]  time = 0.73, size = 587, normalized size = 4.02 \[ \frac {2 \, \sqrt {-c x^{2} e^{2} + c d^{2} - b x e^{2} - b d e} {\left ({\left ({\left (\frac {{\left (16 \, c^{3} d^{3} g e^{3} - 8 \, c^{3} d^{2} f e^{4} - 20 \, b c^{2} d^{2} g e^{4} + 8 \, b c^{2} d f e^{5} + 8 \, b^{2} c d g e^{5} - 2 \, b^{2} c f e^{6} - b^{3} g e^{6}\right )} x}{16 \, c^{4} d^{4} e^{2} - 32 \, b c^{3} d^{3} e^{3} + 24 \, b^{2} c^{2} d^{2} e^{4} - 8 \, b^{3} c d e^{5} + b^{4} e^{6}} + \frac {3 \, {\left (8 \, c^{3} d^{4} g e^{2} - 8 \, b c^{2} d^{3} g e^{3} - 4 \, b c^{2} d^{2} f e^{4} + 2 \, b^{2} c d^{2} g e^{4} + 4 \, b^{2} c d f e^{5} - b^{3} f e^{6}\right )}}{16 \, c^{4} d^{4} e^{2} - 32 \, b c^{3} d^{3} e^{3} + 24 \, b^{2} c^{2} d^{2} e^{4} - 8 \, b^{3} c d e^{5} + b^{4} e^{6}}\right )} x + \frac {3 \, {\left (8 \, c^{3} d^{4} f e^{2} + 4 \, b c^{2} d^{4} g e^{2} - 16 \, b c^{2} d^{3} f e^{3} - 4 \, b^{2} c d^{3} g e^{3} + 10 \, b^{2} c d^{2} f e^{4} + b^{3} d^{2} g e^{4} - 2 \, b^{3} d f e^{5}\right )}}{16 \, c^{4} d^{4} e^{2} - 32 \, b c^{3} d^{3} e^{3} + 24 \, b^{2} c^{2} d^{2} e^{4} - 8 \, b^{3} c d e^{5} + b^{4} e^{6}}\right )} x - \frac {8 \, c^{3} d^{6} g - 16 \, c^{3} d^{5} f e - 16 \, b c^{2} d^{5} g e + 28 \, b c^{2} d^{4} f e^{2} + 10 \, b^{2} c d^{4} g e^{2} - 16 \, b^{2} c d^{3} f e^{3} - 2 \, b^{3} d^{3} g e^{3} + 3 \, b^{3} d^{2} f e^{4}}{16 \, c^{4} d^{4} e^{2} - 32 \, b c^{3} d^{3} e^{3} + 24 \, b^{2} c^{2} d^{2} e^{4} - 8 \, b^{3} c d e^{5} + b^{4} e^{6}}\right )}}{3 \, {\left (c x^{2} e^{2} - c d^{2} + b x e^{2} + b d e\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x, algorithm="giac")

[Out]

2/3*sqrt(-c*x^2*e^2 + c*d^2 - b*x*e^2 - b*d*e)*((((16*c^3*d^3*g*e^3 - 8*c^3*d^2*f*e^4 - 20*b*c^2*d^2*g*e^4 + 8
*b*c^2*d*f*e^5 + 8*b^2*c*d*g*e^5 - 2*b^2*c*f*e^6 - b^3*g*e^6)*x/(16*c^4*d^4*e^2 - 32*b*c^3*d^3*e^3 + 24*b^2*c^
2*d^2*e^4 - 8*b^3*c*d*e^5 + b^4*e^6) + 3*(8*c^3*d^4*g*e^2 - 8*b*c^2*d^3*g*e^3 - 4*b*c^2*d^2*f*e^4 + 2*b^2*c*d^
2*g*e^4 + 4*b^2*c*d*f*e^5 - b^3*f*e^6)/(16*c^4*d^4*e^2 - 32*b*c^3*d^3*e^3 + 24*b^2*c^2*d^2*e^4 - 8*b^3*c*d*e^5
+ b^4*e^6))*x + 3*(8*c^3*d^4*f*e^2 + 4*b*c^2*d^4*g*e^2 - 16*b*c^2*d^3*f*e^3 - 4*b^2*c*d^3*g*e^3 + 10*b^2*c*d^
2*f*e^4 + b^3*d^2*g*e^4 - 2*b^3*d*f*e^5)/(16*c^4*d^4*e^2 - 32*b*c^3*d^3*e^3 + 24*b^2*c^2*d^2*e^4 - 8*b^3*c*d*e
^5 + b^4*e^6))*x - (8*c^3*d^6*g - 16*c^3*d^5*f*e - 16*b*c^2*d^5*g*e + 28*b*c^2*d^4*f*e^2 + 10*b^2*c*d^4*g*e^2
- 16*b^2*c*d^3*f*e^3 - 2*b^3*d^3*g*e^3 + 3*b^3*d^2*f*e^4)/(16*c^4*d^4*e^2 - 32*b*c^3*d^3*e^3 + 24*b^2*c^2*d^2*
e^4 - 8*b^3*c*d*e^5 + b^4*e^6))/(c*x^2*e^2 - c*d^2 + b*x*e^2 + b*d*e)^2

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maple [A]  time = 0.05, size = 128, normalized size = 0.88 \[ -\frac {2 \left (e x +d \right )^{3} \left (c e x +b e -c d \right ) \left (-b \,e^{2} g x +4 c d e g x -2 c \,e^{2} f x +2 b d e g -3 b \,e^{2} f -2 c \,d^{2} g +4 c d e f \right )}{3 \left (b^{2} e^{2}-4 b c d e +4 c^{2} d^{2}\right ) \left (-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}\right )^{\frac {5}{2}} e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(g*x+f)/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-2*c*d>0)', see `assume?` f
or more details)Is b*e-2*c*d zero or nonzero?

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mupad [B]  time = 3.30, size = 107, normalized size = 0.73 \[ -\frac {2\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}\,\left (3\,b\,e^2\,f+2\,c\,d^2\,g+b\,e^2\,g\,x+2\,c\,e^2\,f\,x-2\,b\,d\,e\,g-4\,c\,d\,e\,f-4\,c\,d\,e\,g\,x\right )}{3\,e^2\,{\left (b\,e-2\,c\,d\right )}^2\,{\left (b\,e-c\,d+c\,e\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f + g*x)*(d + e*x)^2)/(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(5/2),x)

[Out]

-(2*(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(1/2)*(3*b*e^2*f + 2*c*d^2*g + b*e^2*g*x + 2*c*e^2*f*x - 2*b*d*e*g -
4*c*d*e*f - 4*c*d*e*g*x))/(3*e^2*(b*e - 2*c*d)^2*(b*e - c*d + c*e*x)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{2} \left (f + g x\right )}{\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2*(g*x+f)/(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(5/2),x)

[Out]

Integral((d + e*x)**2*(f + g*x)/(-(d + e*x)*(b*e - c*d + c*e*x))**(5/2), x)

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