3.2280 \(\int \frac{(d+e x)^{5/2} (f+g x)}{(c d^2-b d e-b e^2 x-c e^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=152 \[ \frac{2 \sqrt{d+e x} (2 b e g-5 c d g+c e f)}{3 c^2 e^2 (2 c d-b e) \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac{2 (d+e x)^{5/2} (-b e g+c d g+c e f)}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.160173, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 46, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.043, Rules used = {788, 648} \[ \frac{2 \sqrt{d+e x} (2 b e g-5 c d g+c e f)}{3 c^2 e^2 (2 c d-b e) \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac{2 (d+e x)^{5/2} (-b e g+c d g+c e f)}{3 c e^2 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

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]

Rule 648

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), 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 + p, 0]

Rubi steps

\begin{align*} \int \frac{(d+e x)^{5/2} (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)^{5/2}}{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{(c e f-5 c d g+2 b e g) \int \frac{(d+e x)^{3/2}}{\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)^{5/2}}{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-5 c d g+2 b e g) \sqrt{d+e x}}{3 c^2 e^2 (2 c d-b e) \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0630239, size = 76, normalized size = 0.5 \[ -\frac{2 \sqrt{d+e x} (2 b e g-2 c d g+c e (f+3 g x))}{3 c^2 e^2 (b e-c d+c e x) \sqrt{(d+e x) (c (d-e x)-b e)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 78, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 2\,cex+2\,be-2\,cd \right ) \left ( 3\,cegx+2\,beg-2\,cdg+cef \right ) }{3\,{c}^{2}{e}^{2}} \left ( ex+d \right ) ^{{\frac{5}{2}}} \left ( -c{e}^{2}{x}^{2}-b{e}^{2}x-bde+c{d}^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.29947, size = 139, normalized size = 0.91 \begin{align*} -\frac{2 \,{\left (3 \, c e x - 2 \, c d + 2 \, b e\right )} g}{3 \,{\left (c^{3} e^{3} x - c^{3} d e^{2} + b c^{2} e^{3}\right )} \sqrt{-c e x + c d - b e}} - \frac{2 \, f}{3 \,{\left (c^{2} e^{2} x - c^{2} d e + b c e^{2}\right )} \sqrt{-c e x + c d - b e}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.29828, size = 305, normalized size = 2.01 \begin{align*} \frac{2 \, \sqrt{-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e}{\left (3 \, c e g x + c e f - 2 \,{\left (c d - b e\right )} g\right )} \sqrt{e x + d}}{3 \,{\left (c^{4} e^{5} x^{3} + c^{4} d^{3} e^{2} - 2 \, b c^{3} d^{2} e^{3} + b^{2} c^{2} d e^{4} -{\left (c^{4} d e^{4} - 2 \, b c^{3} e^{5}\right )} x^{2} -{\left (c^{4} d^{2} e^{3} - b^{2} c^{2} e^{5}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x