3.2203 \(\int \frac{(f+g x) (c d^2-b d e-b e^2 x-c e^2 x^2)^{5/2}}{(d+e x)^6} \, dx\)

Optimal. Leaf size=352 \[ -\frac{c^2 \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2} (5 b e g-12 c d g+2 c e f)}{e^2 (2 c d-b e)}-\frac{c^{3/2} (5 b e g-12 c d g+2 c e f) \tan ^{-1}\left (\frac{e (b+2 c x)}{2 \sqrt{c} \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{2 e^2}-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{5 e^2 (d+e x)^6 (2 c d-b e)}+\frac{2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2} (5 b e g-12 c d g+2 c e f)}{15 e^2 (d+e x)^4 (2 c d-b e)}-\frac{2 c \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2} (5 b e g-12 c d g+2 c e f)}{3 e^2 (d+e x)^2 (2 c d-b e)} \]

[Out]

-((c^2*(2*c*e*f - 12*c*d*g + 5*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*c*(
2*c*e*f - 12*c*d*g + 5*b*e*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(3*e^2*(2*c*d - b*e)*(d + e*x)^2) +
(2*(2*c*e*f - 12*c*d*g + 5*b*e*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(5/2))/(15*e^2*(2*c*d - b*e)*(d + e*x
)^4) - (2*(e*f - d*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(7/2))/(5*e^2*(2*c*d - b*e)*(d + e*x)^6) - (c^(3/2
)*(2*c*e*f - 12*c*d*g + 5*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*e^2)

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Rubi [A]  time = 0.571117, antiderivative size = 352, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 44, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {792, 662, 664, 621, 204} \[ -\frac{c^2 \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2} (5 b e g-12 c d g+2 c e f)}{e^2 (2 c d-b e)}-\frac{c^{3/2} (5 b e g-12 c d g+2 c e f) \tan ^{-1}\left (\frac{e (b+2 c x)}{2 \sqrt{c} \sqrt{d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{2 e^2}-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{5 e^2 (d+e x)^6 (2 c d-b e)}+\frac{2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2} (5 b e g-12 c d g+2 c e f)}{15 e^2 (d+e x)^4 (2 c d-b e)}-\frac{2 c \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2} (5 b e g-12 c d g+2 c e f)}{3 e^2 (d+e x)^2 (2 c d-b e)} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x)*(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(5/2))/(d + e*x)^6,x]

[Out]

-((c^2*(2*c*e*f - 12*c*d*g + 5*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*c*(
2*c*e*f - 12*c*d*g + 5*b*e*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(3*e^2*(2*c*d - b*e)*(d + e*x)^2) +
(2*(2*c*e*f - 12*c*d*g + 5*b*e*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(5/2))/(15*e^2*(2*c*d - b*e)*(d + e*x
)^4) - (2*(e*f - d*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(7/2))/(5*e^2*(2*c*d - b*e)*(d + e*x)^6) - (c^(3/2
)*(2*c*e*f - 12*c*d*g + 5*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*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 662

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 + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

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) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^6} \, dx &=-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{5 e^2 (2 c d-b e) (d+e x)^6}-\frac{(2 c e f-12 c d g+5 b e g) \int \frac{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^5} \, dx}{5 e (2 c d-b e)}\\ &=\frac{2 (2 c e f-12 c d g+5 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{15 e^2 (2 c d-b e) (d+e x)^4}-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{5 e^2 (2 c d-b e) (d+e x)^6}+\frac{(c (2 c e f-12 c d g+5 b e g)) \int \frac{\left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{(d+e x)^3} \, dx}{3 e (2 c d-b e)}\\ &=-\frac{2 c (2 c e f-12 c d g+5 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 e^2 (2 c d-b e) (d+e x)^2}+\frac{2 (2 c e f-12 c d g+5 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{15 e^2 (2 c d-b e) (d+e x)^4}-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{5 e^2 (2 c d-b e) (d+e x)^6}-\frac{\left (c^2 (2 c e f-12 c d g+5 b e g)\right ) \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{c^2 (2 c e f-12 c d g+5 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 c (2 c e f-12 c d g+5 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 e^2 (2 c d-b e) (d+e x)^2}+\frac{2 (2 c e f-12 c d g+5 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{15 e^2 (2 c d-b e) (d+e x)^4}-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{5 e^2 (2 c d-b e) (d+e x)^6}+\frac{\left (c^2 (-2 c d+b e) (2 c e f-12 c d g+5 b e g)\right ) \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{c^2 (2 c e f-12 c d g+5 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 c (2 c e f-12 c d g+5 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 e^2 (2 c d-b e) (d+e x)^2}+\frac{2 (2 c e f-12 c d g+5 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{15 e^2 (2 c d-b e) (d+e x)^4}-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{5 e^2 (2 c d-b e) (d+e x)^6}-\frac{\left (c^2 (2 c e f-12 c d g+5 b e g)\right ) \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}\\ &=-\frac{c^2 (2 c e f-12 c d g+5 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 c (2 c e f-12 c d g+5 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 e^2 (2 c d-b e) (d+e x)^2}+\frac{2 (2 c e f-12 c d g+5 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{15 e^2 (2 c d-b e) (d+e x)^4}-\frac{2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{5 e^2 (2 c d-b e) (d+e x)^6}-\frac{c^{3/2} (2 c e f-12 c d g+5 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 e^2}\\ \end{align*}

Mathematica [C]  time = 0.247877, size = 175, normalized size = 0.5 \[ \frac{2 ((d+e x) (c (d-e x)-b e))^{5/2} \left (\frac{(d+e x) (b e-2 c d)^2 (5 b e g+2 c (e f-6 d g)) \, _2F_1\left (-\frac{5}{2},-\frac{3}{2};-\frac{1}{2};\frac{c (d+e x)}{2 c d-b e}\right )}{\sqrt{\frac{b e-c d+c e x}{b e-2 c d}}}+3 (e f-d g) (b e-c d+c e x)^3\right )}{15 e^2 (d+e x)^5 (2 c d-b e) (b e-c d+c e x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x)*(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(5/2))/(d + e*x)^6,x]

[Out]

(2*((d + e*x)*(-(b*e) + c*(d - e*x)))^(5/2)*(3*(e*f - d*g)*(-(c*d) + b*e + c*e*x)^3 + ((-2*c*d + b*e)^2*(5*b*e
*g + 2*c*(e*f - 6*d*g))*(d + e*x)*Hypergeometric2F1[-5/2, -3/2, -1/2, (c*(d + e*x))/(2*c*d - b*e)])/Sqrt[(-(c*
d) + b*e + c*e*x)/(-2*c*d + b*e)]))/(15*e^2*(2*c*d - b*e)*(d + e*x)^5*(-(c*d) + b*e + c*e*x)^2)

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Maple [B]  time = 0.016, size = 5440, normalized size = 15.5 \begin{align*} \text{output 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)^(5/2)/(e*x+d)^6,x)

[Out]

result too large to display

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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)^(5/2)/(e*x+d)^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 88.5178, size = 1935, normalized size = 5.5 \begin{align*} \text{result too large to display} \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)^(5/2)/(e*x+d)^6,x, algorithm="fricas")

[Out]

[1/60*(15*(2*c^2*d^3*e*f + (2*c^2*e^4*f - (12*c^2*d*e^3 - 5*b*c*e^4)*g)*x^3 + 3*(2*c^2*d*e^3*f - (12*c^2*d^2*e
^2 - 5*b*c*d*e^3)*g)*x^2 - (12*c^2*d^4 - 5*b*c*d^3*e)*g + 3*(2*c^2*d^2*e^2*f - (12*c^2*d^3*e - 5*b*c*d^2*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)*sqrt(-c)) + 4*(15*c^2*e^3*g*x^3 - (46*c^2*e^3*f - 7*(33*c^2*d*e^2 - 10*b*c*e
^3)*g)*x^2 - 2*(13*c^2*d^2*e - b*c*d*e^2 + 3*b^2*e^3)*f + (141*c^2*d^3 - 32*b*c*d^2*e - 4*b^2*d*e^2)*g - (2*(2
4*c^2*d*e^2 + 11*b*c*e^3)*f - (333*c^2*d^2*e - 78*b*c*d*e^2 - 10*b^2*e^3)*g)*x)*sqrt(-c*e^2*x^2 - b*e^2*x + c*
d^2 - b*d*e))/(e^5*x^3 + 3*d*e^4*x^2 + 3*d^2*e^3*x + d^3*e^2), 1/30*(15*(2*c^2*d^3*e*f + (2*c^2*e^4*f - (12*c^
2*d*e^3 - 5*b*c*e^4)*g)*x^3 + 3*(2*c^2*d*e^3*f - (12*c^2*d^2*e^2 - 5*b*c*d*e^3)*g)*x^2 - (12*c^2*d^4 - 5*b*c*d
^3*e)*g + 3*(2*c^2*d^2*e^2*f - (12*c^2*d^3*e - 5*b*c*d^2*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*(15*c^2*e^3*g*x
^3 - (46*c^2*e^3*f - 7*(33*c^2*d*e^2 - 10*b*c*e^3)*g)*x^2 - 2*(13*c^2*d^2*e - b*c*d*e^2 + 3*b^2*e^3)*f + (141*
c^2*d^3 - 32*b*c*d^2*e - 4*b^2*d*e^2)*g - (2*(24*c^2*d*e^2 + 11*b*c*e^3)*f - (333*c^2*d^2*e - 78*b*c*d*e^2 - 1
0*b^2*e^3)*g)*x)*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e))/(e^5*x^3 + 3*d*e^4*x^2 + 3*d^2*e^3*x + d^3*e^2)]

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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((g*x+f)*(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(5/2)/(e*x+d)**6,x)

[Out]

Timed out

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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)^(5/2)/(e*x+d)^6,x, algorithm="giac")

[Out]

Timed out