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

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

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Rubi [A]  time = 0.16, antiderivative size = 152, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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]

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)^{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*}

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Mathematica [A]  time = 0.06, size = 76, normalized size = 0.50 \begin {gather*} -\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)}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 6.22, size = 73, normalized size = 0.48 \begin {gather*} \frac {2 (d+e x)^{3/2} (2 b e g+3 c g (d+e x)-5 c d g+c e f)}{3 c^2 e^2 \left ((d+e x) (2 c d-b e)-c (d+e x)^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.40, size = 154, normalized size = 1.01 \begin {gather*} \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 {gather*}

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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]

Timed out

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maple [A]  time = 0.04, size = 78, normalized size = 0.51 \begin {gather*} -\frac {2 \left (c e x +b e -c d \right ) \left (3 c e g x +2 b e g -2 c d g +c e f \right ) \left (e x +d \right )^{\frac {5}{2}}}{3 \left (-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}\right )^{\frac {5}{2}} c^{2} e^{2}} \end {gather*}

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 = 0.97, size = 103, normalized size = 0.68 \begin {gather*} -\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 {gather*}

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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mupad [B]  time = 2.85, size = 154, normalized size = 1.01 \begin {gather*} \frac {\left (\frac {\sqrt {d+e\,x}\,\left (4\,b\,e\,g-4\,c\,d\,g+2\,c\,e\,f\right )}{3\,c^4\,e^5}+\frac {2\,g\,x\,\sqrt {d+e\,x}}{c^3\,e^4}\right )\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}}{x^3+\frac {x\,\left (3\,b^2\,c^2\,e^5-3\,c^4\,d^2\,e^3\right )}{3\,c^4\,e^5}+\frac {d\,{\left (b\,e-c\,d\right )}^2}{c^2\,e^3}+\frac {x^2\,\left (2\,b\,e-c\,d\right )}{c\,e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((((d + e*x)^(1/2)*(4*b*e*g - 4*c*d*g + 2*c*e*f))/(3*c^4*e^5) + (2*g*x*(d + e*x)^(1/2))/(c^3*e^4))*(c*d^2 - c*
e^2*x^2 - b*d*e - b*e^2*x)^(1/2))/(x^3 + (x*(3*b^2*c^2*e^5 - 3*c^4*d^2*e^3))/(3*c^4*e^5) + (d*(b*e - c*d)^2)/(
c^2*e^3) + (x^2*(2*b*e - c*d))/(c*e))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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