3.21.40 \(\int \frac {(d+e x)^{5/2} (f+g x)}{(c d^2-b d e-b e^2 x-c e^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=217 \[ \frac {4 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-4 b e g+5 c d g+3 c e f)}{3 c^3 e^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-4 b e g+5 c d g+3 c e f)}{3 c^2 e^2 (2 c d-b e)}+\frac {2 (d+e x)^{5/2} (-b e g+c d g+c e f)}{c e^2 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}} \]

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Rubi [A]  time = 0.29, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {788, 656, 648} \begin {gather*} \frac {2 \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-4 b e g+5 c d g+3 c e f)}{3 c^2 e^2 (2 c d-b e)}+\frac {4 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2} (-4 b e g+5 c d g+3 c e f)}{3 c^3 e^2 \sqrt {d+e x}}+\frac {2 (d+e x)^{5/2} (-b e g+c d g+c e f)}{c e^2 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^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)^(3/2),x]

[Out]

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

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 656

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*(m + 2*p + 1)), x] + Dist[(Simplify[m + p]*(2*c*d - b*e))/(c*(m + 2*p + 1)), In
t[(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] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[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 )^{3/2}} \, dx &=\frac {2 (c e f+c d g-b e g) (d+e x)^{5/2}}{c e^2 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}-\frac {(3 c e f+5 c d g-4 b e g) \int \frac {(d+e x)^{3/2}}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{c e (2 c d-b e)}\\ &=\frac {2 (c e f+c d g-b e g) (d+e x)^{5/2}}{c e^2 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {2 (3 c e f+5 c d g-4 b e g) \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{3 c^2 e^2 (2 c d-b e)}-\frac {(2 (3 c e f+5 c d g-4 b e g)) \int \frac {\sqrt {d+e x}}{\sqrt {c d^2-b d e-b e^2 x-c e^2 x^2}} \, dx}{3 c^2 e}\\ &=\frac {2 (c e f+c d g-b e g) (d+e x)^{5/2}}{c e^2 (2 c d-b e) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}+\frac {4 (3 c e f+5 c d g-4 b e g) \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{3 c^3 e^2 \sqrt {d+e x}}+\frac {2 (3 c e f+5 c d g-4 b e g) \sqrt {d+e x} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{3 c^2 e^2 (2 c d-b e)}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 105, normalized size = 0.48 \begin {gather*} -\frac {2 \sqrt {d+e x} \left (-8 b^2 e^2 g+2 b c e (11 d g+3 e f-2 e g x)+c^2 \left (-14 d^2 g+d e (7 g x-9 f)+e^2 x (3 f+g x)\right )\right )}{3 c^3 e^2 \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)^(3/2),x]

[Out]

(-2*Sqrt[d + e*x]*(-8*b^2*e^2*g + 2*b*c*e*(3*e*f + 11*d*g - 2*e*g*x) + c^2*(-14*d^2*g + e^2*x*(3*f + g*x) + d*
e*(-9*f + 7*g*x))))/(3*c^3*e^2*Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))])

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IntegrateAlgebraic [A]  time = 4.55, size = 155, normalized size = 0.71 \begin {gather*} \frac {2 \sqrt {(d+e x) (2 c d-b e)-c (d+e x)^2} \left (-8 b^2 e^2 g-4 b c e g (d+e x)+26 b c d e g+6 b c e^2 f-20 c^2 d^2 g+3 c^2 e f (d+e x)-12 c^2 d e f+c^2 g (d+e x)^2+5 c^2 d g (d+e x)\right )}{3 c^3 e^2 \sqrt {d+e x} (b e+c (d+e x)-2 c d)} \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)^(3/2),x]

[Out]

(2*Sqrt[(2*c*d - b*e)*(d + e*x) - c*(d + e*x)^2]*(-12*c^2*d*e*f + 6*b*c*e^2*f - 20*c^2*d^2*g + 26*b*c*d*e*g -
8*b^2*e^2*g + 3*c^2*e*f*(d + e*x) + 5*c^2*d*g*(d + e*x) - 4*b*c*e*g*(d + e*x) + c^2*g*(d + e*x)^2))/(3*c^3*e^2
*Sqrt[d + e*x]*(-2*c*d + b*e + c*(d + e*x)))

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fricas [A]  time = 0.41, size = 165, normalized size = 0.76 \begin {gather*} \frac {2 \, {\left (c^{2} e^{2} g x^{2} - 3 \, {\left (3 \, c^{2} d e - 2 \, b c e^{2}\right )} f - 2 \, {\left (7 \, c^{2} d^{2} - 11 \, b c d e + 4 \, b^{2} e^{2}\right )} g + {\left (3 \, c^{2} e^{2} f + {\left (7 \, c^{2} d e - 4 \, b c e^{2}\right )} g\right )} x\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {e x + d}}{3 \, {\left (c^{4} e^{4} x^{2} + b c^{3} e^{4} x - c^{4} d^{2} e^{2} + b c^{3} d e^{3}\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)^(3/2),x, algorithm="fricas")

[Out]

2/3*(c^2*e^2*g*x^2 - 3*(3*c^2*d*e - 2*b*c*e^2)*f - 2*(7*c^2*d^2 - 11*b*c*d*e + 4*b^2*e^2)*g + (3*c^2*e^2*f + (
7*c^2*d*e - 4*b*c*e^2)*g)*x)*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(e*x + d)/(c^4*e^4*x^2 + b*c^3*e^4
*x - c^4*d^2*e^2 + b*c^3*d*e^3)

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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)^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.04, size = 139, normalized size = 0.64 \begin {gather*} -\frac {2 \left (c e x +b e -c d \right ) \left (-g \,x^{2} c^{2} e^{2}+4 b c \,e^{2} g x -7 c^{2} d e g x -3 c^{2} e^{2} f x +8 b^{2} e^{2} g -22 b c d e g -6 b c \,e^{2} f +14 c^{2} d^{2} g +9 c^{2} d e f \right ) \left (e x +d \right )^{\frac {3}{2}}}{3 \left (-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}\right )^{\frac {3}{2}} c^{3} 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)^(3/2),x)

[Out]

-2/3*(c*e*x+b*e-c*d)*(-c^2*e^2*g*x^2+4*b*c*e^2*g*x-7*c^2*d*e*g*x-3*c^2*e^2*f*x+8*b^2*e^2*g-22*b*c*d*e*g-6*b*c*
e^2*f+14*c^2*d^2*g+9*c^2*d*e*f)*(e*x+d)^(3/2)/c^3/e^2/(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(3/2)

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maxima [A]  time = 0.72, size = 112, normalized size = 0.52 \begin {gather*} -\frac {2 \, {\left (c e x - 3 \, c d + 2 \, b e\right )} f}{\sqrt {-c e x + c d - b e} c^{2} e} - \frac {2 \, {\left (c^{2} e^{2} x^{2} - 14 \, c^{2} d^{2} + 22 \, b c d e - 8 \, b^{2} e^{2} + {\left (7 \, c^{2} d e - 4 \, b c e^{2}\right )} x\right )} g}{3 \, \sqrt {-c e x + c d - b e} c^{3} e^{2}} \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)^(3/2),x, algorithm="maxima")

[Out]

-2*(c*e*x - 3*c*d + 2*b*e)*f/(sqrt(-c*e*x + c*d - b*e)*c^2*e) - 2/3*(c^2*e^2*x^2 - 14*c^2*d^2 + 22*b*c*d*e - 8
*b^2*e^2 + (7*c^2*d*e - 4*b*c*e^2)*x)*g/(sqrt(-c*e*x + c*d - b*e)*c^3*e^2)

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mupad [B]  time = 2.76, size = 167, normalized size = 0.77 \begin {gather*} \frac {\left (\frac {2\,g\,x^2\,\sqrt {d+e\,x}}{3\,c^2\,e^2}-\frac {\sqrt {d+e\,x}\,\left (16\,g\,b^2\,e^2-44\,g\,b\,c\,d\,e-12\,f\,b\,c\,e^2+28\,g\,c^2\,d^2+18\,f\,c^2\,d\,e\right )}{3\,c^4\,e^4}+\frac {2\,x\,\sqrt {d+e\,x}\,\left (7\,c\,d\,g-4\,b\,e\,g+3\,c\,e\,f\right )}{3\,c^3\,e^3}\right )\,\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}}{x^2+\frac {b\,x}{c}+\frac {d\,\left (b\,e-c\,d\right )}{c\,e^2}} \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)^(3/2),x)

[Out]

(((2*g*x^2*(d + e*x)^(1/2))/(3*c^2*e^2) - ((d + e*x)^(1/2)*(16*b^2*e^2*g + 28*c^2*d^2*g - 12*b*c*e^2*f + 18*c^
2*d*e*f - 44*b*c*d*e*g))/(3*c^4*e^4) + (2*x*(d + e*x)^(1/2)*(7*c*d*g - 4*b*e*g + 3*c*e*f))/(3*c^3*e^3))*(c*d^2
- c*e^2*x^2 - b*d*e - b*e^2*x)^(1/2))/(x^2 + (b*x)/c + (d*(b*e - c*d))/(c*e^2))

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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)**(3/2),x)

[Out]

Timed out

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