3.1599 \(\int \frac{(b+2 c x) (a+b x+c x^2)}{(d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=126 \[ \frac{2 \sqrt{d+e x} \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{e^4}+\frac{2 (2 c d-b e) \left (a e^2-b d e+c d^2\right )}{e^4 \sqrt{d+e x}}-\frac{2 c (d+e x)^{3/2} (2 c d-b e)}{e^4}+\frac{4 c^2 (d+e x)^{5/2}}{5 e^4} \]

[Out]

(2*(2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2))/(e^4*Sqrt[d + e*x]) + (2*(6*c^2*d^2 + b^2*e^2 - 2*c*e*(3*b*d - a*e))
*Sqrt[d + e*x])/e^4 - (2*c*(2*c*d - b*e)*(d + e*x)^(3/2))/e^4 + (4*c^2*(d + e*x)^(5/2))/(5*e^4)

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Rubi [A]  time = 0.0678039, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.038, Rules used = {771} \[ \frac{2 \sqrt{d+e x} \left (-2 c e (3 b d-a e)+b^2 e^2+6 c^2 d^2\right )}{e^4}+\frac{2 (2 c d-b e) \left (a e^2-b d e+c d^2\right )}{e^4 \sqrt{d+e x}}-\frac{2 c (d+e x)^{3/2} (2 c d-b e)}{e^4}+\frac{4 c^2 (d+e x)^{5/2}}{5 e^4} \]

Antiderivative was successfully verified.

[In]

Int[((b + 2*c*x)*(a + b*x + c*x^2))/(d + e*x)^(3/2),x]

[Out]

(2*(2*c*d - b*e)*(c*d^2 - b*d*e + a*e^2))/(e^4*Sqrt[d + e*x]) + (2*(6*c^2*d^2 + b^2*e^2 - 2*c*e*(3*b*d - a*e))
*Sqrt[d + e*x])/e^4 - (2*c*(2*c*d - b*e)*(d + e*x)^(3/2))/e^4 + (4*c^2*(d + e*x)^(5/2))/(5*e^4)

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int \frac{(b+2 c x) \left (a+b x+c x^2\right )}{(d+e x)^{3/2}} \, dx &=\int \left (\frac{(-2 c d+b e) \left (c d^2-b d e+a e^2\right )}{e^3 (d+e x)^{3/2}}+\frac{6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)}{e^3 \sqrt{d+e x}}-\frac{3 c (2 c d-b e) \sqrt{d+e x}}{e^3}+\frac{2 c^2 (d+e x)^{3/2}}{e^3}\right ) \, dx\\ &=\frac{2 (2 c d-b e) \left (c d^2-b d e+a e^2\right )}{e^4 \sqrt{d+e x}}+\frac{2 \left (6 c^2 d^2+b^2 e^2-2 c e (3 b d-a e)\right ) \sqrt{d+e x}}{e^4}-\frac{2 c (2 c d-b e) (d+e x)^{3/2}}{e^4}+\frac{4 c^2 (d+e x)^{5/2}}{5 e^4}\\ \end{align*}

Mathematica [A]  time = 0.108237, size = 106, normalized size = 0.84 \[ \frac{2 \left (5 c e \left (2 a e (2 d+e x)+b \left (-8 d^2-4 d e x+e^2 x^2\right )\right )+5 b e^2 (-a e+2 b d+b e x)+2 c^2 \left (8 d^2 e x+16 d^3-2 d e^2 x^2+e^3 x^3\right )\right )}{5 e^4 \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

Integrate[((b + 2*c*x)*(a + b*x + c*x^2))/(d + e*x)^(3/2),x]

[Out]

(2*(5*b*e^2*(2*b*d - a*e + b*e*x) + 2*c^2*(16*d^3 + 8*d^2*e*x - 2*d*e^2*x^2 + e^3*x^3) + 5*c*e*(2*a*e*(2*d + e
*x) + b*(-8*d^2 - 4*d*e*x + e^2*x^2))))/(5*e^4*Sqrt[d + e*x])

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Maple [A]  time = 0.005, size = 123, normalized size = 1. \begin{align*} -{\frac{-4\,{c}^{2}{x}^{3}{e}^{3}-10\,bc{e}^{3}{x}^{2}+8\,{c}^{2}d{e}^{2}{x}^{2}-20\,ac{e}^{3}x-10\,{b}^{2}{e}^{3}x+40\,bcd{e}^{2}x-32\,{c}^{2}{d}^{2}ex+10\,ab{e}^{3}-40\,acd{e}^{2}-20\,{b}^{2}d{e}^{2}+80\,b{d}^{2}ce-64\,{c}^{2}{d}^{3}}{5\,{e}^{4}}{\frac{1}{\sqrt{ex+d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*c*x+b)*(c*x^2+b*x+a)/(e*x+d)^(3/2),x)

[Out]

-2/5/(e*x+d)^(1/2)*(-2*c^2*e^3*x^3-5*b*c*e^3*x^2+4*c^2*d*e^2*x^2-10*a*c*e^3*x-5*b^2*e^3*x+20*b*c*d*e^2*x-16*c^
2*d^2*e*x+5*a*b*e^3-20*a*c*d*e^2-10*b^2*d*e^2+40*b*c*d^2*e-32*c^2*d^3)/e^4

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Maxima [A]  time = 1.00919, size = 174, normalized size = 1.38 \begin{align*} \frac{2 \,{\left (\frac{2 \,{\left (e x + d\right )}^{\frac{5}{2}} c^{2} - 5 \,{\left (2 \, c^{2} d - b c e\right )}{\left (e x + d\right )}^{\frac{3}{2}} + 5 \,{\left (6 \, c^{2} d^{2} - 6 \, b c d e +{\left (b^{2} + 2 \, a c\right )} e^{2}\right )} \sqrt{e x + d}}{e^{3}} + \frac{5 \,{\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e - a b e^{3} +{\left (b^{2} + 2 \, a c\right )} d e^{2}\right )}}{\sqrt{e x + d} e^{3}}\right )}}{5 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(c*x^2+b*x+a)/(e*x+d)^(3/2),x, algorithm="maxima")

[Out]

2/5*((2*(e*x + d)^(5/2)*c^2 - 5*(2*c^2*d - b*c*e)*(e*x + d)^(3/2) + 5*(6*c^2*d^2 - 6*b*c*d*e + (b^2 + 2*a*c)*e
^2)*sqrt(e*x + d))/e^3 + 5*(2*c^2*d^3 - 3*b*c*d^2*e - a*b*e^3 + (b^2 + 2*a*c)*d*e^2)/(sqrt(e*x + d)*e^3))/e

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Fricas [A]  time = 1.34043, size = 275, normalized size = 2.18 \begin{align*} \frac{2 \,{\left (2 \, c^{2} e^{3} x^{3} + 32 \, c^{2} d^{3} - 40 \, b c d^{2} e - 5 \, a b e^{3} + 10 \,{\left (b^{2} + 2 \, a c\right )} d e^{2} -{\left (4 \, c^{2} d e^{2} - 5 \, b c e^{3}\right )} x^{2} +{\left (16 \, c^{2} d^{2} e - 20 \, b c d e^{2} + 5 \,{\left (b^{2} + 2 \, a c\right )} e^{3}\right )} x\right )} \sqrt{e x + d}}{5 \,{\left (e^{5} x + d e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(c*x^2+b*x+a)/(e*x+d)^(3/2),x, algorithm="fricas")

[Out]

2/5*(2*c^2*e^3*x^3 + 32*c^2*d^3 - 40*b*c*d^2*e - 5*a*b*e^3 + 10*(b^2 + 2*a*c)*d*e^2 - (4*c^2*d*e^2 - 5*b*c*e^3
)*x^2 + (16*c^2*d^2*e - 20*b*c*d*e^2 + 5*(b^2 + 2*a*c)*e^3)*x)*sqrt(e*x + d)/(e^5*x + d*e^4)

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Sympy [A]  time = 19.0123, size = 128, normalized size = 1.02 \begin{align*} \frac{4 c^{2} \left (d + e x\right )^{\frac{5}{2}}}{5 e^{4}} + \frac{\left (d + e x\right )^{\frac{3}{2}} \left (6 b c e - 12 c^{2} d\right )}{3 e^{4}} + \frac{\sqrt{d + e x} \left (4 a c e^{2} + 2 b^{2} e^{2} - 12 b c d e + 12 c^{2} d^{2}\right )}{e^{4}} - \frac{2 \left (b e - 2 c d\right ) \left (a e^{2} - b d e + c d^{2}\right )}{e^{4} \sqrt{d + e x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(c*x**2+b*x+a)/(e*x+d)**(3/2),x)

[Out]

4*c**2*(d + e*x)**(5/2)/(5*e**4) + (d + e*x)**(3/2)*(6*b*c*e - 12*c**2*d)/(3*e**4) + sqrt(d + e*x)*(4*a*c*e**2
+ 2*b**2*e**2 - 12*b*c*d*e + 12*c**2*d**2)/e**4 - 2*(b*e - 2*c*d)*(a*e**2 - b*d*e + c*d**2)/(e**4*sqrt(d + e*
x))

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Giac [A]  time = 1.41537, size = 220, normalized size = 1.75 \begin{align*} \frac{2}{5} \,{\left (2 \,{\left (x e + d\right )}^{\frac{5}{2}} c^{2} e^{16} - 10 \,{\left (x e + d\right )}^{\frac{3}{2}} c^{2} d e^{16} + 30 \, \sqrt{x e + d} c^{2} d^{2} e^{16} + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} b c e^{17} - 30 \, \sqrt{x e + d} b c d e^{17} + 5 \, \sqrt{x e + d} b^{2} e^{18} + 10 \, \sqrt{x e + d} a c e^{18}\right )} e^{\left (-20\right )} + \frac{2 \,{\left (2 \, c^{2} d^{3} - 3 \, b c d^{2} e + b^{2} d e^{2} + 2 \, a c d e^{2} - a b e^{3}\right )} e^{\left (-4\right )}}{\sqrt{x e + d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*c*x+b)*(c*x^2+b*x+a)/(e*x+d)^(3/2),x, algorithm="giac")

[Out]

2/5*(2*(x*e + d)^(5/2)*c^2*e^16 - 10*(x*e + d)^(3/2)*c^2*d*e^16 + 30*sqrt(x*e + d)*c^2*d^2*e^16 + 5*(x*e + d)^
(3/2)*b*c*e^17 - 30*sqrt(x*e + d)*b*c*d*e^17 + 5*sqrt(x*e + d)*b^2*e^18 + 10*sqrt(x*e + d)*a*c*e^18)*e^(-20) +
2*(2*c^2*d^3 - 3*b*c*d^2*e + b^2*d*e^2 + 2*a*c*d*e^2 - a*b*e^3)*e^(-4)/sqrt(x*e + d)