3.2047 \(\int \frac{(a+b x) (a^2+2 a b x+b^2 x^2)}{(d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=94 \[ -\frac{2 b^2 (d+e x)^{3/2} (b d-a e)}{e^4}+\frac{6 b \sqrt{d+e x} (b d-a e)^2}{e^4}+\frac{2 (b d-a e)^3}{e^4 \sqrt{d+e x}}+\frac{2 b^3 (d+e x)^{5/2}}{5 e^4} \]

[Out]

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

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Rubi [A]  time = 0.0331072, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {27, 43} \[ -\frac{2 b^2 (d+e x)^{3/2} (b d-a e)}{e^4}+\frac{6 b \sqrt{d+e x} (b d-a e)^2}{e^4}+\frac{2 (b d-a e)^3}{e^4 \sqrt{d+e x}}+\frac{2 b^3 (d+e x)^{5/2}}{5 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0499887, size = 78, normalized size = 0.83 \[ \frac{2 \left (-5 b^2 (d+e x)^2 (b d-a e)+15 b (d+e x) (b d-a e)^2+5 (b d-a e)^3+b^3 (d+e x)^3\right )}{5 e^4 \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*(b*d - a*e)^3 + 15*b*(b*d - a*e)^2*(d + e*x) - 5*b^2*(b*d - a*e)*(d + e*x)^2 + b^3*(d + e*x)^3))/(5*e^4*
Sqrt[d + e*x])

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Maple [A]  time = 0.006, size = 116, normalized size = 1.2 \begin{align*} -{\frac{-2\,{x}^{3}{b}^{3}{e}^{3}-10\,{x}^{2}a{b}^{2}{e}^{3}+4\,{x}^{2}{b}^{3}d{e}^{2}-30\,x{a}^{2}b{e}^{3}+40\,xa{b}^{2}d{e}^{2}-16\,x{b}^{3}{d}^{2}e+10\,{e}^{3}{a}^{3}-60\,d{e}^{2}{a}^{2}b+80\,a{d}^{2}e{b}^{2}-32\,{d}^{3}{b}^{3}}{5\,{e}^{4}}{\frac{1}{\sqrt{ex+d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(-b^3*e^3*x^3-5*a*b^2*e^3*x^2+2*b^3*d*e^2*x^2-15*a^2*b*e^3*x+20*a*b^2*d*e^2*x-8*b^3*d^2*e*x+5*a^3*e^3-30*
a^2*b*d*e^2+40*a*b^2*d^2*e-16*b^3*d^3)/(e*x+d)^(1/2)/e^4

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Maxima [A]  time = 0.975297, size = 169, normalized size = 1.8 \begin{align*} \frac{2 \,{\left (\frac{{\left (e x + d\right )}^{\frac{5}{2}} b^{3} - 5 \,{\left (b^{3} d - a b^{2} e\right )}{\left (e x + d\right )}^{\frac{3}{2}} + 15 \,{\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )} \sqrt{e x + d}}{e^{3}} + \frac{5 \,{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )}}{\sqrt{e x + d} e^{3}}\right )}}{5 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(((e*x + d)^(5/2)*b^3 - 5*(b^3*d - a*b^2*e)*(e*x + d)^(3/2) + 15*(b^3*d^2 - 2*a*b^2*d*e + a^2*b*e^2)*sqrt(
e*x + d))/e^3 + 5*(b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)/(sqrt(e*x + d)*e^3))/e

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Fricas [A]  time = 1.30665, size = 259, normalized size = 2.76 \begin{align*} \frac{2 \,{\left (b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 40 \, a b^{2} d^{2} e + 30 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3} -{\left (2 \, b^{3} d e^{2} - 5 \, a b^{2} e^{3}\right )} x^{2} +{\left (8 \, b^{3} d^{2} e - 20 \, a b^{2} d e^{2} + 15 \, a^{2} b 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((b*x+a)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^(3/2),x, algorithm="fricas")

[Out]

2/5*(b^3*e^3*x^3 + 16*b^3*d^3 - 40*a*b^2*d^2*e + 30*a^2*b*d*e^2 - 5*a^3*e^3 - (2*b^3*d*e^2 - 5*a*b^2*e^3)*x^2
+ (8*b^3*d^2*e - 20*a*b^2*d*e^2 + 15*a^2*b*e^3)*x)*sqrt(e*x + d)/(e^5*x + d*e^4)

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Sympy [A]  time = 15.3016, size = 109, normalized size = 1.16 \begin{align*} \frac{2 b^{3} \left (d + e x\right )^{\frac{5}{2}}}{5 e^{4}} + \frac{\left (d + e x\right )^{\frac{3}{2}} \left (6 a b^{2} e - 6 b^{3} d\right )}{3 e^{4}} + \frac{\sqrt{d + e x} \left (6 a^{2} b e^{2} - 12 a b^{2} d e + 6 b^{3} d^{2}\right )}{e^{4}} - \frac{2 \left (a e - b d\right )^{3}}{e^{4} \sqrt{d + e x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1434, size = 203, normalized size = 2.16 \begin{align*} \frac{2}{5} \,{\left ({\left (x e + d\right )}^{\frac{5}{2}} b^{3} e^{16} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{3} d e^{16} + 15 \, \sqrt{x e + d} b^{3} d^{2} e^{16} + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{2} e^{17} - 30 \, \sqrt{x e + d} a b^{2} d e^{17} + 15 \, \sqrt{x e + d} a^{2} b e^{18}\right )} e^{\left (-20\right )} + \frac{2 \,{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} 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((b*x+a)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^(3/2),x, algorithm="giac")

[Out]

2/5*((x*e + d)^(5/2)*b^3*e^16 - 5*(x*e + d)^(3/2)*b^3*d*e^16 + 15*sqrt(x*e + d)*b^3*d^2*e^16 + 5*(x*e + d)^(3/
2)*a*b^2*e^17 - 30*sqrt(x*e + d)*a*b^2*d*e^17 + 15*sqrt(x*e + d)*a^2*b*e^18)*e^(-20) + 2*(b^3*d^3 - 3*a*b^2*d^
2*e + 3*a^2*b*d*e^2 - a^3*e^3)*e^(-4)/sqrt(x*e + d)