3.2049 \(\int \frac{(a+b x) (a^2+2 a b x+b^2 x^2)}{(d+e x)^{7/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0324152, 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} \[ \frac{6 b^2 (b d-a e)}{e^4 \sqrt{d+e x}}-\frac{2 b (b d-a e)^2}{e^4 (d+e x)^{3/2}}+\frac{2 (b d-a e)^3}{5 e^4 (d+e x)^{5/2}}+\frac{2 b^3 \sqrt{d+e x}}{e^4} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2))/(d + e*x)^(7/2),x]

[Out]

(2*(b*d - a*e)^3)/(5*e^4*(d + e*x)^(5/2)) - (2*b*(b*d - a*e)^2)/(e^4*(d + e*x)^(3/2)) + (6*b^2*(b*d - a*e))/(e
^4*Sqrt[d + e*x]) + (2*b^3*Sqrt[d + e*x])/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)^{7/2}} \, dx &=\int \frac{(a+b x)^3}{(d+e x)^{7/2}} \, dx\\ &=\int \left (\frac{(-b d+a e)^3}{e^3 (d+e x)^{7/2}}+\frac{3 b (b d-a e)^2}{e^3 (d+e x)^{5/2}}-\frac{3 b^2 (b d-a e)}{e^3 (d+e x)^{3/2}}+\frac{b^3}{e^3 \sqrt{d+e x}}\right ) \, dx\\ &=\frac{2 (b d-a e)^3}{5 e^4 (d+e x)^{5/2}}-\frac{2 b (b d-a e)^2}{e^4 (d+e x)^{3/2}}+\frac{6 b^2 (b d-a e)}{e^4 \sqrt{d+e x}}+\frac{2 b^3 \sqrt{d+e x}}{e^4}\\ \end{align*}

Mathematica [A]  time = 0.0521089, size = 77, normalized size = 0.82 \[ \frac{2 \left (15 b^2 (d+e x)^2 (b d-a e)-5 b (d+e x) (b d-a e)^2+(b d-a e)^3+5 b^3 (d+e x)^3\right )}{5 e^4 (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2))/(d + e*x)^(7/2),x]

[Out]

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

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Maple [A]  time = 0.007, size = 115, normalized size = 1.2 \begin{align*} -{\frac{-10\,{x}^{3}{b}^{3}{e}^{3}+30\,{x}^{2}a{b}^{2}{e}^{3}-60\,{x}^{2}{b}^{3}d{e}^{2}+10\,x{a}^{2}b{e}^{3}+40\,xa{b}^{2}d{e}^{2}-80\,x{b}^{3}{d}^{2}e+2\,{e}^{3}{a}^{3}+4\,d{e}^{2}{a}^{2}b+16\,a{d}^{2}e{b}^{2}-32\,{d}^{3}{b}^{3}}{5\,{e}^{4}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}} \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)^(7/2),x)

[Out]

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

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Maxima [A]  time = 0.977622, size = 163, normalized size = 1.73 \begin{align*} \frac{2 \,{\left (\frac{5 \, \sqrt{e x + d} b^{3}}{e^{3}} + \frac{b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3} + 15 \,{\left (b^{3} d - a b^{2} e\right )}{\left (e x + d\right )}^{2} - 5 \,{\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )}{\left (e x + d\right )}}{{\left (e x + d\right )}^{\frac{5}{2}} 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)^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.2802, size = 298, normalized size = 3.17 \begin{align*} \frac{2 \,{\left (5 \, b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 8 \, a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} - a^{3} e^{3} + 15 \,{\left (2 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 5 \,{\left (8 \, b^{3} d^{2} e - 4 \, a b^{2} d e^{2} - a^{2} b e^{3}\right )} x\right )} \sqrt{e x + d}}{5 \,{\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} 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)^(7/2),x, algorithm="fricas")

[Out]

2/5*(5*b^3*e^3*x^3 + 16*b^3*d^3 - 8*a*b^2*d^2*e - 2*a^2*b*d*e^2 - a^3*e^3 + 15*(2*b^3*d*e^2 - a*b^2*e^3)*x^2 +
5*(8*b^3*d^2*e - 4*a*b^2*d*e^2 - a^2*b*e^3)*x)*sqrt(e*x + d)/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)

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Sympy [A]  time = 3.95412, size = 665, normalized size = 7.07 \begin{align*} \begin{cases} - \frac{2 a^{3} e^{3}}{5 d^{2} e^{4} \sqrt{d + e x} + 10 d e^{5} x \sqrt{d + e x} + 5 e^{6} x^{2} \sqrt{d + e x}} - \frac{4 a^{2} b d e^{2}}{5 d^{2} e^{4} \sqrt{d + e x} + 10 d e^{5} x \sqrt{d + e x} + 5 e^{6} x^{2} \sqrt{d + e x}} - \frac{10 a^{2} b e^{3} x}{5 d^{2} e^{4} \sqrt{d + e x} + 10 d e^{5} x \sqrt{d + e x} + 5 e^{6} x^{2} \sqrt{d + e x}} - \frac{16 a b^{2} d^{2} e}{5 d^{2} e^{4} \sqrt{d + e x} + 10 d e^{5} x \sqrt{d + e x} + 5 e^{6} x^{2} \sqrt{d + e x}} - \frac{40 a b^{2} d e^{2} x}{5 d^{2} e^{4} \sqrt{d + e x} + 10 d e^{5} x \sqrt{d + e x} + 5 e^{6} x^{2} \sqrt{d + e x}} - \frac{30 a b^{2} e^{3} x^{2}}{5 d^{2} e^{4} \sqrt{d + e x} + 10 d e^{5} x \sqrt{d + e x} + 5 e^{6} x^{2} \sqrt{d + e x}} + \frac{32 b^{3} d^{3}}{5 d^{2} e^{4} \sqrt{d + e x} + 10 d e^{5} x \sqrt{d + e x} + 5 e^{6} x^{2} \sqrt{d + e x}} + \frac{80 b^{3} d^{2} e x}{5 d^{2} e^{4} \sqrt{d + e x} + 10 d e^{5} x \sqrt{d + e x} + 5 e^{6} x^{2} \sqrt{d + e x}} + \frac{60 b^{3} d e^{2} x^{2}}{5 d^{2} e^{4} \sqrt{d + e x} + 10 d e^{5} x \sqrt{d + e x} + 5 e^{6} x^{2} \sqrt{d + e x}} + \frac{10 b^{3} e^{3} x^{3}}{5 d^{2} e^{4} \sqrt{d + e x} + 10 d e^{5} x \sqrt{d + e x} + 5 e^{6} x^{2} \sqrt{d + e x}} & \text{for}\: e \neq 0 \\\frac{a^{3} x + \frac{3 a^{2} b x^{2}}{2} + a b^{2} x^{3} + \frac{b^{3} x^{4}}{4}}{d^{\frac{7}{2}}} & \text{otherwise} \end{cases} \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)**(7/2),x)

[Out]

Piecewise((-2*a**3*e**3/(5*d**2*e**4*sqrt(d + e*x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) -
4*a**2*b*d*e**2/(5*d**2*e**4*sqrt(d + e*x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) - 10*a**2*
b*e**3*x/(5*d**2*e**4*sqrt(d + e*x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) - 16*a*b**2*d**2*
e/(5*d**2*e**4*sqrt(d + e*x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) - 40*a*b**2*d*e**2*x/(5*
d**2*e**4*sqrt(d + e*x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) - 30*a*b**2*e**3*x**2/(5*d**2
*e**4*sqrt(d + e*x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) + 32*b**3*d**3/(5*d**2*e**4*sqrt(
d + e*x) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) + 80*b**3*d**2*e*x/(5*d**2*e**4*sqrt(d + e*x
) + 10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) + 60*b**3*d*e**2*x**2/(5*d**2*e**4*sqrt(d + e*x) +
10*d*e**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)) + 10*b**3*e**3*x**3/(5*d**2*e**4*sqrt(d + e*x) + 10*d*e
**5*x*sqrt(d + e*x) + 5*e**6*x**2*sqrt(d + e*x)), Ne(e, 0)), ((a**3*x + 3*a**2*b*x**2/2 + a*b**2*x**3 + b**3*x
**4/4)/d**(7/2), True))

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Giac [A]  time = 1.10535, size = 184, normalized size = 1.96 \begin{align*} 2 \, \sqrt{x e + d} b^{3} e^{\left (-4\right )} + \frac{2 \,{\left (15 \,{\left (x e + d\right )}^{2} b^{3} d - 5 \,{\left (x e + d\right )} b^{3} d^{2} + b^{3} d^{3} - 15 \,{\left (x e + d\right )}^{2} a b^{2} e + 10 \,{\left (x e + d\right )} a b^{2} d e - 3 \, a b^{2} d^{2} e - 5 \,{\left (x e + d\right )} a^{2} b e^{2} + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} e^{\left (-4\right )}}{5 \,{\left (x e + d\right )}^{\frac{5}{2}}} \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)^(7/2),x, algorithm="giac")

[Out]

2*sqrt(x*e + d)*b^3*e^(-4) + 2/5*(15*(x*e + d)^2*b^3*d - 5*(x*e + d)*b^3*d^2 + b^3*d^3 - 15*(x*e + d)^2*a*b^2*
e + 10*(x*e + d)*a*b^2*d*e - 3*a*b^2*d^2*e - 5*(x*e + d)*a^2*b*e^2 + 3*a^2*b*d*e^2 - a^3*e^3)*e^(-4)/(x*e + d)
^(5/2)