3.1636 \(\int \frac{(a^2+2 a b x+b^2 x^2)^2}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=125 \[ -\frac{8 b^3 \sqrt{d+e x} (b d-a e)}{e^5}-\frac{12 b^2 (b d-a e)^2}{e^5 \sqrt{d+e x}}+\frac{8 b (b d-a e)^3}{3 e^5 (d+e x)^{3/2}}-\frac{2 (b d-a e)^4}{5 e^5 (d+e x)^{5/2}}+\frac{2 b^4 (d+e x)^{3/2}}{3 e^5} \]

[Out]

(-2*(b*d - a*e)^4)/(5*e^5*(d + e*x)^(5/2)) + (8*b*(b*d - a*e)^3)/(3*e^5*(d + e*x)^(3/2)) - (12*b^2*(b*d - a*e)
^2)/(e^5*Sqrt[d + e*x]) - (8*b^3*(b*d - a*e)*Sqrt[d + e*x])/e^5 + (2*b^4*(d + e*x)^(3/2))/(3*e^5)

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Rubi [A]  time = 0.0430044, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {27, 43} \[ -\frac{8 b^3 \sqrt{d+e x} (b d-a e)}{e^5}-\frac{12 b^2 (b d-a e)^2}{e^5 \sqrt{d+e x}}+\frac{8 b (b d-a e)^3}{3 e^5 (d+e x)^{3/2}}-\frac{2 (b d-a e)^4}{5 e^5 (d+e x)^{5/2}}+\frac{2 b^4 (d+e x)^{3/2}}{3 e^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*d - a*e)^4)/(5*e^5*(d + e*x)^(5/2)) + (8*b*(b*d - a*e)^3)/(3*e^5*(d + e*x)^(3/2)) - (12*b^2*(b*d - a*e)
^2)/(e^5*Sqrt[d + e*x]) - (8*b^3*(b*d - a*e)*Sqrt[d + e*x])/e^5 + (2*b^4*(d + e*x)^(3/2))/(3*e^5)

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{\left (a^2+2 a b x+b^2 x^2\right )^2}{(d+e x)^{7/2}} \, dx &=\int \frac{(a+b x)^4}{(d+e x)^{7/2}} \, dx\\ &=\int \left (\frac{(-b d+a e)^4}{e^4 (d+e x)^{7/2}}-\frac{4 b (b d-a e)^3}{e^4 (d+e x)^{5/2}}+\frac{6 b^2 (b d-a e)^2}{e^4 (d+e x)^{3/2}}-\frac{4 b^3 (b d-a e)}{e^4 \sqrt{d+e x}}+\frac{b^4 \sqrt{d+e x}}{e^4}\right ) \, dx\\ &=-\frac{2 (b d-a e)^4}{5 e^5 (d+e x)^{5/2}}+\frac{8 b (b d-a e)^3}{3 e^5 (d+e x)^{3/2}}-\frac{12 b^2 (b d-a e)^2}{e^5 \sqrt{d+e x}}-\frac{8 b^3 (b d-a e) \sqrt{d+e x}}{e^5}+\frac{2 b^4 (d+e x)^{3/2}}{3 e^5}\\ \end{align*}

Mathematica [A]  time = 0.0715284, size = 101, normalized size = 0.81 \[ \frac{2 \left (-90 b^2 (d+e x)^2 (b d-a e)^2-60 b^3 (d+e x)^3 (b d-a e)+20 b (d+e x) (b d-a e)^3-3 (b d-a e)^4+5 b^4 (d+e x)^4\right )}{15 e^5 (d+e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-3*(b*d - a*e)^4 + 20*b*(b*d - a*e)^3*(d + e*x) - 90*b^2*(b*d - a*e)^2*(d + e*x)^2 - 60*b^3*(b*d - a*e)*(d
+ e*x)^3 + 5*b^4*(d + e*x)^4))/(15*e^5*(d + e*x)^(5/2))

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Maple [A]  time = 0.046, size = 186, normalized size = 1.5 \begin{align*} -{\frac{-10\,{x}^{4}{b}^{4}{e}^{4}-120\,{x}^{3}a{b}^{3}{e}^{4}+80\,{x}^{3}{b}^{4}d{e}^{3}+180\,{x}^{2}{a}^{2}{b}^{2}{e}^{4}-720\,{x}^{2}a{b}^{3}d{e}^{3}+480\,{x}^{2}{b}^{4}{d}^{2}{e}^{2}+40\,x{a}^{3}b{e}^{4}+240\,x{a}^{2}{b}^{2}d{e}^{3}-960\,xa{b}^{3}{d}^{2}{e}^{2}+640\,x{b}^{4}{d}^{3}e+6\,{a}^{4}{e}^{4}+16\,{a}^{3}bd{e}^{3}+96\,{d}^{2}{e}^{2}{a}^{2}{b}^{2}-384\,a{b}^{3}{d}^{3}e+256\,{b}^{4}{d}^{4}}{15\,{e}^{5}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^(7/2),x)

[Out]

-2/15*(-5*b^4*e^4*x^4-60*a*b^3*e^4*x^3+40*b^4*d*e^3*x^3+90*a^2*b^2*e^4*x^2-360*a*b^3*d*e^3*x^2+240*b^4*d^2*e^2
*x^2+20*a^3*b*e^4*x+120*a^2*b^2*d*e^3*x-480*a*b^3*d^2*e^2*x+320*b^4*d^3*e*x+3*a^4*e^4+8*a^3*b*d*e^3+48*a^2*b^2
*d^2*e^2-192*a*b^3*d^3*e+128*b^4*d^4)/(e*x+d)^(5/2)/e^5

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Maxima [A]  time = 1.21617, size = 255, normalized size = 2.04 \begin{align*} \frac{2 \,{\left (\frac{5 \,{\left ({\left (e x + d\right )}^{\frac{3}{2}} b^{4} - 12 \,{\left (b^{4} d - a b^{3} e\right )} \sqrt{e x + d}\right )}}{e^{4}} - \frac{3 \, b^{4} d^{4} - 12 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 12 \, a^{3} b d e^{3} + 3 \, a^{4} e^{4} + 90 \,{\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )}{\left (e x + d\right )}^{2} - 20 \,{\left (b^{4} d^{3} - 3 \, a b^{3} d^{2} e + 3 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}\right )}{\left (e x + d\right )}}{{\left (e x + d\right )}^{\frac{5}{2}} e^{4}}\right )}}{15 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

2/15*(5*((e*x + d)^(3/2)*b^4 - 12*(b^4*d - a*b^3*e)*sqrt(e*x + d))/e^4 - (3*b^4*d^4 - 12*a*b^3*d^3*e + 18*a^2*
b^2*d^2*e^2 - 12*a^3*b*d*e^3 + 3*a^4*e^4 + 90*(b^4*d^2 - 2*a*b^3*d*e + a^2*b^2*e^2)*(e*x + d)^2 - 20*(b^4*d^3
- 3*a*b^3*d^2*e + 3*a^2*b^2*d*e^2 - a^3*b*e^3)*(e*x + d))/((e*x + d)^(5/2)*e^4))/e

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Fricas [A]  time = 1.49603, size = 447, normalized size = 3.58 \begin{align*} \frac{2 \,{\left (5 \, b^{4} e^{4} x^{4} - 128 \, b^{4} d^{4} + 192 \, a b^{3} d^{3} e - 48 \, a^{2} b^{2} d^{2} e^{2} - 8 \, a^{3} b d e^{3} - 3 \, a^{4} e^{4} - 20 \,{\left (2 \, b^{4} d e^{3} - 3 \, a b^{3} e^{4}\right )} x^{3} - 30 \,{\left (8 \, b^{4} d^{2} e^{2} - 12 \, a b^{3} d e^{3} + 3 \, a^{2} b^{2} e^{4}\right )} x^{2} - 20 \,{\left (16 \, b^{4} d^{3} e - 24 \, a b^{3} d^{2} e^{2} + 6 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \sqrt{e x + d}}{15 \,{\left (e^{8} x^{3} + 3 \, d e^{7} x^{2} + 3 \, d^{2} e^{6} x + d^{3} e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

2/15*(5*b^4*e^4*x^4 - 128*b^4*d^4 + 192*a*b^3*d^3*e - 48*a^2*b^2*d^2*e^2 - 8*a^3*b*d*e^3 - 3*a^4*e^4 - 20*(2*b
^4*d*e^3 - 3*a*b^3*e^4)*x^3 - 30*(8*b^4*d^2*e^2 - 12*a*b^3*d*e^3 + 3*a^2*b^2*e^4)*x^2 - 20*(16*b^4*d^3*e - 24*
a*b^3*d^2*e^2 + 6*a^2*b^2*d*e^3 + a^3*b*e^4)*x)*sqrt(e*x + d)/(e^8*x^3 + 3*d*e^7*x^2 + 3*d^2*e^6*x + d^3*e^5)

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Sympy [A]  time = 4.30085, size = 1008, normalized size = 8.06 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**2/(e*x+d)**(7/2),x)

[Out]

Piecewise((-6*a**4*e**4/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x))
- 16*a**3*b*d*e**3/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) - 40*
a**3*b*e**4*x/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) - 96*a**2*
b**2*d**2*e**2/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) - 240*a**
2*b**2*d*e**3*x/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) - 180*a*
*2*b**2*e**4*x**2/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) + 384*
a*b**3*d**3*e/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) + 960*a*b*
*3*d**2*e**2*x/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) + 720*a*b
**3*d*e**3*x**2/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) + 120*a*
b**3*e**4*x**3/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) - 256*b**
4*d**4/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) - 640*b**4*d**3*e
*x/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) - 480*b**4*d**2*e**2*
x**2/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) - 80*b**4*d*e**3*x*
*3/(15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)) + 10*b**4*e**4*x**4/(
15*d**2*e**5*sqrt(d + e*x) + 30*d*e**6*x*sqrt(d + e*x) + 15*e**7*x**2*sqrt(d + e*x)), Ne(e, 0)), ((a**4*x + 2*
a**3*b*x**2 + 2*a**2*b**2*x**3 + a*b**3*x**4 + b**4*x**5/5)/d**(7/2), True))

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Giac [B]  time = 1.24122, size = 305, normalized size = 2.44 \begin{align*} \frac{2}{3} \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} b^{4} e^{10} - 12 \, \sqrt{x e + d} b^{4} d e^{10} + 12 \, \sqrt{x e + d} a b^{3} e^{11}\right )} e^{\left (-15\right )} - \frac{2 \,{\left (90 \,{\left (x e + d\right )}^{2} b^{4} d^{2} - 20 \,{\left (x e + d\right )} b^{4} d^{3} + 3 \, b^{4} d^{4} - 180 \,{\left (x e + d\right )}^{2} a b^{3} d e + 60 \,{\left (x e + d\right )} a b^{3} d^{2} e - 12 \, a b^{3} d^{3} e + 90 \,{\left (x e + d\right )}^{2} a^{2} b^{2} e^{2} - 60 \,{\left (x e + d\right )} a^{2} b^{2} d e^{2} + 18 \, a^{2} b^{2} d^{2} e^{2} + 20 \,{\left (x e + d\right )} a^{3} b e^{3} - 12 \, a^{3} b d e^{3} + 3 \, a^{4} e^{4}\right )} e^{\left (-5\right )}}{15 \,{\left (x e + d\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

2/3*((x*e + d)^(3/2)*b^4*e^10 - 12*sqrt(x*e + d)*b^4*d*e^10 + 12*sqrt(x*e + d)*a*b^3*e^11)*e^(-15) - 2/15*(90*
(x*e + d)^2*b^4*d^2 - 20*(x*e + d)*b^4*d^3 + 3*b^4*d^4 - 180*(x*e + d)^2*a*b^3*d*e + 60*(x*e + d)*a*b^3*d^2*e
- 12*a*b^3*d^3*e + 90*(x*e + d)^2*a^2*b^2*e^2 - 60*(x*e + d)*a^2*b^2*d*e^2 + 18*a^2*b^2*d^2*e^2 + 20*(x*e + d)
*a^3*b*e^3 - 12*a^3*b*d*e^3 + 3*a^4*e^4)*e^(-5)/(x*e + d)^(5/2)