### 3.1635 $$\int \frac{(a^2+2 a b x+b^2 x^2)^2}{(d+e x)^{5/2}} \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0415078, 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} $-\frac{8 b^3 (d+e x)^{3/2} (b d-a e)}{3 e^5}+\frac{12 b^2 \sqrt{d+e x} (b d-a e)^2}{e^5}+\frac{8 b (b d-a e)^3}{e^5 \sqrt{d+e x}}-\frac{2 (b d-a e)^4}{3 e^5 (d+e x)^{3/2}}+\frac{2 b^4 (d+e x)^{5/2}}{5 e^5}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^2/(d + e*x)^(5/2),x]

[Out]

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

Mathematica [A]  time = 0.0716295, size = 101, normalized size = 0.81 $\frac{2 \left (90 b^2 (d+e x)^2 (b d-a e)^2-20 b^3 (d+e x)^3 (b d-a e)+60 b (d+e x) (b d-a e)^3-5 (b d-a e)^4+3 b^4 (d+e x)^4\right )}{15 e^5 (d+e x)^{3/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^2/(d + e*x)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.049, size = 186, normalized size = 1.5 \begin{align*} -{\frac{-6\,{x}^{4}{b}^{4}{e}^{4}-40\,{x}^{3}a{b}^{3}{e}^{4}+16\,{x}^{3}{b}^{4}d{e}^{3}-180\,{x}^{2}{a}^{2}{b}^{2}{e}^{4}+240\,{x}^{2}a{b}^{3}d{e}^{3}-96\,{x}^{2}{b}^{4}{d}^{2}{e}^{2}+120\,x{a}^{3}b{e}^{4}-720\,x{a}^{2}{b}^{2}d{e}^{3}+960\,xa{b}^{3}{d}^{2}{e}^{2}-384\,x{b}^{4}{d}^{3}e+10\,{a}^{4}{e}^{4}+80\,{a}^{3}bd{e}^{3}-480\,{d}^{2}{e}^{2}{a}^{2}{b}^{2}+640\,a{b}^{3}{d}^{3}e-256\,{b}^{4}{d}^{4}}{15\,{e}^{5}} \left ( ex+d \right ) ^{-{\frac{3}{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)^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.12195, size = 252, normalized size = 2.02 \begin{align*} \frac{2 \,{\left (\frac{3 \,{\left (e x + d\right )}^{\frac{5}{2}} b^{4} - 20 \,{\left (b^{4} d - a b^{3} e\right )}{\left (e x + d\right )}^{\frac{3}{2}} + 90 \,{\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} \sqrt{e x + d}}{e^{4}} - \frac{5 \,{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4} - 12 \,{\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 )}\right )}}{{\left (e x + d\right )}^{\frac{3}{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)^(5/2),x, algorithm="maxima")

[Out]

2/15*((3*(e*x + d)^(5/2)*b^4 - 20*(b^4*d - a*b^3*e)*(e*x + d)^(3/2) + 90*(b^4*d^2 - 2*a*b^3*d*e + a^2*b^2*e^2)
*sqrt(e*x + d))/e^4 - 5*(b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4 - 12*(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)^(3/2)*e^4))/e

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Fricas [A]  time = 1.61883, size = 431, normalized size = 3.45 \begin{align*} \frac{2 \,{\left (3 \, b^{4} e^{4} x^{4} + 128 \, b^{4} d^{4} - 320 \, a b^{3} d^{3} e + 240 \, a^{2} b^{2} d^{2} e^{2} - 40 \, a^{3} b d e^{3} - 5 \, a^{4} e^{4} - 4 \,{\left (2 \, b^{4} d e^{3} - 5 \, a b^{3} e^{4}\right )} x^{3} + 6 \,{\left (8 \, b^{4} d^{2} e^{2} - 20 \, a b^{3} d e^{3} + 15 \, a^{2} b^{2} e^{4}\right )} x^{2} + 12 \,{\left (16 \, b^{4} d^{3} e - 40 \, a b^{3} d^{2} e^{2} + 30 \, a^{2} b^{2} d e^{3} - 5 \, a^{3} b e^{4}\right )} x\right )} \sqrt{e x + d}}{15 \,{\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} 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)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*b^4*e^4*x^4 + 128*b^4*d^4 - 320*a*b^3*d^3*e + 240*a^2*b^2*d^2*e^2 - 40*a^3*b*d*e^3 - 5*a^4*e^4 - 4*(2*
b^4*d*e^3 - 5*a*b^3*e^4)*x^3 + 6*(8*b^4*d^2*e^2 - 20*a*b^3*d*e^3 + 15*a^2*b^2*e^4)*x^2 + 12*(16*b^4*d^3*e - 40
*a*b^3*d^2*e^2 + 30*a^2*b^2*d*e^3 - 5*a^3*b*e^4)*x)*sqrt(e*x + d)/(e^7*x^2 + 2*d*e^6*x + d^2*e^5)

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Sympy [A]  time = 40.3786, size = 136, normalized size = 1.09 \begin{align*} \frac{2 b^{4} \left (d + e x\right )^{\frac{5}{2}}}{5 e^{5}} - \frac{8 b \left (a e - b d\right )^{3}}{e^{5} \sqrt{d + e x}} + \frac{\left (d + e x\right )^{\frac{3}{2}} \left (8 a b^{3} e - 8 b^{4} d\right )}{3 e^{5}} + \frac{\sqrt{d + e x} \left (12 a^{2} b^{2} e^{2} - 24 a b^{3} d e + 12 b^{4} d^{2}\right )}{e^{5}} - \frac{2 \left (a e - b d\right )^{4}}{3 e^{5} \left (d + e x\right )^{\frac{3}{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)**(5/2),x)

[Out]

2*b**4*(d + e*x)**(5/2)/(5*e**5) - 8*b*(a*e - b*d)**3/(e**5*sqrt(d + e*x)) + (d + e*x)**(3/2)*(8*a*b**3*e - 8*
b**4*d)/(3*e**5) + sqrt(d + e*x)*(12*a**2*b**2*e**2 - 24*a*b**3*d*e + 12*b**4*d**2)/e**5 - 2*(a*e - b*d)**4/(3
*e**5*(d + e*x)**(3/2))

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Giac [B]  time = 1.15454, size = 309, normalized size = 2.47 \begin{align*} \frac{2}{15} \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{4} e^{20} - 20 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{4} d e^{20} + 90 \, \sqrt{x e + d} b^{4} d^{2} e^{20} + 20 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{3} e^{21} - 180 \, \sqrt{x e + d} a b^{3} d e^{21} + 90 \, \sqrt{x e + d} a^{2} b^{2} e^{22}\right )} e^{\left (-25\right )} + \frac{2 \,{\left (12 \,{\left (x e + d\right )} b^{4} d^{3} - b^{4} d^{4} - 36 \,{\left (x e + d\right )} a b^{3} d^{2} e + 4 \, a b^{3} d^{3} e + 36 \,{\left (x e + d\right )} a^{2} b^{2} d e^{2} - 6 \, a^{2} b^{2} d^{2} e^{2} - 12 \,{\left (x e + d\right )} a^{3} b e^{3} + 4 \, a^{3} b d e^{3} - a^{4} e^{4}\right )} e^{\left (-5\right )}}{3 \,{\left (x e + d\right )}^{\frac{3}{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)^(5/2),x, algorithm="giac")

[Out]

2/15*(3*(x*e + d)^(5/2)*b^4*e^20 - 20*(x*e + d)^(3/2)*b^4*d*e^20 + 90*sqrt(x*e + d)*b^4*d^2*e^20 + 20*(x*e + d
)^(3/2)*a*b^3*e^21 - 180*sqrt(x*e + d)*a*b^3*d*e^21 + 90*sqrt(x*e + d)*a^2*b^2*e^22)*e^(-25) + 2/3*(12*(x*e +
d)*b^4*d^3 - b^4*d^4 - 36*(x*e + d)*a*b^3*d^2*e + 4*a*b^3*d^3*e + 36*(x*e + d)*a^2*b^2*d*e^2 - 6*a^2*b^2*d^2*e
^2 - 12*(x*e + d)*a^3*b*e^3 + 4*a^3*b*d*e^3 - a^4*e^4)*e^(-5)/(x*e + d)^(3/2)