3.1791 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)}{(d+e x)^{7/2}} \, dx\)

Optimal. Leaf size=124 \[ \frac{2 b (-2 a B e-A b e+3 b B d)}{e^4 \sqrt{d+e x}}-\frac{2 (b d-a e) (-a B e-2 A b e+3 b B d)}{3 e^4 (d+e x)^{3/2}}+\frac{2 (b d-a e)^2 (B d-A e)}{5 e^4 (d+e x)^{5/2}}+\frac{2 b^2 B \sqrt{d+e x}}{e^4} \]

[Out]

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

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Rubi [A]  time = 0.0554085, antiderivative size = 124, 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, 77} \[ \frac{2 b (-2 a B e-A b e+3 b B d)}{e^4 \sqrt{d+e x}}-\frac{2 (b d-a e) (-a B e-2 A b e+3 b B d)}{3 e^4 (d+e x)^{3/2}}+\frac{2 (b d-a e)^2 (B d-A e)}{5 e^4 (d+e x)^{5/2}}+\frac{2 b^2 B \sqrt{d+e x}}{e^4} \]

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)^2*(B*d - A*e))/(5*e^4*(d + e*x)^(5/2)) - (2*(b*d - a*e)*(3*b*B*d - 2*A*b*e - a*B*e))/(3*e^4*(d
+ e*x)^(3/2)) + (2*b*(3*b*B*d - A*b*e - 2*a*B*e))/(e^4*Sqrt[d + e*x]) + (2*b^2*B*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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

Mathematica [A]  time = 0.0990118, size = 107, normalized size = 0.86 \[ \frac{2 \left (15 b (d+e x)^2 (-2 a B e-A b e+3 b B d)-5 (d+e x) (b d-a e) (-a B e-2 A b e+3 b B d)+3 (b d-a e)^2 (B d-A e)+15 b^2 B (d+e x)^3\right )}{15 e^4 (d+e x)^{5/2}} \]

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

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Maple [A]  time = 0.008, size = 169, normalized size = 1.4 \begin{align*} -{\frac{-30\,{b}^{2}B{x}^{3}{e}^{3}+30\,A{b}^{2}{e}^{3}{x}^{2}+60\,Bab{e}^{3}{x}^{2}-180\,B{b}^{2}d{e}^{2}{x}^{2}+20\,Axab{e}^{3}+40\,Ax{b}^{2}d{e}^{2}+10\,Bx{a}^{2}{e}^{3}+80\,Bxabd{e}^{2}-240\,B{b}^{2}{d}^{2}ex+6\,A{a}^{2}{e}^{3}+8\,Aabd{e}^{2}+16\,A{b}^{2}{d}^{2}e+4\,B{a}^{2}d{e}^{2}+32\,Bab{d}^{2}e-96\,{b}^{2}B{d}^{3}}{15\,{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]

-2/15*(-15*B*b^2*e^3*x^3+15*A*b^2*e^3*x^2+30*B*a*b*e^3*x^2-90*B*b^2*d*e^2*x^2+10*A*a*b*e^3*x+20*A*b^2*d*e^2*x+
5*B*a^2*e^3*x+40*B*a*b*d*e^2*x-120*B*b^2*d^2*e*x+3*A*a^2*e^3+4*A*a*b*d*e^2+8*A*b^2*d^2*e+2*B*a^2*d*e^2+16*B*a*
b*d^2*e-48*B*b^2*d^3)/(e*x+d)^(5/2)/e^4

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Maxima [A]  time = 0.985875, size = 221, normalized size = 1.78 \begin{align*} \frac{2 \,{\left (\frac{15 \, \sqrt{e x + d} B b^{2}}{e^{3}} + \frac{3 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} - 3 \,{\left (2 \, B a b + A b^{2}\right )} d^{2} e + 3 \,{\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 15 \,{\left (3 \, B b^{2} d -{\left (2 \, B a b + A b^{2}\right )} e\right )}{\left (e x + d\right )}^{2} - 5 \,{\left (3 \, B b^{2} d^{2} - 2 \,{\left (2 \, B a b + A b^{2}\right )} d e +{\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )}{\left (e x + d\right )}}{{\left (e x + d\right )}^{\frac{5}{2}} e^{3}}\right )}}{15 \, 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]

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

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Fricas [A]  time = 1.32734, size = 400, normalized size = 3.23 \begin{align*} \frac{2 \,{\left (15 \, B b^{2} e^{3} x^{3} + 48 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} - 8 \,{\left (2 \, B a b + A b^{2}\right )} d^{2} e - 2 \,{\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 15 \,{\left (6 \, B b^{2} d e^{2} -{\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 5 \,{\left (24 \, B b^{2} d^{2} e - 4 \,{\left (2 \, B a b + A b^{2}\right )} d e^{2} -{\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x\right )} \sqrt{e x + d}}{15 \,{\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]

2/15*(15*B*b^2*e^3*x^3 + 48*B*b^2*d^3 - 3*A*a^2*e^3 - 8*(2*B*a*b + A*b^2)*d^2*e - 2*(B*a^2 + 2*A*a*b)*d*e^2 +
15*(6*B*b^2*d*e^2 - (2*B*a*b + A*b^2)*e^3)*x^2 + 5*(24*B*b^2*d^2*e - 4*(2*B*a*b + A*b^2)*d*e^2 - (B*a^2 + 2*A*
a*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 = 4.01722, size = 1015, normalized size = 8.19 \begin{align*} \text{result too large to display} \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((-6*A*a**2*e**3/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)
) - 8*A*a*b*d*e**2/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 20*
A*a*b*e**3*x/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 16*A*b**2
*d**2*e/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 40*A*b**2*d*e*
*2*x/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 30*A*b**2*e**3*x*
*2/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 4*B*a**2*d*e**2/(15
*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 10*B*a**2*e**3*x/(15*d**2
*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 32*B*a*b*d**2*e/(15*d**2*e**4*
sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 80*B*a*b*d*e**2*x/(15*d**2*e**4*sqrt
(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 60*B*a*b*e**3*x**2/(15*d**2*e**4*sqrt(d
+ e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) + 96*B*b**2*d**3/(15*d**2*e**4*sqrt(d + e*x)
+ 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) + 240*B*b**2*d**2*e*x/(15*d**2*e**4*sqrt(d + e*x) +
30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) + 180*B*b**2*d*e**2*x**2/(15*d**2*e**4*sqrt(d + e*x) +
30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) + 30*B*b**2*e**3*x**3/(15*d**2*e**4*sqrt(d + e*x) + 3
0*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)), Ne(e, 0)), ((A*a**2*x + A*a*b*x**2 + A*b**2*x**3/3 + B
*a**2*x**2/2 + 2*B*a*b*x**3/3 + B*b**2*x**4/4)/d**(7/2), True))

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Giac [A]  time = 1.13119, size = 273, normalized size = 2.2 \begin{align*} 2 \, \sqrt{x e + d} B b^{2} e^{\left (-4\right )} + \frac{2 \,{\left (45 \,{\left (x e + d\right )}^{2} B b^{2} d - 15 \,{\left (x e + d\right )} B b^{2} d^{2} + 3 \, B b^{2} d^{3} - 30 \,{\left (x e + d\right )}^{2} B a b e - 15 \,{\left (x e + d\right )}^{2} A b^{2} e + 20 \,{\left (x e + d\right )} B a b d e + 10 \,{\left (x e + d\right )} A b^{2} d e - 6 \, B a b d^{2} e - 3 \, A b^{2} d^{2} e - 5 \,{\left (x e + d\right )} B a^{2} e^{2} - 10 \,{\left (x e + d\right )} A a b e^{2} + 3 \, B a^{2} d e^{2} + 6 \, A a b d e^{2} - 3 \, A a^{2} e^{3}\right )} e^{\left (-4\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*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*b^2*e^(-4) + 2/15*(45*(x*e + d)^2*B*b^2*d - 15*(x*e + d)*B*b^2*d^2 + 3*B*b^2*d^3 - 30*(x*e +
d)^2*B*a*b*e - 15*(x*e + d)^2*A*b^2*e + 20*(x*e + d)*B*a*b*d*e + 10*(x*e + d)*A*b^2*d*e - 6*B*a*b*d^2*e - 3*A
*b^2*d^2*e - 5*(x*e + d)*B*a^2*e^2 - 10*(x*e + d)*A*a*b*e^2 + 3*B*a^2*d*e^2 + 6*A*a*b*d*e^2 - 3*A*a^2*e^3)*e^(
-4)/(x*e + d)^(5/2)