3.2482 \(\int \frac{(A+B x) (d+e x)^2}{(a+b x+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=121 \[ -\frac{2 (d+e x)^2 (-2 a B-x (b B-2 A c)+A b)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac{8 (-2 a e+x (2 c d-b e)+b d) (-2 a B e+A b e-2 A c d+b B d)}{3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}} \]

[Out]

(-2*(A*b - 2*a*B - (b*B - 2*A*c)*x)*(d + e*x)^2)/(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) - (8*(b*B*d - 2*A*c
*d + A*b*e - 2*a*B*e)*(b*d - 2*a*e + (2*c*d - b*e)*x))/(3*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2])

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Rubi [A]  time = 0.067421, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {804, 636} \[ -\frac{2 (d+e x)^2 (-2 a B-x (b B-2 A c)+A b)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}-\frac{8 (-2 a e+x (2 c d-b e)+b d) (-2 a B e+A b e-2 A c d+b B d)}{3 \left (b^2-4 a c\right )^2 \sqrt{a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(A*b - 2*a*B - (b*B - 2*A*c)*x)*(d + e*x)^2)/(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) - (8*(b*B*d - 2*A*c
*d + A*b*e - 2*a*B*e)*(b*d - 2*a*e + (2*c*d - b*e)*x))/(3*(b^2 - 4*a*c)^2*Sqrt[a + b*x + c*x^2])

Rule 804

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*(b*f - 2*a*g + (2*c*f - b*g)*x))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[(m
*(b*(e*f + d*g) - 2*(c*d*f + a*e*g)))/((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)
, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0] && LtQ[p, -1]

Rule 636

Int[((d_.) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-2*(b*d - 2*a*e + (2*c*
d - b*e)*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] &&
NeQ[b^2 - 4*a*c, 0]

Rubi steps

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

Mathematica [B]  time = 1.13625, size = 314, normalized size = 2.6 \[ \frac{2 A \left (4 b \left (2 a^2 e^2+3 a c (d-e x)^2+2 c^2 d x^2 (3 d-2 e x)\right )+8 c \left (-2 a^2 d e+a c x \left (3 d^2+e^2 x^2\right )+2 c^2 d^2 x^3\right )+b^2 \left (2 c x \left (3 d^2-12 d e x+e^2 x^2\right )-4 a e (d-3 e x)\right )+b^3 \left (-\left (d^2+6 d e x-3 e^2 x^2\right )\right )\right )-2 B \left (8 a^2 \left (b e (3 e x-2 d)+c \left (d^2+3 e^2 x^2\right )\right )+16 a^3 e^2+2 a \left (b^2 \left (d^2-12 d e x+3 e^2 x^2\right )+6 b c x (d-e x)^2-8 c^2 d e x^3\right )+b x \left (b^2 \left (3 d^2-6 d e x-e^2 x^2\right )+4 b c d x (3 d-e x)+8 c^2 d^2 x^2\right )\right )}{3 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*(-(b^3*(d^2 + 6*d*e*x - 3*e^2*x^2)) + 4*b*(2*a^2*e^2 + 2*c^2*d*x^2*(3*d - 2*e*x) + 3*a*c*(d - e*x)^2) + 8
*c*(-2*a^2*d*e + 2*c^2*d^2*x^3 + a*c*x*(3*d^2 + e^2*x^2)) + b^2*(-4*a*e*(d - 3*e*x) + 2*c*x*(3*d^2 - 12*d*e*x
+ e^2*x^2))) - 2*B*(16*a^3*e^2 + b*x*(8*c^2*d^2*x^2 + 4*b*c*d*x*(3*d - e*x) + b^2*(3*d^2 - 6*d*e*x - e^2*x^2))
+ 8*a^2*(b*e*(-2*d + 3*e*x) + c*(d^2 + 3*e^2*x^2)) + 2*a*(-8*c^2*d*e*x^3 + 6*b*c*x*(d - e*x)^2 + b^2*(d^2 - 1
2*d*e*x + 3*e^2*x^2))))/(3*(b^2 - 4*a*c)^2*(a + x*(b + c*x))^(3/2))

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Maple [B]  time = 0.01, size = 433, normalized size = 3.6 \begin{align*}{\frac{16\,Aa{c}^{2}{e}^{2}{x}^{3}+4\,A{b}^{2}c{e}^{2}{x}^{3}-32\,Ab{c}^{2}de{x}^{3}+32\,A{c}^{3}{d}^{2}{x}^{3}-24\,Babc{e}^{2}{x}^{3}+32\,Ba{c}^{2}de{x}^{3}+2\,B{x}^{3}{b}^{3}{e}^{2}+8\,B{b}^{2}cde{x}^{3}-16\,Bb{c}^{2}{d}^{2}{x}^{3}+24\,Aabc{e}^{2}{x}^{2}+6\,A{x}^{2}{b}^{3}{e}^{2}-48\,A{b}^{2}cde{x}^{2}+48\,Ab{c}^{2}{d}^{2}{x}^{2}-48\,B{a}^{2}c{e}^{2}{x}^{2}-12\,B{x}^{2}a{b}^{2}{e}^{2}+48\,Babcde{x}^{2}+12\,B{x}^{2}{b}^{3}de-24\,B{b}^{2}c{d}^{2}{x}^{2}+24\,Axa{b}^{2}{e}^{2}-48\,Aabcdex+48\,Aa{c}^{2}{d}^{2}x-12\,Ax{b}^{3}de+12\,A{b}^{2}c{d}^{2}x-48\,Bx{a}^{2}b{e}^{2}+48\,Bxa{b}^{2}de-24\,Babc{d}^{2}x-6\,Bx{b}^{3}{d}^{2}+16\,A{a}^{2}b{e}^{2}-32\,A{a}^{2}cde-8\,Aa{b}^{2}de+24\,Aabc{d}^{2}-2\,A{b}^{3}{d}^{2}-32\,B{e}^{2}{a}^{3}+32\,B{a}^{2}bde-16\,B{a}^{2}c{d}^{2}-4\,Ba{b}^{2}{d}^{2}}{48\,{a}^{2}{c}^{2}-24\,a{b}^{2}c+3\,{b}^{4}} \left ( c{x}^{2}+bx+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^2/(c*x^2+b*x+a)^(5/2),x)

[Out]

2/3/(c*x^2+b*x+a)^(3/2)*(8*A*a*c^2*e^2*x^3+2*A*b^2*c*e^2*x^3-16*A*b*c^2*d*e*x^3+16*A*c^3*d^2*x^3-12*B*a*b*c*e^
2*x^3+16*B*a*c^2*d*e*x^3+B*b^3*e^2*x^3+4*B*b^2*c*d*e*x^3-8*B*b*c^2*d^2*x^3+12*A*a*b*c*e^2*x^2+3*A*b^3*e^2*x^2-
24*A*b^2*c*d*e*x^2+24*A*b*c^2*d^2*x^2-24*B*a^2*c*e^2*x^2-6*B*a*b^2*e^2*x^2+24*B*a*b*c*d*e*x^2+6*B*b^3*d*e*x^2-
12*B*b^2*c*d^2*x^2+12*A*a*b^2*e^2*x-24*A*a*b*c*d*e*x+24*A*a*c^2*d^2*x-6*A*b^3*d*e*x+6*A*b^2*c*d^2*x-24*B*a^2*b
*e^2*x+24*B*a*b^2*d*e*x-12*B*a*b*c*d^2*x-3*B*b^3*d^2*x+8*A*a^2*b*e^2-16*A*a^2*c*d*e-4*A*a*b^2*d*e+12*A*a*b*c*d
^2-A*b^3*d^2-16*B*a^3*e^2+16*B*a^2*b*d*e-8*B*a^2*c*d^2-2*B*a*b^2*d^2)/(16*a^2*c^2-8*a*b^2*c+b^4)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^2/(c*x^2+b*x+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 71.1859, size = 984, normalized size = 8.13 \begin{align*} -\frac{2 \,{\left ({\left (8 \,{\left (B b c^{2} - 2 \, A c^{3}\right )} d^{2} - 4 \,{\left (B b^{2} c + 4 \,{\left (B a - A b\right )} c^{2}\right )} d e -{\left (B b^{3} + 8 \, A a c^{2} - 2 \,{\left (6 \, B a b - A b^{2}\right )} c\right )} e^{2}\right )} x^{3} +{\left (2 \, B a b^{2} + A b^{3} + 4 \,{\left (2 \, B a^{2} - 3 \, A a b\right )} c\right )} d^{2} - 4 \,{\left (4 \, B a^{2} b - A a b^{2} - 4 \, A a^{2} c\right )} d e + 8 \,{\left (2 \, B a^{3} - A a^{2} b\right )} e^{2} + 3 \,{\left (4 \,{\left (B b^{2} c - 2 \, A b c^{2}\right )} d^{2} - 2 \,{\left (B b^{3} + 4 \,{\left (B a b - A b^{2}\right )} c\right )} d e +{\left (2 \, B a b^{2} - A b^{3} + 4 \,{\left (2 \, B a^{2} - A a b\right )} c\right )} e^{2}\right )} x^{2} + 3 \,{\left ({\left (B b^{3} - 8 \, A a c^{2} + 2 \,{\left (2 \, B a b - A b^{2}\right )} c\right )} d^{2} - 2 \,{\left (4 \, B a b^{2} - A b^{3} - 4 \, A a b c\right )} d e + 4 \,{\left (2 \, B a^{2} b - A a b^{2}\right )} e^{2}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{3 \,{\left (a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} +{\left (b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}\right )} x^{4} + 2 \,{\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} +{\left (b^{6} - 6 \, a b^{4} c + 32 \, a^{3} c^{3}\right )} x^{2} + 2 \,{\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^2/(c*x^2+b*x+a)^(5/2),x, algorithm="fricas")

[Out]

-2/3*((8*(B*b*c^2 - 2*A*c^3)*d^2 - 4*(B*b^2*c + 4*(B*a - A*b)*c^2)*d*e - (B*b^3 + 8*A*a*c^2 - 2*(6*B*a*b - A*b
^2)*c)*e^2)*x^3 + (2*B*a*b^2 + A*b^3 + 4*(2*B*a^2 - 3*A*a*b)*c)*d^2 - 4*(4*B*a^2*b - A*a*b^2 - 4*A*a^2*c)*d*e
+ 8*(2*B*a^3 - A*a^2*b)*e^2 + 3*(4*(B*b^2*c - 2*A*b*c^2)*d^2 - 2*(B*b^3 + 4*(B*a*b - A*b^2)*c)*d*e + (2*B*a*b^
2 - A*b^3 + 4*(2*B*a^2 - A*a*b)*c)*e^2)*x^2 + 3*((B*b^3 - 8*A*a*c^2 + 2*(2*B*a*b - A*b^2)*c)*d^2 - 2*(4*B*a*b^
2 - A*b^3 - 4*A*a*b*c)*d*e + 4*(2*B*a^2*b - A*a*b^2)*e^2)*x)*sqrt(c*x^2 + b*x + a)/(a^2*b^4 - 8*a^3*b^2*c + 16
*a^4*c^2 + (b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 6*a*
b^4*c + 32*a^3*c^3)*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**2/(c*x**2+b*x+a)**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 1.15544, size = 637, normalized size = 5.26 \begin{align*} -\frac{{\left ({\left (\frac{{\left (8 \, B b c^{2} d^{2} - 16 \, A c^{3} d^{2} - 4 \, B b^{2} c d e - 16 \, B a c^{2} d e + 16 \, A b c^{2} d e - B b^{3} e^{2} + 12 \, B a b c e^{2} - 2 \, A b^{2} c e^{2} - 8 \, A a c^{2} e^{2}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac{3 \,{\left (4 \, B b^{2} c d^{2} - 8 \, A b c^{2} d^{2} - 2 \, B b^{3} d e - 8 \, B a b c d e + 8 \, A b^{2} c d e + 2 \, B a b^{2} e^{2} - A b^{3} e^{2} + 8 \, B a^{2} c e^{2} - 4 \, A a b c e^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{3 \,{\left (B b^{3} d^{2} + 4 \, B a b c d^{2} - 2 \, A b^{2} c d^{2} - 8 \, A a c^{2} d^{2} - 8 \, B a b^{2} d e + 2 \, A b^{3} d e + 8 \, A a b c d e + 8 \, B a^{2} b e^{2} - 4 \, A a b^{2} e^{2}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac{2 \, B a b^{2} d^{2} + A b^{3} d^{2} + 8 \, B a^{2} c d^{2} - 12 \, A a b c d^{2} - 16 \, B a^{2} b d e + 4 \, A a b^{2} d e + 16 \, A a^{2} c d e + 16 \, B a^{3} e^{2} - 8 \, A a^{2} b e^{2}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}}{3 \,{\left (c x^{2} + b x + a\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^2/(c*x^2+b*x+a)^(5/2),x, algorithm="giac")

[Out]

-1/3*((((8*B*b*c^2*d^2 - 16*A*c^3*d^2 - 4*B*b^2*c*d*e - 16*B*a*c^2*d*e + 16*A*b*c^2*d*e - B*b^3*e^2 + 12*B*a*b
*c*e^2 - 2*A*b^2*c*e^2 - 8*A*a*c^2*e^2)*x/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4) + 3*(4*B*b^2*c*d^2 - 8*A*b*c^2*
d^2 - 2*B*b^3*d*e - 8*B*a*b*c*d*e + 8*A*b^2*c*d*e + 2*B*a*b^2*e^2 - A*b^3*e^2 + 8*B*a^2*c*e^2 - 4*A*a*b*c*e^2)
/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x + 3*(B*b^3*d^2 + 4*B*a*b*c*d^2 - 2*A*b^2*c*d^2 - 8*A*a*c^2*d^2 - 8*B*
a*b^2*d*e + 2*A*b^3*d*e + 8*A*a*b*c*d*e + 8*B*a^2*b*e^2 - 4*A*a*b^2*e^2)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))
*x + (2*B*a*b^2*d^2 + A*b^3*d^2 + 8*B*a^2*c*d^2 - 12*A*a*b*c*d^2 - 16*B*a^2*b*d*e + 4*A*a*b^2*d*e + 16*A*a^2*c
*d*e + 16*B*a^3*e^2 - 8*A*a^2*b*e^2)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))/(c*x^2 + b*x + a)^(3/2)