3.1143 \(\int \frac{(a+b x+c x^2)^3}{(b d+2 c d x)^2} \, dx\)

Optimal. Leaf size=94 \[ -\frac{\left (b^2-4 a c\right ) (b+2 c x)^3}{128 c^4 d^2}+\frac{3 x \left (b^2-4 a c\right )^2}{64 c^3 d^2}+\frac{\left (b^2-4 a c\right )^3}{128 c^4 d^2 (b+2 c x)}+\frac{(b+2 c x)^5}{640 c^4 d^2} \]

[Out]

(3*(b^2 - 4*a*c)^2*x)/(64*c^3*d^2) + (b^2 - 4*a*c)^3/(128*c^4*d^2*(b + 2*c*x)) - ((b^2 - 4*a*c)*(b + 2*c*x)^3)
/(128*c^4*d^2) + (b + 2*c*x)^5/(640*c^4*d^2)

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Rubi [A]  time = 0.0945815, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.042, Rules used = {683} \[ -\frac{\left (b^2-4 a c\right ) (b+2 c x)^3}{128 c^4 d^2}+\frac{3 x \left (b^2-4 a c\right )^2}{64 c^3 d^2}+\frac{\left (b^2-4 a c\right )^3}{128 c^4 d^2 (b+2 c x)}+\frac{(b+2 c x)^5}{640 c^4 d^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^2,x]

[Out]

(3*(b^2 - 4*a*c)^2*x)/(64*c^3*d^2) + (b^2 - 4*a*c)^3/(128*c^4*d^2*(b + 2*c*x)) - ((b^2 - 4*a*c)*(b + 2*c*x)^3)
/(128*c^4*d^2) + (b + 2*c*x)^5/(640*c^4*d^2)

Rule 683

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[2*c*d - b*e,
0] && IGtQ[p, 0] &&  !(EqQ[m, 3] && NeQ[p, 1])

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^3}{(b d+2 c d x)^2} \, dx &=\int \left (\frac{3 \left (-b^2+4 a c\right )^2}{64 c^3 d^2}+\frac{\left (-b^2+4 a c\right )^3}{64 c^3 (b d+2 c d x)^2}+\frac{3 \left (-b^2+4 a c\right ) (b d+2 c d x)^2}{64 c^3 d^4}+\frac{(b d+2 c d x)^4}{64 c^3 d^6}\right ) \, dx\\ &=\frac{3 \left (b^2-4 a c\right )^2 x}{64 c^3 d^2}+\frac{\left (b^2-4 a c\right )^3}{128 c^4 d^2 (b+2 c x)}-\frac{\left (b^2-4 a c\right ) (b+2 c x)^3}{128 c^4 d^2}+\frac{(b+2 c x)^5}{640 c^4 d^2}\\ \end{align*}

Mathematica [A]  time = 0.0791392, size = 101, normalized size = 1.07 \[ \frac{\frac{10 x \left (48 a^2 c^2-12 a b^2 c+b^4\right )}{c^3}-\frac{20 b x^2 \left (b^2-12 a c\right )}{c^2}+\frac{5 \left (b^2-4 a c\right )^3}{c^4 (b+2 c x)}+\frac{40 x^3 \left (4 a c+b^2\right )}{c}+80 b x^4+32 c x^5}{640 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^3/(b*d + 2*c*d*x)^2,x]

[Out]

((10*(b^4 - 12*a*b^2*c + 48*a^2*c^2)*x)/c^3 - (20*b*(b^2 - 12*a*c)*x^2)/c^2 + (40*(b^2 + 4*a*c)*x^3)/c + 80*b*
x^4 + 32*c*x^5 + (5*(b^2 - 4*a*c)^3)/(c^4*(b + 2*c*x)))/(640*d^2)

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Maple [A]  time = 0.044, size = 135, normalized size = 1.4 \begin{align*}{\frac{1}{{d}^{2}} \left ({\frac{1}{64\,{c}^{3}} \left ({\frac{16\,{x}^{5}{c}^{4}}{5}}+8\,b{x}^{4}{c}^{3}+16\,{x}^{3}a{c}^{3}+4\,{b}^{2}{c}^{2}{x}^{3}+24\,{x}^{2}ab{c}^{2}-2\,{x}^{2}{b}^{3}c+48\,{a}^{2}{c}^{2}x-12\,ac{b}^{2}x+{b}^{4}x \right ) }-{\frac{64\,{a}^{3}{c}^{3}-48\,{a}^{2}{b}^{2}{c}^{2}+12\,a{b}^{4}c-{b}^{6}}{128\,{c}^{4} \left ( 2\,cx+b \right ) }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^2,x)

[Out]

1/d^2*(1/64/c^3*(16/5*x^5*c^4+8*b*x^4*c^3+16*x^3*a*c^3+4*b^2*c^2*x^3+24*x^2*a*b*c^2-2*x^2*b^3*c+48*a^2*c^2*x-1
2*a*c*b^2*x+b^4*x)-1/128*(64*a^3*c^3-48*a^2*b^2*c^2+12*a*b^4*c-b^6)/c^4/(2*c*x+b))

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Maxima [A]  time = 1.16678, size = 186, normalized size = 1.98 \begin{align*} \frac{b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}}{128 \,{\left (2 \, c^{5} d^{2} x + b c^{4} d^{2}\right )}} + \frac{16 \, c^{4} x^{5} + 40 \, b c^{3} x^{4} + 20 \,{\left (b^{2} c^{2} + 4 \, a c^{3}\right )} x^{3} - 10 \,{\left (b^{3} c - 12 \, a b c^{2}\right )} x^{2} + 5 \,{\left (b^{4} - 12 \, a b^{2} c + 48 \, a^{2} c^{2}\right )} x}{320 \, c^{3} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^2,x, algorithm="maxima")

[Out]

1/128*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)/(2*c^5*d^2*x + b*c^4*d^2) + 1/320*(16*c^4*x^5 + 40*b*c^
3*x^4 + 20*(b^2*c^2 + 4*a*c^3)*x^3 - 10*(b^3*c - 12*a*b*c^2)*x^2 + 5*(b^4 - 12*a*b^2*c + 48*a^2*c^2)*x)/(c^3*d
^2)

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Fricas [A]  time = 1.90991, size = 305, normalized size = 3.24 \begin{align*} \frac{64 \, c^{6} x^{6} + 192 \, b c^{5} x^{5} + 640 \, a b c^{4} x^{3} + 960 \, a^{2} c^{4} x^{2} + 5 \, b^{6} - 60 \, a b^{4} c + 240 \, a^{2} b^{2} c^{2} - 320 \, a^{3} c^{3} + 160 \,{\left (b^{2} c^{4} + 2 \, a c^{5}\right )} x^{4} + 10 \,{\left (b^{5} c - 12 \, a b^{3} c^{2} + 48 \, a^{2} b c^{3}\right )} x}{640 \,{\left (2 \, c^{5} d^{2} x + b c^{4} d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^2,x, algorithm="fricas")

[Out]

1/640*(64*c^6*x^6 + 192*b*c^5*x^5 + 640*a*b*c^4*x^3 + 960*a^2*c^4*x^2 + 5*b^6 - 60*a*b^4*c + 240*a^2*b^2*c^2 -
320*a^3*c^3 + 160*(b^2*c^4 + 2*a*c^5)*x^4 + 10*(b^5*c - 12*a*b^3*c^2 + 48*a^2*b*c^3)*x)/(2*c^5*d^2*x + b*c^4*
d^2)

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Sympy [A]  time = 0.973318, size = 143, normalized size = 1.52 \begin{align*} \frac{b x^{4}}{8 d^{2}} + \frac{c x^{5}}{20 d^{2}} - \frac{64 a^{3} c^{3} - 48 a^{2} b^{2} c^{2} + 12 a b^{4} c - b^{6}}{128 b c^{4} d^{2} + 256 c^{5} d^{2} x} + \frac{x^{3} \left (4 a c + b^{2}\right )}{16 c d^{2}} + \frac{x^{2} \left (12 a b c - b^{3}\right )}{32 c^{2} d^{2}} + \frac{x \left (48 a^{2} c^{2} - 12 a b^{2} c + b^{4}\right )}{64 c^{3} d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**3/(2*c*d*x+b*d)**2,x)

[Out]

b*x**4/(8*d**2) + c*x**5/(20*d**2) - (64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)/(128*b*c**4*d**2
+ 256*c**5*d**2*x) + x**3*(4*a*c + b**2)/(16*c*d**2) + x**2*(12*a*b*c - b**3)/(32*c**2*d**2) + x*(48*a**2*c**2
- 12*a*b**2*c + b**4)/(64*c**3*d**2)

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Giac [B]  time = 1.22794, size = 298, normalized size = 3.17 \begin{align*} \frac{{\left (\frac{15 \, b^{4} d^{4}}{{\left (2 \, c d x + b d\right )}^{4}} - \frac{120 \, a b^{2} c d^{4}}{{\left (2 \, c d x + b d\right )}^{4}} + \frac{240 \, a^{2} c^{2} d^{4}}{{\left (2 \, c d x + b d\right )}^{4}} - \frac{5 \, b^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + \frac{20 \, a c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} + 1\right )}{\left (2 \, c d x + b d\right )}^{5}}{640 \, c^{4} d^{7}} + \frac{\frac{b^{6} c^{5} d^{11}}{2 \, c d x + b d} - \frac{12 \, a b^{4} c^{6} d^{11}}{2 \, c d x + b d} + \frac{48 \, a^{2} b^{2} c^{7} d^{11}}{2 \, c d x + b d} - \frac{64 \, a^{3} c^{8} d^{11}}{2 \, c d x + b d}}{128 \, c^{9} d^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^3/(2*c*d*x+b*d)^2,x, algorithm="giac")

[Out]

1/640*(15*b^4*d^4/(2*c*d*x + b*d)^4 - 120*a*b^2*c*d^4/(2*c*d*x + b*d)^4 + 240*a^2*c^2*d^4/(2*c*d*x + b*d)^4 -
5*b^2*d^2/(2*c*d*x + b*d)^2 + 20*a*c*d^2/(2*c*d*x + b*d)^2 + 1)*(2*c*d*x + b*d)^5/(c^4*d^7) + 1/128*(b^6*c^5*d
^11/(2*c*d*x + b*d) - 12*a*b^4*c^6*d^11/(2*c*d*x + b*d) + 48*a^2*b^2*c^7*d^11/(2*c*d*x + b*d) - 64*a^3*c^8*d^1
1/(2*c*d*x + b*d))/(c^9*d^12)