\(\int \frac {(a+b x+c x^2)^2}{(b d+2 c d x)^2} \, dx\) [1127]

Optimal result

Integrand size = 24, antiderivative size = 72 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^2} \, dx=-\frac {\left (b^2-8 a c\right ) x}{16 c^2 d^2}+\frac {b x^2}{8 c d^2}+\frac {x^3}{12 d^2}-\frac {\left (b^2-4 a c\right )^2}{32 c^3 d^2 (b+2 c x)} \]

[Out]

-1/16*(-8*a*c+b^2)*x/c^2/d^2+1/8*b*x^2/c/d^2+1/12*x^3/d^2-1/32*(-4*a*c+b^2)^2/c^3/d^2/(2*c*x+b)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {697} \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^2} \, dx=-\frac {\left (b^2-4 a c\right )^2}{32 c^3 d^2 (b+2 c x)}-\frac {x \left (b^2-8 a c\right )}{16 c^2 d^2}+\frac {b x^2}{8 c d^2}+\frac {x^3}{12 d^2} \]

[In]

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

[Out]

-1/16*((b^2 - 8*a*c)*x)/(c^2*d^2) + (b*x^2)/(8*c*d^2) + x^3/(12*d^2) - (b^2 - 4*a*c)^2/(32*c^3*d^2*(b + 2*c*x)
)

Rule 697

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*} \text {integral}& = \int \left (\frac {-b^2+8 a c}{16 c^2 d^2}+\frac {b x}{4 c d^2}+\frac {x^2}{4 d^2}+\frac {\left (-b^2+4 a c\right )^2}{16 c^2 d^2 (b+2 c x)^2}\right ) \, dx \\ & = -\frac {\left (b^2-8 a c\right ) x}{16 c^2 d^2}+\frac {b x^2}{8 c d^2}+\frac {x^3}{12 d^2}-\frac {\left (b^2-4 a c\right )^2}{32 c^3 d^2 (b+2 c x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^2} \, dx=\frac {-\frac {6 \left (b^2-8 a c\right ) x}{c^2}+\frac {12 b x^2}{c}+8 x^3-\frac {3 \left (b^2-4 a c\right )^2}{c^3 (b+2 c x)}}{96 d^2} \]

[In]

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

[Out]

((-6*(b^2 - 8*a*c)*x)/c^2 + (12*b*x^2)/c + 8*x^3 - (3*(b^2 - 4*a*c)^2)/(c^3*(b + 2*c*x)))/(96*d^2)

Maple [A] (verified)

Time = 2.68 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.62

method	result	size
gosper	\(-\frac {-x^{4} c^{2}-2 b c \,x^{3}-6 a c \,x^{2}+3 a^{2}}{6 \left (2 c x +b \right ) c \,d^{2}}\)	\(45\)
parallelrisch	\(\frac {b c \,x^{4}+2 b^{2} x^{3}+6 a b \,x^{2}+6 a^{2} x}{6 b \,d^{2} \left (2 c x +b \right )}\)	\(45\)
norman	\(\frac {\frac {x^{2} a}{d}+\frac {a^{2} x}{b d}+\frac {b \,x^{3}}{3 d}+\frac {c \,x^{4}}{6 d}}{d \left (2 c x +b \right )}\)	\(51\)
default	\(\frac {\frac {\frac {4}{3} c^{2} x^{3}+2 c b \,x^{2}+8 a c x -b^{2} x}{16 c^{2}}-\frac {16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}{32 c^{3} \left (2 c x +b \right )}}{d^{2}}\)	\(70\)
risch	\(\frac {x^{3}}{12 d^{2}}+\frac {b \,x^{2}}{8 c \,d^{2}}+\frac {a x}{2 c \,d^{2}}-\frac {b^{2} x}{16 c^{2} d^{2}}-\frac {a^{2}}{2 c \,d^{2} \left (2 c x +b \right )}+\frac {a \,b^{2}}{4 c^{2} d^{2} \left (2 c x +b \right )}-\frac {b^{4}}{32 c^{3} d^{2} \left (2 c x +b \right )}\)	\(102\)

[In]

int((c*x^2+b*x+a)^2/(2*c*d*x+b*d)^2,x,method=_RETURNVERBOSE)

[Out]

-1/6*(-c^2*x^4-2*b*c*x^3-6*a*c*x^2+3*a^2)/(2*c*x+b)/c/d^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.18 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^2} \, dx=\frac {16 \, c^{4} x^{4} + 32 \, b c^{3} x^{3} + 96 \, a c^{3} x^{2} - 3 \, b^{4} + 24 \, a b^{2} c - 48 \, a^{2} c^{2} - 6 \, {\left (b^{3} c - 8 \, a b c^{2}\right )} x}{96 \, {\left (2 \, c^{4} d^{2} x + b c^{3} d^{2}\right )}} \]

[In]

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

[Out]

1/96*(16*c^4*x^4 + 32*b*c^3*x^3 + 96*a*c^3*x^2 - 3*b^4 + 24*a*b^2*c - 48*a^2*c^2 - 6*(b^3*c - 8*a*b*c^2)*x)/(2
*c^4*d^2*x + b*c^3*d^2)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^2} \, dx=\frac {b x^{2}}{8 c d^{2}} + x \left (\frac {a}{2 c d^{2}} - \frac {b^{2}}{16 c^{2} d^{2}}\right ) + \frac {- 16 a^{2} c^{2} + 8 a b^{2} c - b^{4}}{32 b c^{3} d^{2} + 64 c^{4} d^{2} x} + \frac {x^{3}}{12 d^{2}} \]

[In]

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

[Out]

b*x**2/(8*c*d**2) + x*(a/(2*c*d**2) - b**2/(16*c**2*d**2)) + (-16*a**2*c**2 + 8*a*b**2*c - b**4)/(32*b*c**3*d*
*2 + 64*c**4*d**2*x) + x**3/(12*d**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^2} \, dx=-\frac {b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}}{32 \, {\left (2 \, c^{4} d^{2} x + b c^{3} d^{2}\right )}} + \frac {4 \, c^{2} x^{3} + 6 \, b c x^{2} - 3 \, {\left (b^{2} - 8 \, a c\right )} x}{48 \, c^{2} d^{2}} \]

[In]

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

[Out]

-1/32*(b^4 - 8*a*b^2*c + 16*a^2*c^2)/(2*c^4*d^2*x + b*c^3*d^2) + 1/48*(4*c^2*x^3 + 6*b*c*x^2 - 3*(b^2 - 8*a*c)
*x)/(c^2*d^2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (64) = 128\).

Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.86 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^2} \, dx=-\frac {{\left (2 \, c d x + b d\right )}^{3} {\left (\frac {6 \, b^{2} d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} - \frac {24 \, a c d^{2}}{{\left (2 \, c d x + b d\right )}^{2}} - 1\right )}}{96 \, c^{3} d^{5}} - \frac {\frac {b^{4} c^{3} d^{7}}{2 \, c d x + b d} - \frac {8 \, a b^{2} c^{4} d^{7}}{2 \, c d x + b d} + \frac {16 \, a^{2} c^{5} d^{7}}{2 \, c d x + b d}}{32 \, c^{6} d^{8}} \]

[In]

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

[Out]

-1/96*(2*c*d*x + b*d)^3*(6*b^2*d^2/(2*c*d*x + b*d)^2 - 24*a*c*d^2/(2*c*d*x + b*d)^2 - 1)/(c^3*d^5) - 1/32*(b^4
*c^3*d^7/(2*c*d*x + b*d) - 8*a*b^2*c^4*d^7/(2*c*d*x + b*d) + 16*a^2*c^5*d^7/(2*c*d*x + b*d))/(c^6*d^8)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a+b x+c x^2\right )^2}{(b d+2 c d x)^2} \, dx=x\,\left (\frac {b^2+2\,a\,c}{4\,c^2\,d^2}-\frac {5\,b^2}{16\,c^2\,d^2}\right )+\frac {x^3}{12\,d^2}-\frac {16\,a^2\,c^2-8\,a\,b^2\,c+b^4}{2\,c\,\left (32\,x\,c^3\,d^2+16\,b\,c^2\,d^2\right )}+\frac {b\,x^2}{8\,c\,d^2} \]

[In]

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

[Out]

x*((2*a*c + b^2)/(4*c^2*d^2) - (5*b^2)/(16*c^2*d^2)) + x^3/(12*d^2) - (b^4 + 16*a^2*c^2 - 8*a*b^2*c)/(2*c*(16*
b*c^2*d^2 + 32*c^3*d^2*x)) + (b*x^2)/(8*c*d^2)