\(\int \frac {(a+b x+c x^2)^2}{b d+2 c d x} \, dx\) [1126]

Optimal result

Integrand size = 24, antiderivative size = 72 \[ \int \frac {\left (a+b x+c x^2\right )^2}{b d+2 c d x} \, dx=-\frac {\left (b^2-4 a c\right ) (b+2 c x)^2}{32 c^3 d}+\frac {(b+2 c x)^4}{128 c^3 d}+\frac {\left (b^2-4 a c\right )^2 \log (b+2 c x)}{32 c^3 d} \]

[Out]

-1/32*(-4*a*c+b^2)*(2*c*x+b)^2/c^3/d+1/128*(2*c*x+b)^4/c^3/d+1/32*(-4*a*c+b^2)^2*ln(2*c*x+b)/c^3/d

Rubi [A] (verified)

Time = 0.05 (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} \[ \int \frac {\left (a+b x+c x^2\right )^2}{b d+2 c d x} \, dx=-\frac {\left (b^2-4 a c\right ) (b+2 c x)^2}{32 c^3 d}+\frac {\left (b^2-4 a c\right )^2 \log (b+2 c x)}{32 c^3 d}+\frac {(b+2 c x)^4}{128 c^3 d} \]

[In]

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

[Out]

-1/32*((b^2 - 4*a*c)*(b + 2*c*x)^2)/(c^3*d) + (b + 2*c*x)^4/(128*c^3*d) + ((b^2 - 4*a*c)^2*Log[b + 2*c*x])/(32
*c^3*d)

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 {\left (-b^2+4 a c\right )^2}{16 c^2 (b d+2 c d x)}+\frac {\left (-b^2+4 a c\right ) (b d+2 c d x)}{8 c^2 d^2}+\frac {(b d+2 c d x)^3}{16 c^2 d^4}\right ) \, dx \\ & = -\frac {\left (b^2-4 a c\right ) (b+2 c x)^2}{32 c^3 d}+\frac {(b+2 c x)^4}{128 c^3 d}+\frac {\left (b^2-4 a c\right )^2 \log (b+2 c x)}{32 c^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x+c x^2\right )^2}{b d+2 c d x} \, dx=\frac {2 c x (b+c x) \left (-b^2+2 b c x+2 c \left (4 a+c x^2\right )\right )+\left (b^2-4 a c\right )^2 \log (b+2 c x)}{32 c^3 d} \]

[In]

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

[Out]

(2*c*x*(b + c*x)*(-b^2 + 2*b*c*x + 2*c*(4*a + c*x^2)) + (b^2 - 4*a*c)^2*Log[b + 2*c*x])/(32*c^3*d)

Maple [A] (verified)

Time = 2.61 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\frac {\left (4 c^{2} x^{2}+4 b c x +8 a c -b^{2}\right )^{2}}{128 c^{3}}+\frac {\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (2 c x +b \right )}{32 c^{3}}}{d}\)	\(67\)
norman	\(\frac {b \,x^{3}}{4 d}+\frac {c \,x^{4}}{8 d}+\frac {\left (8 a c +b^{2}\right ) x^{2}}{16 c d}+\frac {b \left (8 a c -b^{2}\right ) x}{16 c^{2} d}+\frac {\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (2 c x +b \right )}{32 c^{3} d}\)	\(93\)
parallelrisch	\(\frac {4 c^{4} x^{4}+8 b \,c^{3} x^{3}+16 x^{2} a \,c^{3}+2 b^{2} c^{2} x^{2}+16 \ln \left (\frac {b}{2}+c x \right ) a^{2} c^{2}-8 \ln \left (\frac {b}{2}+c x \right ) a \,b^{2} c +\ln \left (\frac {b}{2}+c x \right ) b^{4}+16 a b \,c^{2} x -2 b^{3} c x}{32 c^{3} d}\)	\(105\)
risch	\(\frac {c \,x^{4}}{8 d}+\frac {b \,x^{3}}{4 d}+\frac {x^{2} a}{2 d}+\frac {b^{2} x^{2}}{16 c d}+\frac {a b x}{2 c d}-\frac {b^{3} x}{16 c^{2} d}+\frac {a^{2}}{2 c d}-\frac {a \,b^{2}}{8 c^{2} d}+\frac {b^{4}}{128 c^{3} d}+\frac {\ln \left (2 c x +b \right ) a^{2}}{2 c d}-\frac {\ln \left (2 c x +b \right ) a \,b^{2}}{4 c^{2} d}+\frac {\ln \left (2 c x +b \right ) b^{4}}{32 c^{3} d}\)	\(155\)

[In]

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

[Out]

1/d*(1/128/c^3*(4*c^2*x^2+4*b*c*x+8*a*c-b^2)^2+1/32*(16*a^2*c^2-8*a*b^2*c+b^4)/c^3*ln(2*c*x+b))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.22 \[ \int \frac {\left (a+b x+c x^2\right )^2}{b d+2 c d x} \, dx=\frac {4 \, c^{4} x^{4} + 8 \, b c^{3} x^{3} + 2 \, {\left (b^{2} c^{2} + 8 \, a c^{3}\right )} x^{2} - 2 \, {\left (b^{3} c - 8 \, a b c^{2}\right )} x + {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left (2 \, c x + b\right )}{32 \, c^{3} d} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x+c x^2\right )^2}{b d+2 c d x} \, dx=\frac {b x^{3}}{4 d} + \frac {c x^{4}}{8 d} + x^{2} \left (\frac {a}{2 d} + \frac {b^{2}}{16 c d}\right ) + x \left (\frac {a b}{2 c d} - \frac {b^{3}}{16 c^{2} d}\right ) + \frac {\left (4 a c - b^{2}\right )^{2} \log {\left (b + 2 c x \right )}}{32 c^{3} d} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a+b x+c x^2\right )^2}{b d+2 c d x} \, dx=\frac {2 \, c^{3} x^{4} + 4 \, b c^{2} x^{3} + {\left (b^{2} c + 8 \, a c^{2}\right )} x^{2} - {\left (b^{3} - 8 \, a b c\right )} x}{16 \, c^{2} d} + \frac {{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left (2 \, c x + b\right )}{32 \, c^{3} d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.61 \[ \int \frac {\left (a+b x+c x^2\right )^2}{b d+2 c d x} \, dx=\frac {{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \log \left ({\left | 2 \, c x + b \right |}\right )}{32 \, c^{3} d} + \frac {2 \, c^{5} d^{3} x^{4} + 4 \, b c^{4} d^{3} x^{3} + b^{2} c^{3} d^{3} x^{2} + 8 \, a c^{4} d^{3} x^{2} - b^{3} c^{2} d^{3} x + 8 \, a b c^{3} d^{3} x}{16 \, c^{4} d^{4}} \]

[In]

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

[Out]

1/32*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*log(abs(2*c*x + b))/(c^3*d) + 1/16*(2*c^5*d^3*x^4 + 4*b*c^4*d^3*x^3 + b^2*
c^3*d^3*x^2 + 8*a*c^4*d^3*x^2 - b^3*c^2*d^3*x + 8*a*b*c^3*d^3*x)/(c^4*d^4)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.85 \[ \int \frac {\left (a+b x+c x^2\right )^2}{b d+2 c d x} \, dx=x^2\,\left (\frac {b^2+2\,a\,c}{4\,c\,d}-\frac {3\,b^2}{16\,c\,d}\right )-x\,\left (\frac {b\,\left (\frac {b^2+2\,a\,c}{2\,c\,d}-\frac {3\,b^2}{8\,c\,d}\right )}{2\,c}-\frac {a\,b}{c\,d}\right )+\frac {b\,x^3}{4\,d}+\frac {c\,x^4}{8\,d}+\frac {\ln \left (b+2\,c\,x\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{32\,c^3\,d} \]

[In]

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

[Out]

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