\(\int \frac {a+b x+c x^2}{(b d+2 c d x)^5} \, dx\) [1117]

Optimal result

Integrand size = 22, antiderivative size = 37 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^5} \, dx=\frac {\left (a+b x+c x^2\right )^2}{2 \left (b^2-4 a c\right ) d^5 (b+2 c x)^4} \]

[Out]

1/2*(c*x^2+b*x+a)^2/(-4*a*c+b^2)/d^5/(2*c*x+b)^4

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {696} \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^5} \, dx=\frac {\left (a+b x+c x^2\right )^2}{2 d^5 \left (b^2-4 a c\right ) (b+2 c x)^4} \]

[In]

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

[Out]

(a + b*x + c*x^2)^2/(2*(b^2 - 4*a*c)*d^5*(b + 2*c*x)^4)

Rule 696

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a+b x+c x^2\right )^2}{2 \left (b^2-4 a c\right ) d^5 (b+2 c x)^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.03 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^5} \, dx=-\frac {b^2+8 b c x+4 c \left (a+2 c x^2\right )}{32 c^2 d^5 (b+2 c x)^4} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.03

method	result	size
gosper	\(-\frac {8 c^{2} x^{2}+8 b c x +4 a c +b^{2}}{32 c^{2} \left (2 c x +b \right )^{4} d^{5}}\)	\(38\)
risch	\(\frac {-\frac {x^{2}}{4}-\frac {b x}{4 c}-\frac {4 a c +b^{2}}{32 c^{2}}}{d^{5} \left (2 c x +b \right )^{4}}\)	\(39\)
default	\(\frac {-\frac {4 a c -b^{2}}{32 c^{2} \left (2 c x +b \right )^{4}}-\frac {1}{16 c^{2} \left (2 c x +b \right )^{2}}}{d^{5}}\)	\(42\)
parallelrisch	\(\frac {4 x^{4} a \,c^{3}+b^{2} c^{2} x^{4}+8 a b \,c^{2} x^{3}+2 b^{3} c \,x^{3}+6 a \,b^{2} c \,x^{2}+b^{4} x^{2}+2 a \,b^{3} x}{2 b^{4} d^{5} \left (2 c x +b \right )^{4}}\)	\(80\)
norman	\(\frac {\frac {a x}{b d}+\frac {c \left (4 a c +b^{2}\right ) x^{3}}{b^{3} d}+\frac {\left (6 a c +b^{2}\right ) x^{2}}{2 b^{2} d}+\frac {c^{2} \left (4 a c +b^{2}\right ) x^{4}}{2 b^{4} d}}{d^{4} \left (2 c x +b \right )^{4}}\)	\(83\)

[In]

int((c*x^2+b*x+a)/(2*c*d*x+b*d)^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (35) = 70\).

Time = 0.33 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.30 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^5} \, dx=-\frac {8 \, c^{2} x^{2} + 8 \, b c x + b^{2} + 4 \, a c}{32 \, {\left (16 \, c^{6} d^{5} x^{4} + 32 \, b c^{5} d^{5} x^{3} + 24 \, b^{2} c^{4} d^{5} x^{2} + 8 \, b^{3} c^{3} d^{5} x + b^{4} c^{2} d^{5}\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (32) = 64\).

Time = 0.34 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.43 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^5} \, dx=\frac {- 4 a c - b^{2} - 8 b c x - 8 c^{2} x^{2}}{32 b^{4} c^{2} d^{5} + 256 b^{3} c^{3} d^{5} x + 768 b^{2} c^{4} d^{5} x^{2} + 1024 b c^{5} d^{5} x^{3} + 512 c^{6} d^{5} x^{4}} \]

[In]

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

[Out]

(-4*a*c - b**2 - 8*b*c*x - 8*c**2*x**2)/(32*b**4*c**2*d**5 + 256*b**3*c**3*d**5*x + 768*b**2*c**4*d**5*x**2 +
1024*b*c**5*d**5*x**3 + 512*c**6*d**5*x**4)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (35) = 70\).

Time = 0.20 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.30 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^5} \, dx=-\frac {8 \, c^{2} x^{2} + 8 \, b c x + b^{2} + 4 \, a c}{32 \, {\left (16 \, c^{6} d^{5} x^{4} + 32 \, b c^{5} d^{5} x^{3} + 24 \, b^{2} c^{4} d^{5} x^{2} + 8 \, b^{3} c^{3} d^{5} x + b^{4} c^{2} d^{5}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.62 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^5} \, dx=\frac {\frac {b^{2}}{{\left (2 \, c d x + b d\right )}^{4} c^{2}} - \frac {4 \, a}{{\left (2 \, c d x + b d\right )}^{4} c} - \frac {2}{{\left (2 \, c d x + b d\right )}^{2} c^{2} d^{2}}}{32 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.30 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^5} \, dx=-\frac {\frac {b^2+4\,a\,c}{32\,c^2}+\frac {x^2}{4}+\frac {b\,x}{4\,c}}{b^4\,d^5+8\,b^3\,c\,d^5\,x+24\,b^2\,c^2\,d^5\,x^2+32\,b\,c^3\,d^5\,x^3+16\,c^4\,d^5\,x^4} \]

[In]

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

[Out]

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