\(\int \frac {a+b x+c x^2}{(b d+2 c d x)^{7/2}} \, dx\) [1267]

Optimal result

Integrand size = 24, antiderivative size = 55 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{7/2}} \, dx=\frac {b^2-4 a c}{20 c^2 d (b d+2 c d x)^{5/2}}-\frac {1}{4 c^2 d^3 \sqrt {b d+2 c d x}} \]

[Out]

1/20*(-4*a*c+b^2)/c^2/d/(2*c*d*x+b*d)^(5/2)-1/4/c^2/d^3/(2*c*d*x+b*d)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, 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} \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{7/2}} \, dx=\frac {b^2-4 a c}{20 c^2 d (b d+2 c d x)^{5/2}}-\frac {1}{4 c^2 d^3 \sqrt {b d+2 c d x}} \]

[In]

Int[(a + b*x + c*x^2)/(b*d + 2*c*d*x)^(7/2),x]

[Out]

(b^2 - 4*a*c)/(20*c^2*d*(b*d + 2*c*d*x)^(5/2)) - 1/(4*c^2*d^3*Sqrt[b*d + 2*c*d*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+4 a c}{4 c (b d+2 c d x)^{7/2}}+\frac {1}{4 c d^2 (b d+2 c d x)^{3/2}}\right ) \, dx \\ & = \frac {b^2-4 a c}{20 c^2 d (b d+2 c d x)^{5/2}}-\frac {1}{4 c^2 d^3 \sqrt {b d+2 c d x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{7/2}} \, dx=\frac {b^2-4 a c-5 (b+2 c x)^2}{20 c^2 d (d (b+2 c x))^{5/2}} \]

[In]

Integrate[(a + b*x + c*x^2)/(b*d + 2*c*d*x)^(7/2),x]

[Out]

(b^2 - 4*a*c - 5*(b + 2*c*x)^2)/(20*c^2*d*(d*(b + 2*c*x))^(5/2))

Maple [A] (verified)

Time = 2.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78

method	result	size
gosper	\(-\frac {\left (2 c x +b \right ) \left (5 c^{2} x^{2}+5 b c x +a c +b^{2}\right )}{5 c^{2} \left (2 c d x +b d \right )^{\frac {7}{2}}}\)	\(43\)
pseudoelliptic	\(-\frac {5 c^{2} x^{2}+\left (5 b x +a \right ) c +b^{2}}{5 \sqrt {d \left (2 c x +b \right )}\, d^{3} \left (2 c x +b \right )^{2} c^{2}}\)	\(47\)
trager	\(-\frac {\left (5 c^{2} x^{2}+5 b c x +a c +b^{2}\right ) \sqrt {2 c d x +b d}}{5 d^{4} \left (2 c x +b \right )^{3} c^{2}}\)	\(48\)
derivativedivides	\(\frac {-\frac {d^{2} \left (4 a c -b^{2}\right )}{5 \left (2 c d x +b d \right )^{\frac {5}{2}}}-\frac {1}{\sqrt {2 c d x +b d}}}{4 c^{2} d^{3}}\)	\(49\)
default	\(\frac {-\frac {d^{2} \left (4 a c -b^{2}\right )}{5 \left (2 c d x +b d \right )^{\frac {5}{2}}}-\frac {1}{\sqrt {2 c d x +b d}}}{4 c^{2} d^{3}}\)	\(49\)

[In]

int((c*x^2+b*x+a)/(2*c*d*x+b*d)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/5*(2*c*x+b)*(5*c^2*x^2+5*b*c*x+a*c+b^2)/c^2/(2*c*d*x+b*d)^(7/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.47 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{7/2}} \, dx=-\frac {{\left (5 \, c^{2} x^{2} + 5 \, b c x + b^{2} + a c\right )} \sqrt {2 \, c d x + b d}}{5 \, {\left (8 \, c^{5} d^{4} x^{3} + 12 \, b c^{4} d^{4} x^{2} + 6 \, b^{2} c^{3} d^{4} x + b^{3} c^{2} d^{4}\right )}} \]

[In]

integrate((c*x^2+b*x+a)/(2*c*d*x+b*d)^(7/2),x, algorithm="fricas")

[Out]

-1/5*(5*c^2*x^2 + 5*b*c*x + b^2 + a*c)*sqrt(2*c*d*x + b*d)/(8*c^5*d^4*x^3 + 12*b*c^4*d^4*x^2 + 6*b^2*c^3*d^4*x
+ b^3*c^2*d^4)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (51) = 102\).

Time = 0.44 (sec) , antiderivative size = 298, normalized size of antiderivative = 5.42 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{7/2}} \, dx=\begin {cases} - \frac {a c \sqrt {b d + 2 c d x}}{5 b^{3} c^{2} d^{4} + 30 b^{2} c^{3} d^{4} x + 60 b c^{4} d^{4} x^{2} + 40 c^{5} d^{4} x^{3}} - \frac {b^{2} \sqrt {b d + 2 c d x}}{5 b^{3} c^{2} d^{4} + 30 b^{2} c^{3} d^{4} x + 60 b c^{4} d^{4} x^{2} + 40 c^{5} d^{4} x^{3}} - \frac {5 b c x \sqrt {b d + 2 c d x}}{5 b^{3} c^{2} d^{4} + 30 b^{2} c^{3} d^{4} x + 60 b c^{4} d^{4} x^{2} + 40 c^{5} d^{4} x^{3}} - \frac {5 c^{2} x^{2} \sqrt {b d + 2 c d x}}{5 b^{3} c^{2} d^{4} + 30 b^{2} c^{3} d^{4} x + 60 b c^{4} d^{4} x^{2} + 40 c^{5} d^{4} x^{3}} & \text {for}\: c \neq 0 \\\frac {a x + \frac {b x^{2}}{2}}{\left (b d\right )^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate((c*x**2+b*x+a)/(2*c*d*x+b*d)**(7/2),x)

[Out]

Piecewise((-a*c*sqrt(b*d + 2*c*d*x)/(5*b**3*c**2*d**4 + 30*b**2*c**3*d**4*x + 60*b*c**4*d**4*x**2 + 40*c**5*d*
*4*x**3) - b**2*sqrt(b*d + 2*c*d*x)/(5*b**3*c**2*d**4 + 30*b**2*c**3*d**4*x + 60*b*c**4*d**4*x**2 + 40*c**5*d*
*4*x**3) - 5*b*c*x*sqrt(b*d + 2*c*d*x)/(5*b**3*c**2*d**4 + 30*b**2*c**3*d**4*x + 60*b*c**4*d**4*x**2 + 40*c**5
*d**4*x**3) - 5*c**2*x**2*sqrt(b*d + 2*c*d*x)/(5*b**3*c**2*d**4 + 30*b**2*c**3*d**4*x + 60*b*c**4*d**4*x**2 +
40*c**5*d**4*x**3), Ne(c, 0)), ((a*x + b*x**2/2)/(b*d)**(7/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{7/2}} \, dx=\frac {{\left (b^{2} - 4 \, a c\right )} d^{2} - 5 \, {\left (2 \, c d x + b d\right )}^{2}}{20 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c^{2} d^{3}} \]

[In]

integrate((c*x^2+b*x+a)/(2*c*d*x+b*d)^(7/2),x, algorithm="maxima")

[Out]

1/20*((b^2 - 4*a*c)*d^2 - 5*(2*c*d*x + b*d)^2)/((2*c*d*x + b*d)^(5/2)*c^2*d^3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{7/2}} \, dx=\frac {b^{2} d^{2} - 4 \, a c d^{2} - 5 \, {\left (2 \, c d x + b d\right )}^{2}}{20 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}} c^{2} d^{3}} \]

[In]

integrate((c*x^2+b*x+a)/(2*c*d*x+b*d)^(7/2),x, algorithm="giac")

[Out]

1/20*(b^2*d^2 - 4*a*c*d^2 - 5*(2*c*d*x + b*d)^2)/((2*c*d*x + b*d)^(5/2)*c^2*d^3)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.67 \[ \int \frac {a+b x+c x^2}{(b d+2 c d x)^{7/2}} \, dx=-\frac {\frac {4\,a\,c}{5}+{\left (b+2\,c\,x\right )}^2-\frac {b^2}{5}}{4\,c^2\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{5/2}} \]

[In]

int((a + b*x + c*x^2)/(b*d + 2*c*d*x)^(7/2),x)

[Out]

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