\(\int \frac {b d+2 c d x}{(a+b x+c x^2)^{5/2}} \, dx\) [1255]

Optimal result

Integrand size = 24, antiderivative size = 19 \[ \int \frac {b d+2 c d x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 d}{3 \left (a+b x+c x^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {643} \[ \int \frac {b d+2 c d x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 d}{3 \left (a+b x+c x^2\right )^{3/2}} \]

[In]

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

[Out]

(-2*d)/(3*(a + b*x + c*x^2)^(3/2))

Rule 643

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 d}{3 \left (a+b x+c x^2\right )^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {b d+2 c d x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 d}{3 (a+x (b+c x))^{3/2}} \]

[In]

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

[Out]

(-2*d)/(3*(a + x*(b + c*x))^(3/2))

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
gosper	\(-\frac {2 d}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\)	\(16\)
default	\(-\frac {2 d}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\)	\(16\)
trager	\(-\frac {2 d}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\)	\(16\)
pseudoelliptic	\(-\frac {2 d}{3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}\)	\(16\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (15) = 30\).

Time = 0.39 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.74 \[ \int \frac {b d+2 c d x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, \sqrt {c x^{2} + b x + a} d}{3 \, {\left (c^{2} x^{4} + 2 \, b c x^{3} + 2 \, a b x + {\left (b^{2} + 2 \, a c\right )} x^{2} + a^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (19) = 38\).

Time = 0.51 (sec) , antiderivative size = 82, normalized size of antiderivative = 4.32 \[ \int \frac {b d+2 c d x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\begin {cases} - \frac {2 d}{3 a \sqrt {a + b x + c x^{2}} + 3 b x \sqrt {a + b x + c x^{2}} + 3 c x^{2} \sqrt {a + b x + c x^{2}}} & \text {for}\: a \neq - x \left (b + c x\right ) \\\tilde {\infty } b d x + \tilde {\infty } c d x^{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-2*d/(3*a*sqrt(a + b*x + c*x**2) + 3*b*x*sqrt(a + b*x + c*x**2) + 3*c*x**2*sqrt(a + b*x + c*x**2)),
Ne(a, -x*(b + c*x))), (zoo*b*d*x + zoo*c*d*x**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {b d+2 c d x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, d}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \]

[In]

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

[Out]

-2/3*d/(c*x^2 + b*x + a)^(3/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {b d+2 c d x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, d^{5}}{3 \, {\left (a d^{2} + {\left (c d x^{2} + b d x\right )} d\right )}^{\frac {3}{2}} {\left | d \right |}} \]

[In]

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

[Out]

-2/3*d^5/((a*d^2 + (c*d*x^2 + b*d*x)*d)^(3/2)*abs(d))

Mupad [B] (verification not implemented)

Time = 9.51 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {b d+2 c d x}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2\,d}{3\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \]

[In]

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

[Out]

-(2*d)/(3*(a + b*x + c*x^2)^(3/2))