3.1266 \(\int \frac{a+b x+c x^2}{(b d+2 c d x)^{5/2}} \, dx\)

Optimal. Leaf size=55 \[ \frac{b^2-4 a c}{12 c^2 d (b d+2 c d x)^{3/2}}+\frac{\sqrt{b d+2 c d x}}{4 c^2 d^3} \]

[Out]

(b^2 - 4*a*c)/(12*c^2*d*(b*d + 2*c*d*x)^(3/2)) + Sqrt[b*d + 2*c*d*x]/(4*c^2*d^3)

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Rubi [A]  time = 0.0216346, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.042, Rules used = {683} \[ \frac{b^2-4 a c}{12 c^2 d (b d+2 c d x)^{3/2}}+\frac{\sqrt{b d+2 c d x}}{4 c^2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2 - 4*a*c)/(12*c^2*d*(b*d + 2*c*d*x)^(3/2)) + Sqrt[b*d + 2*c*d*x]/(4*c^2*d^3)

Rule 683

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*} \int \frac{a+b x+c x^2}{(b d+2 c d x)^{5/2}} \, dx &=\int \left (\frac{-b^2+4 a c}{4 c (b d+2 c d x)^{5/2}}+\frac{1}{4 c d^2 \sqrt{b d+2 c d x}}\right ) \, dx\\ &=\frac{b^2-4 a c}{12 c^2 d (b d+2 c d x)^{3/2}}+\frac{\sqrt{b d+2 c d x}}{4 c^2 d^3}\\ \end{align*}

Mathematica [A]  time = 0.0266623, size = 43, normalized size = 0.78 \[ \frac{c \left (3 c x^2-a\right )+b^2+3 b c x}{3 c^2 d (d (b+2 c x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.042, size = 45, normalized size = 0.8 \begin{align*} -{\frac{ \left ( 2\,cx+b \right ) \left ( -3\,{c}^{2}{x}^{2}-3\,bcx+ac-{b}^{2} \right ) }{3\,{c}^{2}} \left ( 2\,cdx+bd \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.16885, size = 69, normalized size = 1.25 \begin{align*} \frac{\frac{b^{2} - 4 \, a c}{{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} c} + \frac{3 \, \sqrt{2 \, c d x + b d}}{c d^{2}}}{12 \, c d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.03218, size = 142, normalized size = 2.58 \begin{align*} \frac{{\left (3 \, c^{2} x^{2} + 3 \, b c x + b^{2} - a c\right )} \sqrt{2 \, c d x + b d}}{3 \,{\left (4 \, c^{4} d^{3} x^{2} + 4 \, b c^{3} d^{3} x + b^{2} c^{2} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.71848, size = 235, normalized size = 4.27 \begin{align*} \begin{cases} - \frac{a c \sqrt{b d + 2 c d x}}{3 b^{2} c^{2} d^{3} + 12 b c^{3} d^{3} x + 12 c^{4} d^{3} x^{2}} + \frac{b^{2} \sqrt{b d + 2 c d x}}{3 b^{2} c^{2} d^{3} + 12 b c^{3} d^{3} x + 12 c^{4} d^{3} x^{2}} + \frac{3 b c x \sqrt{b d + 2 c d x}}{3 b^{2} c^{2} d^{3} + 12 b c^{3} d^{3} x + 12 c^{4} d^{3} x^{2}} + \frac{3 c^{2} x^{2} \sqrt{b d + 2 c d x}}{3 b^{2} c^{2} d^{3} + 12 b c^{3} d^{3} x + 12 c^{4} d^{3} x^{2}} & \text{for}\: c \neq 0 \\\frac{a x + \frac{b x^{2}}{2}}{\left (b d\right )^{\frac{5}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a*c*sqrt(b*d + 2*c*d*x)/(3*b**2*c**2*d**3 + 12*b*c**3*d**3*x + 12*c**4*d**3*x**2) + b**2*sqrt(b*d
+ 2*c*d*x)/(3*b**2*c**2*d**3 + 12*b*c**3*d**3*x + 12*c**4*d**3*x**2) + 3*b*c*x*sqrt(b*d + 2*c*d*x)/(3*b**2*c**
2*d**3 + 12*b*c**3*d**3*x + 12*c**4*d**3*x**2) + 3*c**2*x**2*sqrt(b*d + 2*c*d*x)/(3*b**2*c**2*d**3 + 12*b*c**3
*d**3*x + 12*c**4*d**3*x**2), Ne(c, 0)), ((a*x + b*x**2/2)/(b*d)**(5/2), True))

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Giac [A]  time = 1.15943, size = 63, normalized size = 1.15 \begin{align*} \frac{b^{2} - 4 \, a c}{12 \,{\left (2 \, c d x + b d\right )}^{\frac{3}{2}} c^{2} d} + \frac{\sqrt{2 \, c d x + b d}}{4 \, c^{2} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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