### 3.1267 $$\int \frac{a+b x+c x^2}{(b d+2 c d x)^{7/2}} \, dx$$

Optimal. Leaf size=55 $\frac{b^2-4 a c}{20 c^2 d (b d+2 c d x)^{5/2}}-\frac{1}{4 c^2 d^3 \sqrt{b d+2 c d x}}$

[Out]

(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])

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Rubi [A]  time = 0.0223381, 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} $\frac{b^2-4 a c}{20 c^2 d (b d+2 c d x)^{5/2}}-\frac{1}{4 c^2 d^3 \sqrt{b d+2 c d x}}$

Antiderivative was successfully veriﬁed.

[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 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)^{7/2}} \, dx &=\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]  time = 0.0293239, size = 44, normalized size = 0.8 $\frac{-c \left (a+5 c x^2\right )-b^2-5 b c x}{5 c^2 d (d (b+2 c x))^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 43, normalized size = 0.8 \begin{align*} -{\frac{ \left ( 2\,cx+b \right ) \left ( 5\,{c}^{2}{x}^{2}+5\,bcx+ac+{b}^{2} \right ) }{5\,{c}^{2}} \left ( 2\,cdx+bd \right ) ^{-{\frac{7}{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)^(7/2),x)

[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)

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Maxima [A]  time = 1.026, size = 61, normalized size = 1.11 \begin{align*} \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}} \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)^(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)

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Fricas [A]  time = 2.01084, size = 171, normalized size = 3.11 \begin{align*} -\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 )}} \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)^(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)

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Sympy [A]  time = 4.4656, size = 298, normalized size = 5.42 \begin{align*} \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} \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)**(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))

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Giac [A]  time = 1.1477, size = 63, normalized size = 1.15 \begin{align*} \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}} \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)^(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)