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

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

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Rubi [A]  time = 0.02, antiderivative size = 55, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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 {gather*}

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*}

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Mathematica [A]  time = 0.03, size = 44, normalized size = 0.80 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.08, size = 46, normalized size = 0.84 \begin {gather*} \frac {-a c-b^2-5 b c x-5 c^2 x^2}{5 c^2 d (b d+2 c d x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 81, normalized size = 1.47 \begin {gather*} -\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 {gather*}

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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giac [A]  time = 0.18, size = 47, normalized size = 0.85 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.04, size = 43, normalized size = 0.78 \begin {gather*} -\frac {\left (2 c x +b \right ) \left (5 c^{2} x^{2}+5 b c x +a c +b^{2}\right )}{5 \left (2 c d x +b d \right )^{\frac {7}{2}} c^{2}} \end {gather*}

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 = 1.42, size = 45, normalized size = 0.82 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.04, size = 37, normalized size = 0.67 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 3.02, size = 298, normalized size = 5.42 \begin {gather*} \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 {gather*}

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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