3.13.66 \(\int \frac {a+b x+c x^2}{(b d+2 c d x)^{5/2}} \, dx\) [1266]

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]

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

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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 = {697} \begin {gather*} \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 {gather*}

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 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*} \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*}

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Mathematica [A]
time = 0.04, size = 40, normalized size = 0.73 \begin {gather*} \frac {b^2-4 a c+3 (b+2 c x)^2}{12 c^2 d (d (b+2 c x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.45, size = 47, normalized size = 0.85

method	result	size
gosper	\(-\frac {\left (2 c x +b \right ) \left (-3 c^{2} x^{2}-3 b c x +a c -b^{2}\right )}{3 c^{2} \left (2 c d x +b d \right )^{\frac {5}{2}}}\)	\(45\)
derivativedivides	\(\frac {\sqrt {2 c d x +b d}-\frac {d^{2} \left (4 a c -b^{2}\right )}{3 \left (2 c d x +b d \right )^{\frac {3}{2}}}}{4 c^{2} d^{3}}\)	\(47\)
default	\(\frac {\sqrt {2 c d x +b d}-\frac {d^{2} \left (4 a c -b^{2}\right )}{3 \left (2 c d x +b d \right )^{\frac {3}{2}}}}{4 c^{2} d^{3}}\)	\(47\)
trager	\(-\frac {\left (-3 c^{2} x^{2}-3 b c x +a c -b^{2}\right ) \sqrt {2 c d x +b d}}{3 d^{3} c^{2} \left (2 c x +b \right )^{2}}\)	\(50\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 51, normalized size = 0.93 \begin {gather*} \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 {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)^(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 = 2.95, size = 68, normalized size = 1.24 \begin {gather*} \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 {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)^(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 [B] Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (49) = 98\).
time = 0.38, size = 235, normalized size = 4.27 \begin {gather*} \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 {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)**(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 = 1.40, size = 47, normalized size = 0.85 \begin {gather*} \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 {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)^(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)

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Mupad [B]
time = 0.48, size = 37, normalized size = 0.67 \begin {gather*} \frac {3\,{\left (b+2\,c\,x\right )}^2-4\,a\,c+b^2}{12\,c^2\,d\,{\left (b\,d+2\,c\,d\,x\right )}^{3/2}} \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)^(5/2),x)

[Out]

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

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