3.1264 \(\int \frac {a+b x+c x^2}{\sqrt {b d+2 c d x}} \, dx\)

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

[Out]

1/20*(2*c*d*x+b*d)^(5/2)/c^2/d^3-1/4*(-4*a*c+b^2)*(2*c*d*x+b*d)^(1/2)/c^2/d

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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} \[ \frac {(b d+2 c d x)^{5/2}}{20 c^2 d^3}-\frac {\left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}{4 c^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2 - 4*a*c)*Sqrt[b*d + 2*c*d*x])/(4*c^2*d) + (b*d + 2*c*d*x)^(5/2)/(20*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}{\sqrt {b d+2 c d x}} \, dx &=\int \left (\frac {-b^2+4 a c}{4 c \sqrt {b d+2 c d x}}+\frac {(b d+2 c d x)^{3/2}}{4 c d^2}\right ) \, dx\\ &=-\frac {\left (b^2-4 a c\right ) \sqrt {b d+2 c d x}}{4 c^2 d}+\frac {(b d+2 c d x)^{5/2}}{20 c^2 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 43, normalized size = 0.78 \[ \frac {\left (c \left (5 a+c x^2\right )-b^2+b c x\right ) \sqrt {d (b+2 c x)}}{5 c^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.83, size = 40, normalized size = 0.73 \[ \frac {{\left (c^{2} x^{2} + b c x - b^{2} + 5 \, a c\right )} \sqrt {2 \, c d x + b d}}{5 \, c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(2*c*d*x+b*d)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.20, size = 116, normalized size = 2.11 \[ \frac {60 \, \sqrt {2 \, c d x + b d} a - \frac {10 \, {\left (3 \, \sqrt {2 \, c d x + b d} b d - {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}\right )} b}{c d} + \frac {15 \, \sqrt {2 \, c d x + b d} b^{2} d^{2} - 10 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b d + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}}{c d^{2}}}{60 \, c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(2*c*d*x+b*d)^(1/2),x, algorithm="giac")

[Out]

1/60*(60*sqrt(2*c*d*x + b*d)*a - 10*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*b/(c*d) + (15*sqrt(2*c
*d*x + b*d)*b^2*d^2 - 10*(2*c*d*x + b*d)^(3/2)*b*d + 3*(2*c*d*x + b*d)^(5/2))/(c*d^2))/(c*d)

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maple [A]  time = 0.04, size = 44, normalized size = 0.80 \[ \frac {\left (2 c x +b \right ) \left (c^{2} x^{2}+b c x +5 a c -b^{2}\right )}{5 \sqrt {2 c d x +b d}\, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)/(2*c*d*x+b*d)^(1/2),x)

[Out]

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

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maxima [B]  time = 1.36, size = 116, normalized size = 2.11 \[ \frac {60 \, \sqrt {2 \, c d x + b d} a - \frac {10 \, {\left (3 \, \sqrt {2 \, c d x + b d} b d - {\left (2 \, c d x + b d\right )}^{\frac {3}{2}}\right )} b}{c d} + \frac {15 \, \sqrt {2 \, c d x + b d} b^{2} d^{2} - 10 \, {\left (2 \, c d x + b d\right )}^{\frac {3}{2}} b d + 3 \, {\left (2 \, c d x + b d\right )}^{\frac {5}{2}}}{c d^{2}}}{60 \, c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(2*c*d*x+b*d)^(1/2),x, algorithm="maxima")

[Out]

1/60*(60*sqrt(2*c*d*x + b*d)*a - 10*(3*sqrt(2*c*d*x + b*d)*b*d - (2*c*d*x + b*d)^(3/2))*b/(c*d) + (15*sqrt(2*c
*d*x + b*d)*b^2*d^2 - 10*(2*c*d*x + b*d)^(3/2)*b*d + 3*(2*c*d*x + b*d)^(5/2))/(c*d^2))/(c*d)

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mupad [B]  time = 0.04, size = 37, normalized size = 0.67 \[ \frac {\sqrt {b\,d+2\,c\,d\,x}\,\left (20\,a\,c+{\left (b+2\,c\,x\right )}^2-5\,b^2\right )}{20\,c^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)/(b*d + 2*c*d*x)^(1/2),x)

[Out]

((b*d + 2*c*d*x)^(1/2)*(20*a*c + (b + 2*c*x)^2 - 5*b^2))/(20*c^2*d)

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sympy [A]  time = 10.54, size = 258, normalized size = 4.69 \[ \begin {cases} \frac {- \frac {a b}{\sqrt {b d + 2 c d x}} - \frac {a \left (- \frac {b d}{\sqrt {b d + 2 c d x}} - \sqrt {b d + 2 c d x}\right )}{d} - \frac {b^{2} \left (- \frac {b d}{\sqrt {b d + 2 c d x}} - \sqrt {b d + 2 c d x}\right )}{2 c d} - \frac {3 b \left (\frac {b^{2} d^{2}}{\sqrt {b d + 2 c d x}} + 2 b d \sqrt {b d + 2 c d x} - \frac {\left (b d + 2 c d x\right )^{\frac {3}{2}}}{3}\right )}{4 c d^{2}} - \frac {- \frac {b^{3} d^{3}}{\sqrt {b d + 2 c d x}} - 3 b^{2} d^{2} \sqrt {b d + 2 c d x} + b d \left (b d + 2 c d x\right )^{\frac {3}{2}} - \frac {\left (b d + 2 c d x\right )^{\frac {5}{2}}}{5}}{4 c d^{3}}}{c} & \text {for}\: c \neq 0 \\\frac {a x + \frac {b x^{2}}{2}}{\sqrt {b d}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)/(2*c*d*x+b*d)**(1/2),x)

[Out]

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

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