3.9.22 \(\int \frac {a+b x+c x^2}{\sqrt {f+g x}} \, dx\) [822]

Optimal. Leaf size=73 \[ \frac {2 \left (c f^2-b f g+a g^2\right ) \sqrt {f+g x}}{g^3}-\frac {2 (2 c f-b g) (f+g x)^{3/2}}{3 g^3}+\frac {2 c (f+g x)^{5/2}}{5 g^3} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {712} \begin {gather*} \frac {2 \sqrt {f+g x} \left (a g^2-b f g+c f^2\right )}{g^3}-\frac {2 (f+g x)^{3/2} (2 c f-b g)}{3 g^3}+\frac {2 c (f+g x)^{5/2}}{5 g^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(c*f^2 - b*f*g + a*g^2)*Sqrt[f + g*x])/g^3 - (2*(2*c*f - b*g)*(f + g*x)^(3/2))/(3*g^3) + (2*c*(f + g*x)^(5/
2))/(5*g^3)

Rule 712

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] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 54, normalized size = 0.74 \begin {gather*} \frac {2 \sqrt {f+g x} \left (5 g (-2 b f+3 a g+b g x)+c \left (8 f^2-4 f g x+3 g^2 x^2\right )\right )}{15 g^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[f + g*x]*(5*g*(-2*b*f + 3*a*g + b*g*x) + c*(8*f^2 - 4*f*g*x + 3*g^2*x^2)))/(15*g^3)

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Maple [A]
time = 0.08, size = 75, normalized size = 1.03

method	result	size
gosper	\(\frac {2 \sqrt {g x +f}\, \left (3 c \,x^{2} g^{2}+5 b \,g^{2} x -4 c f g x +15 a \,g^{2}-10 b f g +8 c \,f^{2}\right )}{15 g^{3}}\)	\(53\)
trager	\(\frac {2 \sqrt {g x +f}\, \left (3 c \,x^{2} g^{2}+5 b \,g^{2} x -4 c f g x +15 a \,g^{2}-10 b f g +8 c \,f^{2}\right )}{15 g^{3}}\)	\(53\)
risch	\(\frac {2 \sqrt {g x +f}\, \left (3 c \,x^{2} g^{2}+5 b \,g^{2} x -4 c f g x +15 a \,g^{2}-10 b f g +8 c \,f^{2}\right )}{15 g^{3}}\)	\(53\)
derivativedivides	\(\frac {\frac {2 c \left (g x +f \right )^{\frac {5}{2}}}{5}+\frac {2 b g \left (g x +f \right )^{\frac {3}{2}}}{3}-\frac {4 c f \left (g x +f \right )^{\frac {3}{2}}}{3}+2 a \,g^{2} \sqrt {g x +f}-2 b f g \sqrt {g x +f}+2 c \,f^{2} \sqrt {g x +f}}{g^{3}}\)	\(75\)
default	\(\frac {\frac {2 c \left (g x +f \right )^{\frac {5}{2}}}{5}+\frac {2 b g \left (g x +f \right )^{\frac {3}{2}}}{3}-\frac {4 c f \left (g x +f \right )^{\frac {3}{2}}}{3}+2 a \,g^{2} \sqrt {g x +f}-2 b f g \sqrt {g x +f}+2 c \,f^{2} \sqrt {g x +f}}{g^{3}}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)/(g*x+f)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 77, normalized size = 1.05 \begin {gather*} \frac {2 \, {\left (15 \, \sqrt {g x + f} a + \frac {5 \, {\left ({\left (g x + f\right )}^{\frac {3}{2}} - 3 \, \sqrt {g x + f} f\right )} b}{g} + \frac {{\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} - 10 \, {\left (g x + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {g x + f} f^{2}\right )} c}{g^{2}}\right )}}{15 \, g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*sqrt(g*x + f)*a + 5*((g*x + f)^(3/2) - 3*sqrt(g*x + f)*f)*b/g + (3*(g*x + f)^(5/2) - 10*(g*x + f)^(3/
2)*f + 15*sqrt(g*x + f)*f^2)*c/g^2)/g

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Fricas [A]
time = 2.45, size = 54, normalized size = 0.74 \begin {gather*} \frac {2 \, {\left (3 \, c g^{2} x^{2} + 8 \, c f^{2} - 10 \, b f g + 15 \, a g^{2} - {\left (4 \, c f g - 5 \, b g^{2}\right )} x\right )} \sqrt {g x + f}}{15 \, g^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*c*g^2*x^2 + 8*c*f^2 - 10*b*f*g + 15*a*g^2 - (4*c*f*g - 5*b*g^2)*x)*sqrt(g*x + f)/g^3

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (73) = 146\).
time = 4.76, size = 223, normalized size = 3.05 \begin {gather*} \begin {cases} \frac {- \frac {2 a f}{\sqrt {f + g x}} - 2 a \left (- \frac {f}{\sqrt {f + g x}} - \sqrt {f + g x}\right ) - \frac {2 b f \left (- \frac {f}{\sqrt {f + g x}} - \sqrt {f + g x}\right )}{g} - \frac {2 b \left (\frac {f^{2}}{\sqrt {f + g x}} + 2 f \sqrt {f + g x} - \frac {\left (f + g x\right )^{\frac {3}{2}}}{3}\right )}{g} - \frac {2 c f \left (\frac {f^{2}}{\sqrt {f + g x}} + 2 f \sqrt {f + g x} - \frac {\left (f + g x\right )^{\frac {3}{2}}}{3}\right )}{g^{2}} - \frac {2 c \left (- \frac {f^{3}}{\sqrt {f + g x}} - 3 f^{2} \sqrt {f + g x} + f \left (f + g x\right )^{\frac {3}{2}} - \frac {\left (f + g x\right )^{\frac {5}{2}}}{5}\right )}{g^{2}}}{g} & \text {for}\: g \neq 0 \\\frac {a x + \frac {b x^{2}}{2} + \frac {c x^{3}}{3}}{\sqrt {f}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.96, size = 77, normalized size = 1.05 \begin {gather*} \frac {2 \, {\left (15 \, \sqrt {g x + f} a + \frac {5 \, {\left ({\left (g x + f\right )}^{\frac {3}{2}} - 3 \, \sqrt {g x + f} f\right )} b}{g} + \frac {{\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} - 10 \, {\left (g x + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {g x + f} f^{2}\right )} c}{g^{2}}\right )}}{15 \, g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*sqrt(g*x + f)*a + 5*((g*x + f)^(3/2) - 3*sqrt(g*x + f)*f)*b/g + (3*(g*x + f)^(5/2) - 10*(g*x + f)^(3/
2)*f + 15*sqrt(g*x + f)*f^2)*c/g^2)/g

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Mupad [B]
time = 3.12, size = 58, normalized size = 0.79 \begin {gather*} \frac {2\,\sqrt {f+g\,x}\,\left (3\,c\,{\left (f+g\,x\right )}^2+15\,a\,g^2+15\,c\,f^2+5\,b\,g\,\left (f+g\,x\right )-10\,c\,f\,\left (f+g\,x\right )-15\,b\,f\,g\right )}{15\,g^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(f + g*x)^(1/2)*(3*c*(f + g*x)^2 + 15*a*g^2 + 15*c*f^2 + 5*b*g*(f + g*x) - 10*c*f*(f + g*x) - 15*b*f*g))/(1
5*g^3)

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