3.6.72 \(\int \frac {a+b x+c x^2}{(f+g x)^{3/2}} \, dx\)

Optimal. Leaf size=71 \[ -\frac {2 \left (a g^2-b f g+c f^2\right )}{g^3 \sqrt {f+g x}}-\frac {2 \sqrt {f+g x} (2 c f-b g)}{g^3}+\frac {2 c (f+g x)^{3/2}}{3 g^3} \]

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Rubi [A]  time = 0.04, antiderivative size = 71, 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 = {698} \begin {gather*} -\frac {2 \left (a g^2-b f g+c f^2\right )}{g^3 \sqrt {f+g x}}-\frac {2 \sqrt {f+g x} (2 c f-b g)}{g^3}+\frac {2 c (f+g x)^{3/2}}{3 g^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)/(f + g*x)^(3/2),x]

[Out]

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

Rule 698

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

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Mathematica [A]  time = 0.06, size = 54, normalized size = 0.76 \begin {gather*} \frac {6 g (-a g+2 b f+b g x)+2 c \left (-8 f^2-4 f g x+g^2 x^2\right )}{3 g^3 \sqrt {f+g x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)/(f + g*x)^(3/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.04, size = 61, normalized size = 0.86 \begin {gather*} \frac {2 \left (-3 a g^2+3 b g (f+g x)+3 b f g-3 c f^2-6 c f (f+g x)+c (f+g x)^2\right )}{3 g^3 \sqrt {f+g x}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x + c*x^2)/(f + g*x)^(3/2),x]

[Out]

(2*(-3*c*f^2 + 3*b*f*g - 3*a*g^2 - 6*c*f*(f + g*x) + 3*b*g*(f + g*x) + c*(f + g*x)^2))/(3*g^3*Sqrt[f + g*x])

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fricas [A]  time = 0.40, size = 63, normalized size = 0.89 \begin {gather*} \frac {2 \, {\left (c g^{2} x^{2} - 8 \, c f^{2} + 6 \, b f g - 3 \, a g^{2} - {\left (4 \, c f g - 3 \, b g^{2}\right )} x\right )} \sqrt {g x + f}}{3 \, {\left (g^{4} x + f g^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(g*x+f)^(3/2),x, algorithm="fricas")

[Out]

2/3*(c*g^2*x^2 - 8*c*f^2 + 6*b*f*g - 3*a*g^2 - (4*c*f*g - 3*b*g^2)*x)*sqrt(g*x + f)/(g^4*x + f*g^3)

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giac [A]  time = 0.15, size = 74, normalized size = 1.04 \begin {gather*} -\frac {2 \, {\left (c f^{2} - b f g + a g^{2}\right )}}{\sqrt {g x + f} g^{3}} + \frac {2 \, {\left ({\left (g x + f\right )}^{\frac {3}{2}} c g^{6} - 6 \, \sqrt {g x + f} c f g^{6} + 3 \, \sqrt {g x + f} b g^{7}\right )}}{3 \, g^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(g*x+f)^(3/2),x, algorithm="giac")

[Out]

-2*(c*f^2 - b*f*g + a*g^2)/(sqrt(g*x + f)*g^3) + 2/3*((g*x + f)^(3/2)*c*g^6 - 6*sqrt(g*x + f)*c*f*g^6 + 3*sqrt
(g*x + f)*b*g^7)/g^9

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maple [A]  time = 0.00, size = 53, normalized size = 0.75 \begin {gather*} -\frac {2 \left (-c \,x^{2} g^{2}-3 b \,g^{2} x +4 c f g x +3 a \,g^{2}-6 b f g +8 c \,f^{2}\right )}{3 \sqrt {g x +f}\, g^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)/(g*x+f)^(3/2),x)

[Out]

-2/3/(g*x+f)^(1/2)*(-c*g^2*x^2-3*b*g^2*x+4*c*f*g*x+3*a*g^2-6*b*f*g+8*c*f^2)/g^3

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maxima [A]  time = 0.44, size = 66, normalized size = 0.93 \begin {gather*} \frac {2 \, {\left (\frac {{\left (g x + f\right )}^{\frac {3}{2}} c - 3 \, {\left (2 \, c f - b g\right )} \sqrt {g x + f}}{g^{2}} - \frac {3 \, {\left (c f^{2} - b f g + a g^{2}\right )}}{\sqrt {g x + f} g^{2}}\right )}}{3 \, g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(g*x+f)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.06, size = 58, normalized size = 0.82 \begin {gather*} \frac {2\,c\,{\left (f+g\,x\right )}^2-6\,a\,g^2-6\,c\,f^2+6\,b\,g\,\left (f+g\,x\right )-12\,c\,f\,\left (f+g\,x\right )+6\,b\,f\,g}{3\,g^3\,\sqrt {f+g\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)/(f + g*x)^(3/2),x)

[Out]

(2*c*(f + g*x)^2 - 6*a*g^2 - 6*c*f^2 + 6*b*g*(f + g*x) - 12*c*f*(f + g*x) + 6*b*f*g)/(3*g^3*(f + g*x)^(1/2))

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sympy [A]  time = 13.12, size = 70, normalized size = 0.99 \begin {gather*} \frac {2 c \left (f + g x\right )^{\frac {3}{2}}}{3 g^{3}} + \frac {\sqrt {f + g x} \left (2 b g - 4 c f\right )}{g^{3}} - \frac {2 \left (a g^{2} - b f g + c f^{2}\right )}{g^{3} \sqrt {f + g x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)/(g*x+f)**(3/2),x)

[Out]

2*c*(f + g*x)**(3/2)/(3*g**3) + sqrt(f + g*x)*(2*b*g - 4*c*f)/g**3 - 2*(a*g**2 - b*f*g + c*f**2)/(g**3*sqrt(f
+ g*x))

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