3.6.64 \(\int \frac {(d+e x) (a+b x+c x^2)}{\sqrt {f+g x}} \, dx\)

Optimal. Leaf size=137 \[ -\frac {2 \sqrt {f+g x} (e f-d g) \left (a g^2-b f g+c f^2\right )}{g^4}+\frac {2 (f+g x)^{3/2} (c f (3 e f-2 d g)-g (-a e g-b d g+2 b e f))}{3 g^4}-\frac {2 (f+g x)^{5/2} (-b e g-c d g+3 c e f)}{5 g^4}+\frac {2 c e (f+g x)^{7/2}}{7 g^4} \]

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Rubi [A]  time = 0.11, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {771} \begin {gather*} -\frac {2 \sqrt {f+g x} (e f-d g) \left (a g^2-b f g+c f^2\right )}{g^4}+\frac {2 (f+g x)^{3/2} (c f (3 e f-2 d g)-g (-a e g-b d g+2 b e f))}{3 g^4}-\frac {2 (f+g x)^{5/2} (-b e g-c d g+3 c e f)}{5 g^4}+\frac {2 c e (f+g x)^{7/2}}{7 g^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(e*f - d*g)*(c*f^2 - b*f*g + a*g^2)*Sqrt[f + g*x])/g^4 + (2*(c*f*(3*e*f - 2*d*g) - g*(2*b*e*f - b*d*g - a*
e*g))*(f + g*x)^(3/2))/(3*g^4) - (2*(3*c*e*f - c*d*g - b*e*g)*(f + g*x)^(5/2))/(5*g^4) + (2*c*e*(f + g*x)^(7/2
))/(7*g^4)

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {align*} \int \frac {(d+e x) \left (a+b x+c x^2\right )}{\sqrt {f+g x}} \, dx &=\int \left (\frac {(-e f+d g) \left (c f^2-b f g+a g^2\right )}{g^3 \sqrt {f+g x}}+\frac {(c f (3 e f-2 d g)-g (2 b e f-b d g-a e g)) \sqrt {f+g x}}{g^3}+\frac {(-3 c e f+c d g+b e g) (f+g x)^{3/2}}{g^3}+\frac {c e (f+g x)^{5/2}}{g^3}\right ) \, dx\\ &=-\frac {2 (e f-d g) \left (c f^2-b f g+a g^2\right ) \sqrt {f+g x}}{g^4}+\frac {2 (c f (3 e f-2 d g)-g (2 b e f-b d g-a e g)) (f+g x)^{3/2}}{3 g^4}-\frac {2 (3 c e f-c d g-b e g) (f+g x)^{5/2}}{5 g^4}+\frac {2 c e (f+g x)^{7/2}}{7 g^4}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 131, normalized size = 0.96 \begin {gather*} \frac {2 \sqrt {f+g x} \left (7 g \left (5 a g (3 d g-2 e f+e g x)+5 b d g (g x-2 f)+b e \left (8 f^2-4 f g x+3 g^2 x^2\right )\right )+c \left (7 d g \left (8 f^2-4 f g x+3 g^2 x^2\right )-3 e \left (16 f^3-8 f^2 g x+6 f g^2 x^2-5 g^3 x^3\right )\right )\right )}{105 g^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[f + g*x]*(7*g*(5*b*d*g*(-2*f + g*x) + 5*a*g*(-2*e*f + 3*d*g + e*g*x) + b*e*(8*f^2 - 4*f*g*x + 3*g^2*x^
2)) + c*(7*d*g*(8*f^2 - 4*f*g*x + 3*g^2*x^2) - 3*e*(16*f^3 - 8*f^2*g*x + 6*f*g^2*x^2 - 5*g^3*x^3))))/(105*g^4)

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IntegrateAlgebraic [A]  time = 0.13, size = 168, normalized size = 1.23 \begin {gather*} \frac {2 \sqrt {f+g x} \left (105 a d g^3+35 a e g^2 (f+g x)-105 a e f g^2+35 b d g^2 (f+g x)-105 b d f g^2+105 b e f^2 g-70 b e f g (f+g x)+21 b e g (f+g x)^2+105 c d f^2 g-70 c d f g (f+g x)+21 c d g (f+g x)^2-105 c e f^3+105 c e f^2 (f+g x)-63 c e f (f+g x)^2+15 c e (f+g x)^3\right )}{105 g^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[f + g*x]*(-105*c*e*f^3 + 105*c*d*f^2*g + 105*b*e*f^2*g - 105*b*d*f*g^2 - 105*a*e*f*g^2 + 105*a*d*g^3 +
105*c*e*f^2*(f + g*x) - 70*c*d*f*g*(f + g*x) - 70*b*e*f*g*(f + g*x) + 35*b*d*g^2*(f + g*x) + 35*a*e*g^2*(f +
g*x) - 63*c*e*f*(f + g*x)^2 + 21*c*d*g*(f + g*x)^2 + 21*b*e*g*(f + g*x)^2 + 15*c*e*(f + g*x)^3))/(105*g^4)

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fricas [A]  time = 0.56, size = 125, normalized size = 0.91 \begin {gather*} \frac {2 \, {\left (15 \, c e g^{3} x^{3} - 48 \, c e f^{3} + 105 \, a d g^{3} + 56 \, {\left (c d + b e\right )} f^{2} g - 70 \, {\left (b d + a e\right )} f g^{2} - 3 \, {\left (6 \, c e f g^{2} - 7 \, {\left (c d + b e\right )} g^{3}\right )} x^{2} + {\left (24 \, c e f^{2} g - 28 \, {\left (c d + b e\right )} f g^{2} + 35 \, {\left (b d + a e\right )} g^{3}\right )} x\right )} \sqrt {g x + f}}{105 \, g^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*c*e*g^3*x^3 - 48*c*e*f^3 + 105*a*d*g^3 + 56*(c*d + b*e)*f^2*g - 70*(b*d + a*e)*f*g^2 - 3*(6*c*e*f*g^
2 - 7*(c*d + b*e)*g^3)*x^2 + (24*c*e*f^2*g - 28*(c*d + b*e)*f*g^2 + 35*(b*d + a*e)*g^3)*x)*sqrt(g*x + f)/g^4

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giac [A]  time = 0.19, size = 199, normalized size = 1.45 \begin {gather*} \frac {2 \, {\left (105 \, \sqrt {g x + f} a d + \frac {35 \, {\left ({\left (g x + f\right )}^{\frac {3}{2}} - 3 \, \sqrt {g x + f} f\right )} b d}{g} + \frac {35 \, {\left ({\left (g x + f\right )}^{\frac {3}{2}} - 3 \, \sqrt {g x + f} f\right )} a e}{g} + \frac {7 \, {\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 d}{g^{2}} + \frac {7 \, {\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 )} b e}{g^{2}} + \frac {3 \, {\left (5 \, {\left (g x + f\right )}^{\frac {7}{2}} - 21 \, {\left (g x + f\right )}^{\frac {5}{2}} f + 35 \, {\left (g x + f\right )}^{\frac {3}{2}} f^{2} - 35 \, \sqrt {g x + f} f^{3}\right )} c e}{g^{3}}\right )}}{105 \, g} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(105*sqrt(g*x + f)*a*d + 35*((g*x + f)^(3/2) - 3*sqrt(g*x + f)*f)*b*d/g + 35*((g*x + f)^(3/2) - 3*sqrt(g
*x + f)*f)*a*e/g + 7*(3*(g*x + f)^(5/2) - 10*(g*x + f)^(3/2)*f + 15*sqrt(g*x + f)*f^2)*c*d/g^2 + 7*(3*(g*x + f
)^(5/2) - 10*(g*x + f)^(3/2)*f + 15*sqrt(g*x + f)*f^2)*b*e/g^2 + 3*(5*(g*x + f)^(7/2) - 21*(g*x + f)^(5/2)*f +
35*(g*x + f)^(3/2)*f^2 - 35*sqrt(g*x + f)*f^3)*c*e/g^3)/g

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maple [A]  time = 0.01, size = 144, normalized size = 1.05 \begin {gather*} \frac {2 \sqrt {g x +f}\, \left (15 c e \,x^{3} g^{3}+21 b e \,g^{3} x^{2}+21 c d \,g^{3} x^{2}-18 c e f \,g^{2} x^{2}+35 a e \,g^{3} x +35 b d \,g^{3} x -28 b e f \,g^{2} x -28 c d f \,g^{2} x +24 c e \,f^{2} g x +105 a d \,g^{3}-70 a e f \,g^{2}-70 b d f \,g^{2}+56 b e \,f^{2} g +56 c d \,f^{2} g -48 c e \,f^{3}\right )}{105 g^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(g*x+f)^(1/2)*(15*c*e*g^3*x^3+21*b*e*g^3*x^2+21*c*d*g^3*x^2-18*c*e*f*g^2*x^2+35*a*e*g^3*x+35*b*d*g^3*x-2
8*b*e*f*g^2*x-28*c*d*f*g^2*x+24*c*e*f^2*g*x+105*a*d*g^3-70*a*e*f*g^2-70*b*d*f*g^2+56*b*e*f^2*g+56*c*d*f^2*g-48
*c*e*f^3)/g^4

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maxima [A]  time = 0.46, size = 129, normalized size = 0.94 \begin {gather*} \frac {2 \, {\left (15 \, {\left (g x + f\right )}^{\frac {7}{2}} c e - 21 \, {\left (3 \, c e f - {\left (c d + b e\right )} g\right )} {\left (g x + f\right )}^{\frac {5}{2}} + 35 \, {\left (3 \, c e f^{2} - 2 \, {\left (c d + b e\right )} f g + {\left (b d + a e\right )} g^{2}\right )} {\left (g x + f\right )}^{\frac {3}{2}} - 105 \, {\left (c e f^{3} - a d g^{3} - {\left (c d + b e\right )} f^{2} g + {\left (b d + a e\right )} f g^{2}\right )} \sqrt {g x + f}\right )}}{105 \, g^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*(g*x + f)^(7/2)*c*e - 21*(3*c*e*f - (c*d + b*e)*g)*(g*x + f)^(5/2) + 35*(3*c*e*f^2 - 2*(c*d + b*e)*f
*g + (b*d + a*e)*g^2)*(g*x + f)^(3/2) - 105*(c*e*f^3 - a*d*g^3 - (c*d + b*e)*f^2*g + (b*d + a*e)*f*g^2)*sqrt(g
*x + f))/g^4

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mupad [B]  time = 0.08, size = 125, normalized size = 0.91 \begin {gather*} \frac {{\left (f+g\,x\right )}^{5/2}\,\left (2\,b\,e\,g+2\,c\,d\,g-6\,c\,e\,f\right )}{5\,g^4}+\frac {{\left (f+g\,x\right )}^{3/2}\,\left (2\,a\,e\,g^2+2\,b\,d\,g^2+6\,c\,e\,f^2-4\,b\,e\,f\,g-4\,c\,d\,f\,g\right )}{3\,g^4}+\frac {2\,\sqrt {f+g\,x}\,\left (d\,g-e\,f\right )\,\left (c\,f^2-b\,f\,g+a\,g^2\right )}{g^4}+\frac {2\,c\,e\,{\left (f+g\,x\right )}^{7/2}}{7\,g^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((f + g*x)^(5/2)*(2*b*e*g + 2*c*d*g - 6*c*e*f))/(5*g^4) + ((f + g*x)^(3/2)*(2*a*e*g^2 + 2*b*d*g^2 + 6*c*e*f^2
- 4*b*e*f*g - 4*c*d*f*g))/(3*g^4) + (2*(f + g*x)^(1/2)*(d*g - e*f)*(a*g^2 + c*f^2 - b*f*g))/g^4 + (2*c*e*(f +
g*x)^(7/2))/(7*g^4)

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sympy [A]  time = 55.83, size = 549, normalized size = 4.01 \begin {gather*} \begin {cases} \frac {- \frac {2 a d f}{\sqrt {f + g x}} - 2 a d \left (- \frac {f}{\sqrt {f + g x}} - \sqrt {f + g x}\right ) - \frac {2 a e f \left (- \frac {f}{\sqrt {f + g x}} - \sqrt {f + g x}\right )}{g} - \frac {2 a e \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 b d f \left (- \frac {f}{\sqrt {f + g x}} - \sqrt {f + g x}\right )}{g} - \frac {2 b d \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 b e 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 b e \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}} - \frac {2 c d 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 d \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}} - \frac {2 c e f \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^{3}} - \frac {2 c e \left (\frac {f^{4}}{\sqrt {f + g x}} + 4 f^{3} \sqrt {f + g x} - 2 f^{2} \left (f + g x\right )^{\frac {3}{2}} + \frac {4 f \left (f + g x\right )^{\frac {5}{2}}}{5} - \frac {\left (f + g x\right )^{\frac {7}{2}}}{7}\right )}{g^{3}}}{g} & \text {for}\: g \neq 0 \\\frac {a d x + \frac {c e x^{4}}{4} + \frac {x^{3} \left (b e + c d\right )}{3} + \frac {x^{2} \left (a e + b d\right )}{2}}{\sqrt {f}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-2*a*d*f/sqrt(f + g*x) - 2*a*d*(-f/sqrt(f + g*x) - sqrt(f + g*x)) - 2*a*e*f*(-f/sqrt(f + g*x) - sq
rt(f + g*x))/g - 2*a*e*(f**2/sqrt(f + g*x) + 2*f*sqrt(f + g*x) - (f + g*x)**(3/2)/3)/g - 2*b*d*f*(-f/sqrt(f +
g*x) - sqrt(f + g*x))/g - 2*b*d*(f**2/sqrt(f + g*x) + 2*f*sqrt(f + g*x) - (f + g*x)**(3/2)/3)/g - 2*b*e*f*(f**
2/sqrt(f + g*x) + 2*f*sqrt(f + g*x) - (f + g*x)**(3/2)/3)/g**2 - 2*b*e*(-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 - 2*c*d*f*(f**2/sqrt(f + g*x) + 2*f*sqrt(f + g*x) - (f +
g*x)**(3/2)/3)/g**2 - 2*c*d*(-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 - 2*c*e*f*(-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**
3 - 2*c*e*(f**4/sqrt(f + g*x) + 4*f**3*sqrt(f + g*x) - 2*f**2*(f + g*x)**(3/2) + 4*f*(f + g*x)**(5/2)/5 - (f +
g*x)**(7/2)/7)/g**3)/g, Ne(g, 0)), ((a*d*x + c*e*x**4/4 + x**3*(b*e + c*d)/3 + x**2*(a*e + b*d)/2)/sqrt(f), T
rue))

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