\(\int \frac {(d+e x) (a+b x+c x^2)}{\sqrt {f+g x}} \, dx\) [821]

Optimal result

Integrand size = 25, antiderivative size = 137 \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{\sqrt {f+g x}} \, dx=-\frac {2 (e f-d g) \left (c f^2-b f g+a g^2\right ) \sqrt {f+g x}}{g^4}+\frac {2 (c f (3 e f-2 d g)-g (2 b e f-b d g-a e g)) (f+g x)^{3/2}}{3 g^4}-\frac {2 (3 c e f-c d g-b e g) (f+g x)^{5/2}}{5 g^4}+\frac {2 c e (f+g x)^{7/2}}{7 g^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {785} \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{\sqrt {f+g x}} \, dx=-\frac {2 \sqrt {f+g x} (e f-d g) \left (a g^2-b f g+c f^2\right )}{g^4}+\frac {2 (f+g x)^{3/2} (c f (3 e f-2 d g)-g (-a e g-b d g+2 b e f))}{3 g^4}-\frac {2 (f+g x)^{5/2} (-b e g-c d g+3 c e f)}{5 g^4}+\frac {2 c e (f+g x)^{7/2}}{7 g^4} \]

[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 785

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.96 \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{\sqrt {f+g x}} \, dx=\frac {2 \sqrt {f+g x} \left (7 g \left (5 b d g (-2 f+g x)+5 a g (-2 e f+3 d g+e g x)+b e \left (8 f^2-4 f g x+3 g^2 x^2\right )\right )+c \left (7 d g \left (8 f^2-4 f g x+3 g^2 x^2\right )-3 e \left (16 f^3-8 f^2 g x+6 f g^2 x^2-5 g^3 x^3\right )\right )\right )}{105 g^4} \]

[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)

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(\frac {2 \sqrt {g x +f}\, \left (\left (\frac {\left (\frac {3}{7} c \,x^{2}+\frac {3}{5} b x +a \right ) x e}{3}+d \left (\frac {1}{5} c \,x^{2}+\frac {1}{3} b x +a \right )\right ) g^{3}-\frac {2 f \left (\left (\frac {9}{35} c \,x^{2}+\frac {2}{5} b x +a \right ) e +d \left (\frac {2 c x}{5}+b \right )\right ) g^{2}}{3}+\frac {8 \left (\left (\frac {3 c x}{7}+b \right ) e +c d \right ) f^{2} g}{15}-\frac {16 c e \,f^{3}}{35}\right )}{g^{4}}\)	\(103\)
derivativedivides	\(\frac {\frac {2 c e \left (g x +f \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (d g -e f \right ) c +e \left (b g -2 c f \right )\right ) \left (g x +f \right )^{\frac {5}{2}}}{5}+\frac {2 \left (\left (d g -e f \right ) \left (b g -2 c f \right )+e \left (a \,g^{2}-b f g +c \,f^{2}\right )\right ) \left (g x +f \right )^{\frac {3}{2}}}{3}+2 \left (d g -e f \right ) \left (a \,g^{2}-b f g +c \,f^{2}\right ) \sqrt {g x +f}}{g^{4}}\)	\(125\)
default	\(\frac {\frac {2 c e \left (g x +f \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (d g -e f \right ) c +e \left (b g -2 c f \right )\right ) \left (g x +f \right )^{\frac {5}{2}}}{5}+\frac {2 \left (\left (d g -e f \right ) \left (b g -2 c f \right )+e \left (a \,g^{2}-b f g +c \,f^{2}\right )\right ) \left (g x +f \right )^{\frac {3}{2}}}{3}+2 \left (d g -e f \right ) \left (a \,g^{2}-b f g +c \,f^{2}\right ) \sqrt {g x +f}}{g^{4}}\)	\(125\)
gosper	\(\frac {2 \sqrt {g x +f}\, \left (15 c e \,x^{3} g^{3}+21 b e \,g^{3} x^{2}+21 c d \,g^{3} x^{2}-18 c e f \,g^{2} x^{2}+35 a e \,g^{3} x +35 b d \,g^{3} x -28 b e f \,g^{2} x -28 c d f \,g^{2} x +24 c e \,f^{2} g x +105 a d \,g^{3}-70 a e f \,g^{2}-70 b d f \,g^{2}+56 b e \,f^{2} g +56 d \,f^{2} g c -48 c e \,f^{3}\right )}{105 g^{4}}\)	\(144\)
trager	\(\frac {2 \sqrt {g x +f}\, \left (15 c e \,x^{3} g^{3}+21 b e \,g^{3} x^{2}+21 c d \,g^{3} x^{2}-18 c e f \,g^{2} x^{2}+35 a e \,g^{3} x +35 b d \,g^{3} x -28 b e f \,g^{2} x -28 c d f \,g^{2} x +24 c e \,f^{2} g x +105 a d \,g^{3}-70 a e f \,g^{2}-70 b d f \,g^{2}+56 b e \,f^{2} g +56 d \,f^{2} g c -48 c e \,f^{3}\right )}{105 g^{4}}\)	\(144\)
risch	\(\frac {2 \sqrt {g x +f}\, \left (15 c e \,x^{3} g^{3}+21 b e \,g^{3} x^{2}+21 c d \,g^{3} x^{2}-18 c e f \,g^{2} x^{2}+35 a e \,g^{3} x +35 b d \,g^{3} x -28 b e f \,g^{2} x -28 c d f \,g^{2} x +24 c e \,f^{2} g x +105 a d \,g^{3}-70 a e f \,g^{2}-70 b d f \,g^{2}+56 b e \,f^{2} g +56 d \,f^{2} g c -48 c e \,f^{3}\right )}{105 g^{4}}\)	\(144\)

[In]

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

[Out]

2*(g*x+f)^(1/2)*((1/3*(3/7*c*x^2+3/5*b*x+a)*x*e+d*(1/5*c*x^2+1/3*b*x+a))*g^3-2/3*f*((9/35*c*x^2+2/5*b*x+a)*e+d
*(2/5*c*x+b))*g^2+8/15*((3/7*c*x+b)*e+c*d)*f^2*g-16/35*c*e*f^3)/g^4

Fricas [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.91 \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{\sqrt {f+g x}} \, dx=\frac {2 \, {\left (15 \, c e g^{3} x^{3} - 48 \, c e f^{3} + 105 \, a d g^{3} + 56 \, {\left (c d + b e\right )} f^{2} g - 70 \, {\left (b d + a e\right )} f g^{2} - 3 \, {\left (6 \, c e f g^{2} - 7 \, {\left (c d + b e\right )} g^{3}\right )} x^{2} + {\left (24 \, c e f^{2} g - 28 \, {\left (c d + b e\right )} f g^{2} + 35 \, {\left (b d + a e\right )} g^{3}\right )} x\right )} \sqrt {g x + f}}{105 \, g^{4}} \]

[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

Sympy [A] (verification not implemented)

Time = 0.85 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.53 \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{\sqrt {f+g x}} \, dx=\begin {cases} \frac {2 \left (\frac {c e \left (f + g x\right )^{\frac {7}{2}}}{7 g^{3}} + \frac {\left (f + g x\right )^{\frac {5}{2}} \left (b e g + c d g - 3 c e f\right )}{5 g^{3}} + \frac {\left (f + g x\right )^{\frac {3}{2}} \left (a e g^{2} + b d g^{2} - 2 b e f g - 2 c d f g + 3 c e f^{2}\right )}{3 g^{3}} + \frac {\sqrt {f + g x} \left (a d g^{3} - a e f g^{2} - b d f g^{2} + b e f^{2} g + c d f^{2} g - c e f^{3}\right )}{g^{3}}\right )}{g} & \text {for}\: g \neq 0 \\\frac {a d x + \frac {c e x^{4}}{4} + \frac {x^{3} \left (b e + c d\right )}{3} + \frac {x^{2} \left (a e + b d\right )}{2}}{\sqrt {f}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(c*e*(f + g*x)**(7/2)/(7*g**3) + (f + g*x)**(5/2)*(b*e*g + c*d*g - 3*c*e*f)/(5*g**3) + (f + g*x)*
*(3/2)*(a*e*g**2 + b*d*g**2 - 2*b*e*f*g - 2*c*d*f*g + 3*c*e*f**2)/(3*g**3) + sqrt(f + g*x)*(a*d*g**3 - a*e*f*g
**2 - b*d*f*g**2 + b*e*f**2*g + c*d*f**2*g - c*e*f**3)/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), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.94 \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{\sqrt {f+g x}} \, dx=\frac {2 \, {\left (15 \, {\left (g x + f\right )}^{\frac {7}{2}} c e - 21 \, {\left (3 \, c e f - {\left (c d + b e\right )} g\right )} {\left (g x + f\right )}^{\frac {5}{2}} + 35 \, {\left (3 \, c e f^{2} - 2 \, {\left (c d + b e\right )} f g + {\left (b d + a e\right )} g^{2}\right )} {\left (g x + f\right )}^{\frac {3}{2}} - 105 \, {\left (c e f^{3} - a d g^{3} - {\left (c d + b e\right )} f^{2} g + {\left (b d + a e\right )} f g^{2}\right )} \sqrt {g x + f}\right )}}{105 \, g^{4}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.43 \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{\sqrt {f+g x}} \, dx=\frac {2 \, {\left (105 \, \sqrt {g x + f} a d + \frac {35 \, {\left ({\left (g x + f\right )}^{\frac {3}{2}} - 3 \, \sqrt {g x + f} f\right )} b d}{g} + \frac {35 \, {\left ({\left (g x + f\right )}^{\frac {3}{2}} - 3 \, \sqrt {g x + f} f\right )} a e}{g} + \frac {7 \, {\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} - 10 \, {\left (g x + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {g x + f} f^{2}\right )} c d}{g^{2}} + \frac {7 \, {\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} - 10 \, {\left (g x + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {g x + f} f^{2}\right )} b e}{g^{2}} + \frac {3 \, {\left (5 \, {\left (g x + f\right )}^{\frac {7}{2}} - 21 \, {\left (g x + f\right )}^{\frac {5}{2}} f + 35 \, {\left (g x + f\right )}^{\frac {3}{2}} f^{2} - 35 \, \sqrt {g x + f} f^{3}\right )} c e}{g^{3}}\right )}}{105 \, g} \]

[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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.91 \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{\sqrt {f+g x}} \, dx=\frac {{\left (f+g\,x\right )}^{5/2}\,\left (2\,b\,e\,g+2\,c\,d\,g-6\,c\,e\,f\right )}{5\,g^4}+\frac {{\left (f+g\,x\right )}^{3/2}\,\left (2\,a\,e\,g^2+2\,b\,d\,g^2+6\,c\,e\,f^2-4\,b\,e\,f\,g-4\,c\,d\,f\,g\right )}{3\,g^4}+\frac {2\,\sqrt {f+g\,x}\,\left (d\,g-e\,f\right )\,\left (c\,f^2-b\,f\,g+a\,g^2\right )}{g^4}+\frac {2\,c\,e\,{\left (f+g\,x\right )}^{7/2}}{7\,g^4} \]

[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)