\(\int \frac {(d+e x) (a+b x+c x^2)}{(f+g x)^{3/2}} \, dx\) [828]

Optimal result

Integrand size = 25, antiderivative size = 135 \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{(f+g x)^{3/2}} \, dx=\frac {2 (e f-d g) \left (c f^2-b f g+a g^2\right )}{g^4 \sqrt {f+g x}}+\frac {2 (c f (3 e f-2 d g)-g (2 b e f-b d g-a e g)) \sqrt {f+g x}}{g^4}-\frac {2 (3 c e f-c d g-b e g) (f+g x)^{3/2}}{3 g^4}+\frac {2 c e (f+g x)^{5/2}}{5 g^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 135, 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} \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{(f+g x)^{3/2}} \, dx=\frac {2 (e f-d g) \left (a g^2-b f g+c f^2\right )}{g^4 \sqrt {f+g x}}+\frac {2 \sqrt {f+g x} (c f (3 e f-2 d g)-g (-a e g-b d g+2 b e f))}{g^4}-\frac {2 (f+g x)^{3/2} (-b e g-c d g+3 c e f)}{3 g^4}+\frac {2 c e (f+g x)^{5/2}}{5 g^4} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.95 \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{(f+g x)^{3/2}} \, dx=\frac {2 \left (5 g \left (3 b d g (2 f+g x)+3 a g (2 e f-d g+e g x)+b e \left (-8 f^2-4 f g x+g^2 x^2\right )\right )+c \left (5 d g \left (-8 f^2-4 f g x+g^2 x^2\right )+3 e \left (16 f^3+8 f^2 g x-2 f g^2 x^2+g^3 x^3\right )\right )\right )}{15 g^4 \sqrt {f+g x}} \]

[In]

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

[Out]

(2*(5*g*(3*b*d*g*(2*f + g*x) + 3*a*g*(2*e*f - d*g + e*g*x) + b*e*(-8*f^2 - 4*f*g*x + g^2*x^2)) + c*(5*d*g*(-8*
f^2 - 4*f*g*x + g^2*x^2) + 3*e*(16*f^3 + 8*f^2*g*x - 2*f*g^2*x^2 + g^3*x^3))))/(15*g^4*Sqrt[f + g*x])

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(\frac {\left (\left (6 c \,x^{3}+10 b \,x^{2}+30 a x \right ) e -30 d \left (a -\frac {1}{3} c \,x^{2}-b x \right )\right ) g^{3}+60 \left (\left (-\frac {1}{5} c \,x^{2}-\frac {2}{3} b x +a \right ) e +d \left (-\frac {2 c x}{3}+b \right )\right ) f \,g^{2}-80 \left (\left (-\frac {3 c x}{5}+b \right ) e +c d \right ) f^{2} g +96 c e \,f^{3}}{15 \sqrt {g x +f}\, g^{4}}\)	\(107\)
risch	\(\frac {2 \left (3 c e \,x^{2} g^{2}+5 b e x \,g^{2}+5 c d \,g^{2} x -9 c e f g x +15 a e \,g^{2}+15 b d \,g^{2}-25 b e f g -25 c d f g +33 c e \,f^{2}\right ) \sqrt {g x +f}}{15 g^{4}}-\frac {2 \left (a d \,g^{3}-a e f \,g^{2}-b d f \,g^{2}+b e \,f^{2} g +d \,f^{2} g c -c e \,f^{3}\right )}{g^{4} \sqrt {g x +f}}\)	\(137\)
gosper	\(-\frac {2 \left (-3 c e \,x^{3} g^{3}-5 b e \,g^{3} x^{2}-5 c d \,g^{3} x^{2}+6 c e f \,g^{2} x^{2}-15 a e \,g^{3} x -15 b d \,g^{3} x +20 b e f \,g^{2} x +20 c d f \,g^{2} x -24 c e \,f^{2} g x +15 a d \,g^{3}-30 a e f \,g^{2}-30 b d f \,g^{2}+40 b e \,f^{2} g +40 d \,f^{2} g c -48 c e \,f^{3}\right )}{15 \sqrt {g x +f}\, g^{4}}\)	\(144\)
trager	\(-\frac {2 \left (-3 c e \,x^{3} g^{3}-5 b e \,g^{3} x^{2}-5 c d \,g^{3} x^{2}+6 c e f \,g^{2} x^{2}-15 a e \,g^{3} x -15 b d \,g^{3} x +20 b e f \,g^{2} x +20 c d f \,g^{2} x -24 c e \,f^{2} g x +15 a d \,g^{3}-30 a e f \,g^{2}-30 b d f \,g^{2}+40 b e \,f^{2} g +40 d \,f^{2} g c -48 c e \,f^{3}\right )}{15 \sqrt {g x +f}\, g^{4}}\)	\(144\)
derivativedivides	\(\frac {\frac {2 c e \left (g x +f \right )^{\frac {5}{2}}}{5}+\frac {2 b e g \left (g x +f \right )^{\frac {3}{2}}}{3}+\frac {2 c d g \left (g x +f \right )^{\frac {3}{2}}}{3}-2 c e f \left (g x +f \right )^{\frac {3}{2}}+2 a e \,g^{2} \sqrt {g x +f}+2 b d \,g^{2} \sqrt {g x +f}-4 b e f g \sqrt {g x +f}-4 c d f g \sqrt {g x +f}+6 c e \,f^{2} \sqrt {g x +f}-\frac {2 \left (a d \,g^{3}-a e f \,g^{2}-b d f \,g^{2}+b e \,f^{2} g +d \,f^{2} g c -c e \,f^{3}\right )}{\sqrt {g x +f}}}{g^{4}}\)	\(173\)
default	\(\frac {\frac {2 c e \left (g x +f \right )^{\frac {5}{2}}}{5}+\frac {2 b e g \left (g x +f \right )^{\frac {3}{2}}}{3}+\frac {2 c d g \left (g x +f \right )^{\frac {3}{2}}}{3}-2 c e f \left (g x +f \right )^{\frac {3}{2}}+2 a e \,g^{2} \sqrt {g x +f}+2 b d \,g^{2} \sqrt {g x +f}-4 b e f g \sqrt {g x +f}-4 c d f g \sqrt {g x +f}+6 c e \,f^{2} \sqrt {g x +f}-\frac {2 \left (a d \,g^{3}-a e f \,g^{2}-b d f \,g^{2}+b e \,f^{2} g +d \,f^{2} g c -c e \,f^{3}\right )}{\sqrt {g x +f}}}{g^{4}}\)	\(173\)

[In]

int((e*x+d)*(c*x^2+b*x+a)/(g*x+f)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/15*(((6*c*x^3+10*b*x^2+30*a*x)*e-30*d*(a-1/3*c*x^2-b*x))*g^3+60*((-1/5*c*x^2-2/3*b*x+a)*e+d*(-2/3*c*x+b))*f*
g^2-80*((-3/5*c*x+b)*e+c*d)*f^2*g+96*c*e*f^3)/(g*x+f)^(1/2)/g^4

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{(f+g x)^{3/2}} \, dx=\frac {2 \, {\left (3 \, c e g^{3} x^{3} + 48 \, c e f^{3} - 15 \, a d g^{3} - 40 \, {\left (c d + b e\right )} f^{2} g + 30 \, {\left (b d + a e\right )} f g^{2} - {\left (6 \, c e f g^{2} - 5 \, {\left (c d + b e\right )} g^{3}\right )} x^{2} + {\left (24 \, c e f^{2} g - 20 \, {\left (c d + b e\right )} f g^{2} + 15 \, {\left (b d + a e\right )} g^{3}\right )} x\right )} \sqrt {g x + f}}{15 \, {\left (g^{5} x + f g^{4}\right )}} \]

[In]

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

[Out]

2/15*(3*c*e*g^3*x^3 + 48*c*e*f^3 - 15*a*d*g^3 - 40*(c*d + b*e)*f^2*g + 30*(b*d + a*e)*f*g^2 - (6*c*e*f*g^2 - 5
*(c*d + b*e)*g^3)*x^2 + (24*c*e*f^2*g - 20*(c*d + b*e)*f*g^2 + 15*(b*d + a*e)*g^3)*x)*sqrt(g*x + f)/(g^5*x + f
*g^4)

Sympy [A] (verification not implemented)

Time = 3.25 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.35 \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{(f+g x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {c e \left (f + g x\right )^{\frac {5}{2}}}{5 g^{3}} + \frac {\left (f + g x\right )^{\frac {3}{2}} \left (b e g + c d g - 3 c e f\right )}{3 g^{3}} + \frac {\sqrt {f + g x} \left (a e g^{2} + b d g^{2} - 2 b e f g - 2 c d f g + 3 c e f^{2}\right )}{g^{3}} - \frac {\left (d g - e f\right ) \left (a g^{2} - b f g + c f^{2}\right )}{g^{3} \sqrt {f + g x}}\right )}{g} & \text {for}\: g \neq 0 \\\frac {a d x + \frac {c e x^{4}}{4} + \frac {x^{3} \left (b e + c d\right )}{3} + \frac {x^{2} \left (a e + b d\right )}{2}}{f^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01 \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{(f+g x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {3 \, {\left (g x + f\right )}^{\frac {5}{2}} c e - 5 \, {\left (3 \, c e f - {\left (c d + b e\right )} g\right )} {\left (g x + f\right )}^{\frac {3}{2}} + 15 \, {\left (3 \, c e f^{2} - 2 \, {\left (c d + b e\right )} f g + {\left (b d + a e\right )} g^{2}\right )} \sqrt {g x + f}}{g^{3}} + \frac {15 \, {\left (c e f^{3} - a d g^{3} - {\left (c d + b e\right )} f^{2} g + {\left (b d + a e\right )} f g^{2}\right )}}{\sqrt {g x + f} g^{3}}\right )}}{15 \, g} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.44 \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{(f+g x)^{3/2}} \, dx=\frac {2 \, {\left (c e f^{3} - c d f^{2} g - b e f^{2} g + b d f g^{2} + a e f g^{2} - a d g^{3}\right )}}{\sqrt {g x + f} g^{4}} + \frac {2 \, {\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} c e g^{16} - 15 \, {\left (g x + f\right )}^{\frac {3}{2}} c e f g^{16} + 45 \, \sqrt {g x + f} c e f^{2} g^{16} + 5 \, {\left (g x + f\right )}^{\frac {3}{2}} c d g^{17} + 5 \, {\left (g x + f\right )}^{\frac {3}{2}} b e g^{17} - 30 \, \sqrt {g x + f} c d f g^{17} - 30 \, \sqrt {g x + f} b e f g^{17} + 15 \, \sqrt {g x + f} b d g^{18} + 15 \, \sqrt {g x + f} a e g^{18}\right )}}{15 \, g^{20}} \]

[In]

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

[Out]

2*(c*e*f^3 - c*d*f^2*g - b*e*f^2*g + b*d*f*g^2 + a*e*f*g^2 - a*d*g^3)/(sqrt(g*x + f)*g^4) + 2/15*(3*(g*x + f)^
(5/2)*c*e*g^16 - 15*(g*x + f)^(3/2)*c*e*f*g^16 + 45*sqrt(g*x + f)*c*e*f^2*g^16 + 5*(g*x + f)^(3/2)*c*d*g^17 +
5*(g*x + f)^(3/2)*b*e*g^17 - 30*sqrt(g*x + f)*c*d*f*g^17 - 30*sqrt(g*x + f)*b*e*f*g^17 + 15*sqrt(g*x + f)*b*d*
g^18 + 15*sqrt(g*x + f)*a*e*g^18)/g^20

Mupad [B] (verification not implemented)

Time = 11.75 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.09 \[ \int \frac {(d+e x) \left (a+b x+c x^2\right )}{(f+g x)^{3/2}} \, dx=\frac {{\left (f+g\,x\right )}^{3/2}\,\left (2\,b\,e\,g+2\,c\,d\,g-6\,c\,e\,f\right )}{3\,g^4}-\frac {2\,a\,d\,g^3-2\,c\,e\,f^3-2\,a\,e\,f\,g^2-2\,b\,d\,f\,g^2+2\,b\,e\,f^2\,g+2\,c\,d\,f^2\,g}{g^4\,\sqrt {f+g\,x}}+\frac {\sqrt {f+g\,x}\,\left (2\,a\,e\,g^2+2\,b\,d\,g^2+6\,c\,e\,f^2-4\,b\,e\,f\,g-4\,c\,d\,f\,g\right )}{g^4}+\frac {2\,c\,e\,{\left (f+g\,x\right )}^{5/2}}{5\,g^4} \]

[In]

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

[Out]

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