\(\int \frac {(d+e x)^2 (a+c x^2)}{\sqrt {f+g x}} \, dx\) [590]

Optimal result

Integrand size = 24, antiderivative size = 175 \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{\sqrt {f+g x}} \, dx=\frac {2 (e f-d g)^2 \left (c f^2+a g^2\right ) \sqrt {f+g x}}{g^5}-\frac {4 (e f-d g) \left (a e g^2+c f (2 e f-d g)\right ) (f+g x)^{3/2}}{3 g^5}+\frac {2 \left (a e^2 g^2+c \left (6 e^2 f^2-6 d e f g+d^2 g^2\right )\right ) (f+g x)^{5/2}}{5 g^5}-\frac {4 c e (2 e f-d g) (f+g x)^{7/2}}{7 g^5}+\frac {2 c e^2 (f+g x)^{9/2}}{9 g^5} \]

[Out]

-4/3*(-d*g+e*f)*(a*e*g^2+c*f*(-d*g+2*e*f))*(g*x+f)^(3/2)/g^5+2/5*(a*e^2*g^2+c*(d^2*g^2-6*d*e*f*g+6*e^2*f^2))*(
g*x+f)^(5/2)/g^5-4/7*c*e*(-d*g+2*e*f)*(g*x+f)^(7/2)/g^5+2/9*c*e^2*(g*x+f)^(9/2)/g^5+2*(-d*g+e*f)^2*(a*g^2+c*f^
2)*(g*x+f)^(1/2)/g^5

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {912, 1167} \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{\sqrt {f+g x}} \, dx=\frac {2 (f+g x)^{5/2} \left (a e^2 g^2+c \left (d^2 g^2-6 d e f g+6 e^2 f^2\right )\right )}{5 g^5}+\frac {2 \sqrt {f+g x} \left (a g^2+c f^2\right ) (e f-d g)^2}{g^5}-\frac {4 (f+g x)^{3/2} (e f-d g) \left (a e g^2+c f (2 e f-d g)\right )}{3 g^5}-\frac {4 c e (f+g x)^{7/2} (2 e f-d g)}{7 g^5}+\frac {2 c e^2 (f+g x)^{9/2}}{9 g^5} \]

[In]

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

[Out]

(2*(e*f - d*g)^2*(c*f^2 + a*g^2)*Sqrt[f + g*x])/g^5 - (4*(e*f - d*g)*(a*e*g^2 + c*f*(2*e*f - d*g))*(f + g*x)^(
3/2))/(3*g^5) + (2*(a*e^2*g^2 + c*(6*e^2*f^2 - 6*d*e*f*g + d^2*g^2))*(f + g*x)^(5/2))/(5*g^5) - (4*c*e*(2*e*f
- d*g)*(f + g*x)^(7/2))/(7*g^5) + (2*c*e^2*(f + g*x)^(9/2))/(9*g^5)

Rule 912

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 + a*e^2)/e^2 - 2*c*
d*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]

Rule 1167

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \text {Subst}\left (\int \left (\frac {-e f+d g}{g}+\frac {e x^2}{g}\right )^2 \left (\frac {c f^2+a g^2}{g^2}-\frac {2 c f x^2}{g^2}+\frac {c x^4}{g^2}\right ) \, dx,x,\sqrt {f+g x}\right )}{g} \\ & = \frac {2 \text {Subst}\left (\int \left (\frac {(-e f+d g)^2 \left (c f^2+a g^2\right )}{g^4}+\frac {2 (e f-d g) \left (-a e g^2-c f (2 e f-d g)\right ) x^2}{g^4}+\frac {\left (a e^2 g^2+c \left (6 e^2 f^2-6 d e f g+d^2 g^2\right )\right ) x^4}{g^4}-\frac {2 c e (2 e f-d g) x^6}{g^4}+\frac {c e^2 x^8}{g^4}\right ) \, dx,x,\sqrt {f+g x}\right )}{g} \\ & = \frac {2 (e f-d g)^2 \left (c f^2+a g^2\right ) \sqrt {f+g x}}{g^5}-\frac {4 (e f-d g) \left (a e g^2+c f (2 e f-d g)\right ) (f+g x)^{3/2}}{3 g^5}+\frac {2 \left (a e^2 g^2+c \left (6 e^2 f^2-6 d e f g+d^2 g^2\right )\right ) (f+g x)^{5/2}}{5 g^5}-\frac {4 c e (2 e f-d g) (f+g x)^{7/2}}{7 g^5}+\frac {2 c e^2 (f+g x)^{9/2}}{9 g^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.01 \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{\sqrt {f+g x}} \, dx=\frac {2 \sqrt {f+g x} \left (21 a g^2 \left (15 d^2 g^2+10 d e g (-2 f+g x)+e^2 \left (8 f^2-4 f g x+3 g^2 x^2\right )\right )+c \left (21 d^2 g^2 \left (8 f^2-4 f g x+3 g^2 x^2\right )+18 d e g \left (-16 f^3+8 f^2 g x-6 f g^2 x^2+5 g^3 x^3\right )+e^2 \left (128 f^4-64 f^3 g x+48 f^2 g^2 x^2-40 f g^3 x^3+35 g^4 x^4\right )\right )\right )}{315 g^5} \]

[In]

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

[Out]

(2*Sqrt[f + g*x]*(21*a*g^2*(15*d^2*g^2 + 10*d*e*g*(-2*f + g*x) + e^2*(8*f^2 - 4*f*g*x + 3*g^2*x^2)) + c*(21*d^
2*g^2*(8*f^2 - 4*f*g*x + 3*g^2*x^2) + 18*d*e*g*(-16*f^3 + 8*f^2*g*x - 6*f*g^2*x^2 + 5*g^3*x^3) + e^2*(128*f^4
- 64*f^3*g*x + 48*f^2*g^2*x^2 - 40*f*g^3*x^3 + 35*g^4*x^4))))/(315*g^5)

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.89

method	result	size
pseudoelliptic	\(\frac {2 \sqrt {g x +f}\, \left (\left (\frac {x^{2} \left (\frac {5 c \,x^{2}}{9}+a \right ) e^{2}}{5}+\frac {2 x d \left (\frac {3 c \,x^{2}}{7}+a \right ) e}{3}+d^{2} \left (\frac {c \,x^{2}}{5}+a \right )\right ) g^{4}-\frac {4 \left (\left (\frac {2}{21} c \,x^{3}+\frac {1}{5} a x \right ) e^{2}+d \left (\frac {9 c \,x^{2}}{35}+a \right ) e +\frac {c \,d^{2} x}{5}\right ) f \,g^{3}}{3}+\frac {8 f^{2} \left (\left (\frac {2 c \,x^{2}}{7}+a \right ) e^{2}+\frac {6 c d e x}{7}+c \,d^{2}\right ) g^{2}}{15}-\frac {32 e c \,f^{3} \left (\frac {2 e x}{9}+d \right ) g}{35}+\frac {128 c \,e^{2} f^{4}}{315}\right )}{g^{5}}\)	\(155\)
derivativedivides	\(\frac {\frac {2 c \,e^{2} \left (g x +f \right )^{\frac {9}{2}}}{9}+\frac {2 \left (2 \left (d g -e f \right ) e c -2 c \,e^{2} f \right ) \left (g x +f \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (d g -e f \right )^{2} c -4 \left (d g -e f \right ) e c f +e^{2} \left (a \,g^{2}+c \,f^{2}\right )\right ) \left (g x +f \right )^{\frac {5}{2}}}{5}+\frac {2 \left (-2 \left (d g -e f \right )^{2} c f +2 \left (d g -e f \right ) e \left (a \,g^{2}+c \,f^{2}\right )\right ) \left (g x +f \right )^{\frac {3}{2}}}{3}+2 \left (d g -e f \right )^{2} \left (a \,g^{2}+c \,f^{2}\right ) \sqrt {g x +f}}{g^{5}}\)	\(174\)
default	\(\frac {\frac {2 c \,e^{2} \left (g x +f \right )^{\frac {9}{2}}}{9}+\frac {2 \left (2 \left (d g -e f \right ) e c -2 c \,e^{2} f \right ) \left (g x +f \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (d g -e f \right )^{2} c -4 \left (d g -e f \right ) e c f +e^{2} \left (a \,g^{2}+c \,f^{2}\right )\right ) \left (g x +f \right )^{\frac {5}{2}}}{5}+\frac {2 \left (-2 \left (d g -e f \right )^{2} c f +2 \left (d g -e f \right ) e \left (a \,g^{2}+c \,f^{2}\right )\right ) \left (g x +f \right )^{\frac {3}{2}}}{3}+2 \left (d g -e f \right )^{2} \left (a \,g^{2}+c \,f^{2}\right ) \sqrt {g x +f}}{g^{5}}\)	\(174\)
gosper	\(\frac {2 \sqrt {g x +f}\, \left (35 c \,e^{2} x^{4} g^{4}+90 c d e \,g^{4} x^{3}-40 c \,e^{2} f \,g^{3} x^{3}+63 a \,e^{2} g^{4} x^{2}+63 c \,d^{2} g^{4} x^{2}-108 c d e f \,g^{3} x^{2}+48 c \,e^{2} f^{2} g^{2} x^{2}+210 a d e \,g^{4} x -84 a \,e^{2} f \,g^{3} x -84 c \,d^{2} f \,g^{3} x +144 c d e \,f^{2} g^{2} x -64 c \,e^{2} f^{3} g x +315 a \,d^{2} g^{4}-420 a d e f \,g^{3}+168 a \,e^{2} f^{2} g^{2}+168 c \,d^{2} f^{2} g^{2}-288 c d e \,f^{3} g +128 c \,e^{2} f^{4}\right )}{315 g^{5}}\)	\(215\)
trager	\(\frac {2 \sqrt {g x +f}\, \left (35 c \,e^{2} x^{4} g^{4}+90 c d e \,g^{4} x^{3}-40 c \,e^{2} f \,g^{3} x^{3}+63 a \,e^{2} g^{4} x^{2}+63 c \,d^{2} g^{4} x^{2}-108 c d e f \,g^{3} x^{2}+48 c \,e^{2} f^{2} g^{2} x^{2}+210 a d e \,g^{4} x -84 a \,e^{2} f \,g^{3} x -84 c \,d^{2} f \,g^{3} x +144 c d e \,f^{2} g^{2} x -64 c \,e^{2} f^{3} g x +315 a \,d^{2} g^{4}-420 a d e f \,g^{3}+168 a \,e^{2} f^{2} g^{2}+168 c \,d^{2} f^{2} g^{2}-288 c d e \,f^{3} g +128 c \,e^{2} f^{4}\right )}{315 g^{5}}\)	\(215\)
risch	\(\frac {2 \sqrt {g x +f}\, \left (35 c \,e^{2} x^{4} g^{4}+90 c d e \,g^{4} x^{3}-40 c \,e^{2} f \,g^{3} x^{3}+63 a \,e^{2} g^{4} x^{2}+63 c \,d^{2} g^{4} x^{2}-108 c d e f \,g^{3} x^{2}+48 c \,e^{2} f^{2} g^{2} x^{2}+210 a d e \,g^{4} x -84 a \,e^{2} f \,g^{3} x -84 c \,d^{2} f \,g^{3} x +144 c d e \,f^{2} g^{2} x -64 c \,e^{2} f^{3} g x +315 a \,d^{2} g^{4}-420 a d e f \,g^{3}+168 a \,e^{2} f^{2} g^{2}+168 c \,d^{2} f^{2} g^{2}-288 c d e \,f^{3} g +128 c \,e^{2} f^{4}\right )}{315 g^{5}}\)	\(215\)

[In]

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

[Out]

2*(g*x+f)^(1/2)*((1/5*x^2*(5/9*c*x^2+a)*e^2+2/3*x*d*(3/7*c*x^2+a)*e+d^2*(1/5*c*x^2+a))*g^4-4/3*((2/21*c*x^3+1/
5*a*x)*e^2+d*(9/35*c*x^2+a)*e+1/5*c*d^2*x)*f*g^3+8/15*f^2*((2/7*c*x^2+a)*e^2+6/7*c*d*e*x+c*d^2)*g^2-32/35*e*c*
f^3*(2/9*e*x+d)*g+128/315*c*e^2*f^4)/g^5

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.13 \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{\sqrt {f+g x}} \, dx=\frac {2 \, {\left (35 \, c e^{2} g^{4} x^{4} + 128 \, c e^{2} f^{4} - 288 \, c d e f^{3} g - 420 \, a d e f g^{3} + 315 \, a d^{2} g^{4} + 168 \, {\left (c d^{2} + a e^{2}\right )} f^{2} g^{2} - 10 \, {\left (4 \, c e^{2} f g^{3} - 9 \, c d e g^{4}\right )} x^{3} + 3 \, {\left (16 \, c e^{2} f^{2} g^{2} - 36 \, c d e f g^{3} + 21 \, {\left (c d^{2} + a e^{2}\right )} g^{4}\right )} x^{2} - 2 \, {\left (32 \, c e^{2} f^{3} g - 72 \, c d e f^{2} g^{2} - 105 \, a d e g^{4} + 42 \, {\left (c d^{2} + a e^{2}\right )} f g^{3}\right )} x\right )} \sqrt {g x + f}}{315 \, g^{5}} \]

[In]

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

[Out]

2/315*(35*c*e^2*g^4*x^4 + 128*c*e^2*f^4 - 288*c*d*e*f^3*g - 420*a*d*e*f*g^3 + 315*a*d^2*g^4 + 168*(c*d^2 + a*e
^2)*f^2*g^2 - 10*(4*c*e^2*f*g^3 - 9*c*d*e*g^4)*x^3 + 3*(16*c*e^2*f^2*g^2 - 36*c*d*e*f*g^3 + 21*(c*d^2 + a*e^2)
*g^4)*x^2 - 2*(32*c*e^2*f^3*g - 72*c*d*e*f^2*g^2 - 105*a*d*e*g^4 + 42*(c*d^2 + a*e^2)*f*g^3)*x)*sqrt(g*x + f)/
g^5

Sympy [A] (verification not implemented)

Time = 0.86 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.78 \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{\sqrt {f+g x}} \, dx=\begin {cases} \frac {2 \left (\frac {c e^{2} \left (f + g x\right )^{\frac {9}{2}}}{9 g^{4}} + \frac {\left (f + g x\right )^{\frac {7}{2}} \cdot \left (2 c d e g - 4 c e^{2} f\right )}{7 g^{4}} + \frac {\left (f + g x\right )^{\frac {5}{2}} \left (a e^{2} g^{2} + c d^{2} g^{2} - 6 c d e f g + 6 c e^{2} f^{2}\right )}{5 g^{4}} + \frac {\left (f + g x\right )^{\frac {3}{2}} \cdot \left (2 a d e g^{3} - 2 a e^{2} f g^{2} - 2 c d^{2} f g^{2} + 6 c d e f^{2} g - 4 c e^{2} f^{3}\right )}{3 g^{4}} + \frac {\sqrt {f + g x} \left (a d^{2} g^{4} - 2 a d e f g^{3} + a e^{2} f^{2} g^{2} + c d^{2} f^{2} g^{2} - 2 c d e f^{3} g + c e^{2} f^{4}\right )}{g^{4}}\right )}{g} & \text {for}\: g \neq 0 \\\frac {a d^{2} x + a d e x^{2} + \frac {c d e x^{4}}{2} + \frac {c e^{2} x^{5}}{5} + \frac {x^{3} \left (a e^{2} + c d^{2}\right )}{3}}{\sqrt {f}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(c*e**2*(f + g*x)**(9/2)/(9*g**4) + (f + g*x)**(7/2)*(2*c*d*e*g - 4*c*e**2*f)/(7*g**4) + (f + g*x
)**(5/2)*(a*e**2*g**2 + c*d**2*g**2 - 6*c*d*e*f*g + 6*c*e**2*f**2)/(5*g**4) + (f + g*x)**(3/2)*(2*a*d*e*g**3 -
2*a*e**2*f*g**2 - 2*c*d**2*f*g**2 + 6*c*d*e*f**2*g - 4*c*e**2*f**3)/(3*g**4) + sqrt(f + g*x)*(a*d**2*g**4 - 2
*a*d*e*f*g**3 + a*e**2*f**2*g**2 + c*d**2*f**2*g**2 - 2*c*d*e*f**3*g + c*e**2*f**4)/g**4)/g, Ne(g, 0)), ((a*d*
*2*x + a*d*e*x**2 + c*d*e*x**4/2 + c*e**2*x**5/5 + x**3*(a*e**2 + c*d**2)/3)/sqrt(f), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.13 \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{\sqrt {f+g x}} \, dx=\frac {2 \, {\left (35 \, {\left (g x + f\right )}^{\frac {9}{2}} c e^{2} - 90 \, {\left (2 \, c e^{2} f - c d e g\right )} {\left (g x + f\right )}^{\frac {7}{2}} + 63 \, {\left (6 \, c e^{2} f^{2} - 6 \, c d e f g + {\left (c d^{2} + a e^{2}\right )} g^{2}\right )} {\left (g x + f\right )}^{\frac {5}{2}} - 210 \, {\left (2 \, c e^{2} f^{3} - 3 \, c d e f^{2} g - a d e g^{3} + {\left (c d^{2} + a e^{2}\right )} f g^{2}\right )} {\left (g x + f\right )}^{\frac {3}{2}} + 315 \, {\left (c e^{2} f^{4} - 2 \, c d e f^{3} g - 2 \, a d e f g^{3} + a d^{2} g^{4} + {\left (c d^{2} + a e^{2}\right )} f^{2} g^{2}\right )} \sqrt {g x + f}\right )}}{315 \, g^{5}} \]

[In]

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

[Out]

2/315*(35*(g*x + f)^(9/2)*c*e^2 - 90*(2*c*e^2*f - c*d*e*g)*(g*x + f)^(7/2) + 63*(6*c*e^2*f^2 - 6*c*d*e*f*g + (
c*d^2 + a*e^2)*g^2)*(g*x + f)^(5/2) - 210*(2*c*e^2*f^3 - 3*c*d*e*f^2*g - a*d*e*g^3 + (c*d^2 + a*e^2)*f*g^2)*(g
*x + f)^(3/2) + 315*(c*e^2*f^4 - 2*c*d*e*f^3*g - 2*a*d*e*f*g^3 + a*d^2*g^4 + (c*d^2 + a*e^2)*f^2*g^2)*sqrt(g*x
+ f))/g^5

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.39 \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{\sqrt {f+g x}} \, dx=\frac {2 \, {\left (315 \, \sqrt {g x + f} a d^{2} + \frac {210 \, {\left ({\left (g x + f\right )}^{\frac {3}{2}} - 3 \, \sqrt {g x + f} f\right )} a d e}{g} + \frac {21 \, {\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} - 10 \, {\left (g x + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {g x + f} f^{2}\right )} c d^{2}}{g^{2}} + \frac {21 \, {\left (3 \, {\left (g x + f\right )}^{\frac {5}{2}} - 10 \, {\left (g x + f\right )}^{\frac {3}{2}} f + 15 \, \sqrt {g x + f} f^{2}\right )} a e^{2}}{g^{2}} + \frac {18 \, {\left (5 \, {\left (g x + f\right )}^{\frac {7}{2}} - 21 \, {\left (g x + f\right )}^{\frac {5}{2}} f + 35 \, {\left (g x + f\right )}^{\frac {3}{2}} f^{2} - 35 \, \sqrt {g x + f} f^{3}\right )} c d e}{g^{3}} + \frac {{\left (35 \, {\left (g x + f\right )}^{\frac {9}{2}} - 180 \, {\left (g x + f\right )}^{\frac {7}{2}} f + 378 \, {\left (g x + f\right )}^{\frac {5}{2}} f^{2} - 420 \, {\left (g x + f\right )}^{\frac {3}{2}} f^{3} + 315 \, \sqrt {g x + f} f^{4}\right )} c e^{2}}{g^{4}}\right )}}{315 \, g} \]

[In]

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

[Out]

2/315*(315*sqrt(g*x + f)*a*d^2 + 210*((g*x + f)^(3/2) - 3*sqrt(g*x + f)*f)*a*d*e/g + 21*(3*(g*x + f)^(5/2) - 1
0*(g*x + f)^(3/2)*f + 15*sqrt(g*x + f)*f^2)*c*d^2/g^2 + 21*(3*(g*x + f)^(5/2) - 10*(g*x + f)^(3/2)*f + 15*sqrt
(g*x + f)*f^2)*a*e^2/g^2 + 18*(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*d*e/g^3 + (35*(g*x + f)^(9/2) - 180*(g*x + f)^(7/2)*f + 378*(g*x + f)^(5/2)*f^2 - 420*(g*x + f)^(
3/2)*f^3 + 315*sqrt(g*x + f)*f^4)*c*e^2/g^4)/g

Mupad [B] (verification not implemented)

Time = 12.19 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.91 \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{\sqrt {f+g x}} \, dx=\frac {{\left (f+g\,x\right )}^{5/2}\,\left (2\,c\,d^2\,g^2-12\,c\,d\,e\,f\,g+12\,c\,e^2\,f^2+2\,a\,e^2\,g^2\right )}{5\,g^5}+\frac {2\,\sqrt {f+g\,x}\,\left (c\,f^2+a\,g^2\right )\,{\left (d\,g-e\,f\right )}^2}{g^5}+\frac {4\,{\left (f+g\,x\right )}^{3/2}\,\left (d\,g-e\,f\right )\,\left (2\,c\,e\,f^2-c\,d\,f\,g+a\,e\,g^2\right )}{3\,g^5}+\frac {2\,c\,e^2\,{\left (f+g\,x\right )}^{9/2}}{9\,g^5}+\frac {4\,c\,e\,{\left (f+g\,x\right )}^{7/2}\,\left (d\,g-2\,e\,f\right )}{7\,g^5} \]

[In]

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

[Out]

((f + g*x)^(5/2)*(2*a*e^2*g^2 + 2*c*d^2*g^2 + 12*c*e^2*f^2 - 12*c*d*e*f*g))/(5*g^5) + (2*(f + g*x)^(1/2)*(a*g^
2 + c*f^2)*(d*g - e*f)^2)/g^5 + (4*(f + g*x)^(3/2)*(d*g - e*f)*(a*e*g^2 + 2*c*e*f^2 - c*d*f*g))/(3*g^5) + (2*c
*e^2*(f + g*x)^(9/2))/(9*g^5) + (4*c*e*(f + g*x)^(7/2)*(d*g - 2*e*f))/(7*g^5)