\(\int \frac {(d+e x)^2 (a+c x^2)}{(f+g x)^{3/2}} \, dx\) [597]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 24, antiderivative size = 173 \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{(f+g x)^{3/2}} \, dx=-\frac {2 (e f-d g)^2 \left (c f^2+a g^2\right )}{g^5 \sqrt {f+g x}}-\frac {4 (e f-d g) \left (a e g^2+c f (2 e f-d g)\right ) \sqrt {f+g x}}{g^5}+\frac {2 \left (a e^2 g^2+c \left (6 e^2 f^2-6 d e f g+d^2 g^2\right )\right ) (f+g x)^{3/2}}{3 g^5}-\frac {4 c e (2 e f-d g) (f+g x)^{5/2}}{5 g^5}+\frac {2 c e^2 (f+g x)^{7/2}}{7 g^5} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 173, 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, 1275} \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{(f+g x)^{3/2}} \, dx=\frac {2 (f+g x)^{3/2} \left (a e^2 g^2+c \left (d^2 g^2-6 d e f g+6 e^2 f^2\right )\right )}{3 g^5}-\frac {2 \left (a g^2+c f^2\right ) (e f-d g)^2}{g^5 \sqrt {f+g x}}-\frac {4 \sqrt {f+g x} (e f-d g) \left (a e g^2+c f (2 e f-d g)\right )}{g^5}-\frac {4 c e (f+g x)^{5/2} (2 e f-d g)}{5 g^5}+\frac {2 c e^2 (f+g x)^{7/2}}{7 g^5} \]

[In]

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

[Out]

(-2*(e*f - d*g)^2*(c*f^2 + a*g^2))/(g^5*Sqrt[f + g*x]) - (4*(e*f - d*g)*(a*e*g^2 + c*f*(2*e*f - d*g))*Sqrt[f +
 g*x])/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)^(3/2))/(3*g^5) - (4*c*e*(2*e*f - d
*g)*(f + g*x)^(5/2))/(5*g^5) + (2*c*e^2*(f + g*x)^(7/2))/(7*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 1275

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

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.02 \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{(f+g x)^{3/2}} \, dx=\frac {-70 a g^2 \left (3 d^2 g^2-6 d e g (2 f+g x)+e^2 \left (8 f^2+4 f g x-g^2 x^2\right )\right )+2 c \left (35 d^2 g^2 \left (-8 f^2-4 f g x+g^2 x^2\right )+42 d e g \left (16 f^3+8 f^2 g x-2 f g^2 x^2+g^3 x^3\right )-3 e^2 \left (128 f^4+64 f^3 g x-16 f^2 g^2 x^2+8 f g^3 x^3-5 g^4 x^4\right )\right )}{105 g^5 \sqrt {f+g x}} \]

[In]

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

[Out]

(-70*a*g^2*(3*d^2*g^2 - 6*d*e*g*(2*f + g*x) + e^2*(8*f^2 + 4*f*g*x - g^2*x^2)) + 2*c*(35*d^2*g^2*(-8*f^2 - 4*f
*g*x + g^2*x^2) + 42*d*e*g*(16*f^3 + 8*f^2*g*x - 2*f*g^2*x^2 + g^3*x^3) - 3*e^2*(128*f^4 + 64*f^3*g*x - 16*f^2
*g^2*x^2 + 8*f*g^3*x^3 - 5*g^4*x^4)))/(105*g^5*Sqrt[f + g*x])

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.89

method result size
pseudoelliptic \(-\frac {2 \left (\left (-\frac {x^{2} \left (\frac {3 c \,x^{2}}{7}+a \right ) e^{2}}{3}-2 \left (\frac {c \,x^{2}}{5}+a \right ) x d e +d^{2} \left (-\frac {c \,x^{2}}{3}+a \right )\right ) g^{4}-4 \left (-\frac {x \left (\frac {6 c \,x^{2}}{35}+a \right ) e^{2}}{3}+d \left (-\frac {c \,x^{2}}{5}+a \right ) e -\frac {c \,d^{2} x}{3}\right ) f \,g^{3}+\frac {8 f^{2} \left (\left (-\frac {6 c \,x^{2}}{35}+a \right ) e^{2}-\frac {6 c d e x}{5}+c \,d^{2}\right ) g^{2}}{3}-\frac {32 e \left (-\frac {2 e x}{7}+d \right ) c \,f^{3} g}{5}+\frac {128 c \,e^{2} f^{4}}{35}\right )}{\sqrt {g x +f}\, g^{5}}\) \(154\)
risch \(\frac {2 \left (15 c \,e^{2} x^{3} g^{3}+42 c d e \,g^{3} x^{2}-39 c \,e^{2} f \,g^{2} x^{2}+35 a \,e^{2} g^{3} x +35 c \,d^{2} g^{3} x -126 c d e f \,g^{2} x +87 c \,e^{2} f^{2} g x +210 a d e \,g^{3}-175 a \,e^{2} f \,g^{2}-175 c \,d^{2} f \,g^{2}+462 c d e \,f^{2} g -279 c \,e^{2} f^{3}\right ) \sqrt {g x +f}}{105 g^{5}}-\frac {2 \left (a \,d^{2} g^{4}-2 a d e f \,g^{3}+a \,e^{2} f^{2} g^{2}+c \,d^{2} f^{2} g^{2}-2 c d e \,f^{3} g +c \,e^{2} f^{4}\right )}{g^{5} \sqrt {g x +f}}\) \(207\)
gosper \(-\frac {2 \left (-15 c \,e^{2} x^{4} g^{4}-42 c d e \,g^{4} x^{3}+24 c \,e^{2} f \,g^{3} x^{3}-35 a \,e^{2} g^{4} x^{2}-35 c \,d^{2} g^{4} x^{2}+84 c d e f \,g^{3} x^{2}-48 c \,e^{2} f^{2} g^{2} x^{2}-210 a d e \,g^{4} x +140 a \,e^{2} f \,g^{3} x +140 c \,d^{2} f \,g^{3} x -336 c d e \,f^{2} g^{2} x +192 c \,e^{2} f^{3} g x +105 a \,d^{2} g^{4}-420 a d e f \,g^{3}+280 a \,e^{2} f^{2} g^{2}+280 c \,d^{2} f^{2} g^{2}-672 c d e \,f^{3} g +384 c \,e^{2} f^{4}\right )}{105 \sqrt {g x +f}\, g^{5}}\) \(215\)
trager \(-\frac {2 \left (-15 c \,e^{2} x^{4} g^{4}-42 c d e \,g^{4} x^{3}+24 c \,e^{2} f \,g^{3} x^{3}-35 a \,e^{2} g^{4} x^{2}-35 c \,d^{2} g^{4} x^{2}+84 c d e f \,g^{3} x^{2}-48 c \,e^{2} f^{2} g^{2} x^{2}-210 a d e \,g^{4} x +140 a \,e^{2} f \,g^{3} x +140 c \,d^{2} f \,g^{3} x -336 c d e \,f^{2} g^{2} x +192 c \,e^{2} f^{3} g x +105 a \,d^{2} g^{4}-420 a d e f \,g^{3}+280 a \,e^{2} f^{2} g^{2}+280 c \,d^{2} f^{2} g^{2}-672 c d e \,f^{3} g +384 c \,e^{2} f^{4}\right )}{105 \sqrt {g x +f}\, g^{5}}\) \(215\)
derivativedivides \(\frac {\frac {2 c \,e^{2} \left (g x +f \right )^{\frac {7}{2}}}{7}+\frac {4 c d e g \left (g x +f \right )^{\frac {5}{2}}}{5}-\frac {8 c \,e^{2} f \left (g x +f \right )^{\frac {5}{2}}}{5}+\frac {2 a \,e^{2} g^{2} \left (g x +f \right )^{\frac {3}{2}}}{3}+\frac {2 c \,d^{2} g^{2} \left (g x +f \right )^{\frac {3}{2}}}{3}-4 c d e f g \left (g x +f \right )^{\frac {3}{2}}+4 c \,e^{2} f^{2} \left (g x +f \right )^{\frac {3}{2}}+4 a d e \,g^{3} \sqrt {g x +f}-4 a \,e^{2} f \,g^{2} \sqrt {g x +f}-4 c \,d^{2} f \,g^{2} \sqrt {g x +f}+12 c d e \,f^{2} g \sqrt {g x +f}-8 c \,e^{2} f^{3} \sqrt {g x +f}-\frac {2 \left (a \,d^{2} g^{4}-2 a d e f \,g^{3}+a \,e^{2} f^{2} g^{2}+c \,d^{2} f^{2} g^{2}-2 c d e \,f^{3} g +c \,e^{2} f^{4}\right )}{\sqrt {g x +f}}}{g^{5}}\) \(256\)
default \(\frac {\frac {2 c \,e^{2} \left (g x +f \right )^{\frac {7}{2}}}{7}+\frac {4 c d e g \left (g x +f \right )^{\frac {5}{2}}}{5}-\frac {8 c \,e^{2} f \left (g x +f \right )^{\frac {5}{2}}}{5}+\frac {2 a \,e^{2} g^{2} \left (g x +f \right )^{\frac {3}{2}}}{3}+\frac {2 c \,d^{2} g^{2} \left (g x +f \right )^{\frac {3}{2}}}{3}-4 c d e f g \left (g x +f \right )^{\frac {3}{2}}+4 c \,e^{2} f^{2} \left (g x +f \right )^{\frac {3}{2}}+4 a d e \,g^{3} \sqrt {g x +f}-4 a \,e^{2} f \,g^{2} \sqrt {g x +f}-4 c \,d^{2} f \,g^{2} \sqrt {g x +f}+12 c d e \,f^{2} g \sqrt {g x +f}-8 c \,e^{2} f^{3} \sqrt {g x +f}-\frac {2 \left (a \,d^{2} g^{4}-2 a d e f \,g^{3}+a \,e^{2} f^{2} g^{2}+c \,d^{2} f^{2} g^{2}-2 c d e \,f^{3} g +c \,e^{2} f^{4}\right )}{\sqrt {g x +f}}}{g^{5}}\) \(256\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.19 \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{(f+g x)^{3/2}} \, dx=\frac {2 \, {\left (15 \, c e^{2} g^{4} x^{4} - 384 \, c e^{2} f^{4} + 672 \, c d e f^{3} g + 420 \, a d e f g^{3} - 105 \, a d^{2} g^{4} - 280 \, {\left (c d^{2} + a e^{2}\right )} f^{2} g^{2} - 6 \, {\left (4 \, c e^{2} f g^{3} - 7 \, c d e g^{4}\right )} x^{3} + {\left (48 \, c e^{2} f^{2} g^{2} - 84 \, c d e f g^{3} + 35 \, {\left (c d^{2} + a e^{2}\right )} g^{4}\right )} x^{2} - 2 \, {\left (96 \, c e^{2} f^{3} g - 168 \, c d e f^{2} g^{2} - 105 \, a d e g^{4} + 70 \, {\left (c d^{2} + a e^{2}\right )} f g^{3}\right )} x\right )} \sqrt {g x + f}}{105 \, {\left (g^{6} x + f g^{5}\right )}} \]

[In]

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

[Out]

2/105*(15*c*e^2*g^4*x^4 - 384*c*e^2*f^4 + 672*c*d*e*f^3*g + 420*a*d*e*f*g^3 - 105*a*d^2*g^4 - 280*(c*d^2 + a*e
^2)*f^2*g^2 - 6*(4*c*e^2*f*g^3 - 7*c*d*e*g^4)*x^3 + (48*c*e^2*f^2*g^2 - 84*c*d*e*f*g^3 + 35*(c*d^2 + a*e^2)*g^
4)*x^2 - 2*(96*c*e^2*f^3*g - 168*c*d*e*f^2*g^2 - 105*a*d*e*g^4 + 70*(c*d^2 + a*e^2)*f*g^3)*x)*sqrt(g*x + f)/(g
^6*x + f*g^5)

Sympy [A] (verification not implemented)

Time = 4.82 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.53 \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{(f+g x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {c e^{2} \left (f + g x\right )^{\frac {7}{2}}}{7 g^{4}} + \frac {\left (f + g x\right )^{\frac {5}{2}} \cdot \left (2 c d e g - 4 c e^{2} f\right )}{5 g^{4}} + \frac {\left (f + g x\right )^{\frac {3}{2}} \left (a e^{2} g^{2} + c d^{2} g^{2} - 6 c d e f g + 6 c e^{2} f^{2}\right )}{3 g^{4}} + \frac {\sqrt {f + g x} \left (2 a d e g^{3} - 2 a e^{2} f g^{2} - 2 c d^{2} f g^{2} + 6 c d e f^{2} g - 4 c e^{2} f^{3}\right )}{g^{4}} - \frac {\left (a g^{2} + c f^{2}\right ) \left (d g - e f\right )^{2}}{g^{4} \sqrt {f + g x}}\right )}{g} & \text {for}\: g \neq 0 \\\frac {a d^{2} x + a d e x^{2} + \frac {c d e x^{4}}{2} + \frac {c e^{2} x^{5}}{5} + \frac {x^{3} \left (a e^{2} + c d^{2}\right )}{3}}{f^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.18 \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{(f+g x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {15 \, {\left (g x + f\right )}^{\frac {7}{2}} c e^{2} - 42 \, {\left (2 \, c e^{2} f - c d e g\right )} {\left (g x + f\right )}^{\frac {5}{2}} + 35 \, {\left (6 \, c e^{2} f^{2} - 6 \, c d e f g + {\left (c d^{2} + a e^{2}\right )} g^{2}\right )} {\left (g x + f\right )}^{\frac {3}{2}} - 210 \, {\left (2 \, c e^{2} f^{3} - 3 \, c d e f^{2} g - a d e g^{3} + {\left (c d^{2} + a e^{2}\right )} f g^{2}\right )} \sqrt {g x + f}}{g^{4}} - \frac {105 \, {\left (c e^{2} f^{4} - 2 \, c d e f^{3} g - 2 \, a d e f g^{3} + a d^{2} g^{4} + {\left (c d^{2} + a e^{2}\right )} f^{2} g^{2}\right )}}{\sqrt {g x + f} g^{4}}\right )}}{105 \, g} \]

[In]

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

[Out]

2/105*((15*(g*x + f)^(7/2)*c*e^2 - 42*(2*c*e^2*f - c*d*e*g)*(g*x + f)^(5/2) + 35*(6*c*e^2*f^2 - 6*c*d*e*f*g +
(c*d^2 + a*e^2)*g^2)*(g*x + f)^(3/2) - 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)*s
qrt(g*x + f))/g^4 - 105*(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)/(sqr
t(g*x + f)*g^4))/g

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.60 \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{(f+g x)^{3/2}} \, dx=-\frac {2 \, {\left (c e^{2} f^{4} - 2 \, c d e f^{3} g + c d^{2} f^{2} g^{2} + a e^{2} f^{2} g^{2} - 2 \, a d e f g^{3} + a d^{2} g^{4}\right )}}{\sqrt {g x + f} g^{5}} + \frac {2 \, {\left (15 \, {\left (g x + f\right )}^{\frac {7}{2}} c e^{2} g^{30} - 84 \, {\left (g x + f\right )}^{\frac {5}{2}} c e^{2} f g^{30} + 210 \, {\left (g x + f\right )}^{\frac {3}{2}} c e^{2} f^{2} g^{30} - 420 \, \sqrt {g x + f} c e^{2} f^{3} g^{30} + 42 \, {\left (g x + f\right )}^{\frac {5}{2}} c d e g^{31} - 210 \, {\left (g x + f\right )}^{\frac {3}{2}} c d e f g^{31} + 630 \, \sqrt {g x + f} c d e f^{2} g^{31} + 35 \, {\left (g x + f\right )}^{\frac {3}{2}} c d^{2} g^{32} + 35 \, {\left (g x + f\right )}^{\frac {3}{2}} a e^{2} g^{32} - 210 \, \sqrt {g x + f} c d^{2} f g^{32} - 210 \, \sqrt {g x + f} a e^{2} f g^{32} + 210 \, \sqrt {g x + f} a d e g^{33}\right )}}{105 \, g^{35}} \]

[In]

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

[Out]

-2*(c*e^2*f^4 - 2*c*d*e*f^3*g + c*d^2*f^2*g^2 + a*e^2*f^2*g^2 - 2*a*d*e*f*g^3 + a*d^2*g^4)/(sqrt(g*x + f)*g^5)
 + 2/105*(15*(g*x + f)^(7/2)*c*e^2*g^30 - 84*(g*x + f)^(5/2)*c*e^2*f*g^30 + 210*(g*x + f)^(3/2)*c*e^2*f^2*g^30
 - 420*sqrt(g*x + f)*c*e^2*f^3*g^30 + 42*(g*x + f)^(5/2)*c*d*e*g^31 - 210*(g*x + f)^(3/2)*c*d*e*f*g^31 + 630*s
qrt(g*x + f)*c*d*e*f^2*g^31 + 35*(g*x + f)^(3/2)*c*d^2*g^32 + 35*(g*x + f)^(3/2)*a*e^2*g^32 - 210*sqrt(g*x + f
)*c*d^2*f*g^32 - 210*sqrt(g*x + f)*a*e^2*f*g^32 + 210*sqrt(g*x + f)*a*d*e*g^33)/g^35

Mupad [B] (verification not implemented)

Time = 11.82 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.15 \[ \int \frac {(d+e x)^2 \left (a+c x^2\right )}{(f+g x)^{3/2}} \, dx=\frac {{\left (f+g\,x\right )}^{3/2}\,\left (2\,c\,d^2\,g^2-12\,c\,d\,e\,f\,g+12\,c\,e^2\,f^2+2\,a\,e^2\,g^2\right )}{3\,g^5}-\frac {2\,c\,d^2\,f^2\,g^2+2\,a\,d^2\,g^4-4\,c\,d\,e\,f^3\,g-4\,a\,d\,e\,f\,g^3+2\,c\,e^2\,f^4+2\,a\,e^2\,f^2\,g^2}{g^5\,\sqrt {f+g\,x}}+\frac {4\,\sqrt {f+g\,x}\,\left (d\,g-e\,f\right )\,\left (2\,c\,e\,f^2-c\,d\,f\,g+a\,e\,g^2\right )}{g^5}+\frac {2\,c\,e^2\,{\left (f+g\,x\right )}^{7/2}}{7\,g^5}+\frac {4\,c\,e\,{\left (f+g\,x\right )}^{5/2}\,\left (d\,g-2\,e\,f\right )}{5\,g^5} \]

[In]

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

[Out]

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