\(\int \frac {(d+e x)^{3/2} (f+g x)^3}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\) [665]

Optimal result

Integrand size = 46, antiderivative size = 257 \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {2 \sqrt {d+e x} (f+g x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {16 g (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^4 d^4 e \sqrt {d+e x}}+\frac {16 g^2 (c d f-a e g) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^3 d^3 e}+\frac {12 g (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^2 d^2 \sqrt {d+e x}} \]

[Out]

-2*(g*x+f)^3*(e*x+d)^(1/2)/c/d/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-16/5*g*(-a*e*g+c*d*f)*(2*a*e^2*g-c*d*(-
d*g+3*e*f))*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c^4/d^4/e/(e*x+d)^(1/2)+12/5*g*(g*x+f)^2*(a*d*e+(a*e^2+c*d
^2)*x+c*d*e*x^2)^(1/2)/c^2/d^2/(e*x+d)^(1/2)+16/5*g^2*(-a*e*g+c*d*f)*(e*x+d)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*
e*x^2)^(1/2)/c^3/d^3/e

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {880, 884, 808, 662} \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {16 g \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right )}{5 c^4 d^4 e \sqrt {d+e x}}+\frac {16 g^2 \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} (c d f-a e g)}{5 c^3 d^3 e}+\frac {12 g (f+g x)^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 c^2 d^2 \sqrt {d+e x}}-\frac {2 \sqrt {d+e x} (f+g x)^3}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

[In]

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

[Out]

(-2*Sqrt[d + e*x]*(f + g*x)^3)/(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (16*g*(c*d*f - a*e*g)*(2*a*
e^2*g - c*d*(3*e*f - d*g))*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(5*c^4*d^4*e*Sqrt[d + e*x]) + (16*g^2*
(c*d*f - a*e*g)*Sqrt[d + e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(5*c^3*d^3*e) + (12*g*(f + g*x)^2*S
qrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(5*c^2*d^2*Sqrt[d + e*x])

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rule 808

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rule 880

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Simp[e*(d + e*x)^(m - 1)*(f + g*x)^n*((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] - Dist[e*g*(n/(c*(p + 1))), I
nt[(d + e*x)^(m - 1)*(f + g*x)^(n - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &&
NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0] &
& LtQ[p, -1] && GtQ[n, 0]

Rule 884

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Simp[(-e)*(d + e*x)^(m - 1)*(f + g*x)^n*((a + b*x + c*x^2)^(p + 1)/(c*(m - n - 1))), x] - Dist[n*((c*e*f + c*d
*g - b*e*g)/(c*e*(m - n - 1))), Int[(d + e*x)^m*(f + g*x)^(n - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b,
c, d, e, f, g, m, p}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !Int
egerQ[p] && EqQ[m + p, 0] && GtQ[n, 0] && NeQ[m - n - 1, 0] && (IntegerQ[2*p] || IntegerQ[n])

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {d+e x} (f+g x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {(6 g) \int \frac {\sqrt {d+e x} (f+g x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{c d} \\ & = -\frac {2 \sqrt {d+e x} (f+g x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {12 g (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^2 d^2 \sqrt {d+e x}}+\frac {(24 g (c d f-a e g)) \int \frac {\sqrt {d+e x} (f+g x)}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{5 c^2 d^2} \\ & = -\frac {2 \sqrt {d+e x} (f+g x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {16 g^2 (c d f-a e g) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^3 d^3 e}+\frac {12 g (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^2 d^2 \sqrt {d+e x}}-\frac {\left (8 g (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right )\right ) \int \frac {\sqrt {d+e x}}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{5 c^3 d^3 e} \\ & = -\frac {2 \sqrt {d+e x} (f+g x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {16 g (c d f-a e g) \left (2 a e^2 g-c d (3 e f-d g)\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^4 d^4 e \sqrt {d+e x}}+\frac {16 g^2 (c d f-a e g) \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^3 d^3 e}+\frac {12 g (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{5 c^2 d^2 \sqrt {d+e x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.52 \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {d+e x} \left (16 a^3 e^3 g^3+8 a^2 c d e^2 g^2 (-5 f+g x)-2 a c^2 d^2 e g \left (-15 f^2+10 f g x+g^2 x^2\right )+c^3 d^3 \left (-5 f^3+15 f^2 g x+5 f g^2 x^2+g^3 x^3\right )\right )}{5 c^4 d^4 \sqrt {(a e+c d x) (d+e x)}} \]

[In]

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

[Out]

(2*Sqrt[d + e*x]*(16*a^3*e^3*g^3 + 8*a^2*c*d*e^2*g^2*(-5*f + g*x) - 2*a*c^2*d^2*e*g*(-15*f^2 + 10*f*g*x + g^2*
x^2) + c^3*d^3*(-5*f^3 + 15*f^2*g*x + 5*f*g^2*x^2 + g^3*x^3)))/(5*c^4*d^4*Sqrt[(a*e + c*d*x)*(d + e*x)])

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.70

method	result	size
default	\(\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (g^{3} x^{3} c^{3} d^{3}-2 a \,c^{2} d^{2} e \,g^{3} x^{2}+5 c^{3} d^{3} f \,g^{2} x^{2}+8 a^{2} c d \,e^{2} g^{3} x -20 a \,c^{2} d^{2} e f \,g^{2} x +15 c^{3} d^{3} f^{2} g x +16 a^{3} e^{3} g^{3}-40 a^{2} c d \,e^{2} f \,g^{2}+30 a \,c^{2} d^{2} e \,f^{2} g -5 f^{3} c^{3} d^{3}\right )}{5 \sqrt {e x +d}\, \left (c d x +a e \right ) c^{4} d^{4}}\)	\(179\)
gosper	\(\frac {2 \left (c d x +a e \right ) \left (g^{3} x^{3} c^{3} d^{3}-2 a \,c^{2} d^{2} e \,g^{3} x^{2}+5 c^{3} d^{3} f \,g^{2} x^{2}+8 a^{2} c d \,e^{2} g^{3} x -20 a \,c^{2} d^{2} e f \,g^{2} x +15 c^{3} d^{3} f^{2} g x +16 a^{3} e^{3} g^{3}-40 a^{2} c d \,e^{2} f \,g^{2}+30 a \,c^{2} d^{2} e \,f^{2} g -5 f^{3} c^{3} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}}{5 c^{4} d^{4} \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}}\)	\(187\)

[In]

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

[Out]

2/5/(e*x+d)^(1/2)*((c*d*x+a*e)*(e*x+d))^(1/2)*(c^3*d^3*g^3*x^3-2*a*c^2*d^2*e*g^3*x^2+5*c^3*d^3*f*g^2*x^2+8*a^2
*c*d*e^2*g^3*x-20*a*c^2*d^2*e*f*g^2*x+15*c^3*d^3*f^2*g*x+16*a^3*e^3*g^3-40*a^2*c*d*e^2*f*g^2+30*a*c^2*d^2*e*f^
2*g-5*c^3*d^3*f^3)/(c*d*x+a*e)/c^4/d^4

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.84 \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (c^{3} d^{3} g^{3} x^{3} - 5 \, c^{3} d^{3} f^{3} + 30 \, a c^{2} d^{2} e f^{2} g - 40 \, a^{2} c d e^{2} f g^{2} + 16 \, a^{3} e^{3} g^{3} + {\left (5 \, c^{3} d^{3} f g^{2} - 2 \, a c^{2} d^{2} e g^{3}\right )} x^{2} + {\left (15 \, c^{3} d^{3} f^{2} g - 20 \, a c^{2} d^{2} e f g^{2} + 8 \, a^{2} c d e^{2} g^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{5 \, {\left (c^{5} d^{5} e x^{2} + a c^{4} d^{5} e + {\left (c^{5} d^{6} + a c^{4} d^{4} e^{2}\right )} x\right )}} \]

[In]

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

[Out]

2/5*(c^3*d^3*g^3*x^3 - 5*c^3*d^3*f^3 + 30*a*c^2*d^2*e*f^2*g - 40*a^2*c*d*e^2*f*g^2 + 16*a^3*e^3*g^3 + (5*c^3*d
^3*f*g^2 - 2*a*c^2*d^2*e*g^3)*x^2 + (15*c^3*d^3*f^2*g - 20*a*c^2*d^2*e*f*g^2 + 8*a^2*c*d*e^2*g^3)*x)*sqrt(c*d*
e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)/(c^5*d^5*e*x^2 + a*c^4*d^5*e + (c^5*d^6 + a*c^4*d^4*e^2)*x)

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.64 \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {2 \, f^{3}}{\sqrt {c d x + a e} c d} + \frac {6 \, {\left (c d x + 2 \, a e\right )} f^{2} g}{\sqrt {c d x + a e} c^{2} d^{2}} + \frac {2 \, {\left (c^{2} d^{2} x^{2} - 4 \, a c d e x - 8 \, a^{2} e^{2}\right )} f g^{2}}{\sqrt {c d x + a e} c^{3} d^{3}} + \frac {2 \, {\left (c^{3} d^{3} x^{3} - 2 \, a c^{2} d^{2} e x^{2} + 8 \, a^{2} c d e^{2} x + 16 \, a^{3} e^{3}\right )} g^{3}}{5 \, \sqrt {c d x + a e} c^{4} d^{4}} \]

[In]

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

[Out]

-2*f^3/(sqrt(c*d*x + a*e)*c*d) + 6*(c*d*x + 2*a*e)*f^2*g/(sqrt(c*d*x + a*e)*c^2*d^2) + 2*(c^2*d^2*x^2 - 4*a*c*
d*e*x - 8*a^2*e^2)*f*g^2/(sqrt(c*d*x + a*e)*c^3*d^3) + 2/5*(c^3*d^3*x^3 - 2*a*c^2*d^2*e*x^2 + 8*a^2*c*d*e^2*x
+ 16*a^3*e^3)*g^3/(sqrt(c*d*x + a*e)*c^4*d^4)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 513 vs. \(2 (235) = 470\).

Time = 0.31 (sec) , antiderivative size = 513, normalized size of antiderivative = 2.00 \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (c^{3} d^{3} e^{2} f^{3} - 3 \, a c^{2} d^{2} e^{3} f^{2} g + 3 \, a^{2} c d e^{4} f g^{2} - a^{3} e^{5} g^{3}\right )}}{\sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} c^{4} d^{4} {\left | e \right |}} + \frac {2 \, {\left (5 \, c^{3} d^{3} e^{3} f^{3} + 15 \, c^{3} d^{4} e^{2} f^{2} g - 30 \, a c^{2} d^{2} e^{4} f^{2} g - 5 \, c^{3} d^{5} e f g^{2} - 20 \, a c^{2} d^{3} e^{3} f g^{2} + 40 \, a^{2} c d e^{5} f g^{2} + c^{3} d^{6} g^{3} + 2 \, a c^{2} d^{4} e^{2} g^{3} + 8 \, a^{2} c d^{2} e^{4} g^{3} - 16 \, a^{3} e^{6} g^{3}\right )}}{5 \, \sqrt {-c d^{2} e + a e^{3}} c^{4} d^{4} e {\left | e \right |}} + \frac {2 \, {\left (15 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} c^{18} d^{18} e^{24} f^{2} g - 30 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} a c^{17} d^{17} e^{25} f g^{2} + 15 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} a^{2} c^{16} d^{16} e^{26} g^{3} + 5 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c^{17} d^{17} e^{22} f g^{2} - 5 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a c^{16} d^{16} e^{23} g^{3} + {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}} c^{16} d^{16} e^{20} g^{3}\right )}}{5 \, c^{20} d^{20} e^{24} {\left | e \right |}} \]

[In]

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

[Out]

-2*(c^3*d^3*e^2*f^3 - 3*a*c^2*d^2*e^3*f^2*g + 3*a^2*c*d*e^4*f*g^2 - a^3*e^5*g^3)/(sqrt((e*x + d)*c*d*e - c*d^2
*e + a*e^3)*c^4*d^4*abs(e)) + 2/5*(5*c^3*d^3*e^3*f^3 + 15*c^3*d^4*e^2*f^2*g - 30*a*c^2*d^2*e^4*f^2*g - 5*c^3*d
^5*e*f*g^2 - 20*a*c^2*d^3*e^3*f*g^2 + 40*a^2*c*d*e^5*f*g^2 + c^3*d^6*g^3 + 2*a*c^2*d^4*e^2*g^3 + 8*a^2*c*d^2*e
^4*g^3 - 16*a^3*e^6*g^3)/(sqrt(-c*d^2*e + a*e^3)*c^4*d^4*e*abs(e)) + 2/5*(15*sqrt((e*x + d)*c*d*e - c*d^2*e +
a*e^3)*c^18*d^18*e^24*f^2*g - 30*sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3)*a*c^17*d^17*e^25*f*g^2 + 15*sqrt((e*x
+ d)*c*d*e - c*d^2*e + a*e^3)*a^2*c^16*d^16*e^26*g^3 + 5*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*c^17*d^17*
e^22*f*g^2 - 5*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a*c^16*d^16*e^23*g^3 + ((e*x + d)*c*d*e - c*d^2*e + a
*e^3)^(5/2)*c^16*d^16*e^20*g^3)/(c^20*d^20*e^24*abs(e))

Mupad [B] (verification not implemented)

Time = 12.73 (sec) , antiderivative size = 252, normalized size of antiderivative = 0.98 \[ \int \frac {(d+e x)^{3/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {\sqrt {d+e\,x}\,\left (32\,a^3\,e^3\,g^3-80\,a^2\,c\,d\,e^2\,f\,g^2+60\,a\,c^2\,d^2\,e\,f^2\,g-10\,c^3\,d^3\,f^3\right )}{5\,c^5\,d^5\,e}+\frac {2\,g^3\,x^3\,\sqrt {d+e\,x}}{5\,c^2\,d^2\,e}-\frac {2\,g^2\,x^2\,\left (2\,a\,e\,g-5\,c\,d\,f\right )\,\sqrt {d+e\,x}}{5\,c^3\,d^3\,e}+\frac {2\,g\,x\,\sqrt {d+e\,x}\,\left (8\,a^2\,e^2\,g^2-20\,a\,c\,d\,e\,f\,g+15\,c^2\,d^2\,f^2\right )}{5\,c^4\,d^4\,e}\right )}{\frac {a}{c}+x^2+\frac {x\,\left (5\,c^5\,d^6+5\,a\,c^4\,d^4\,e^2\right )}{5\,c^5\,d^5\,e}} \]

[In]

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

[Out]

((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*(((d + e*x)^(1/2)*(32*a^3*e^3*g^3 - 10*c^3*d^3*f^3 + 60*a*c^2*d
^2*e*f^2*g - 80*a^2*c*d*e^2*f*g^2))/(5*c^5*d^5*e) + (2*g^3*x^3*(d + e*x)^(1/2))/(5*c^2*d^2*e) - (2*g^2*x^2*(2*
a*e*g - 5*c*d*f)*(d + e*x)^(1/2))/(5*c^3*d^3*e) + (2*g*x*(d + e*x)^(1/2)*(8*a^2*e^2*g^2 + 15*c^2*d^2*f^2 - 20*
a*c*d*e*f*g))/(5*c^4*d^4*e)))/(a/c + x^2 + (x*(5*c^5*d^6 + 5*a*c^4*d^4*e^2))/(5*c^5*d^5*e))