\(\int \frac {(d+e x)^{5/2} (f+g x)^2}{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx\) [673]

Optimal result

Integrand size = 46, antiderivative size = 211 \[ \int \frac {(d+e x)^{5/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=-\frac {2 (d+e x)^{3/2} (f+g x)^2}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {8 g (c d f-a e g) (d+e x)^{3/2}}{3 c^2 d^2 \left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {8 g \left (2 a e^2 g-c d (e f+d g)\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c^3 d^3 \left (c d^2-a e^2\right ) \sqrt {d+e x}} \]

[Out]

-2/3*(e*x+d)^(3/2)*(g*x+f)^2/c/d/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-8/3*g*(-a*e*g+c*d*f)*(e*x+d)^(3/2)/c^
2/d^2/(-a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-8/3*g*(2*a*e^2*g-c*d*(d*g+e*f))*(a*d*e+(a*e^2+c*d
^2)*x+c*d*e*x^2)^(1/2)/c^3/d^3/(-a*e^2+c*d^2)/(e*x+d)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {880, 802, 662} \[ \int \frac {(d+e x)^{5/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=-\frac {8 g \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (2 a e^2 g-c d (d g+e f)\right )}{3 c^3 d^3 \sqrt {d+e x} \left (c d^2-a e^2\right )}-\frac {8 g (d+e x)^{3/2} (c d f-a e g)}{3 c^2 d^2 \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 (d+e x)^{3/2} (f+g x)^2}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \]

[In]

Int[((d + e*x)^(5/2)*(f + g*x)^2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2),x]

[Out]

(-2*(d + e*x)^(3/2)*(f + g*x)^2)/(3*c*d*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)) - (8*g*(c*d*f - a*e*g)*
(d + e*x)^(3/2))/(3*c^2*d^2*(c*d^2 - a*e^2)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (8*g*(2*a*e^2*g - c
*d*(e*f + d*g))*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(3*c^3*d^3*(c*d^2 - a*e^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 802

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.41 \[ \int \frac {(d+e x)^{5/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {2 (d+e x)^{3/2} \left (8 a^2 e^2 g^2-4 a c d e g (f-3 g x)-c^2 d^2 \left (f^2+6 f g x-3 g^2 x^2\right )\right )}{3 c^3 d^3 ((a e+c d x) (d+e x))^{3/2}} \]

[In]

Integrate[((d + e*x)^(5/2)*(f + g*x)^2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.51

method	result	size
default	\(\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (3 g^{2} x^{2} c^{2} d^{2}+12 a c d e \,g^{2} x -6 c^{2} d^{2} f g x +8 a^{2} e^{2} g^{2}-4 a c d e f g -c^{2} d^{2} f^{2}\right )}{3 \sqrt {e x +d}\, \left (c d x +a e \right )^{2} c^{3} d^{3}}\)	\(108\)
gosper	\(\frac {2 \left (c d x +a e \right ) \left (3 g^{2} x^{2} c^{2} d^{2}+12 a c d e \,g^{2} x -6 c^{2} d^{2} f g x +8 a^{2} e^{2} g^{2}-4 a c d e f g -c^{2} d^{2} f^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{3 c^{3} d^{3} \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {5}{2}}}\)	\(116\)

[In]

int((e*x+d)^(5/2)*(g*x+f)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/3/(e*x+d)^(1/2)*((c*d*x+a*e)*(e*x+d))^(1/2)*(3*c^2*d^2*g^2*x^2+12*a*c*d*e*g^2*x-6*c^2*d^2*f*g*x+8*a^2*e^2*g^
2-4*a*c*d*e*f*g-c^2*d^2*f^2)/(c*d*x+a*e)^2/c^3/d^3

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.85 \[ \int \frac {(d+e x)^{5/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (3 \, c^{2} d^{2} g^{2} x^{2} - c^{2} d^{2} f^{2} - 4 \, a c d e f g + 8 \, a^{2} e^{2} g^{2} - 6 \, {\left (c^{2} d^{2} f g - 2 \, a c d e g^{2}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{3 \, {\left (c^{5} d^{5} e x^{3} + a^{2} c^{3} d^{4} e^{2} + {\left (c^{5} d^{6} + 2 \, a c^{4} d^{4} e^{2}\right )} x^{2} + {\left (2 \, a c^{4} d^{5} e + a^{2} c^{3} d^{3} e^{3}\right )} x\right )}} \]

[In]

integrate((e*x+d)^(5/2)*(g*x+f)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{5/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((e*x+d)**(5/2)*(g*x+f)**2/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.65 \[ \int \frac {(d+e x)^{5/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=-\frac {4 \, {\left (3 \, c d x + 2 \, a e\right )} f g}{3 \, {\left (c^{3} d^{3} x + a c^{2} d^{2} e\right )} \sqrt {c d x + a e}} + \frac {2 \, {\left (3 \, c^{2} d^{2} x^{2} + 12 \, a c d e x + 8 \, a^{2} e^{2}\right )} g^{2}}{3 \, {\left (c^{4} d^{4} x + a c^{3} d^{3} e\right )} \sqrt {c d x + a e}} - \frac {2 \, f^{2}}{3 \, {\left (c^{2} d^{2} x + a c d e\right )} \sqrt {c d x + a e}} \]

[In]

integrate((e*x+d)^(5/2)*(g*x+f)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="maxima")

[Out]

-4/3*(3*c*d*x + 2*a*e)*f*g/((c^3*d^3*x + a*c^2*d^2*e)*sqrt(c*d*x + a*e)) + 2/3*(3*c^2*d^2*x^2 + 12*a*c*d*e*x +
8*a^2*e^2)*g^2/((c^4*d^4*x + a*c^3*d^3*e)*sqrt(c*d*x + a*e)) - 2/3*f^2/((c^2*d^2*x + a*c*d*e)*sqrt(c*d*x + a*
e))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.44 \[ \int \frac {(d+e x)^{5/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (c^{2} d^{2} e^{3} f^{2} - 6 \, c^{2} d^{3} e^{2} f g + 4 \, a c d e^{4} f g - 3 \, c^{2} d^{4} e g^{2} + 12 \, a c d^{2} e^{3} g^{2} - 8 \, a^{2} e^{5} g^{2}\right )}}{3 \, {\left (\sqrt {-c d^{2} e + a e^{3}} c^{4} d^{5} {\left | e \right |} - \sqrt {-c d^{2} e + a e^{3}} a c^{3} d^{3} e^{2} {\left | e \right |}\right )}} + \frac {2 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} g^{2}}{c^{3} d^{3} {\left | e \right |}} - \frac {2 \, {\left (c^{2} d^{2} e^{4} f^{2} - 2 \, a c d e^{5} f g + a^{2} e^{6} g^{2} + 6 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )} c d e^{2} f g - 6 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )} a e^{3} g^{2}\right )}}{3 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c^{3} d^{3} {\left | e \right |}} \]

[In]

integrate((e*x+d)^(5/2)*(g*x+f)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="giac")

[Out]

-2/3*(c^2*d^2*e^3*f^2 - 6*c^2*d^3*e^2*f*g + 4*a*c*d*e^4*f*g - 3*c^2*d^4*e*g^2 + 12*a*c*d^2*e^3*g^2 - 8*a^2*e^5
*g^2)/(sqrt(-c*d^2*e + a*e^3)*c^4*d^5*abs(e) - sqrt(-c*d^2*e + a*e^3)*a*c^3*d^3*e^2*abs(e)) + 2*sqrt((e*x + d)
*c*d*e - c*d^2*e + a*e^3)*g^2/(c^3*d^3*abs(e)) - 2/3*(c^2*d^2*e^4*f^2 - 2*a*c*d*e^5*f*g + a^2*e^6*g^2 + 6*((e*
x + d)*c*d*e - c*d^2*e + a*e^3)*c*d*e^2*f*g - 6*((e*x + d)*c*d*e - c*d^2*e + a*e^3)*a*e^3*g^2)/(((e*x + d)*c*d
*e - c*d^2*e + a*e^3)^(3/2)*c^3*d^3*abs(e))

Mupad [B] (verification not implemented)

Time = 12.73 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.98 \[ \int \frac {(d+e x)^{5/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {2\,g^2\,x^2\,\sqrt {d+e\,x}}{c^3\,d^3\,e}-\frac {\sqrt {d+e\,x}\,\left (-16\,a^2\,e^2\,g^2+8\,a\,c\,d\,e\,f\,g+2\,c^2\,d^2\,f^2\right )}{3\,c^5\,d^5\,e}+\frac {4\,g\,x\,\left (2\,a\,e\,g-c\,d\,f\right )\,\sqrt {d+e\,x}}{c^4\,d^4\,e}\right )}{x^3+\frac {a^2\,e}{c^2\,d}+\frac {a\,x\,\left (2\,c\,d^2+a\,e^2\right )}{c^2\,d^2}+\frac {x^2\,\left (3\,c^5\,d^6+6\,a\,c^4\,d^4\,e^2\right )}{3\,c^5\,d^5\,e}} \]

[In]

int(((f + g*x)^2*(d + e*x)^(5/2))/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2),x)

[Out]

((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*((2*g^2*x^2*(d + e*x)^(1/2))/(c^3*d^3*e) - ((d + e*x)^(1/2)*(2*
c^2*d^2*f^2 - 16*a^2*e^2*g^2 + 8*a*c*d*e*f*g))/(3*c^5*d^5*e) + (4*g*x*(2*a*e*g - c*d*f)*(d + e*x)^(1/2))/(c^4*
d^4*e)))/(x^3 + (a^2*e)/(c^2*d) + (a*x*(a*e^2 + 2*c*d^2))/(c^2*d^2) + (x^2*(3*c^5*d^6 + 6*a*c^4*d^4*e^2))/(3*c
^5*d^5*e))