Integrand size = 46, antiderivative size = 239 \[ \int \frac {(d+e x)^{5/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=-\frac {2 (d+e x)^{3/2} (f+g x)^3}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {4 g \sqrt {d+e x} (f+g x)^2}{c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {16 g^2 \left (2 a e^2 g-c d (3 e f-d g)\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c^4 d^4 e \sqrt {d+e x}}+\frac {16 g^3 \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c^3 d^3 e} \]
-2/3*(e*x+d)^(3/2)*(g*x+f)^3/c/d/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-4 *g*(g*x+f)^2*(e*x+d)^(1/2)/c^2/d^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2) -16/3*g^2*(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)+16/3*g^3*(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
Time = 0.12 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.55 \[ \int \frac {(d+e x)^{5/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {2 (d+e x)^{3/2} \left (-16 a^3 e^3 g^3+24 a^2 c d e^2 g^2 (f-g x)-6 a c^2 d^2 e g \left (f^2-6 f g x+g^2 x^2\right )+c^3 d^3 \left (-f^3-9 f^2 g x+9 f g^2 x^2+g^3 x^3\right )\right )}{3 c^4 d^4 ((a e+c d x) (d+e x))^{3/2}} \]
(2*(d + e*x)^(3/2)*(-16*a^3*e^3*g^3 + 24*a^2*c*d*e^2*g^2*(f - g*x) - 6*a*c ^2*d^2*e*g*(f^2 - 6*f*g*x + g^2*x^2) + c^3*d^3*(-f^3 - 9*f^2*g*x + 9*f*g^2 *x^2 + g^3*x^3)))/(3*c^4*d^4*((a*e + c*d*x)*(d + e*x))^(3/2))
Time = 0.48 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1251, 1251, 1221, 1122}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(d+e x)^{5/2} (f+g x)^3}{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1251 |
| \(\displaystyle \frac {2 g \int \frac {(d+e x)^{3/2} (f+g x)^2}{\left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right )^{3/2}}dx}{c d}-\frac {2 (d+e x)^{3/2} (f+g x)^3}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1251 |
| \(\displaystyle \frac {2 g \left (\frac {4 g \int \frac {\sqrt {d+e x} (f+g x)}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx}{c d}-\frac {2 \sqrt {d+e x} (f+g x)^2}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{c d}-\frac {2 (d+e x)^{3/2} (f+g x)^3}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1221 |
| \(\displaystyle \frac {2 g \left (\frac {4 g \left (\frac {1}{3} \left (-\frac {2 a e g}{c d}-\frac {d g}{e}+3 f\right ) \int \frac {\sqrt {d+e x}}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx+\frac {2 g \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d e}\right )}{c d}-\frac {2 \sqrt {d+e x} (f+g x)^2}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{c d}-\frac {2 (d+e x)^{3/2} (f+g x)^3}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1122 |
| \(\displaystyle \frac {2 g \left (\frac {4 g \left (\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (-\frac {2 a e g}{c d}-\frac {d g}{e}+3 f\right )}{3 c d \sqrt {d+e x}}+\frac {2 g \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d e}\right )}{c d}-\frac {2 \sqrt {d+e x} (f+g x)^2}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{c d}-\frac {2 (d+e x)^{3/2} (f+g x)^3}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}\) |
(-2*(d + e*x)^(3/2)*(f + g*x)^3)/(3*c*d*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e *x^2)^(3/2)) + (2*g*((-2*Sqrt[d + e*x]*(f + g*x)^2)/(c*d*Sqrt[a*d*e + (c*d ^2 + a*e^2)*x + c*d*e*x^2]) + (4*g*((2*(3*f - (d*g)/e - (2*a*e*g)/(c*d))*S qrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(3*c*d*Sqrt[d + e*x]) + (2*g*S qrt[d + e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(3*c*d*e)))/(c*d )))/(c*d)
3.7.72.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> 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] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[m + p, 0]
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] + Simp[(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] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0]
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] - Simp[e*g*(n/(c*(p + 1))) Int[( d + e*x)^(m - 1)*(f + g*x)^(n - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; Fre eQ[{a, b, c, d, e, f, g}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[m + p, 0] && LtQ[p, -1] && GtQ[n, 0]
Time = 0.54 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.75
| method | result | size |
| default | \(-\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (-g^{3} x^{3} c^{3} d^{3}+6 a \,c^{2} d^{2} e \,g^{3} x^{2}-9 c^{3} d^{3} f \,g^{2} x^{2}+24 a^{2} c d \,e^{2} g^{3} x -36 a \,c^{2} d^{2} e f \,g^{2} x +9 c^{3} d^{3} f^{2} g x +16 a^{3} e^{3} g^{3}-24 a^{2} c d \,e^{2} f \,g^{2}+6 a \,c^{2} d^{2} e \,f^{2} g +f^{3} c^{3} d^{3}\right )}{3 \sqrt {e x +d}\, \left (c d x +a e \right )^{2} c^{4} d^{4}}\) | \(179\) |
| gosper | \(-\frac {2 \left (c d x +a e \right ) \left (-g^{3} x^{3} c^{3} d^{3}+6 a \,c^{2} d^{2} e \,g^{3} x^{2}-9 c^{3} d^{3} f \,g^{2} x^{2}+24 a^{2} c d \,e^{2} g^{3} x -36 a \,c^{2} d^{2} e f \,g^{2} x +9 c^{3} d^{3} f^{2} g x +16 a^{3} e^{3} g^{3}-24 a^{2} c d \,e^{2} f \,g^{2}+6 a \,c^{2} d^{2} e \,f^{2} g +f^{3} c^{3} d^{3}\right ) \left (e x +d \right )^{\frac {5}{2}}}{3 c^{4} d^{4} \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {5}{2}}}\) | \(187\) |
-2/3/(e*x+d)^(1/2)*((c*d*x+a*e)*(e*x+d))^(1/2)*(-c^3*d^3*g^3*x^3+6*a*c^2*d ^2*e*g^3*x^2-9*c^3*d^3*f*g^2*x^2+24*a^2*c*d*e^2*g^3*x-36*a*c^2*d^2*e*f*g^2 *x+9*c^3*d^3*f^2*g*x+16*a^3*e^3*g^3-24*a^2*c*d*e^2*f*g^2+6*a*c^2*d^2*e*f^2 *g+c^3*d^3*f^3)/(c*d*x+a*e)^2/c^4/d^4
Time = 0.42 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.05 \[ \int \frac {(d+e x)^{5/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (c^{3} d^{3} g^{3} x^{3} - c^{3} d^{3} f^{3} - 6 \, a c^{2} d^{2} e f^{2} g + 24 \, a^{2} c d e^{2} f g^{2} - 16 \, a^{3} e^{3} g^{3} + 3 \, {\left (3 \, c^{3} d^{3} f g^{2} - 2 \, a c^{2} d^{2} e g^{3}\right )} x^{2} - 3 \, {\left (3 \, c^{3} d^{3} f^{2} g - 12 \, a c^{2} d^{2} e f g^{2} + 8 \, a^{2} c d e^{2} g^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{3 \, {\left (c^{6} d^{6} e x^{3} + a^{2} c^{4} d^{5} e^{2} + {\left (c^{6} d^{7} + 2 \, a c^{5} d^{5} e^{2}\right )} x^{2} + {\left (2 \, a c^{5} d^{6} e + a^{2} c^{4} d^{4} e^{3}\right )} x\right )}} \]
integrate((e*x+d)^(5/2)*(g*x+f)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2), x, algorithm="fricas")
2/3*(c^3*d^3*g^3*x^3 - c^3*d^3*f^3 - 6*a*c^2*d^2*e*f^2*g + 24*a^2*c*d*e^2* f*g^2 - 16*a^3*e^3*g^3 + 3*(3*c^3*d^3*f*g^2 - 2*a*c^2*d^2*e*g^3)*x^2 - 3*( 3*c^3*d^3*f^2*g - 12*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^6*d^6*e*x^3 + a^2*c^4*d^ 5*e^2 + (c^6*d^7 + 2*a*c^5*d^5*e^2)*x^2 + (2*a*c^5*d^6*e + a^2*c^4*d^4*e^3 )*x)
Timed out. \[ \int \frac {(d+e x)^{5/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\text {Timed out} \]
Time = 0.25 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.92 \[ \int \frac {(d+e x)^{5/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (3 \, c d x + 2 \, a e\right )} f^{2} g}{{\left (c^{3} d^{3} x + a c^{2} d^{2} e\right )} \sqrt {c d x + a e}} + \frac {2 \, {\left (3 \, c^{2} d^{2} x^{2} + 12 \, a c d e x + 8 \, a^{2} e^{2}\right )} f g^{2}}{{\left (c^{4} d^{4} x + a c^{3} d^{3} e\right )} \sqrt {c d x + a e}} + \frac {2 \, {\left (c^{3} d^{3} x^{3} - 6 \, a c^{2} d^{2} e x^{2} - 24 \, a^{2} c d e^{2} x - 16 \, a^{3} e^{3}\right )} g^{3}}{3 \, {\left (c^{5} d^{5} x + a c^{4} d^{4} e\right )} \sqrt {c d x + a e}} - \frac {2 \, f^{3}}{3 \, {\left (c^{2} d^{2} x + a c d e\right )} \sqrt {c d x + a e}} \]
integrate((e*x+d)^(5/2)*(g*x+f)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2), x, algorithm="maxima")
-2*(3*c*d*x + 2*a*e)*f^2*g/((c^3*d^3*x + a*c^2*d^2*e)*sqrt(c*d*x + a*e)) + 2*(3*c^2*d^2*x^2 + 12*a*c*d*e*x + 8*a^2*e^2)*f*g^2/((c^4*d^4*x + a*c^3*d^ 3*e)*sqrt(c*d*x + a*e)) + 2/3*(c^3*d^3*x^3 - 6*a*c^2*d^2*e*x^2 - 24*a^2*c* d*e^2*x - 16*a^3*e^3)*g^3/((c^5*d^5*x + a*c^4*d^4*e)*sqrt(c*d*x + a*e)) - 2/3*f^3/((c^2*d^2*x + a*c*d*e)*sqrt(c*d*x + a*e))
Leaf count of result is larger than twice the leaf count of optimal. 524 vs. \(2 (217) = 434\).
Time = 0.34 (sec) , antiderivative size = 524, normalized size of antiderivative = 2.19 \[ \int \frac {(d+e x)^{5/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (c^{3} d^{3} e^{3} f^{3} - 9 \, c^{3} d^{4} e^{2} f^{2} g + 6 \, a c^{2} d^{2} e^{4} f^{2} g - 9 \, c^{3} d^{5} e f g^{2} + 36 \, a c^{2} d^{3} e^{3} f g^{2} - 24 \, a^{2} c d e^{5} f g^{2} + c^{3} d^{6} g^{3} + 6 \, a c^{2} d^{4} e^{2} g^{3} - 24 \, a^{2} c d^{2} e^{4} g^{3} + 16 \, a^{3} e^{6} g^{3}\right )}}{3 \, {\left (\sqrt {-c d^{2} e + a e^{3}} c^{5} d^{6} {\left | e \right |} - \sqrt {-c d^{2} e + a e^{3}} a c^{4} d^{4} e^{2} {\left | e \right |}\right )}} - \frac {2 \, {\left (c^{3} d^{3} e^{4} f^{3} - 3 \, a c^{2} d^{2} e^{5} f^{2} g + 3 \, a^{2} c d e^{6} f g^{2} - a^{3} e^{7} g^{3} + 9 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )} c^{2} d^{2} e^{2} f^{2} g - 18 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )} a c d e^{3} f g^{2} + 9 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )} a^{2} e^{4} g^{3}\right )}}{3 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c^{4} d^{4} {\left | e \right |}} + \frac {2 \, {\left (9 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} c^{9} d^{9} e^{8} f g^{2} - 9 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} a c^{8} d^{8} e^{9} g^{3} + {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c^{8} d^{8} e^{6} g^{3}\right )}}{3 \, c^{12} d^{12} e^{8} {\left | e \right |}} \]
integrate((e*x+d)^(5/2)*(g*x+f)^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2), x, algorithm="giac")
-2/3*(c^3*d^3*e^3*f^3 - 9*c^3*d^4*e^2*f^2*g + 6*a*c^2*d^2*e^4*f^2*g - 9*c^ 3*d^5*e*f*g^2 + 36*a*c^2*d^3*e^3*f*g^2 - 24*a^2*c*d*e^5*f*g^2 + c^3*d^6*g^ 3 + 6*a*c^2*d^4*e^2*g^3 - 24*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^5*d^6*abs(e) - sqrt(-c*d^2*e + a*e^3)*a*c^4*d^4*e^2*abs(e )) - 2/3*(c^3*d^3*e^4*f^3 - 3*a*c^2*d^2*e^5*f^2*g + 3*a^2*c*d*e^6*f*g^2 - a^3*e^7*g^3 + 9*((e*x + d)*c*d*e - c*d^2*e + a*e^3)*c^2*d^2*e^2*f^2*g - 18 *((e*x + d)*c*d*e - c*d^2*e + a*e^3)*a*c*d*e^3*f*g^2 + 9*((e*x + d)*c*d*e - c*d^2*e + a*e^3)*a^2*e^4*g^3)/(((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2) *c^4*d^4*abs(e)) + 2/3*(9*sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3)*c^9*d^9* e^8*f*g^2 - 9*sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3)*a*c^8*d^8*e^9*g^3 + ((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*c^8*d^8*e^6*g^3)/(c^12*d^12*e^8* abs(e))
Time = 12.94 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.16 \[ \int \frac {(d+e x)^{5/2} (f+g x)^3}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {\sqrt {d+e\,x}\,\left (\frac {32\,a^3\,e^3\,g^3}{3}-16\,a^2\,c\,d\,e^2\,f\,g^2+4\,a\,c^2\,d^2\,e\,f^2\,g+\frac {2\,c^3\,d^3\,f^3}{3}\right )}{c^6\,d^6\,e}-\frac {2\,g^3\,x^3\,\sqrt {d+e\,x}}{3\,c^3\,d^3\,e}+\frac {g^2\,x^2\,\left (4\,a\,e\,g-6\,c\,d\,f\right )\,\sqrt {d+e\,x}}{c^4\,d^4\,e}+\frac {2\,g\,x\,\sqrt {d+e\,x}\,\left (8\,a^2\,e^2\,g^2-12\,a\,c\,d\,e\,f\,g+3\,c^2\,d^2\,f^2\right )}{c^5\,d^5\,e}\right )}{x^3+\frac {a^2\,e}{c^2\,d}+\frac {a\,x\,\left (2\,c\,d^2+a\,e^2\right )}{c^2\,d^2}+\frac {x^2\,\left (c^6\,d^7+2\,a\,c^5\,d^5\,e^2\right )}{c^6\,d^6\,e}} \]
-((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)/3 + (2*c^3*d^3*f^3)/3 + 4*a*c^2*d^2*e*f^2*g - 16*a^2*c*d*e^2*f*g ^2))/(c^6*d^6*e) - (2*g^3*x^3*(d + e*x)^(1/2))/(3*c^3*d^3*e) + (g^2*x^2*(4 *a*e*g - 6*c*d*f)*(d + e*x)^(1/2))/(c^4*d^4*e) + (2*g*x*(d + e*x)^(1/2)*(8 *a^2*e^2*g^2 + 3*c^2*d^2*f^2 - 12*a*c*d*e*f*g))/(c^5*d^5*e)))/(x^3 + (a^2* e)/(c^2*d) + (a*x*(a*e^2 + 2*c*d^2))/(c^2*d^2) + (x^2*(c^6*d^7 + 2*a*c^5*d ^5*e^2))/(c^6*d^6*e))