Integrand size = 46, antiderivative size = 269 \[ \int \frac {\sqrt {d+e x} (f+g x)^3}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=-\frac {16 (c d f-a e g)^2 \left (2 a e^2 g-c d (3 e f-d g)\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{35 c^4 d^4 e \sqrt {d+e x}}+\frac {16 g (c d f-a e g)^2 \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{35 c^3 d^3 e}+\frac {12 (c d f-a e g) (f+g x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{35 c^2 d^2 \sqrt {d+e x}}+\frac {2 (f+g x)^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{7 c d \sqrt {d+e x}} \]
-16/35*(-a*e*g+c*d*f)^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)+12/35*(-a*e*g+c*d*f)*(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)+2/7*(g*x+f)^ 3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c/d/(e*x+d)^(1/2)+16/35*g*(-a*e* g+c*d*f)^2*(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 = 0.13 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.51 \[ \int \frac {\sqrt {d+e x} (f+g x)^3}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \sqrt {(a e+c d x) (d+e x)} \left (-16 a^3 e^3 g^3+8 a^2 c d e^2 g^2 (7 f+g x)-2 a c^2 d^2 e g \left (35 f^2+14 f g x+3 g^2 x^2\right )+c^3 d^3 \left (35 f^3+35 f^2 g x+21 f g^2 x^2+5 g^3 x^3\right )\right )}{35 c^4 d^4 \sqrt {d+e x}} \]
(2*Sqrt[(a*e + c*d*x)*(d + e*x)]*(-16*a^3*e^3*g^3 + 8*a^2*c*d*e^2*g^2*(7*f + g*x) - 2*a*c^2*d^2*e*g*(35*f^2 + 14*f*g*x + 3*g^2*x^2) + c^3*d^3*(35*f^ 3 + 35*f^2*g*x + 21*f*g^2*x^2 + 5*g^3*x^3)))/(35*c^4*d^4*Sqrt[d + e*x])
Time = 0.53 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1253, 1253, 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.
| \(\displaystyle \int \frac {\sqrt {d+e x} (f+g x)^3}{\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \, dx\) |
| \(\Big \downarrow \) 1253 |
| \(\displaystyle \frac {6 (c d f-a e g) \int \frac {\sqrt {d+e x} (f+g x)^2}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx}{7 c d}+\frac {2 (f+g x)^3 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{7 c d \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 1253 |
| \(\displaystyle \frac {6 (c d f-a e g) \left (\frac {4 (c d f-a e g) \int \frac {\sqrt {d+e x} (f+g x)}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx}{5 c d}+\frac {2 (f+g x)^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 c d \sqrt {d+e x}}\right )}{7 c d}+\frac {2 (f+g x)^3 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{7 c d \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 1221 |
| \(\displaystyle \frac {6 (c d f-a e g) \left (\frac {4 (c d f-a e g) \left (\frac {1}{3} \left (-\frac {2 a e g}{c d}-\frac {d g}{e}+3 f\right ) \int \frac {\sqrt {d+e x}}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx+\frac {2 g \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d e}\right )}{5 c d}+\frac {2 (f+g x)^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 c d \sqrt {d+e x}}\right )}{7 c d}+\frac {2 (f+g x)^3 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{7 c d \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 1122 |
| \(\displaystyle \frac {2 (f+g x)^3 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{7 c d \sqrt {d+e x}}+\frac {6 (c d f-a e g) \left (\frac {2 (f+g x)^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{5 c d \sqrt {d+e x}}+\frac {4 (c d f-a e g) \left (\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (-\frac {2 a e g}{c d}-\frac {d g}{e}+3 f\right )}{3 c d \sqrt {d+e x}}+\frac {2 g \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d e}\right )}{5 c d}\right )}{7 c d}\) |
(2*(f + g*x)^3*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(7*c*d*Sqrt[d + e*x]) + (6*(c*d*f - a*e*g)*((2*(f + g*x)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)* x + c*d*e*x^2])/(5*c*d*Sqrt[d + e*x]) + (4*(c*d*f - a*e*g)*((2*(3*f - (d*g )/e - (2*a*e*g)/(c*d))*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(3*c*d *Sqrt[d + e*x]) + (2*g*Sqrt[d + e*x]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d* e*x^2])/(3*c*d*e)))/(5*c*d)))/(7*c*d)
3.7.57.3.1 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*(m - n - 1))), x] - Simp[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] && EqQ[c*d^2 - b*d* e + a*e^2, 0] && EqQ[m + p, 0] && GtQ[n, 0] && NeQ[m - n - 1, 0] && (Intege rQ[2*p] || IntegerQ[n])
Time = 0.51 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.63
| method | result | size |
| default | \(-\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (-5 g^{3} x^{3} c^{3} d^{3}+6 a \,c^{2} d^{2} e \,g^{3} x^{2}-21 c^{3} d^{3} f \,g^{2} x^{2}-8 a^{2} c d \,e^{2} g^{3} x +28 a \,c^{2} d^{2} e f \,g^{2} x -35 c^{3} d^{3} f^{2} g x +16 a^{3} e^{3} g^{3}-56 a^{2} c d \,e^{2} f \,g^{2}+70 a \,c^{2} d^{2} e \,f^{2} g -35 f^{3} c^{3} d^{3}\right )}{35 \sqrt {e x +d}\, c^{4} d^{4}}\) | \(170\) |
| gosper | \(-\frac {2 \left (c d x +a e \right ) \left (-5 g^{3} x^{3} c^{3} d^{3}+6 a \,c^{2} d^{2} e \,g^{3} x^{2}-21 c^{3} d^{3} f \,g^{2} x^{2}-8 a^{2} c d \,e^{2} g^{3} x +28 a \,c^{2} d^{2} e f \,g^{2} x -35 c^{3} d^{3} f^{2} g x +16 a^{3} e^{3} g^{3}-56 a^{2} c d \,e^{2} f \,g^{2}+70 a \,c^{2} d^{2} e \,f^{2} g -35 f^{3} c^{3} d^{3}\right ) \sqrt {e x +d}}{35 c^{4} d^{4} \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}\) | \(188\) |
-2/35/(e*x+d)^(1/2)*((c*d*x+a*e)*(e*x+d))^(1/2)*(-5*c^3*d^3*g^3*x^3+6*a*c^ 2*d^2*e*g^3*x^2-21*c^3*d^3*f*g^2*x^2-8*a^2*c*d*e^2*g^3*x+28*a*c^2*d^2*e*f* g^2*x-35*c^3*d^3*f^2*g*x+16*a^3*e^3*g^3-56*a^2*c*d*e^2*f*g^2+70*a*c^2*d^2* e*f^2*g-35*c^3*d^3*f^3)/c^4/d^4
Time = 0.30 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {d+e x} (f+g x)^3}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, {\left (5 \, c^{3} d^{3} g^{3} x^{3} + 35 \, c^{3} d^{3} f^{3} - 70 \, a c^{2} d^{2} e f^{2} g + 56 \, a^{2} c d e^{2} f g^{2} - 16 \, a^{3} e^{3} g^{3} + 3 \, {\left (7 \, c^{3} d^{3} f g^{2} - 2 \, a c^{2} d^{2} e g^{3}\right )} x^{2} + {\left (35 \, c^{3} d^{3} f^{2} g - 28 \, a c^{2} d^{2} e f g^{2} + 8 \, a^{2} c d e^{2} g^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{35 \, {\left (c^{4} d^{4} e x + c^{4} d^{5}\right )}} \]
integrate((g*x+f)^3*(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2), x, algorithm="fricas")
2/35*(5*c^3*d^3*g^3*x^3 + 35*c^3*d^3*f^3 - 70*a*c^2*d^2*e*f^2*g + 56*a^2*c *d*e^2*f*g^2 - 16*a^3*e^3*g^3 + 3*(7*c^3*d^3*f*g^2 - 2*a*c^2*d^2*e*g^3)*x^ 2 + (35*c^3*d^3*f^2*g - 28*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^4*d^4*e*x + c^4*d^ 5)
\[ \int \frac {\sqrt {d+e x} (f+g x)^3}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\int \frac {\sqrt {d + e x} \left (f + g x\right )^{3}}{\sqrt {\left (d + e x\right ) \left (a e + c d x\right )}}\, dx \]
Time = 0.24 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {d+e x} (f+g x)^3}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, \sqrt {c d x + a e} f^{3}}{c d} + \frac {2 \, {\left (c^{2} d^{2} x^{2} - a c d e x - 2 \, a^{2} e^{2}\right )} f^{2} g}{\sqrt {c d x + a e} c^{2} d^{2}} + \frac {2 \, {\left (3 \, c^{3} d^{3} x^{3} - a c^{2} d^{2} e x^{2} + 4 \, a^{2} c d e^{2} x + 8 \, a^{3} e^{3}\right )} f g^{2}}{5 \, \sqrt {c d x + a e} c^{3} d^{3}} + \frac {2 \, {\left (5 \, c^{4} d^{4} x^{4} - a c^{3} d^{3} e x^{3} + 2 \, a^{2} c^{2} d^{2} e^{2} x^{2} - 8 \, a^{3} c d e^{3} x - 16 \, a^{4} e^{4}\right )} g^{3}}{35 \, \sqrt {c d x + a e} c^{4} d^{4}} \]
integrate((g*x+f)^3*(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2), x, algorithm="maxima")
2*sqrt(c*d*x + a*e)*f^3/(c*d) + 2*(c^2*d^2*x^2 - a*c*d*e*x - 2*a^2*e^2)*f^ 2*g/(sqrt(c*d*x + a*e)*c^2*d^2) + 2/5*(3*c^3*d^3*x^3 - a*c^2*d^2*e*x^2 + 4 *a^2*c*d*e^2*x + 8*a^3*e^3)*f*g^2/(sqrt(c*d*x + a*e)*c^3*d^3) + 2/35*(5*c^ 4*d^4*x^4 - a*c^3*d^3*e*x^3 + 2*a^2*c^2*d^2*e^2*x^2 - 8*a^3*c*d*e^3*x - 16 *a^4*e^4)*g^3/(sqrt(c*d*x + a*e)*c^4*d^4)
Leaf count of result is larger than twice the leaf count of optimal. 613 vs. \(2 (245) = 490\).
Time = 0.31 (sec) , antiderivative size = 613, normalized size of antiderivative = 2.28 \[ \int \frac {\sqrt {d+e x} (f+g x)^3}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \, e {\left (\frac {35 \, {\left (c^{3} d^{3} f^{3} - 3 \, a c^{2} d^{2} e f^{2} g + 3 \, a^{2} c d e^{2} f g^{2} - a^{3} e^{3} g^{3}\right )} \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}}}{c^{4} d^{4} e} - \frac {35 \, \sqrt {-c d^{2} e + a e^{3}} c^{3} d^{3} e^{3} f^{3} - 35 \, \sqrt {-c d^{2} e + a e^{3}} c^{3} d^{4} e^{2} f^{2} g - 70 \, \sqrt {-c d^{2} e + a e^{3}} a c^{2} d^{2} e^{4} f^{2} g + 21 \, \sqrt {-c d^{2} e + a e^{3}} c^{3} d^{5} e f g^{2} + 28 \, \sqrt {-c d^{2} e + a e^{3}} a c^{2} d^{3} e^{3} f g^{2} + 56 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} c d e^{5} f g^{2} - 5 \, \sqrt {-c d^{2} e + a e^{3}} c^{3} d^{6} g^{3} - 6 \, \sqrt {-c d^{2} e + a e^{3}} a c^{2} d^{4} e^{2} g^{3} - 8 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} c d^{2} e^{4} g^{3} - 16 \, \sqrt {-c d^{2} e + a e^{3}} a^{3} e^{6} g^{3}}{c^{4} d^{4} e^{4}} + \frac {35 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c^{2} d^{2} e^{4} f^{2} g - 70 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a c d e^{5} f g^{2} + 35 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a^{2} e^{6} g^{3} + 21 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}} c d e^{2} f g^{2} - 21 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}} a e^{3} g^{3} + 5 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {7}{2}} g^{3}}{c^{4} d^{4} e^{7}}\right )}}{35 \, {\left | e \right |}} \]
integrate((g*x+f)^3*(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2), x, algorithm="giac")
2/35*e*(35*(c^3*d^3*f^3 - 3*a*c^2*d^2*e*f^2*g + 3*a^2*c*d*e^2*f*g^2 - a^3* e^3*g^3)*sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3)/(c^4*d^4*e) - (35*sqrt(-c *d^2*e + a*e^3)*c^3*d^3*e^3*f^3 - 35*sqrt(-c*d^2*e + a*e^3)*c^3*d^4*e^2*f^ 2*g - 70*sqrt(-c*d^2*e + a*e^3)*a*c^2*d^2*e^4*f^2*g + 21*sqrt(-c*d^2*e + a *e^3)*c^3*d^5*e*f*g^2 + 28*sqrt(-c*d^2*e + a*e^3)*a*c^2*d^3*e^3*f*g^2 + 56 *sqrt(-c*d^2*e + a*e^3)*a^2*c*d*e^5*f*g^2 - 5*sqrt(-c*d^2*e + a*e^3)*c^3*d ^6*g^3 - 6*sqrt(-c*d^2*e + a*e^3)*a*c^2*d^4*e^2*g^3 - 8*sqrt(-c*d^2*e + a* e^3)*a^2*c*d^2*e^4*g^3 - 16*sqrt(-c*d^2*e + a*e^3)*a^3*e^6*g^3)/(c^4*d^4*e ^4) + (35*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*c^2*d^2*e^4*f^2*g - 70 *((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a*c*d*e^5*f*g^2 + 35*((e*x + d) *c*d*e - c*d^2*e + a*e^3)^(3/2)*a^2*e^6*g^3 + 21*((e*x + d)*c*d*e - c*d^2* e + a*e^3)^(5/2)*c*d*e^2*f*g^2 - 21*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(5 /2)*a*e^3*g^3 + 5*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(7/2)*g^3)/(c^4*d^4* e^7))/abs(e)
Time = 12.61 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {d+e x} (f+g x)^3}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=-\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {\sqrt {d+e\,x}\,\left (32\,a^3\,e^3\,g^3-112\,a^2\,c\,d\,e^2\,f\,g^2+140\,a\,c^2\,d^2\,e\,f^2\,g-70\,c^3\,d^3\,f^3\right )}{35\,c^4\,d^4\,e}-\frac {2\,g^3\,x^3\,\sqrt {d+e\,x}}{7\,c\,d\,e}+\frac {6\,g^2\,x^2\,\left (2\,a\,e\,g-7\,c\,d\,f\right )\,\sqrt {d+e\,x}}{35\,c^2\,d^2\,e}-\frac {2\,g\,x\,\sqrt {d+e\,x}\,\left (8\,a^2\,e^2\,g^2-28\,a\,c\,d\,e\,f\,g+35\,c^2\,d^2\,f^2\right )}{35\,c^3\,d^3\,e}\right )}{x+\frac {d}{e}} \]
-((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 - 70*c^3*d^3*f^3 + 140*a*c^2*d^2*e*f^2*g - 112*a^2*c*d*e^2*f*g^2)) /(35*c^4*d^4*e) - (2*g^3*x^3*(d + e*x)^(1/2))/(7*c*d*e) + (6*g^2*x^2*(2*a* e*g - 7*c*d*f)*(d + e*x)^(1/2))/(35*c^2*d^2*e) - (2*g*x*(d + e*x)^(1/2)*(8 *a^2*e^2*g^2 + 35*c^2*d^2*f^2 - 28*a*c*d*e*f*g))/(35*c^3*d^3*e)))/(x + d/e )