Integrand size = 46, antiderivative size = 181 \[ \int \frac {(d+e x)^{3/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {2 \sqrt {d+e x} (f+g x)^2}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {8 g \left (2 a e^2 g-c d (3 e f-d g)\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c^3 d^3 e \sqrt {d+e x}}+\frac {8 g^2 \sqrt {d+e x} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c^2 d^2 e} \]
-2*(g*x+f)^2*(e*x+d)^(1/2)/c/d/(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+3*e*f))*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c^ 3/d^3/e/(e*x+d)^(1/2)+8/3*g^2*(e*x+d)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x ^2)^(1/2)/c^2/d^2/e
Time = 0.07 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.49 \[ \int \frac {(d+e x)^{3/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {d+e x} \left (-8 a^2 e^2 g^2-4 a c d e g (-3 f+g x)+c^2 d^2 \left (-3 f^2+6 f g x+g^2 x^2\right )\right )}{3 c^3 d^3 \sqrt {(a e+c d x) (d+e x)}} \]
(2*Sqrt[d + e*x]*(-8*a^2*e^2*g^2 - 4*a*c*d*e*g*(-3*f + g*x) + c^2*d^2*(-3* f^2 + 6*f*g*x + g^2*x^2)))/(3*c^3*d^3*Sqrt[(a*e + c*d*x)*(d + e*x)])
Time = 0.37 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {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.
\(\displaystyle \int \frac {(d+e x)^{3/2} (f+g x)^2}{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1251 |
\(\displaystyle \frac {4 g \int \frac {\sqrt {d+e x} (f+g x)}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx}{c d}-\frac {2 \sqrt {d+e x} (f+g x)^2}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\) |
\(\Big \downarrow \) 1221 |
\(\displaystyle \frac {4 g \left (\frac {1}{3} \left (-\frac {2 a e g}{c d}-\frac {d g}{e}+3 f\right ) \int \frac {\sqrt {d+e x}}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx+\frac {2 g \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d e}\right )}{c d}-\frac {2 \sqrt {d+e x} (f+g x)^2}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle \frac {4 g \left (\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (-\frac {2 a e g}{c d}-\frac {d g}{e}+3 f\right )}{3 c d \sqrt {d+e x}}+\frac {2 g \sqrt {d+e x} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d e}\right )}{c d}-\frac {2 \sqrt {d+e x} (f+g x)^2}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\) |
(-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))*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)))/(c*d)
3.7.66.3.1 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 = 0.53 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.60
method | result | size |
default | \(-\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (-g^{2} x^{2} c^{2} d^{2}+4 a c d e \,g^{2} x -6 c^{2} d^{2} f g x +8 a^{2} e^{2} g^{2}-12 a c d e f g +3 c^{2} d^{2} f^{2}\right )}{3 \sqrt {e x +d}\, \left (c d x +a e \right ) c^{3} d^{3}}\) | \(108\) |
gosper | \(-\frac {2 \left (c d x +a e \right ) \left (-g^{2} x^{2} c^{2} d^{2}+4 a c d e \,g^{2} x -6 c^{2} d^{2} f g x +8 a^{2} e^{2} g^{2}-12 a c d e f g +3 c^{2} d^{2} f^{2}\right ) \left (e x +d \right )^{\frac {3}{2}}}{3 c^{3} d^{3} \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}}\) | \(116\) |
-2/3/(e*x+d)^(1/2)*((c*d*x+a*e)*(e*x+d))^(1/2)*(-c^2*d^2*g^2*x^2+4*a*c*d*e *g^2*x-6*c^2*d^2*f*g*x+8*a^2*e^2*g^2-12*a*c*d*e*f*g+3*c^2*d^2*f^2)/(c*d*x+ a*e)/c^3/d^3
Time = 0.27 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.81 \[ \int \frac {(d+e x)^{3/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (c^{2} d^{2} g^{2} x^{2} - 3 \, c^{2} d^{2} f^{2} + 12 \, a c d e f g - 8 \, a^{2} e^{2} g^{2} + 2 \, {\left (3 \, c^{2} d^{2} f g - 2 \, a c d e g^{2}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{3 \, {\left (c^{4} d^{4} e x^{2} + a c^{3} d^{4} e + {\left (c^{4} d^{5} + a c^{3} d^{3} e^{2}\right )} x\right )}} \]
integrate((e*x+d)^(3/2)*(g*x+f)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2), x, algorithm="fricas")
2/3*(c^2*d^2*g^2*x^2 - 3*c^2*d^2*f^2 + 12*a*c*d*e*f*g - 8*a^2*e^2*g^2 + 2* (3*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^4*d^4*e*x^2 + a*c^3*d^4*e + (c^4*d^5 + a*c^3*d^3*e^2 )*x)
\[ \int \frac {(d+e x)^{3/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{\frac {3}{2}} \left (f + g x\right )^{2}}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {3}{2}}}\, dx \]
Time = 0.24 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.54 \[ \int \frac {(d+e x)^{3/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {2 \, f^{2}}{\sqrt {c d x + a e} c d} + \frac {4 \, {\left (c d x + 2 \, a e\right )} f g}{\sqrt {c d x + a e} c^{2} d^{2}} + \frac {2 \, {\left (c^{2} d^{2} x^{2} - 4 \, a c d e x - 8 \, a^{2} e^{2}\right )} g^{2}}{3 \, \sqrt {c d x + a e} c^{3} d^{3}} \]
integrate((e*x+d)^(3/2)*(g*x+f)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2), x, algorithm="maxima")
-2*f^2/(sqrt(c*d*x + a*e)*c*d) + 4*(c*d*x + 2*a*e)*f*g/(sqrt(c*d*x + a*e)* c^2*d^2) + 2/3*(c^2*d^2*x^2 - 4*a*c*d*e*x - 8*a^2*e^2)*g^2/(sqrt(c*d*x + a *e)*c^3*d^3)
Time = 0.30 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.65 \[ \int \frac {(d+e x)^{3/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (c^{2} d^{2} e^{2} f^{2} - 2 \, a c d e^{3} f g + a^{2} e^{4} g^{2}\right )}}{\sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} c^{3} d^{3} {\left | e \right |}} + \frac {2 \, {\left (3 \, c^{2} d^{2} e^{2} f^{2} + 6 \, c^{2} d^{3} e f g - 12 \, a c d e^{3} f g - c^{2} d^{4} g^{2} - 4 \, a c d^{2} e^{2} g^{2} + 8 \, a^{2} e^{4} g^{2}\right )}}{3 \, \sqrt {-c d^{2} e + a e^{3}} c^{3} d^{3} {\left | e \right |}} + \frac {2 \, {\left (6 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} c^{7} d^{7} e^{8} f g - 6 \, \sqrt {{\left (e x + d\right )} c d e - c d^{2} e + a e^{3}} a c^{6} d^{6} e^{9} g^{2} + {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} c^{6} d^{6} e^{6} g^{2}\right )}}{3 \, c^{9} d^{9} e^{8} {\left | e \right |}} \]
integrate((e*x+d)^(3/2)*(g*x+f)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2), x, algorithm="giac")
-2*(c^2*d^2*e^2*f^2 - 2*a*c*d*e^3*f*g + a^2*e^4*g^2)/(sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3)*c^3*d^3*abs(e)) + 2/3*(3*c^2*d^2*e^2*f^2 + 6*c^2*d^3*e *f*g - 12*a*c*d*e^3*f*g - c^2*d^4*g^2 - 4*a*c*d^2*e^2*g^2 + 8*a^2*e^4*g^2) /(sqrt(-c*d^2*e + a*e^3)*c^3*d^3*abs(e)) + 2/3*(6*sqrt((e*x + d)*c*d*e - c *d^2*e + a*e^3)*c^7*d^7*e^8*f*g - 6*sqrt((e*x + d)*c*d*e - c*d^2*e + a*e^3 )*a*c^6*d^6*e^9*g^2 + ((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*c^6*d^6*e^ 6*g^2)/(c^9*d^9*e^8*abs(e))
Time = 12.58 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.98 \[ \int \frac {(d+e x)^{3/2} (f+g x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {\sqrt {d+e\,x}\,\left (16\,a^2\,e^2\,g^2-24\,a\,c\,d\,e\,f\,g+6\,c^2\,d^2\,f^2\right )}{3\,c^4\,d^4\,e}-\frac {2\,g^2\,x^2\,\sqrt {d+e\,x}}{3\,c^2\,d^2\,e}+\frac {4\,g\,x\,\left (2\,a\,e\,g-3\,c\,d\,f\right )\,\sqrt {d+e\,x}}{3\,c^3\,d^3\,e}\right )}{\frac {a}{c}+x^2+\frac {x\,\left (3\,c^4\,d^5+3\,a\,c^3\,d^3\,e^2\right )}{3\,c^4\,d^4\,e}} \]
-((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*(((d + e*x)^(1/2)*(16*a^2* e^2*g^2 + 6*c^2*d^2*f^2 - 24*a*c*d*e*f*g))/(3*c^4*d^4*e) - (2*g^2*x^2*(d + e*x)^(1/2))/(3*c^2*d^2*e) + (4*g*x*(2*a*e*g - 3*c*d*f)*(d + e*x)^(1/2))/( 3*c^3*d^3*e)))/(a/c + x^2 + (x*(3*c^4*d^5 + 3*a*c^3*d^3*e^2))/(3*c^4*d^4*e ))