Integrand size = 44, antiderivative size = 125 \[ \int \frac {(f+g x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=-\frac {2 \left (2 a e^2 g-c d (7 e f-5 d g)\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{35 c^2 d^2 e (d+e x)^{5/2}}+\frac {2 g \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{7 c d e (d+e x)^{3/2}} \]
-2/35*(2*a*e^2*g-c*d*(-5*d*g+7*e*f))*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/ 2)/c^2/d^2/e/(e*x+d)^(5/2)+2/7*g*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2)/c /d/e/(e*x+d)^(3/2)
Time = 0.06 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.43 \[ \int \frac {(f+g x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\frac {2 ((a e+c d x) (d+e x))^{5/2} (-2 a e g+c d (7 f+5 g x))}{35 c^2 d^2 (d+e x)^{5/2}} \]
(2*((a*e + c*d*x)*(d + e*x))^(5/2)*(-2*a*e*g + c*d*(7*f + 5*g*x)))/(35*c^2 *d^2*(d + e*x)^(5/2))
Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {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.
\(\displaystyle \int \frac {(f+g x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 1221 |
\(\displaystyle \frac {1}{7} \left (-\frac {2 a e g}{c d}-\frac {5 d g}{e}+7 f\right ) \int \frac {\left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right )^{3/2}}{(d+e x)^{3/2}}dx+\frac {2 g \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{7 c d e (d+e x)^{3/2}}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle \frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2} \left (-\frac {2 a e g}{c d}-\frac {5 d g}{e}+7 f\right )}{35 c d (d+e x)^{5/2}}+\frac {2 g \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{7 c d e (d+e x)^{3/2}}\) |
(2*(7*f - (5*d*g)/e - (2*a*e*g)/(c*d))*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e* x^2)^(5/2))/(35*c*d*(d + e*x)^(5/2)) + (2*g*(a*d*e + (c*d^2 + a*e^2)*x + c *d*e*x^2)^(5/2))/(7*c*d*e*(d + e*x)^(3/2))
3.7.92.3.1 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]
Time = 0.58 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.47
method | result | size |
default | \(-\frac {2 \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}\, \left (c d x +a e \right )^{2} \left (-5 c d g x +2 a e g -7 c d f \right )}{35 \sqrt {e x +d}\, c^{2} d^{2}}\) | \(59\) |
gosper | \(-\frac {2 \left (c d x +a e \right ) \left (-5 c d g x +2 a e g -7 c d f \right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}}{35 c^{2} d^{2} \left (e x +d \right )^{\frac {3}{2}}}\) | \(67\) |
-2/35*((c*d*x+a*e)*(e*x+d))^(1/2)/(e*x+d)^(1/2)*(c*d*x+a*e)^2*(-5*c*d*g*x+ 2*a*e*g-7*c*d*f)/c^2/d^2
Time = 0.30 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.10 \[ \int \frac {(f+g x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (5 \, c^{3} d^{3} g x^{3} + 7 \, a^{2} c d e^{2} f - 2 \, a^{3} e^{3} g + {\left (7 \, c^{3} d^{3} f + 8 \, a c^{2} d^{2} e g\right )} x^{2} + {\left (14 \, a c^{2} d^{2} e f + a^{2} c d e^{2} g\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{35 \, {\left (c^{2} d^{2} e x + c^{2} d^{3}\right )}} \]
integrate((g*x+f)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(3/2),x, algorithm="fricas")
2/35*(5*c^3*d^3*g*x^3 + 7*a^2*c*d*e^2*f - 2*a^3*e^3*g + (7*c^3*d^3*f + 8*a *c^2*d^2*e*g)*x^2 + (14*a*c^2*d^2*e*f + a^2*c*d*e^2*g)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d)/(c^2*d^2*e*x + c^2*d^3)
\[ \int \frac {(f+g x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\int \frac {\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {3}{2}} \left (f + g x\right )}{\left (d + e x\right )^{\frac {3}{2}}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.86 \[ \int \frac {(f+g x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (c^{2} d^{2} x^{2} + 2 \, a c d e x + a^{2} e^{2}\right )} \sqrt {c d x + a e} f}{5 \, c d} + \frac {2 \, {\left (5 \, c^{3} d^{3} x^{3} + 8 \, a c^{2} d^{2} e x^{2} + a^{2} c d e^{2} x - 2 \, a^{3} e^{3}\right )} \sqrt {c d x + a e} g}{35 \, c^{2} d^{2}} \]
integrate((g*x+f)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^(3/2),x, algorithm="maxima")
2/5*(c^2*d^2*x^2 + 2*a*c*d*e*x + a^2*e^2)*sqrt(c*d*x + a*e)*f/(c*d) + 2/35 *(5*c^3*d^3*x^3 + 8*a*c^2*d^2*e*x^2 + a^2*c*d*e^2*x - 2*a^3*e^3)*sqrt(c*d* x + a*e)*g/(c^2*d^2)
Leaf count of result is larger than twice the leaf count of optimal. 632 vs. \(2 (113) = 226\).
Time = 0.30 (sec) , antiderivative size = 632, normalized size of antiderivative = 5.06 \[ \int \frac {(f+g x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {35 \, a f {\left (\frac {\sqrt {-c d^{2} e + a e^{3}} c d^{2} - \sqrt {-c d^{2} e + a e^{3}} a e^{2}}{c d} + \frac {{\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}}}{c d e}\right )} {\left | e \right |}}{e} + \frac {c d g {\left (\frac {15 \, \sqrt {-c d^{2} e + a e^{3}} c^{3} d^{6} - 3 \, \sqrt {-c d^{2} e + a e^{3}} a c^{2} d^{4} e^{2} - 4 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} c d^{2} e^{4} - 8 \, \sqrt {-c d^{2} e + a e^{3}} a^{3} e^{6}}{c^{3} d^{3} e^{2}} + \frac {35 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a^{2} e^{6} - 42 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}} a e^{3} + 15 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {7}{2}}}{c^{3} d^{3} e^{5}}\right )} {\left | e \right |}}{e^{2}} - \frac {7 \, c d f {\left (\frac {3 \, \sqrt {-c d^{2} e + a e^{3}} c^{2} d^{4} - \sqrt {-c d^{2} e + a e^{3}} a c d^{2} e^{2} - 2 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} e^{4}}{c^{2} d^{2}} + \frac {5 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a e^{3} - 3 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}}}{c^{2} d^{2} e^{2}}\right )} {\left | e \right |}}{e^{3}} - \frac {7 \, a g {\left (\frac {3 \, \sqrt {-c d^{2} e + a e^{3}} c^{2} d^{4} - \sqrt {-c d^{2} e + a e^{3}} a c d^{2} e^{2} - 2 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} e^{4}}{c^{2} d^{2}} + \frac {5 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a e^{3} - 3 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}}}{c^{2} d^{2} e^{2}}\right )} {\left | e \right |}}{e^{2}}\right )}}{105 \, e} \]
2/105*(35*a*f*((sqrt(-c*d^2*e + a*e^3)*c*d^2 - sqrt(-c*d^2*e + a*e^3)*a*e^ 2)/(c*d) + ((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)/(c*d*e))*abs(e)/e + c *d*g*((15*sqrt(-c*d^2*e + a*e^3)*c^3*d^6 - 3*sqrt(-c*d^2*e + a*e^3)*a*c^2* d^4*e^2 - 4*sqrt(-c*d^2*e + a*e^3)*a^2*c*d^2*e^4 - 8*sqrt(-c*d^2*e + a*e^3 )*a^3*e^6)/(c^3*d^3*e^2) + (35*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a ^2*e^6 - 42*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(5/2)*a*e^3 + 15*((e*x + d )*c*d*e - c*d^2*e + a*e^3)^(7/2))/(c^3*d^3*e^5))*abs(e)/e^2 - 7*c*d*f*((3* sqrt(-c*d^2*e + a*e^3)*c^2*d^4 - sqrt(-c*d^2*e + a*e^3)*a*c*d^2*e^2 - 2*sq rt(-c*d^2*e + a*e^3)*a^2*e^4)/(c^2*d^2) + (5*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a*e^3 - 3*((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(5/2))/(c^2*d^2 *e^2))*abs(e)/e^3 - 7*a*g*((3*sqrt(-c*d^2*e + a*e^3)*c^2*d^4 - sqrt(-c*d^2 *e + a*e^3)*a*c*d^2*e^2 - 2*sqrt(-c*d^2*e + a*e^3)*a^2*e^4)/(c^2*d^2) + (5 *((e*x + d)*c*d*e - c*d^2*e + a*e^3)^(3/2)*a*e^3 - 3*((e*x + d)*c*d*e - c* d^2*e + a*e^3)^(5/2))/(c^2*d^2*e^2))*abs(e)/e^2)/e
Time = 11.95 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.87 \[ \int \frac {(f+g x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx=\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (x^2\,\left (\frac {16\,a\,e\,g}{35}+\frac {2\,c\,d\,f}{5}\right )-\frac {4\,a^3\,e^3\,g-14\,a^2\,c\,d\,e^2\,f}{35\,c^2\,d^2}+\frac {2\,c\,d\,g\,x^3}{7}+\frac {2\,a\,e\,x\,\left (a\,e\,g+14\,c\,d\,f\right )}{35\,c\,d}\right )}{\sqrt {d+e\,x}} \]