Integrand size = 39, antiderivative size = 171 \[ \int (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {16 \left (c d^2-a e^2\right )^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{105 c^3 d^3 (d+e x)^{3/2}}+\frac {8 \left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{35 c^2 d^2 \sqrt {d+e x}}+\frac {2 \sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{7 c d} \]
16/105*(-a*e^2+c*d^2)^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/c^3/d^3/(e *x+d)^(3/2)+8/35*(-a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/c^ 2/d^2/(e*x+d)^(1/2)+2/7*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)*(e*x+d)^(1 /2)/c/d
Time = 0.06 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.51 \[ \int (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2 ((a e+c d x) (d+e x))^{3/2} \left (8 a^2 e^4-4 a c d e^2 (7 d+3 e x)+c^2 d^2 \left (35 d^2+42 d e x+15 e^2 x^2\right )\right )}{105 c^3 d^3 (d+e x)^{3/2}} \]
(2*((a*e + c*d*x)*(d + e*x))^(3/2)*(8*a^2*e^4 - 4*a*c*d*e^2*(7*d + 3*e*x) + c^2*d^2*(35*d^2 + 42*d*e*x + 15*e^2*x^2)))/(105*c^3*d^3*(d + e*x)^(3/2))
Time = 0.31 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1128, 1128, 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.
\(\displaystyle \int (d+e x)^{3/2} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \, dx\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {4 \left (d^2-\frac {a e^2}{c}\right ) \int \sqrt {d+e x} \sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}dx}{7 d}+\frac {2 \sqrt {d+e x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{7 c d}\) |
\(\Big \downarrow \) 1128 |
\(\displaystyle \frac {4 \left (d^2-\frac {a e^2}{c}\right ) \left (\frac {2 \left (d^2-\frac {a e^2}{c}\right ) \int \frac {\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}{\sqrt {d+e x}}dx}{5 d}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{5 c d \sqrt {d+e x}}\right )}{7 d}+\frac {2 \sqrt {d+e x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{7 c d}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle \frac {2 \sqrt {d+e x} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{7 c d}+\frac {4 \left (d^2-\frac {a e^2}{c}\right ) \left (\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{5 c d \sqrt {d+e x}}+\frac {4 \left (d^2-\frac {a e^2}{c}\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{15 c d^2 (d+e x)^{3/2}}\right )}{7 d}\) |
(2*Sqrt[d + e*x]*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(7*c*d) + (4*(d^2 - (a*e^2)/c)*((4*(d^2 - (a*e^2)/c)*(a*d*e + (c*d^2 + a*e^2)*x + c* d*e*x^2)^(3/2))/(15*c*d^2*(d + e*x)^(3/2)) + (2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(5*c*d*Sqrt[d + e*x])))/(7*d)
3.21.29.3.1 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_)*((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*(m + 2*p + 1))), x] + Simp[Simplify[m + p]*((2*c*d - b*e)/(c*(m + 2*p + 1))) Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[Simplify[m + p], 0]
Time = 2.45 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.58
method | result | size |
default | \(\frac {2 \left (c d x +a e \right ) \left (15 x^{2} c^{2} d^{2} e^{2}-12 x a c d \,e^{3}+42 x \,c^{2} d^{3} e +8 a^{2} e^{4}-28 a c \,d^{2} e^{2}+35 c^{2} d^{4}\right ) \sqrt {\left (c d x +a e \right ) \left (e x +d \right )}}{105 c^{3} d^{3} \sqrt {e x +d}}\) | \(100\) |
gosper | \(\frac {2 \left (c d x +a e \right ) \left (15 x^{2} c^{2} d^{2} e^{2}-12 x a c d \,e^{3}+42 x \,c^{2} d^{3} e +8 a^{2} e^{4}-28 a c \,d^{2} e^{2}+35 c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{105 c^{3} d^{3} \sqrt {e x +d}}\) | \(110\) |
2/105*(c*d*x+a*e)*(15*c^2*d^2*e^2*x^2-12*a*c*d*e^3*x+42*c^2*d^3*e*x+8*a^2* e^4-28*a*c*d^2*e^2+35*c^2*d^4)*((c*d*x+a*e)*(e*x+d))^(1/2)/c^3/d^3/(e*x+d) ^(1/2)
Time = 0.71 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.93 \[ \int (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2 \, {\left (15 \, c^{3} d^{3} e^{2} x^{3} + 35 \, a c^{2} d^{4} e - 28 \, a^{2} c d^{2} e^{3} + 8 \, a^{3} e^{5} + 3 \, {\left (14 \, c^{3} d^{4} e + a c^{2} d^{2} e^{3}\right )} x^{2} + {\left (35 \, c^{3} d^{5} + 14 \, a c^{2} d^{3} e^{2} - 4 \, a^{2} c d e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {e x + d}}{105 \, {\left (c^{3} d^{3} e x + c^{3} d^{4}\right )}} \]
2/105*(15*c^3*d^3*e^2*x^3 + 35*a*c^2*d^4*e - 28*a^2*c*d^2*e^3 + 8*a^3*e^5 + 3*(14*c^3*d^4*e + a*c^2*d^2*e^3)*x^2 + (35*c^3*d^5 + 14*a*c^2*d^3*e^2 - 4*a^2*c*d*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e*x + d )/(c^3*d^3*e*x + c^3*d^4)
\[ \int (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\int \sqrt {\left (d + e x\right ) \left (a e + c d x\right )} \left (d + e x\right )^{\frac {3}{2}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.82 \[ \int (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2 \, {\left (15 \, c^{3} d^{3} e^{2} x^{3} + 35 \, a c^{2} d^{4} e - 28 \, a^{2} c d^{2} e^{3} + 8 \, a^{3} e^{5} + 3 \, {\left (14 \, c^{3} d^{4} e + a c^{2} d^{2} e^{3}\right )} x^{2} + {\left (35 \, c^{3} d^{5} + 14 \, a c^{2} d^{3} e^{2} - 4 \, a^{2} c d e^{4}\right )} x\right )} \sqrt {c d x + a e} {\left (e x + d\right )}}{105 \, {\left (c^{3} d^{3} e x + c^{3} d^{4}\right )}} \]
2/105*(15*c^3*d^3*e^2*x^3 + 35*a*c^2*d^4*e - 28*a^2*c*d^2*e^3 + 8*a^3*e^5 + 3*(14*c^3*d^4*e + a*c^2*d^2*e^3)*x^2 + (35*c^3*d^5 + 14*a*c^2*d^3*e^2 - 4*a^2*c*d*e^4)*x)*sqrt(c*d*x + a*e)*(e*x + d)/(c^3*d^3*e*x + c^3*d^4)
Leaf count of result is larger than twice the leaf count of optimal. 469 vs. \(2 (153) = 306\).
Time = 0.31 (sec) , antiderivative size = 469, normalized size of antiderivative = 2.74 \[ \int (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2 \, {\left (\frac {35 \, d^{2} {\left (\frac {\sqrt {-c d^{2} e + a e^{3}} c d^{2} - \sqrt {-c d^{2} e + a e^{3}} a e^{2}}{c d} + \frac {{\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}}}{c d e}\right )} {\left | e \right |}}{e^{2}} + {\left (\frac {15 \, \sqrt {-c d^{2} e + a e^{3}} c^{3} d^{6} - 3 \, \sqrt {-c d^{2} e + a e^{3}} a c^{2} d^{4} e^{2} - 4 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} c d^{2} e^{4} - 8 \, \sqrt {-c d^{2} e + a e^{3}} a^{3} e^{6}}{c^{3} d^{3} e^{2}} + \frac {35 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a^{2} e^{6} - 42 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}} a e^{3} + 15 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {7}{2}}}{c^{3} d^{3} e^{5}}\right )} {\left | e \right |} - \frac {14 \, d {\left (\frac {3 \, \sqrt {-c d^{2} e + a e^{3}} c^{2} d^{4} - \sqrt {-c d^{2} e + a e^{3}} a c d^{2} e^{2} - 2 \, \sqrt {-c d^{2} e + a e^{3}} a^{2} e^{4}}{c^{2} d^{2}} + \frac {5 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {3}{2}} a e^{3} - 3 \, {\left ({\left (e x + d\right )} c d e - c d^{2} e + a e^{3}\right )}^{\frac {5}{2}}}{c^{2} d^{2} e^{2}}\right )} {\left | e \right |}}{e^{2}}\right )}}{105 \, e} \]
2/105*(35*d^2*((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^2 + ((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) - 14*d*((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 = 10.20 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.05 \[ \int (d+e x)^{3/2} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (\frac {2\,e\,x^3\,\sqrt {d+e\,x}}{7}+\frac {\sqrt {d+e\,x}\,\left (16\,a^3\,e^5-56\,a^2\,c\,d^2\,e^3+70\,a\,c^2\,d^4\,e\right )}{105\,c^3\,d^3\,e}+\frac {2\,x^2\,\left (14\,c\,d^2+a\,e^2\right )\,\sqrt {d+e\,x}}{35\,c\,d}+\frac {x\,\sqrt {d+e\,x}\,\left (-8\,a^2\,c\,d\,e^4+28\,a\,c^2\,d^3\,e^2+70\,c^3\,d^5\right )}{105\,c^3\,d^3\,e}\right )}{x+\frac {d}{e}} \]
((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*((2*e*x^3*(d + e*x)^(1/2))/ 7 + ((d + e*x)^(1/2)*(16*a^3*e^5 - 56*a^2*c*d^2*e^3 + 70*a*c^2*d^4*e))/(10 5*c^3*d^3*e) + (2*x^2*(a*e^2 + 14*c*d^2)*(d + e*x)^(1/2))/(35*c*d) + (x*(d + e*x)^(1/2)*(70*c^3*d^5 + 28*a*c^2*d^3*e^2 - 8*a^2*c*d*e^4))/(105*c^3*d^ 3*e)))/(x + d/e)