Integrand size = 46, antiderivative size = 193 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=-\frac {4 (2 c d-b e) (9 c e f-c d g-4 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{315 c^3 e^2 (d+e x)^{5/2}}-\frac {2 (9 c e f-c d g-4 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{63 c^2 e^2 (d+e x)^{3/2}}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{9 c e^2 \sqrt {d+e x}} \]
-4/315*(-b*e+2*c*d)*(-4*b*e*g-c*d*g+9*c*e*f)*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x ^2)^(5/2)/c^3/e^2/(e*x+d)^(5/2)-2/63*(-4*b*e*g-c*d*g+9*c*e*f)*(d*(-b*e+c*d )-b*e^2*x-c*e^2*x^2)^(5/2)/c^2/e^2/(e*x+d)^(3/2)-2/9*g*(d*(-b*e+c*d)-b*e^2 *x-c*e^2*x^2)^(5/2)/c/e^2/(e*x+d)^(1/2)
Time = 0.10 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.63 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=-\frac {2 (-c d+b e+c e x)^2 \sqrt {(d+e x) (-b e+c (d-e x))} \left (8 b^2 e^2 g-2 b c e (9 e f+17 d g+10 e g x)+c^2 \left (26 d^2 g+5 e^2 x (9 f+7 g x)+d e (81 f+65 g x)\right )\right )}{315 c^3 e^2 \sqrt {d+e x}} \]
(-2*(-(c*d) + b*e + c*e*x)^2*Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))]*(8*b^2 *e^2*g - 2*b*c*e*(9*e*f + 17*d*g + 10*e*g*x) + c^2*(26*d^2*g + 5*e^2*x*(9* f + 7*g*x) + d*e*(81*f + 65*g*x))))/(315*c^3*e^2*Sqrt[d + e*x])
Time = 0.41 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {1221, 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.
| \(\displaystyle \int \frac {(f+g x) \left (-b d e-b e^2 x+c d^2-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx\) |
| \(\Big \downarrow \) 1221 |
| \(\displaystyle \frac {(-4 b e g-c d g+9 c e f) \int \frac {\left (-c x^2 e^2-b x e^2+d (c d-b e)\right )^{3/2}}{\sqrt {d+e x}}dx}{9 c e}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{9 c e^2 \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 1128 |
| \(\displaystyle \frac {(-4 b e g-c d g+9 c e f) \left (\frac {2 (2 c d-b e) \int \frac {\left (-c x^2 e^2-b x e^2+d (c d-b e)\right )^{3/2}}{(d+e x)^{3/2}}dx}{7 c}-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{7 c e (d+e x)^{3/2}}\right )}{9 c e}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{9 c e^2 \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 1122 |
| \(\displaystyle \frac {\left (-\frac {4 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{35 c^2 e (d+e x)^{5/2}}-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{7 c e (d+e x)^{3/2}}\right ) (-4 b e g-c d g+9 c e f)}{9 c e}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{9 c e^2 \sqrt {d+e x}}\) |
(-2*g*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(5/2))/(9*c*e^2*Sqrt[d + e*x]) + ((9*c*e*f - c*d*g - 4*b*e*g)*((-4*(2*c*d - b*e)*(d*(c*d - b*e) - b*e^2* x - c*e^2*x^2)^(5/2))/(35*c^2*e*(d + e*x)^(5/2)) - (2*(d*(c*d - b*e) - b*e ^2*x - c*e^2*x^2)^(5/2))/(7*c*e*(d + e*x)^(3/2))))/(9*c*e)
3.23.43.3.1 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]
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 = 0.35 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.69
| method | result | size |
| default | \(-\frac {2 \sqrt {-\left (e x +d \right ) \left (x c e +b e -c d \right )}\, \left (x c e +b e -c d \right )^{2} \left (35 g \,x^{2} c^{2} e^{2}-20 b c \,e^{2} g x +65 c^{2} d e g x +45 c^{2} e^{2} f x +8 b^{2} e^{2} g -34 b c d e g -18 b c \,e^{2} f +26 c^{2} d^{2} g +81 c^{2} d e f \right )}{315 \sqrt {e x +d}\, c^{3} e^{2}}\) | \(133\) |
| gosper | \(\frac {2 \left (x c e +b e -c d \right ) \left (35 g \,x^{2} c^{2} e^{2}-20 b c \,e^{2} g x +65 c^{2} d e g x +45 c^{2} e^{2} f x +8 b^{2} e^{2} g -34 b c d e g -18 b c \,e^{2} f +26 c^{2} d^{2} g +81 c^{2} d e f \right ) \left (-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}\right )^{\frac {3}{2}}}{315 c^{3} e^{2} \left (e x +d \right )^{\frac {3}{2}}}\) | \(139\) |
-2/315*(-(e*x+d)*(c*e*x+b*e-c*d))^(1/2)/(e*x+d)^(1/2)*(c*e*x+b*e-c*d)^2*(3 5*c^2*e^2*g*x^2-20*b*c*e^2*g*x+65*c^2*d*e*g*x+45*c^2*e^2*f*x+8*b^2*e^2*g-3 4*b*c*d*e*g-18*b*c*e^2*f+26*c^2*d^2*g+81*c^2*d*e*f)/c^3/e^2
Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (175) = 350\).
Time = 0.38 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.83 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=-\frac {2 \, {\left (35 \, c^{4} e^{4} g x^{4} + 5 \, {\left (9 \, c^{4} e^{4} f - {\left (c^{4} d e^{3} - 10 \, b c^{3} e^{4}\right )} g\right )} x^{3} - 3 \, {\left (3 \, {\left (c^{4} d e^{3} - 8 \, b c^{3} e^{4}\right )} f + {\left (23 \, c^{4} d^{2} e^{2} - 22 \, b c^{3} d e^{3} - b^{2} c^{2} e^{4}\right )} g\right )} x^{2} + 9 \, {\left (9 \, c^{4} d^{3} e - 20 \, b c^{3} d^{2} e^{2} + 13 \, b^{2} c^{2} d e^{3} - 2 \, b^{3} c e^{4}\right )} f + 2 \, {\left (13 \, c^{4} d^{4} - 43 \, b c^{3} d^{3} e + 51 \, b^{2} c^{2} d^{2} e^{2} - 25 \, b^{3} c d e^{3} + 4 \, b^{4} e^{4}\right )} g - {\left (9 \, {\left (13 \, c^{4} d^{2} e^{2} - 12 \, b c^{3} d e^{3} - b^{2} c^{2} e^{4}\right )} f - {\left (13 \, c^{4} d^{3} e - 30 \, b c^{3} d^{2} e^{2} + 21 \, b^{2} c^{2} d e^{3} - 4 \, b^{3} c e^{4}\right )} g\right )} x\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {e x + d}}{315 \, {\left (c^{3} e^{3} x + c^{3} d e^{2}\right )}} \]
-2/315*(35*c^4*e^4*g*x^4 + 5*(9*c^4*e^4*f - (c^4*d*e^3 - 10*b*c^3*e^4)*g)* x^3 - 3*(3*(c^4*d*e^3 - 8*b*c^3*e^4)*f + (23*c^4*d^2*e^2 - 22*b*c^3*d*e^3 - b^2*c^2*e^4)*g)*x^2 + 9*(9*c^4*d^3*e - 20*b*c^3*d^2*e^2 + 13*b^2*c^2*d*e ^3 - 2*b^3*c*e^4)*f + 2*(13*c^4*d^4 - 43*b*c^3*d^3*e + 51*b^2*c^2*d^2*e^2 - 25*b^3*c*d*e^3 + 4*b^4*e^4)*g - (9*(13*c^4*d^2*e^2 - 12*b*c^3*d*e^3 - b^ 2*c^2*e^4)*f - (13*c^4*d^3*e - 30*b*c^3*d^2*e^2 + 21*b^2*c^2*d*e^3 - 4*b^3 *c*e^4)*g)*x)*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*sqrt(e*x + d)/(c^ 3*e^3*x + c^3*d*e^2)
\[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=\int \frac {\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {3}{2}} \left (f + g x\right )}{\sqrt {d + e x}}\, dx \]
Time = 0.23 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.66 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=-\frac {2 \, {\left (5 \, c^{3} e^{3} x^{3} + 9 \, c^{3} d^{3} - 20 \, b c^{2} d^{2} e + 13 \, b^{2} c d e^{2} - 2 \, b^{3} e^{3} - {\left (c^{3} d e^{2} - 8 \, b c^{2} e^{3}\right )} x^{2} - {\left (13 \, c^{3} d^{2} e - 12 \, b c^{2} d e^{2} - b^{2} c e^{3}\right )} x\right )} \sqrt {-c e x + c d - b e} f}{35 \, c^{2} e} - \frac {2 \, {\left (35 \, c^{4} e^{4} x^{4} + 26 \, c^{4} d^{4} - 86 \, b c^{3} d^{3} e + 102 \, b^{2} c^{2} d^{2} e^{2} - 50 \, b^{3} c d e^{3} + 8 \, b^{4} e^{4} - 5 \, {\left (c^{4} d e^{3} - 10 \, b c^{3} e^{4}\right )} x^{3} - 3 \, {\left (23 \, c^{4} d^{2} e^{2} - 22 \, b c^{3} d e^{3} - b^{2} c^{2} e^{4}\right )} x^{2} + {\left (13 \, c^{4} d^{3} e - 30 \, b c^{3} d^{2} e^{2} + 21 \, b^{2} c^{2} d e^{3} - 4 \, b^{3} c e^{4}\right )} x\right )} \sqrt {-c e x + c d - b e} g}{315 \, c^{3} e^{2}} \]
-2/35*(5*c^3*e^3*x^3 + 9*c^3*d^3 - 20*b*c^2*d^2*e + 13*b^2*c*d*e^2 - 2*b^3 *e^3 - (c^3*d*e^2 - 8*b*c^2*e^3)*x^2 - (13*c^3*d^2*e - 12*b*c^2*d*e^2 - b^ 2*c*e^3)*x)*sqrt(-c*e*x + c*d - b*e)*f/(c^2*e) - 2/315*(35*c^4*e^4*x^4 + 2 6*c^4*d^4 - 86*b*c^3*d^3*e + 102*b^2*c^2*d^2*e^2 - 50*b^3*c*d*e^3 + 8*b^4* e^4 - 5*(c^4*d*e^3 - 10*b*c^3*e^4)*x^3 - 3*(23*c^4*d^2*e^2 - 22*b*c^3*d*e^ 3 - b^2*c^2*e^4)*x^2 + (13*c^4*d^3*e - 30*b*c^3*d^2*e^2 + 21*b^2*c^2*d*e^3 - 4*b^3*c*e^4)*x)*sqrt(-c*e*x + c*d - b*e)*g/(c^3*e^2)
Leaf count of result is larger than twice the leaf count of optimal. 1702 vs. \(2 (175) = 350\).
Time = 0.34 (sec) , antiderivative size = 1702, normalized size of antiderivative = 8.82 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=\text {Too large to display} \]
-2/315*(105*c*d^2*f*((-(e*x + d)*c + 2*c*d - b*e)^(3/2)/c - (2*sqrt(2*c*d - b*e)*c*d - sqrt(2*c*d - b*e)*b*e)/c) - 105*b*d*e*f*((-(e*x + d)*c + 2*c* d - b*e)^(3/2)/c - (2*sqrt(2*c*d - b*e)*c*d - sqrt(2*c*d - b*e)*b*e)/c) + 3*c*e^2*f*((22*sqrt(2*c*d - b*e)*c^3*d^3 - 19*sqrt(2*c*d - b*e)*b*c^2*d^2* e + 20*sqrt(2*c*d - b*e)*b^2*c*d*e^2 - 8*sqrt(2*c*d - b*e)*b^3*e^3)/(c^3*e ^2) - (35*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*c^2*d^2 - 70*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*b*c*d*e + 35*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*b^2*e^2 - 42*((e*x + d)*c - 2*c*d + b*e)^2*sqrt(-(e*x + d)*c + 2*c*d - b*e)*c*d + 42*((e*x + d)*c - 2*c*d + b*e)^2*sqrt(-(e*x + d)*c + 2*c*d - b*e)*b*e - 1 5*((e*x + d)*c - 2*c*d + b*e)^3*sqrt(-(e*x + d)*c + 2*c*d - b*e))/(c^3*e^2 )) + 3*b*e^2*g*((22*sqrt(2*c*d - b*e)*c^3*d^3 - 19*sqrt(2*c*d - b*e)*b*c^2 *d^2*e + 20*sqrt(2*c*d - b*e)*b^2*c*d*e^2 - 8*sqrt(2*c*d - b*e)*b^3*e^3)/( c^3*e^2) - (35*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*c^2*d^2 - 70*(-(e*x + d) *c + 2*c*d - b*e)^(3/2)*b*c*d*e + 35*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*b^ 2*e^2 - 42*((e*x + d)*c - 2*c*d + b*e)^2*sqrt(-(e*x + d)*c + 2*c*d - b*e)* c*d + 42*((e*x + d)*c - 2*c*d + b*e)^2*sqrt(-(e*x + d)*c + 2*c*d - b*e)*b* e - 15*((e*x + d)*c - 2*c*d + b*e)^3*sqrt(-(e*x + d)*c + 2*c*d - b*e))/(c^ 3*e^2)) - c*e^2*g*((26*sqrt(2*c*d - b*e)*c^4*d^4 + 47*sqrt(2*c*d - b*e)*b* c^3*d^3*e - 78*sqrt(2*c*d - b*e)*b^2*c^2*d^2*e^2 + 56*sqrt(2*c*d - b*e)*b^ 3*c*d*e^3 - 16*sqrt(2*c*d - b*e)*b^4*e^4)/(c^4*e^3) + (105*(-(e*x + d)*...
Time = 11.63 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.24 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{3/2}}{\sqrt {d+e x}} \, dx=-\frac {\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}\,\left (\frac {2\,e\,x^3\,\left (10\,b\,e\,g-c\,d\,g+9\,c\,e\,f\right )}{63}+\frac {2\,x^2\,\left (g\,b^2\,e^2+22\,g\,b\,c\,d\,e+24\,f\,b\,c\,e^2-23\,g\,c^2\,d^2-3\,f\,c^2\,d\,e\right )}{105\,c}+\frac {2\,c\,e^2\,g\,x^4}{9}+\frac {2\,{\left (b\,e-c\,d\right )}^2\,\left (8\,g\,b^2\,e^2-34\,g\,b\,c\,d\,e-18\,f\,b\,c\,e^2+26\,g\,c^2\,d^2+81\,f\,c^2\,d\,e\right )}{315\,c^3\,e^2}+\frac {2\,x\,\left (b\,e-c\,d\right )\,\left (-4\,g\,b^2\,e^2+17\,g\,b\,c\,d\,e+9\,f\,b\,c\,e^2-13\,g\,c^2\,d^2+117\,f\,c^2\,d\,e\right )}{315\,c^2\,e}\right )}{\sqrt {d+e\,x}} \]
-((c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(1/2)*((2*e*x^3*(10*b*e*g - c*d*g + 9*c*e*f))/63 + (2*x^2*(b^2*e^2*g - 23*c^2*d^2*g + 24*b*c*e^2*f - 3*c^2*d *e*f + 22*b*c*d*e*g))/(105*c) + (2*c*e^2*g*x^4)/9 + (2*(b*e - c*d)^2*(8*b^ 2*e^2*g + 26*c^2*d^2*g - 18*b*c*e^2*f + 81*c^2*d*e*f - 34*b*c*d*e*g))/(315 *c^3*e^2) + (2*x*(b*e - c*d)*(9*b*c*e^2*f - 13*c^2*d^2*g - 4*b^2*e^2*g + 1 17*c^2*d*e*f + 17*b*c*d*e*g))/(315*c^2*e)))/(d + e*x)^(1/2)