Integrand size = 46, antiderivative size = 118 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^{5/2}} \, dx=-\frac {2 (9 c e f-5 c d g-2 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{63 c^2 e^2 (d+e x)^{7/2}}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{9 c e^2 (d+e x)^{5/2}} \]
-2/63*(-2*b*e*g-5*c*d*g+9*c*e*f)*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(7/2)/c^ 2/e^2/(e*x+d)^(7/2)-2/9*g*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(7/2)/c/e^2/(e* x+d)^(5/2)
Time = 0.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.66 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^{5/2}} \, dx=\frac {2 (-c d+b e+c e x)^3 \sqrt {(d+e x) (-b e+c (d-e x))} (-2 b e g+c (9 e f+2 d g+7 e g x))}{63 c^2 e^2 \sqrt {d+e x}} \]
(2*(-(c*d) + b*e + c*e*x)^3*Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))]*(-2*b*e *g + c*(9*e*f + 2*d*g + 7*e*g*x)))/(63*c^2*e^2*Sqrt[d + e*x])
Time = 0.33 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, 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.
\(\displaystyle \int \frac {(f+g x) \left (-b d e-b e^2 x+c d^2-c e^2 x^2\right )^{5/2}}{(d+e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 1221 |
\(\displaystyle \frac {(-2 b e g-5 c d g+9 c e f) \int \frac {\left (-c x^2 e^2-b x e^2+d (c d-b e)\right )^{5/2}}{(d+e x)^{5/2}}dx}{9 c e}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{9 c e^2 (d+e x)^{5/2}}\) |
\(\Big \downarrow \) 1122 |
\(\displaystyle -\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2} (-2 b e g-5 c d g+9 c e f)}{63 c^2 e^2 (d+e x)^{7/2}}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{9 c e^2 (d+e x)^{5/2}}\) |
(-2*(9*c*e*f - 5*c*d*g - 2*b*e*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(7 /2))/(63*c^2*e^2*(d + e*x)^(7/2)) - (2*g*(d*(c*d - b*e) - b*e^2*x - c*e^2* x^2)^(7/2))/(9*c*e^2*(d + e*x)^(5/2))
3.23.55.3.1 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 = 0.40 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62
method | result | size |
default | \(-\frac {2 \sqrt {-\left (e x +d \right ) \left (x c e +b e -c d \right )}\, \left (x c e +b e -c d \right )^{3} \left (-7 c e g x +2 b e g -2 c d g -9 c e f \right )}{63 \sqrt {e x +d}\, c^{2} e^{2}}\) | \(73\) |
gosper | \(-\frac {2 \left (x c e +b e -c d \right ) \left (-7 c e g x +2 b e g -2 c d g -9 c e f \right ) \left (-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}\right )^{\frac {5}{2}}}{63 c^{2} e^{2} \left (e x +d \right )^{\frac {5}{2}}}\) | \(79\) |
-2/63*(-(e*x+d)*(c*e*x+b*e-c*d))^(1/2)/(e*x+d)^(1/2)*(c*e*x+b*e-c*d)^3*(-7 *c*e*g*x+2*b*e*g-2*c*d*g-9*c*e*f)/c^2/e^2
Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (106) = 212\).
Time = 0.33 (sec) , antiderivative size = 345, normalized size of antiderivative = 2.92 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (7 \, c^{4} e^{4} g x^{4} + {\left (9 \, c^{4} e^{4} f - 19 \, {\left (c^{4} d e^{3} - b c^{3} e^{4}\right )} g\right )} x^{3} - 3 \, {\left (9 \, {\left (c^{4} d e^{3} - b c^{3} e^{4}\right )} f - 5 \, {\left (c^{4} d^{2} e^{2} - 2 \, b c^{3} d e^{3} + b^{2} c^{2} e^{4}\right )} g\right )} x^{2} - 9 \, {\left (c^{4} d^{3} e - 3 \, b c^{3} d^{2} e^{2} + 3 \, b^{2} c^{2} d e^{3} - b^{3} c e^{4}\right )} f - 2 \, {\left (c^{4} d^{4} - 4 \, b c^{3} d^{3} e + 6 \, b^{2} c^{2} d^{2} e^{2} - 4 \, b^{3} c d e^{3} + b^{4} e^{4}\right )} g + {\left (27 \, {\left (c^{4} d^{2} e^{2} - 2 \, b c^{3} d e^{3} + b^{2} c^{2} e^{4}\right )} f - {\left (c^{4} d^{3} e - 3 \, b c^{3} d^{2} e^{2} + 3 \, b^{2} c^{2} d e^{3} - b^{3} c e^{4}\right )} g\right )} x\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {e x + d}}{63 \, {\left (c^{2} e^{3} x + c^{2} d e^{2}\right )}} \]
2/63*(7*c^4*e^4*g*x^4 + (9*c^4*e^4*f - 19*(c^4*d*e^3 - b*c^3*e^4)*g)*x^3 - 3*(9*(c^4*d*e^3 - b*c^3*e^4)*f - 5*(c^4*d^2*e^2 - 2*b*c^3*d*e^3 + b^2*c^2 *e^4)*g)*x^2 - 9*(c^4*d^3*e - 3*b*c^3*d^2*e^2 + 3*b^2*c^2*d*e^3 - b^3*c*e^ 4)*f - 2*(c^4*d^4 - 4*b*c^3*d^3*e + 6*b^2*c^2*d^2*e^2 - 4*b^3*c*d*e^3 + b^ 4*e^4)*g + (27*(c^4*d^2*e^2 - 2*b*c^3*d*e^3 + b^2*c^2*e^4)*f - (c^4*d^3*e - 3*b*c^3*d^2*e^2 + 3*b^2*c^2*d*e^3 - 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^2*e^3*x + c^2*d*e^2)
\[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^{5/2}} \, dx=\int \frac {\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {5}{2}} \left (f + g x\right )}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (106) = 212\).
Time = 0.28 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.66 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (c^{3} e^{3} x^{3} - c^{3} d^{3} + 3 \, b c^{2} d^{2} e - 3 \, b^{2} c d e^{2} + b^{3} e^{3} - 3 \, {\left (c^{3} d e^{2} - b c^{2} e^{3}\right )} x^{2} + 3 \, {\left (c^{3} d^{2} e - 2 \, b c^{2} d e^{2} + b^{2} c e^{3}\right )} x\right )} \sqrt {-c e x + c d - b e} f}{7 \, c e} + \frac {2 \, {\left (7 \, c^{4} e^{4} x^{4} - 2 \, c^{4} d^{4} + 8 \, b c^{3} d^{3} e - 12 \, b^{2} c^{2} d^{2} e^{2} + 8 \, b^{3} c d e^{3} - 2 \, b^{4} e^{4} - 19 \, {\left (c^{4} d e^{3} - b c^{3} e^{4}\right )} x^{3} + 15 \, {\left (c^{4} d^{2} e^{2} - 2 \, b c^{3} d e^{3} + b^{2} c^{2} e^{4}\right )} x^{2} - {\left (c^{4} d^{3} e - 3 \, b c^{3} d^{2} e^{2} + 3 \, b^{2} c^{2} d e^{3} - b^{3} c e^{4}\right )} x\right )} \sqrt {-c e x + c d - b e} g}{63 \, c^{2} e^{2}} \]
2/7*(c^3*e^3*x^3 - c^3*d^3 + 3*b*c^2*d^2*e - 3*b^2*c*d*e^2 + b^3*e^3 - 3*( c^3*d*e^2 - b*c^2*e^3)*x^2 + 3*(c^3*d^2*e - 2*b*c^2*d*e^2 + b^2*c*e^3)*x)* sqrt(-c*e*x + c*d - b*e)*f/(c*e) + 2/63*(7*c^4*e^4*x^4 - 2*c^4*d^4 + 8*b*c ^3*d^3*e - 12*b^2*c^2*d^2*e^2 + 8*b^3*c*d*e^3 - 2*b^4*e^4 - 19*(c^4*d*e^3 - b*c^3*e^4)*x^3 + 15*(c^4*d^2*e^2 - 2*b*c^3*d*e^3 + b^2*c^2*e^4)*x^2 - (c ^4*d^3*e - 3*b*c^3*d^2*e^2 + 3*b^2*c^2*d*e^3 - b^3*c*e^4)*x)*sqrt(-c*e*x + c*d - b*e)*g/(c^2*e^2)
Leaf count of result is larger than twice the leaf count of optimal. 2399 vs. \(2 (106) = 212\).
Time = 0.39 (sec) , antiderivative size = 2399, normalized size of antiderivative = 20.33 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^{5/2}} \, dx=\text {Too large to display} \]
-2/315*(105*c^2*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) - 210*b*c*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 ) + 105*b^2*e^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) - 3*c^2*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 - 15*((e*x + d)*c - 2*c*d + b*e)^3*s qrt(-(e*x + d)*c + 2*c*d - b*e))/(c^3*e^2)) + 6*c^2*d*e*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)) - 6*b*c*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*...
Time = 11.65 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.44 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^{5/2}} \, dx=\frac {\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}\,\left (\frac {2\,x^2\,\left (b\,e-c\,d\right )\,\left (5\,b\,e\,g-5\,c\,d\,g+9\,c\,e\,f\right )}{21}+\frac {2\,c\,e\,x^3\,\left (19\,b\,e\,g-19\,c\,d\,g+9\,c\,e\,f\right )}{63}+\frac {2\,c^2\,e^2\,g\,x^4}{9}+\frac {2\,{\left (b\,e-c\,d\right )}^3\,\left (2\,c\,d\,g-2\,b\,e\,g+9\,c\,e\,f\right )}{63\,c^2\,e^2}+\frac {2\,x\,{\left (b\,e-c\,d\right )}^2\,\left (b\,e\,g-c\,d\,g+27\,c\,e\,f\right )}{63\,c\,e}\right )}{\sqrt {d+e\,x}} \]
((c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(1/2)*((2*x^2*(b*e - c*d)*(5*b*e*g - 5*c*d*g + 9*c*e*f))/21 + (2*c*e*x^3*(19*b*e*g - 19*c*d*g + 9*c*e*f))/63 + (2*c^2*e^2*g*x^4)/9 + (2*(b*e - c*d)^3*(2*c*d*g - 2*b*e*g + 9*c*e*f))/(6 3*c^2*e^2) + (2*x*(b*e - c*d)^2*(b*e*g - c*d*g + 27*c*e*f))/(63*c*e)))/(d + e*x)^(1/2)