Integrand size = 46, antiderivative size = 267 \[ \int (d+e x)^{3/2} (f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2} \, dx=-\frac {16 (2 c d-b e)^2 (3 c e f+c d g-2 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{315 c^4 e^2 (d+e x)^{3/2}}-\frac {8 (2 c d-b e) (3 c e f+c d g-2 b e g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{105 c^3 e^2 \sqrt {d+e x}}-\frac {2 (3 c e f+c d g-2 b e g) \sqrt {d+e x} \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{21 c^2 e^2}-\frac {2 g (d+e x)^{3/2} \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{9 c e^2} \]
-16/315*(-b*e+2*c*d)^2*(-2*b*e*g+c*d*g+3*c*e*f)*(d*(-b*e+c*d)-b*e^2*x-c*e^ 2*x^2)^(3/2)/c^4/e^2/(e*x+d)^(3/2)-2/9*g*(e*x+d)^(3/2)*(d*(-b*e+c*d)-b*e^2 *x-c*e^2*x^2)^(3/2)/c/e^2-8/105*(-b*e+2*c*d)*(-2*b*e*g+c*d*g+3*c*e*f)*(d*( -b*e+c*d)-b*e^2*x-c*e^2*x^2)^(3/2)/c^3/e^2/(e*x+d)^(1/2)-2/21*(-2*b*e*g+c* d*g+3*c*e*f)*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(3/2)*(e*x+d)^(1/2)/c^2/e^2
Time = 0.13 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.67 \[ \int (d+e x)^{3/2} (f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2} \, dx=\frac {2 (-c d+b e+c e x) \sqrt {(d+e x) (-b e+c (d-e x))} \left (-16 b^3 e^3 g+24 b^2 c e^2 (4 d g+e (f+g x))-6 b c^2 e \left (31 d^2 g+e^2 x (6 f+5 g x)+d e (22 f+20 g x)\right )+c^3 \left (106 d^3 g+5 e^3 x^2 (9 f+7 g x)+6 d e^2 x (27 f+20 g x)+3 d^2 e (71 f+53 g x)\right )\right )}{315 c^4 e^2 \sqrt {d+e x}} \]
(2*(-(c*d) + b*e + c*e*x)*Sqrt[(d + e*x)*(-(b*e) + c*(d - e*x))]*(-16*b^3* e^3*g + 24*b^2*c*e^2*(4*d*g + e*(f + g*x)) - 6*b*c^2*e*(31*d^2*g + e^2*x*( 6*f + 5*g*x) + d*e*(22*f + 20*g*x)) + c^3*(106*d^3*g + 5*e^3*x^2*(9*f + 7* g*x) + 6*d*e^2*x*(27*f + 20*g*x) + 3*d^2*e*(71*f + 53*g*x))))/(315*c^4*e^2 *Sqrt[d + e*x])
Time = 0.48 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1221, 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.
| \(\displaystyle \int (d+e x)^{3/2} (f+g x) \sqrt {-b d e-b e^2 x+c d^2-c e^2 x^2} \, dx\) |
| \(\Big \downarrow \) 1221 |
| \(\displaystyle \frac {(-2 b e g+c d g+3 c e f) \int (d+e x)^{3/2} \sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}dx}{3 c e}-\frac {2 g (d+e x)^{3/2} \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{9 c e^2}\) |
| \(\Big \downarrow \) 1128 |
| \(\displaystyle \frac {(-2 b e g+c d g+3 c e f) \left (\frac {4 (2 c d-b e) \int \sqrt {d+e x} \sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}dx}{7 c}-\frac {2 \sqrt {d+e x} \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{7 c e}\right )}{3 c e}-\frac {2 g (d+e x)^{3/2} \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{9 c e^2}\) |
| \(\Big \downarrow \) 1128 |
| \(\displaystyle \frac {(-2 b e g+c d g+3 c e f) \left (\frac {4 (2 c d-b e) \left (\frac {2 (2 c d-b e) \int \frac {\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}{\sqrt {d+e x}}dx}{5 c}-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{5 c e \sqrt {d+e x}}\right )}{7 c}-\frac {2 \sqrt {d+e x} \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{7 c e}\right )}{3 c e}-\frac {2 g (d+e x)^{3/2} \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{9 c e^2}\) |
| \(\Big \downarrow \) 1122 |
| \(\displaystyle \frac {\left (\frac {4 (2 c d-b e) \left (-\frac {4 (2 c d-b e) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{15 c^2 e (d+e x)^{3/2}}-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{5 c e \sqrt {d+e x}}\right )}{7 c}-\frac {2 \sqrt {d+e x} \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{7 c e}\right ) (-2 b e g+c d g+3 c e f)}{3 c e}-\frac {2 g (d+e x)^{3/2} \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{9 c e^2}\) |
(-2*g*(d + e*x)^(3/2)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(9*c*e^ 2) + ((3*c*e*f + c*d*g - 2*b*e*g)*((-2*Sqrt[d + e*x]*(d*(c*d - b*e) - b*e^ 2*x - c*e^2*x^2)^(3/2))/(7*c*e) + (4*(2*c*d - b*e)*((-4*(2*c*d - b*e)*(d*( c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(15*c^2*e*(d + e*x)^(3/2)) - (2*( d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/(5*c*e*Sqrt[d + e*x])))/(7*c)) )/(3*c*e)
3.23.32.3.1 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 = 227, normalized size of antiderivative = 0.85
| method | result | size |
| default | \(-\frac {2 \left (x c e +b e -c d \right ) \left (-35 g \,e^{3} x^{3} c^{3}+30 b \,c^{2} e^{3} g \,x^{2}-120 c^{3} d \,e^{2} g \,x^{2}-45 c^{3} e^{3} f \,x^{2}-24 b^{2} c \,e^{3} g x +120 b \,c^{2} d \,e^{2} g x +36 b \,c^{2} e^{3} f x -159 c^{3} d^{2} e g x -162 c^{3} d \,e^{2} f x +16 b^{3} e^{3} g -96 b^{2} c d \,e^{2} g -24 b^{2} c \,e^{3} f +186 b \,c^{2} d^{2} e g +132 b \,c^{2} d \,e^{2} f -106 c^{3} d^{3} g -213 d^{2} f \,c^{3} e \right ) \sqrt {-\left (e x +d \right ) \left (x c e +b e -c d \right )}}{315 c^{4} e^{2} \sqrt {e x +d}}\) | \(227\) |
| gosper | \(-\frac {2 \left (x c e +b e -c d \right ) \left (-35 g \,e^{3} x^{3} c^{3}+30 b \,c^{2} e^{3} g \,x^{2}-120 c^{3} d \,e^{2} g \,x^{2}-45 c^{3} e^{3} f \,x^{2}-24 b^{2} c \,e^{3} g x +120 b \,c^{2} d \,e^{2} g x +36 b \,c^{2} e^{3} f x -159 c^{3} d^{2} e g x -162 c^{3} d \,e^{2} f x +16 b^{3} e^{3} g -96 b^{2} c d \,e^{2} g -24 b^{2} c \,e^{3} f +186 b \,c^{2} d^{2} e g +132 b \,c^{2} d \,e^{2} f -106 c^{3} d^{3} g -213 d^{2} f \,c^{3} e \right ) \sqrt {-c \,e^{2} x^{2}-b \,e^{2} x -b d e +c \,d^{2}}}{315 c^{4} e^{2} \sqrt {e x +d}}\) | \(235\) |
-2/315*(c*e*x+b*e-c*d)*(-35*c^3*e^3*g*x^3+30*b*c^2*e^3*g*x^2-120*c^3*d*e^2 *g*x^2-45*c^3*e^3*f*x^2-24*b^2*c*e^3*g*x+120*b*c^2*d*e^2*g*x+36*b*c^2*e^3* f*x-159*c^3*d^2*e*g*x-162*c^3*d*e^2*f*x+16*b^3*e^3*g-96*b^2*c*d*e^2*g-24*b ^2*c*e^3*f+186*b*c^2*d^2*e*g+132*b*c^2*d*e^2*f-106*c^3*d^3*g-213*c^3*d^2*e *f)*(-(e*x+d)*(c*e*x+b*e-c*d))^(1/2)/c^4/e^2/(e*x+d)^(1/2)
Time = 0.28 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.32 \[ \int (d+e x)^{3/2} (f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2} \, dx=\frac {2 \, {\left (35 \, c^{4} e^{4} g x^{4} + 5 \, {\left (9 \, c^{4} e^{4} f + {\left (17 \, c^{4} d e^{3} + b c^{3} e^{4}\right )} g\right )} x^{3} + 3 \, {\left (3 \, {\left (13 \, c^{4} d e^{3} + b c^{3} e^{4}\right )} f + {\left (13 \, c^{4} d^{2} e^{2} + 10 \, b c^{3} d e^{3} - 2 \, b^{2} c^{2} e^{4}\right )} g\right )} x^{2} - 3 \, {\left (71 \, c^{4} d^{3} e - 115 \, b c^{3} d^{2} e^{2} + 52 \, b^{2} c^{2} d e^{3} - 8 \, b^{3} c e^{4}\right )} f - 2 \, {\left (53 \, c^{4} d^{4} - 146 \, b c^{3} d^{3} e + 141 \, b^{2} c^{2} d^{2} e^{2} - 56 \, b^{3} c d e^{3} + 8 \, b^{4} e^{4}\right )} g + {\left (3 \, {\left (17 \, c^{4} d^{2} e^{2} + 22 \, b c^{3} d e^{3} - 4 \, b^{2} c^{2} e^{4}\right )} f - {\left (53 \, c^{4} d^{3} e - 93 \, b c^{3} d^{2} e^{2} + 48 \, b^{2} c^{2} d e^{3} - 8 \, b^{3} c e^{4}\right )} g\right )} x\right )} \sqrt {-c e^{2} x^{2} - b e^{2} x + c d^{2} - b d e} \sqrt {e x + d}}{315 \, {\left (c^{4} e^{3} x + c^{4} d e^{2}\right )}} \]
2/315*(35*c^4*e^4*g*x^4 + 5*(9*c^4*e^4*f + (17*c^4*d*e^3 + b*c^3*e^4)*g)*x ^3 + 3*(3*(13*c^4*d*e^3 + b*c^3*e^4)*f + (13*c^4*d^2*e^2 + 10*b*c^3*d*e^3 - 2*b^2*c^2*e^4)*g)*x^2 - 3*(71*c^4*d^3*e - 115*b*c^3*d^2*e^2 + 52*b^2*c^2 *d*e^3 - 8*b^3*c*e^4)*f - 2*(53*c^4*d^4 - 146*b*c^3*d^3*e + 141*b^2*c^2*d^ 2*e^2 - 56*b^3*c*d*e^3 + 8*b^4*e^4)*g + (3*(17*c^4*d^2*e^2 + 22*b*c^3*d*e^ 3 - 4*b^2*c^2*e^4)*f - (53*c^4*d^3*e - 93*b*c^3*d^2*e^2 + 48*b^2*c^2*d*e^3 - 8*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^4*e^3*x + c^4*d*e^2)
\[ \int (d+e x)^{3/2} (f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2} \, dx=\int \sqrt {- \left (d + e x\right ) \left (b e - c d + c e x\right )} \left (d + e x\right )^{\frac {3}{2}} \left (f + g x\right )\, dx \]
Time = 0.22 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.33 \[ \int (d+e x)^{3/2} (f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2} \, dx=\frac {2 \, {\left (15 \, c^{3} e^{3} x^{3} - 71 \, c^{3} d^{3} + 115 \, b c^{2} d^{2} e - 52 \, b^{2} c d e^{2} + 8 \, b^{3} e^{3} + 3 \, {\left (13 \, c^{3} d e^{2} + b c^{2} e^{3}\right )} x^{2} + {\left (17 \, c^{3} d^{2} e + 22 \, b c^{2} d e^{2} - 4 \, b^{2} c e^{3}\right )} x\right )} \sqrt {-c e x + c d - b e} {\left (e x + d\right )} f}{105 \, {\left (c^{3} e^{2} x + c^{3} d e\right )}} + \frac {2 \, {\left (35 \, c^{4} e^{4} x^{4} - 106 \, c^{4} d^{4} + 292 \, b c^{3} d^{3} e - 282 \, b^{2} c^{2} d^{2} e^{2} + 112 \, b^{3} c d e^{3} - 16 \, b^{4} e^{4} + 5 \, {\left (17 \, c^{4} d e^{3} + b c^{3} e^{4}\right )} x^{3} + 3 \, {\left (13 \, c^{4} d^{2} e^{2} + 10 \, b c^{3} d e^{3} - 2 \, b^{2} c^{2} e^{4}\right )} x^{2} - {\left (53 \, c^{4} d^{3} e - 93 \, b c^{3} d^{2} e^{2} + 48 \, b^{2} c^{2} d e^{3} - 8 \, b^{3} c e^{4}\right )} x\right )} \sqrt {-c e x + c d - b e} {\left (e x + d\right )} g}{315 \, {\left (c^{4} e^{3} x + c^{4} d e^{2}\right )}} \]
2/105*(15*c^3*e^3*x^3 - 71*c^3*d^3 + 115*b*c^2*d^2*e - 52*b^2*c*d*e^2 + 8* b^3*e^3 + 3*(13*c^3*d*e^2 + b*c^2*e^3)*x^2 + (17*c^3*d^2*e + 22*b*c^2*d*e^ 2 - 4*b^2*c*e^3)*x)*sqrt(-c*e*x + c*d - b*e)*(e*x + d)*f/(c^3*e^2*x + c^3* d*e) + 2/315*(35*c^4*e^4*x^4 - 106*c^4*d^4 + 292*b*c^3*d^3*e - 282*b^2*c^2 *d^2*e^2 + 112*b^3*c*d*e^3 - 16*b^4*e^4 + 5*(17*c^4*d*e^3 + b*c^3*e^4)*x^3 + 3*(13*c^4*d^2*e^2 + 10*b*c^3*d*e^3 - 2*b^2*c^2*e^4)*x^2 - (53*c^4*d^3*e - 93*b*c^3*d^2*e^2 + 48*b^2*c^2*d*e^3 - 8*b^3*c*e^4)*x)*sqrt(-c*e*x + c*d - b*e)*(e*x + d)*g/(c^4*e^3*x + c^4*d*e^2)
Leaf count of result is larger than twice the leaf count of optimal. 1473 vs. \(2 (243) = 486\).
Time = 0.32 (sec) , antiderivative size = 1473, normalized size of antiderivative = 5.52 \[ \int (d+e x)^{3/2} (f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2} \, dx=\text {Too large to display} \]
-2/315*(105*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) - 3*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*s qrt(-(e*x + d)*c + 2*c*d - b*e)*c*d + 42*((e*x + d)*c - 2*c*d + b*e)^2*sqr t(-(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*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*sq rt(-(e*x + d)*c + 2*c*d - b*e))/(c^3*e^2)) + 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)*c + 2*c*d - b*e)^(3/2)*c^3*d^3 - 315*(-(e*x + d)*c + 2*c*d - b*e)^(3/2)*b*c^2*d^2*e + 315*(-(e*x + d)*c + 2*c*d - b...
Time = 11.76 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.26 \[ \int (d+e x)^{3/2} (f+g x) \sqrt {c d^2-b d e-b e^2 x-c e^2 x^2} \, dx=\frac {\sqrt {c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x}\,\left (\frac {2\,x^3\,\sqrt {d+e\,x}\,\left (b\,e\,g+17\,c\,d\,g+9\,c\,e\,f\right )}{63\,c}+\frac {2\,e\,g\,x^4\,\sqrt {d+e\,x}}{9}+\frac {2\,x^2\,\sqrt {d+e\,x}\,\left (-2\,g\,b^2\,e^2+10\,g\,b\,c\,d\,e+3\,f\,b\,c\,e^2+13\,g\,c^2\,d^2+39\,f\,c^2\,d\,e\right )}{105\,c^2\,e}+\frac {2\,\left (b\,e-c\,d\right )\,\sqrt {d+e\,x}\,\left (-16\,g\,b^3\,e^3+96\,g\,b^2\,c\,d\,e^2+24\,f\,b^2\,c\,e^3-186\,g\,b\,c^2\,d^2\,e-132\,f\,b\,c^2\,d\,e^2+106\,g\,c^3\,d^3+213\,f\,c^3\,d^2\,e\right )}{315\,c^4\,e^3}+\frac {x\,\sqrt {d+e\,x}\,\left (16\,g\,b^3\,c\,e^4-96\,g\,b^2\,c^2\,d\,e^3-24\,f\,b^2\,c^2\,e^4+186\,g\,b\,c^3\,d^2\,e^2+132\,f\,b\,c^3\,d\,e^3-106\,g\,c^4\,d^3\,e+102\,f\,c^4\,d^2\,e^2\right )}{315\,c^4\,e^3}\right )}{x+\frac {d}{e}} \]
((c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(1/2)*((2*x^3*(d + e*x)^(1/2)*(b*e* g + 17*c*d*g + 9*c*e*f))/(63*c) + (2*e*g*x^4*(d + e*x)^(1/2))/9 + (2*x^2*( d + e*x)^(1/2)*(13*c^2*d^2*g - 2*b^2*e^2*g + 3*b*c*e^2*f + 39*c^2*d*e*f + 10*b*c*d*e*g))/(105*c^2*e) + (2*(b*e - c*d)*(d + e*x)^(1/2)*(106*c^3*d^3*g - 16*b^3*e^3*g + 24*b^2*c*e^3*f + 213*c^3*d^2*e*f - 132*b*c^2*d*e^2*f - 1 86*b*c^2*d^2*e*g + 96*b^2*c*d*e^2*g))/(315*c^4*e^3) + (x*(d + e*x)^(1/2)*( 102*c^4*d^2*e^2*f - 24*b^2*c^2*e^4*f + 16*b^3*c*e^4*g - 106*c^4*d^3*e*g + 132*b*c^3*d*e^3*f + 186*b*c^3*d^2*e^2*g - 96*b^2*c^2*d*e^3*g))/(315*c^4*e^ 3)))/(x + d/e)