Integrand size = 19, antiderivative size = 200 \[ \int \frac {\left (a+c x^2\right )^3}{\sqrt {d+e x}} \, dx=\frac {2 \left (c d^2+a e^2\right )^3 \sqrt {d+e x}}{e^7}-\frac {4 c d \left (c d^2+a e^2\right )^2 (d+e x)^{3/2}}{e^7}+\frac {6 c \left (c d^2+a e^2\right ) \left (5 c d^2+a e^2\right ) (d+e x)^{5/2}}{5 e^7}-\frac {8 c^2 d \left (5 c d^2+3 a e^2\right ) (d+e x)^{7/2}}{7 e^7}+\frac {2 c^2 \left (5 c d^2+a e^2\right ) (d+e x)^{9/2}}{3 e^7}-\frac {12 c^3 d (d+e x)^{11/2}}{11 e^7}+\frac {2 c^3 (d+e x)^{13/2}}{13 e^7} \]
-4*c*d*(a*e^2+c*d^2)^2*(e*x+d)^(3/2)/e^7+6/5*c*(a*e^2+c*d^2)*(a*e^2+5*c*d^ 2)*(e*x+d)^(5/2)/e^7-8/7*c^2*d*(3*a*e^2+5*c*d^2)*(e*x+d)^(7/2)/e^7+2/3*c^2 *(a*e^2+5*c*d^2)*(e*x+d)^(9/2)/e^7-12/11*c^3*d*(e*x+d)^(11/2)/e^7+2/13*c^3 *(e*x+d)^(13/2)/e^7+2*(a*e^2+c*d^2)^3*(e*x+d)^(1/2)/e^7
Time = 0.14 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.86 \[ \int \frac {\left (a+c x^2\right )^3}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {d+e x} \left (15015 a^3 e^6+3003 a^2 c e^4 \left (8 d^2-4 d e x+3 e^2 x^2\right )+143 a c^2 e^2 \left (128 d^4-64 d^3 e x+48 d^2 e^2 x^2-40 d e^3 x^3+35 e^4 x^4\right )+5 c^3 \left (1024 d^6-512 d^5 e x+384 d^4 e^2 x^2-320 d^3 e^3 x^3+280 d^2 e^4 x^4-252 d e^5 x^5+231 e^6 x^6\right )\right )}{15015 e^7} \]
(2*Sqrt[d + e*x]*(15015*a^3*e^6 + 3003*a^2*c*e^4*(8*d^2 - 4*d*e*x + 3*e^2* x^2) + 143*a*c^2*e^2*(128*d^4 - 64*d^3*e*x + 48*d^2*e^2*x^2 - 40*d*e^3*x^3 + 35*e^4*x^4) + 5*c^3*(1024*d^6 - 512*d^5*e*x + 384*d^4*e^2*x^2 - 320*d^3 *e^3*x^3 + 280*d^2*e^4*x^4 - 252*d*e^5*x^5 + 231*e^6*x^6)))/(15015*e^7)
Time = 0.32 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {476, 2009}
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.
| \(\displaystyle \int \frac {\left (a+c x^2\right )^3}{\sqrt {d+e x}} \, dx\) |
| \(\Big \downarrow \) 476 |
| \(\displaystyle \int \left (\frac {3 c^2 (d+e x)^{7/2} \left (a e^2+5 c d^2\right )}{e^6}-\frac {4 c^2 d (d+e x)^{5/2} \left (3 a e^2+5 c d^2\right )}{e^6}+\frac {3 c (d+e x)^{3/2} \left (a e^2+c d^2\right ) \left (a e^2+5 c d^2\right )}{e^6}-\frac {6 c d \sqrt {d+e x} \left (a e^2+c d^2\right )^2}{e^6}+\frac {\left (a e^2+c d^2\right )^3}{e^6 \sqrt {d+e x}}+\frac {c^3 (d+e x)^{11/2}}{e^6}-\frac {6 c^3 d (d+e x)^{9/2}}{e^6}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 c^2 (d+e x)^{9/2} \left (a e^2+5 c d^2\right )}{3 e^7}-\frac {8 c^2 d (d+e x)^{7/2} \left (3 a e^2+5 c d^2\right )}{7 e^7}+\frac {6 c (d+e x)^{5/2} \left (a e^2+c d^2\right ) \left (a e^2+5 c d^2\right )}{5 e^7}-\frac {4 c d (d+e x)^{3/2} \left (a e^2+c d^2\right )^2}{e^7}+\frac {2 \sqrt {d+e x} \left (a e^2+c d^2\right )^3}{e^7}+\frac {2 c^3 (d+e x)^{13/2}}{13 e^7}-\frac {12 c^3 d (d+e x)^{11/2}}{11 e^7}\) |
(2*(c*d^2 + a*e^2)^3*Sqrt[d + e*x])/e^7 - (4*c*d*(c*d^2 + a*e^2)^2*(d + e* x)^(3/2))/e^7 + (6*c*(c*d^2 + a*e^2)*(5*c*d^2 + a*e^2)*(d + e*x)^(5/2))/(5 *e^7) - (8*c^2*d*(5*c*d^2 + 3*a*e^2)*(d + e*x)^(7/2))/(7*e^7) + (2*c^2*(5* c*d^2 + a*e^2)*(d + e*x)^(9/2))/(3*e^7) - (12*c^3*d*(d + e*x)^(11/2))/(11* e^7) + (2*c^3*(d + e*x)^(13/2))/(13*e^7)
3.7.8.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 0]
Time = 2.20 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.81
| method | result | size |
| pseudoelliptic | \(\frac {2 \left (\frac {\left (e^{6} x^{6}-\frac {12}{11} d \,e^{5} x^{5}+\frac {40}{33} d^{2} e^{4} x^{4}-\frac {320}{231} x^{3} d^{3} e^{3}+\frac {128}{77} d^{4} e^{2} x^{2}-\frac {512}{231} d^{5} e x +\frac {1024}{231} d^{6}\right ) c^{3}}{13}+\frac {128 \left (\frac {35}{128} e^{4} x^{4}-\frac {5}{16} d \,e^{3} x^{3}+\frac {3}{8} d^{2} e^{2} x^{2}-\frac {1}{2} d^{3} e x +d^{4}\right ) e^{2} a \,c^{2}}{105}+\frac {8 \left (\frac {3}{8} x^{2} e^{2}-\frac {1}{2} d e x +d^{2}\right ) e^{4} a^{2} c}{5}+e^{6} a^{3}\right ) \sqrt {e x +d}}{e^{7}}\) | \(162\) |
| gosper | \(\frac {2 \sqrt {e x +d}\, \left (1155 x^{6} c^{3} e^{6}-1260 x^{5} c^{3} d \,e^{5}+5005 x^{4} a \,c^{2} e^{6}+1400 x^{4} c^{3} d^{2} e^{4}-5720 x^{3} a \,c^{2} d \,e^{5}-1600 x^{3} c^{3} d^{3} e^{3}+9009 x^{2} a^{2} c \,e^{6}+6864 x^{2} a \,c^{2} d^{2} e^{4}+1920 x^{2} c^{3} d^{4} e^{2}-12012 x \,a^{2} c d \,e^{5}-9152 x a \,c^{2} d^{3} e^{3}-2560 x \,c^{3} d^{5} e +15015 e^{6} a^{3}+24024 d^{2} e^{4} a^{2} c +18304 d^{4} e^{2} c^{2} a +5120 c^{3} d^{6}\right )}{15015 e^{7}}\) | \(205\) |
| trager | \(\frac {2 \sqrt {e x +d}\, \left (1155 x^{6} c^{3} e^{6}-1260 x^{5} c^{3} d \,e^{5}+5005 x^{4} a \,c^{2} e^{6}+1400 x^{4} c^{3} d^{2} e^{4}-5720 x^{3} a \,c^{2} d \,e^{5}-1600 x^{3} c^{3} d^{3} e^{3}+9009 x^{2} a^{2} c \,e^{6}+6864 x^{2} a \,c^{2} d^{2} e^{4}+1920 x^{2} c^{3} d^{4} e^{2}-12012 x \,a^{2} c d \,e^{5}-9152 x a \,c^{2} d^{3} e^{3}-2560 x \,c^{3} d^{5} e +15015 e^{6} a^{3}+24024 d^{2} e^{4} a^{2} c +18304 d^{4} e^{2} c^{2} a +5120 c^{3} d^{6}\right )}{15015 e^{7}}\) | \(205\) |
| risch | \(\frac {2 \sqrt {e x +d}\, \left (1155 x^{6} c^{3} e^{6}-1260 x^{5} c^{3} d \,e^{5}+5005 x^{4} a \,c^{2} e^{6}+1400 x^{4} c^{3} d^{2} e^{4}-5720 x^{3} a \,c^{2} d \,e^{5}-1600 x^{3} c^{3} d^{3} e^{3}+9009 x^{2} a^{2} c \,e^{6}+6864 x^{2} a \,c^{2} d^{2} e^{4}+1920 x^{2} c^{3} d^{4} e^{2}-12012 x \,a^{2} c d \,e^{5}-9152 x a \,c^{2} d^{3} e^{3}-2560 x \,c^{3} d^{5} e +15015 e^{6} a^{3}+24024 d^{2} e^{4} a^{2} c +18304 d^{4} e^{2} c^{2} a +5120 c^{3} d^{6}\right )}{15015 e^{7}}\) | \(205\) |
| derivativedivides | \(\frac {\frac {2 c^{3} \left (e x +d \right )^{\frac {13}{2}}}{13}-\frac {12 c^{3} d \left (e x +d \right )^{\frac {11}{2}}}{11}+\frac {2 \left (\left (e^{2} a +c \,d^{2}\right ) c^{2}+8 c^{3} d^{2}+c \left (2 \left (e^{2} a +c \,d^{2}\right ) c +4 c^{2} d^{2}\right )\right ) \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (-8 \left (e^{2} a +c \,d^{2}\right ) c^{2} d -2 c d \left (2 \left (e^{2} a +c \,d^{2}\right ) c +4 c^{2} d^{2}\right )\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (e^{2} a +c \,d^{2}\right ) \left (2 \left (e^{2} a +c \,d^{2}\right ) c +4 c^{2} d^{2}\right )+8 c^{2} d^{2} \left (e^{2} a +c \,d^{2}\right )+c \left (e^{2} a +c \,d^{2}\right )^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}-4 \left (e^{2} a +c \,d^{2}\right )^{2} c d \left (e x +d \right )^{\frac {3}{2}}+2 \left (e^{2} a +c \,d^{2}\right )^{3} \sqrt {e x +d}}{e^{7}}\) | \(268\) |
| default | \(\frac {\frac {2 c^{3} \left (e x +d \right )^{\frac {13}{2}}}{13}-\frac {12 c^{3} d \left (e x +d \right )^{\frac {11}{2}}}{11}+\frac {2 \left (\left (e^{2} a +c \,d^{2}\right ) c^{2}+8 c^{3} d^{2}+c \left (2 \left (e^{2} a +c \,d^{2}\right ) c +4 c^{2} d^{2}\right )\right ) \left (e x +d \right )^{\frac {9}{2}}}{9}+\frac {2 \left (-8 \left (e^{2} a +c \,d^{2}\right ) c^{2} d -2 c d \left (2 \left (e^{2} a +c \,d^{2}\right ) c +4 c^{2} d^{2}\right )\right ) \left (e x +d \right )^{\frac {7}{2}}}{7}+\frac {2 \left (\left (e^{2} a +c \,d^{2}\right ) \left (2 \left (e^{2} a +c \,d^{2}\right ) c +4 c^{2} d^{2}\right )+8 c^{2} d^{2} \left (e^{2} a +c \,d^{2}\right )+c \left (e^{2} a +c \,d^{2}\right )^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}{5}-4 \left (e^{2} a +c \,d^{2}\right )^{2} c d \left (e x +d \right )^{\frac {3}{2}}+2 \left (e^{2} a +c \,d^{2}\right )^{3} \sqrt {e x +d}}{e^{7}}\) | \(268\) |
2*(1/13*(e^6*x^6-12/11*d*e^5*x^5+40/33*d^2*e^4*x^4-320/231*x^3*d^3*e^3+128 /77*d^4*e^2*x^2-512/231*d^5*e*x+1024/231*d^6)*c^3+128/105*(35/128*e^4*x^4- 5/16*d*e^3*x^3+3/8*d^2*e^2*x^2-1/2*d^3*e*x+d^4)*e^2*a*c^2+8/5*(3/8*x^2*e^2 -1/2*d*e*x+d^2)*e^4*a^2*c+e^6*a^3)*(e*x+d)^(1/2)/e^7
Time = 0.36 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.01 \[ \int \frac {\left (a+c x^2\right )^3}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (1155 \, c^{3} e^{6} x^{6} - 1260 \, c^{3} d e^{5} x^{5} + 5120 \, c^{3} d^{6} + 18304 \, a c^{2} d^{4} e^{2} + 24024 \, a^{2} c d^{2} e^{4} + 15015 \, a^{3} e^{6} + 35 \, {\left (40 \, c^{3} d^{2} e^{4} + 143 \, a c^{2} e^{6}\right )} x^{4} - 40 \, {\left (40 \, c^{3} d^{3} e^{3} + 143 \, a c^{2} d e^{5}\right )} x^{3} + 3 \, {\left (640 \, c^{3} d^{4} e^{2} + 2288 \, a c^{2} d^{2} e^{4} + 3003 \, a^{2} c e^{6}\right )} x^{2} - 4 \, {\left (640 \, c^{3} d^{5} e + 2288 \, a c^{2} d^{3} e^{3} + 3003 \, a^{2} c d e^{5}\right )} x\right )} \sqrt {e x + d}}{15015 \, e^{7}} \]
2/15015*(1155*c^3*e^6*x^6 - 1260*c^3*d*e^5*x^5 + 5120*c^3*d^6 + 18304*a*c^ 2*d^4*e^2 + 24024*a^2*c*d^2*e^4 + 15015*a^3*e^6 + 35*(40*c^3*d^2*e^4 + 143 *a*c^2*e^6)*x^4 - 40*(40*c^3*d^3*e^3 + 143*a*c^2*d*e^5)*x^3 + 3*(640*c^3*d ^4*e^2 + 2288*a*c^2*d^2*e^4 + 3003*a^2*c*e^6)*x^2 - 4*(640*c^3*d^5*e + 228 8*a*c^2*d^3*e^3 + 3003*a^2*c*d*e^5)*x)*sqrt(e*x + d)/e^7
Time = 0.68 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.52 \[ \int \frac {\left (a+c x^2\right )^3}{\sqrt {d+e x}} \, dx=\begin {cases} \frac {2 \left (- \frac {6 c^{3} d \left (d + e x\right )^{\frac {11}{2}}}{11 e^{6}} + \frac {c^{3} \left (d + e x\right )^{\frac {13}{2}}}{13 e^{6}} + \frac {\left (d + e x\right )^{\frac {9}{2}} \cdot \left (3 a c^{2} e^{2} + 15 c^{3} d^{2}\right )}{9 e^{6}} + \frac {\left (d + e x\right )^{\frac {7}{2}} \left (- 12 a c^{2} d e^{2} - 20 c^{3} d^{3}\right )}{7 e^{6}} + \frac {\left (d + e x\right )^{\frac {5}{2}} \cdot \left (3 a^{2} c e^{4} + 18 a c^{2} d^{2} e^{2} + 15 c^{3} d^{4}\right )}{5 e^{6}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (- 6 a^{2} c d e^{4} - 12 a c^{2} d^{3} e^{2} - 6 c^{3} d^{5}\right )}{3 e^{6}} + \frac {\sqrt {d + e x} \left (a^{3} e^{6} + 3 a^{2} c d^{2} e^{4} + 3 a c^{2} d^{4} e^{2} + c^{3} d^{6}\right )}{e^{6}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {a^{3} x + a^{2} c x^{3} + \frac {3 a c^{2} x^{5}}{5} + \frac {c^{3} x^{7}}{7}}{\sqrt {d}} & \text {otherwise} \end {cases} \]
Piecewise((2*(-6*c**3*d*(d + e*x)**(11/2)/(11*e**6) + c**3*(d + e*x)**(13/ 2)/(13*e**6) + (d + e*x)**(9/2)*(3*a*c**2*e**2 + 15*c**3*d**2)/(9*e**6) + (d + e*x)**(7/2)*(-12*a*c**2*d*e**2 - 20*c**3*d**3)/(7*e**6) + (d + e*x)** (5/2)*(3*a**2*c*e**4 + 18*a*c**2*d**2*e**2 + 15*c**3*d**4)/(5*e**6) + (d + e*x)**(3/2)*(-6*a**2*c*d*e**4 - 12*a*c**2*d**3*e**2 - 6*c**3*d**5)/(3*e** 6) + sqrt(d + e*x)*(a**3*e**6 + 3*a**2*c*d**2*e**4 + 3*a*c**2*d**4*e**2 + c**3*d**6)/e**6)/e, Ne(e, 0)), ((a**3*x + a**2*c*x**3 + 3*a*c**2*x**5/5 + c**3*x**7/7)/sqrt(d), True))
Time = 0.20 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+c x^2\right )^3}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15015 \, \sqrt {e x + d} a^{3} + \frac {3003 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a^{2} c}{e^{2}} + \frac {143 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} a c^{2}}{e^{4}} + \frac {5 \, {\left (231 \, {\left (e x + d\right )}^{\frac {13}{2}} - 1638 \, {\left (e x + d\right )}^{\frac {11}{2}} d + 5005 \, {\left (e x + d\right )}^{\frac {9}{2}} d^{2} - 8580 \, {\left (e x + d\right )}^{\frac {7}{2}} d^{3} + 9009 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{4} - 6006 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{5} + 3003 \, \sqrt {e x + d} d^{6}\right )} c^{3}}{e^{6}}\right )}}{15015 \, e} \]
2/15015*(15015*sqrt(e*x + d)*a^3 + 3003*(3*(e*x + d)^(5/2) - 10*(e*x + d)^ (3/2)*d + 15*sqrt(e*x + d)*d^2)*a^2*c/e^2 + 143*(35*(e*x + d)^(9/2) - 180* (e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 31 5*sqrt(e*x + d)*d^4)*a*c^2/e^4 + 5*(231*(e*x + d)^(13/2) - 1638*(e*x + d)^ (11/2)*d + 5005*(e*x + d)^(9/2)*d^2 - 8580*(e*x + d)^(7/2)*d^3 + 9009*(e*x + d)^(5/2)*d^4 - 6006*(e*x + d)^(3/2)*d^5 + 3003*sqrt(e*x + d)*d^6)*c^3/e ^6)/e
Time = 0.28 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+c x^2\right )^3}{\sqrt {d+e x}} \, dx=\frac {2 \, {\left (15015 \, \sqrt {e x + d} a^{3} + \frac {3003 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x + d} d^{2}\right )} a^{2} c}{e^{2}} + \frac {143 \, {\left (35 \, {\left (e x + d\right )}^{\frac {9}{2}} - 180 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 378 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 420 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 315 \, \sqrt {e x + d} d^{4}\right )} a c^{2}}{e^{4}} + \frac {5 \, {\left (231 \, {\left (e x + d\right )}^{\frac {13}{2}} - 1638 \, {\left (e x + d\right )}^{\frac {11}{2}} d + 5005 \, {\left (e x + d\right )}^{\frac {9}{2}} d^{2} - 8580 \, {\left (e x + d\right )}^{\frac {7}{2}} d^{3} + 9009 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{4} - 6006 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{5} + 3003 \, \sqrt {e x + d} d^{6}\right )} c^{3}}{e^{6}}\right )}}{15015 \, e} \]
2/15015*(15015*sqrt(e*x + d)*a^3 + 3003*(3*(e*x + d)^(5/2) - 10*(e*x + d)^ (3/2)*d + 15*sqrt(e*x + d)*d^2)*a^2*c/e^2 + 143*(35*(e*x + d)^(9/2) - 180* (e*x + d)^(7/2)*d + 378*(e*x + d)^(5/2)*d^2 - 420*(e*x + d)^(3/2)*d^3 + 31 5*sqrt(e*x + d)*d^4)*a*c^2/e^4 + 5*(231*(e*x + d)^(13/2) - 1638*(e*x + d)^ (11/2)*d + 5005*(e*x + d)^(9/2)*d^2 - 8580*(e*x + d)^(7/2)*d^3 + 9009*(e*x + d)^(5/2)*d^4 - 6006*(e*x + d)^(3/2)*d^5 + 3003*sqrt(e*x + d)*d^6)*c^3/e ^6)/e
Time = 0.05 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.94 \[ \int \frac {\left (a+c x^2\right )^3}{\sqrt {d+e x}} \, dx=\frac {\left (30\,c^3\,d^2+6\,a\,c^2\,e^2\right )\,{\left (d+e\,x\right )}^{9/2}}{9\,e^7}+\frac {{\left (d+e\,x\right )}^{5/2}\,\left (6\,a^2\,c\,e^4+36\,a\,c^2\,d^2\,e^2+30\,c^3\,d^4\right )}{5\,e^7}+\frac {2\,c^3\,{\left (d+e\,x\right )}^{13/2}}{13\,e^7}+\frac {2\,{\left (c\,d^2+a\,e^2\right )}^3\,\sqrt {d+e\,x}}{e^7}-\frac {\left (40\,c^3\,d^3+24\,a\,c^2\,d\,e^2\right )\,{\left (d+e\,x\right )}^{7/2}}{7\,e^7}-\frac {12\,c^3\,d\,{\left (d+e\,x\right )}^{11/2}}{11\,e^7}-\frac {4\,c\,d\,{\left (c\,d^2+a\,e^2\right )}^2\,{\left (d+e\,x\right )}^{3/2}}{e^7} \]
((30*c^3*d^2 + 6*a*c^2*e^2)*(d + e*x)^(9/2))/(9*e^7) + ((d + e*x)^(5/2)*(3 0*c^3*d^4 + 6*a^2*c*e^4 + 36*a*c^2*d^2*e^2))/(5*e^7) + (2*c^3*(d + e*x)^(1 3/2))/(13*e^7) + (2*(a*e^2 + c*d^2)^3*(d + e*x)^(1/2))/e^7 - ((40*c^3*d^3 + 24*a*c^2*d*e^2)*(d + e*x)^(7/2))/(7*e^7) - (12*c^3*d*(d + e*x)^(11/2))/( 11*e^7) - (4*c*d*(a*e^2 + c*d^2)^2*(d + e*x)^(3/2))/e^7