Integrand size = 31, antiderivative size = 124 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx=\frac {2 (b d-a e)^2 (B d-A e)}{5 e^4 (d+e x)^{5/2}}-\frac {2 (b d-a e) (3 b B d-2 A b e-a B e)}{3 e^4 (d+e x)^{3/2}}+\frac {2 b (3 b B d-A b e-2 a B e)}{e^4 \sqrt {d+e x}}+\frac {2 b^2 B \sqrt {d+e x}}{e^4} \] Output:
2/5*(-a*e+b*d)^2*(-A*e+B*d)/e^4/(e*x+d)^(5/2)-2/3*(-a*e+b*d)*(-2*A*b*e-B*a *e+3*B*b*d)/e^4/(e*x+d)^(3/2)+2*b*(-A*b*e-2*B*a*e+3*B*b*d)/e^4/(e*x+d)^(1/ 2)+2*b^2*B*(e*x+d)^(1/2)/e^4
Time = 0.20 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.10 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx=-\frac {2 \left (a^2 e^2 (2 B d+3 A e+5 B e x)+2 a b e \left (A e (2 d+5 e x)+B \left (8 d^2+20 d e x+15 e^2 x^2\right )\right )+b^2 \left (A e \left (8 d^2+20 d e x+15 e^2 x^2\right )-3 B \left (16 d^3+40 d^2 e x+30 d e^2 x^2+5 e^3 x^3\right )\right )\right )}{15 e^4 (d+e x)^{5/2}} \] Input:
Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2))/(d + e*x)^(7/2),x]
Output:
(-2*(a^2*e^2*(2*B*d + 3*A*e + 5*B*e*x) + 2*a*b*e*(A*e*(2*d + 5*e*x) + B*(8 *d^2 + 20*d*e*x + 15*e^2*x^2)) + b^2*(A*e*(8*d^2 + 20*d*e*x + 15*e^2*x^2) - 3*B*(16*d^3 + 40*d^2*e*x + 30*d*e^2*x^2 + 5*e^3*x^3))))/(15*e^4*(d + e*x )^(5/2))
Time = 0.48 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1184, 27, 86, 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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x+b^2 x^2\right ) (A+B x)}{(d+e x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1184 |
| \(\displaystyle \frac {\int \frac {b^2 (a+b x)^2 (A+B x)}{(d+e x)^{7/2}}dx}{b^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{7/2}}dx\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \int \left (\frac {b (2 a B e+A b e-3 b B d)}{e^3 (d+e x)^{3/2}}+\frac {(a e-b d) (a B e+2 A b e-3 b B d)}{e^3 (d+e x)^{5/2}}+\frac {(a e-b d)^2 (A e-B d)}{e^3 (d+e x)^{7/2}}+\frac {b^2 B}{e^3 \sqrt {d+e x}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 b (-2 a B e-A b e+3 b B d)}{e^4 \sqrt {d+e x}}-\frac {2 (b d-a e) (-a B e-2 A b e+3 b B d)}{3 e^4 (d+e x)^{3/2}}+\frac {2 (b d-a e)^2 (B d-A e)}{5 e^4 (d+e x)^{5/2}}+\frac {2 b^2 B \sqrt {d+e x}}{e^4}\) |
Input:
Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2))/(d + e*x)^(7/2),x]
Output:
(2*(b*d - a*e)^2*(B*d - A*e))/(5*e^4*(d + e*x)^(5/2)) - (2*(b*d - a*e)*(3* b*B*d - 2*A*b*e - a*B*e))/(3*e^4*(d + e*x)^(3/2)) + (2*b*(3*b*B*d - A*b*e - 2*a*B*e))/(e^4*Sqrt[d + e*x]) + (2*b^2*B*Sqrt[d + e*x])/e^4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 1.13 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.95
| method | result | size |
| pseudoelliptic | \(\frac {\left (-30 x^{2} \left (-B x +A \right ) b^{2}-20 a x \left (3 B x +A \right ) b -6 a^{2} \left (\frac {5 B x}{3}+A \right )\right ) e^{3}-8 d \left (5 x \left (-\frac {9 B x}{2}+A \right ) b^{2}+a \left (10 B x +A \right ) b +\frac {a^{2} B}{2}\right ) e^{2}-16 b \,d^{2} \left (\left (-15 B x +A \right ) b +2 B a \right ) e +96 B \,b^{2} d^{3}}{15 \left (e x +d \right )^{\frac {5}{2}} e^{4}}\) | \(118\) |
| derivativedivides | \(\frac {2 B \,b^{2} \sqrt {e x +d}-\frac {2 \left (a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -B \,b^{2} d^{3}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 b \left (A b e +2 B a e -3 B b d \right )}{\sqrt {e x +d}}-\frac {2 \left (2 A a b \,e^{2}-2 A \,b^{2} d e +B \,e^{2} a^{2}-4 B a b d e +3 B \,b^{2} d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{4}}\) | \(158\) |
| default | \(\frac {2 B \,b^{2} \sqrt {e x +d}-\frac {2 \left (a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -B \,b^{2} d^{3}\right )}{5 \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 b \left (A b e +2 B a e -3 B b d \right )}{\sqrt {e x +d}}-\frac {2 \left (2 A a b \,e^{2}-2 A \,b^{2} d e +B \,e^{2} a^{2}-4 B a b d e +3 B \,b^{2} d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}}}}{e^{4}}\) | \(158\) |
| gosper | \(-\frac {2 \left (-15 B \,b^{2} x^{3} e^{3}+15 A \,b^{2} e^{3} x^{2}+30 B \,x^{2} a b \,e^{3}-90 B \,x^{2} b^{2} d \,e^{2}+10 A x a b \,e^{3}+20 A x \,b^{2} d \,e^{2}+5 B x \,a^{2} e^{3}+40 B x a b d \,e^{2}-120 B x \,b^{2} d^{2} e +3 a^{2} A \,e^{3}+4 A a b d \,e^{2}+8 A \,b^{2} d^{2} e +2 B \,a^{2} d \,e^{2}+16 B a b \,d^{2} e -48 B \,b^{2} d^{3}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{4}}\) | \(169\) |
| trager | \(-\frac {2 \left (-15 B \,b^{2} x^{3} e^{3}+15 A \,b^{2} e^{3} x^{2}+30 B \,x^{2} a b \,e^{3}-90 B \,x^{2} b^{2} d \,e^{2}+10 A x a b \,e^{3}+20 A x \,b^{2} d \,e^{2}+5 B x \,a^{2} e^{3}+40 B x a b d \,e^{2}-120 B x \,b^{2} d^{2} e +3 a^{2} A \,e^{3}+4 A a b d \,e^{2}+8 A \,b^{2} d^{2} e +2 B \,a^{2} d \,e^{2}+16 B a b \,d^{2} e -48 B \,b^{2} d^{3}\right )}{15 \left (e x +d \right )^{\frac {5}{2}} e^{4}}\) | \(169\) |
| risch | \(\frac {2 b^{2} B \sqrt {e x +d}}{e^{4}}-\frac {2 \left (15 A \,b^{2} e^{3} x^{2}+30 B \,x^{2} a b \,e^{3}-45 B \,x^{2} b^{2} d \,e^{2}+10 A x a b \,e^{3}+20 A x \,b^{2} d \,e^{2}+5 B x \,a^{2} e^{3}+40 B x a b d \,e^{2}-75 B x \,b^{2} d^{2} e +3 a^{2} A \,e^{3}+4 A a b d \,e^{2}+8 A \,b^{2} d^{2} e +2 B \,a^{2} d \,e^{2}+16 B a b \,d^{2} e -33 B \,b^{2} d^{3}\right )}{15 e^{4} \sqrt {e x +d}\, \left (e^{2} x^{2}+2 d e x +d^{2}\right )}\) | \(192\) |
| orering | \(-\frac {2 \left (-15 B \,b^{2} x^{3} e^{3}+15 A \,b^{2} e^{3} x^{2}+30 B \,x^{2} a b \,e^{3}-90 B \,x^{2} b^{2} d \,e^{2}+10 A x a b \,e^{3}+20 A x \,b^{2} d \,e^{2}+5 B x \,a^{2} e^{3}+40 B x a b d \,e^{2}-120 B x \,b^{2} d^{2} e +3 a^{2} A \,e^{3}+4 A a b d \,e^{2}+8 A \,b^{2} d^{2} e +2 B \,a^{2} d \,e^{2}+16 B a b \,d^{2} e -48 B \,b^{2} d^{3}\right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )}{15 e^{4} \left (b x +a \right )^{2} \left (e x +d \right )^{\frac {5}{2}}}\) | \(192\) |
Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/15*((-30*x^2*(-B*x+A)*b^2-20*a*x*(3*B*x+A)*b-6*a^2*(5/3*B*x+A))*e^3-8*d* (5*x*(-9/2*B*x+A)*b^2+a*(10*B*x+A)*b+1/2*a^2*B)*e^2-16*b*d^2*((-15*B*x+A)* b+2*B*a)*e+96*B*b^2*d^3)/(e*x+d)^(5/2)/e^4
Time = 0.08 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.52 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (15 \, B b^{2} e^{3} x^{3} + 48 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} - 8 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e - 2 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 15 \, {\left (6 \, B b^{2} d e^{2} - {\left (2 \, B a b + A b^{2}\right )} e^{3}\right )} x^{2} + 5 \, {\left (24 \, B b^{2} d^{2} e - 4 \, {\left (2 \, B a b + A b^{2}\right )} d e^{2} - {\left (B a^{2} + 2 \, A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^(7/2),x, algorithm="fricas ")
Output:
2/15*(15*B*b^2*e^3*x^3 + 48*B*b^2*d^3 - 3*A*a^2*e^3 - 8*(2*B*a*b + A*b^2)* d^2*e - 2*(B*a^2 + 2*A*a*b)*d*e^2 + 15*(6*B*b^2*d*e^2 - (2*B*a*b + A*b^2)* e^3)*x^2 + 5*(24*B*b^2*d^2*e - 4*(2*B*a*b + A*b^2)*d*e^2 - (B*a^2 + 2*A*a* b)*e^3)*x)*sqrt(e*x + d)/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)
Leaf count of result is larger than twice the leaf count of optimal. 1015 vs. \(2 (126) = 252\).
Time = 0.55 (sec) , antiderivative size = 1015, normalized size of antiderivative = 8.19 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx =\text {Too large to display} \] Input:
integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)/(e*x+d)**(7/2),x)
Output:
Piecewise((-6*A*a**2*e**3/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 8*A*a*b*d*e**2/(15*d**2*e**4*sqrt( d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 20*A* a*b*e**3*x/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e* *6*x**2*sqrt(d + e*x)) - 16*A*b**2*d**2*e/(15*d**2*e**4*sqrt(d + e*x) + 30 *d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 40*A*b**2*d*e**2*x /(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sq rt(d + e*x)) - 30*A*b**2*e**3*x**2/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5 *x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 4*B*a**2*d*e**2/(15*d**2* e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x )) - 10*B*a**2*e**3*x/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e *x) + 15*e**6*x**2*sqrt(d + e*x)) - 32*B*a*b*d**2*e/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) - 80*B*a*b *d*e**2*x/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e** 6*x**2*sqrt(d + e*x)) - 60*B*a*b*e**3*x**2/(15*d**2*e**4*sqrt(d + e*x) + 3 0*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) + 96*B*b**2*d**3/(1 5*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt( d + e*x)) + 240*B*b**2*d**2*e*x/(15*d**2*e**4*sqrt(d + e*x) + 30*d*e**5*x* sqrt(d + e*x) + 15*e**6*x**2*sqrt(d + e*x)) + 180*B*b**2*d*e**2*x**2/(15*d **2*e**4*sqrt(d + e*x) + 30*d*e**5*x*sqrt(d + e*x) + 15*e**6*x**2*sqrt(...
Time = 0.04 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.32 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx=\frac {2 \, {\left (\frac {15 \, \sqrt {e x + d} B b^{2}}{e^{3}} + \frac {3 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} - 3 \, {\left (2 \, B a b + A b^{2}\right )} d^{2} e + 3 \, {\left (B a^{2} + 2 \, A a b\right )} d e^{2} + 15 \, {\left (3 \, B b^{2} d - {\left (2 \, B a b + A b^{2}\right )} e\right )} {\left (e x + d\right )}^{2} - 5 \, {\left (3 \, B b^{2} d^{2} - 2 \, {\left (2 \, B a b + A b^{2}\right )} d e + {\left (B a^{2} + 2 \, A a b\right )} e^{2}\right )} {\left (e x + d\right )}}{{\left (e x + d\right )}^{\frac {5}{2}} e^{3}}\right )}}{15 \, e} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^(7/2),x, algorithm="maxima ")
Output:
2/15*(15*sqrt(e*x + d)*B*b^2/e^3 + (3*B*b^2*d^3 - 3*A*a^2*e^3 - 3*(2*B*a*b + A*b^2)*d^2*e + 3*(B*a^2 + 2*A*a*b)*d*e^2 + 15*(3*B*b^2*d - (2*B*a*b + A *b^2)*e)*(e*x + d)^2 - 5*(3*B*b^2*d^2 - 2*(2*B*a*b + A*b^2)*d*e + (B*a^2 + 2*A*a*b)*e^2)*(e*x + d))/((e*x + d)^(5/2)*e^3))/e
Time = 0.17 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.56 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx=\frac {2 \, \sqrt {e x + d} B b^{2}}{e^{4}} + \frac {2 \, {\left (45 \, {\left (e x + d\right )}^{2} B b^{2} d - 15 \, {\left (e x + d\right )} B b^{2} d^{2} + 3 \, B b^{2} d^{3} - 30 \, {\left (e x + d\right )}^{2} B a b e - 15 \, {\left (e x + d\right )}^{2} A b^{2} e + 20 \, {\left (e x + d\right )} B a b d e + 10 \, {\left (e x + d\right )} A b^{2} d e - 6 \, B a b d^{2} e - 3 \, A b^{2} d^{2} e - 5 \, {\left (e x + d\right )} B a^{2} e^{2} - 10 \, {\left (e x + d\right )} A a b e^{2} + 3 \, B a^{2} d e^{2} + 6 \, A a b d e^{2} - 3 \, A a^{2} e^{3}\right )}}{15 \, {\left (e x + d\right )}^{\frac {5}{2}} e^{4}} \] Input:
integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^(7/2),x, algorithm="giac")
Output:
2*sqrt(e*x + d)*B*b^2/e^4 + 2/15*(45*(e*x + d)^2*B*b^2*d - 15*(e*x + d)*B* b^2*d^2 + 3*B*b^2*d^3 - 30*(e*x + d)^2*B*a*b*e - 15*(e*x + d)^2*A*b^2*e + 20*(e*x + d)*B*a*b*d*e + 10*(e*x + d)*A*b^2*d*e - 6*B*a*b*d^2*e - 3*A*b^2* d^2*e - 5*(e*x + d)*B*a^2*e^2 - 10*(e*x + d)*A*a*b*e^2 + 3*B*a^2*d*e^2 + 6 *A*a*b*d*e^2 - 3*A*a^2*e^3)/((e*x + d)^(5/2)*e^4)
Time = 11.44 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.35 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx=-\frac {2\,\left (2\,B\,a^2\,d\,e^2+5\,B\,a^2\,e^3\,x+3\,A\,a^2\,e^3+16\,B\,a\,b\,d^2\,e+40\,B\,a\,b\,d\,e^2\,x+4\,A\,a\,b\,d\,e^2+30\,B\,a\,b\,e^3\,x^2+10\,A\,a\,b\,e^3\,x-48\,B\,b^2\,d^3-120\,B\,b^2\,d^2\,e\,x+8\,A\,b^2\,d^2\,e-90\,B\,b^2\,d\,e^2\,x^2+20\,A\,b^2\,d\,e^2\,x-15\,B\,b^2\,e^3\,x^3+15\,A\,b^2\,e^3\,x^2\right )}{15\,e^4\,{\left (d+e\,x\right )}^{5/2}} \] Input:
int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x))/(d + e*x)^(7/2),x)
Output:
-(2*(3*A*a^2*e^3 - 48*B*b^2*d^3 + 8*A*b^2*d^2*e + 2*B*a^2*d*e^2 + 5*B*a^2* e^3*x + 15*A*b^2*e^3*x^2 - 15*B*b^2*e^3*x^3 + 30*B*a*b*e^3*x^2 + 20*A*b^2* d*e^2*x - 120*B*b^2*d^2*e*x - 90*B*b^2*d*e^2*x^2 + 4*A*a*b*d*e^2 + 16*B*a* b*d^2*e + 10*A*a*b*e^3*x + 40*B*a*b*d*e^2*x))/(15*e^4*(d + e*x)^(5/2))
Time = 0.21 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )}{(d+e x)^{7/2}} \, dx=\frac {2 b^{3} e^{3} x^{3}-6 a \,b^{2} e^{3} x^{2}+12 b^{3} d \,e^{2} x^{2}-2 a^{2} b \,e^{3} x -8 a \,b^{2} d \,e^{2} x +16 b^{3} d^{2} e x -\frac {2}{5} a^{3} e^{3}-\frac {4}{5} a^{2} b d \,e^{2}-\frac {16}{5} a \,b^{2} d^{2} e +\frac {32}{5} b^{3} d^{3}}{\sqrt {e x +d}\, e^{4} \left (e^{2} x^{2}+2 d e x +d^{2}\right )} \] Input:
int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^(7/2),x)
Output:
(2*( - a**3*e**3 - 2*a**2*b*d*e**2 - 5*a**2*b*e**3*x - 8*a*b**2*d**2*e - 2 0*a*b**2*d*e**2*x - 15*a*b**2*e**3*x**2 + 16*b**3*d**3 + 40*b**3*d**2*e*x + 30*b**3*d*e**2*x**2 + 5*b**3*e**3*x**3))/(5*sqrt(d + e*x)*e**4*(d**2 + 2 *d*e*x + e**2*x**2))