Integrand size = 36, antiderivative size = 297 \[ \int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{15/2}} \, dx=-\frac {2 c \left (b c^2-b d^2 x^2\right )^{3/2}}{13 b e (e x)^{13/2} (c+d x)^{3/2}}-\frac {72 d \left (b c^2-b d^2 x^2\right )^{3/2}}{143 b e^2 (e x)^{11/2} (c+d x)^{3/2}}-\frac {862 d^2 \left (b c^2-b d^2 x^2\right )^{3/2}}{1287 b c e^3 (e x)^{9/2} (c+d x)^{3/2}}-\frac {1724 d^3 \left (b c^2-b d^2 x^2\right )^{3/2}}{3003 b c^2 e^4 (e x)^{7/2} (c+d x)^{3/2}}-\frac {6896 d^4 \left (b c^2-b d^2 x^2\right )^{3/2}}{15015 b c^3 e^5 (e x)^{5/2} (c+d x)^{3/2}}-\frac {13792 d^5 \left (b c^2-b d^2 x^2\right )^{3/2}}{45045 b c^4 e^6 (e x)^{3/2} (c+d x)^{3/2}} \] Output:
-2/13*c*(-b*d^2*x^2+b*c^2)^(3/2)/b/e/(e*x)^(13/2)/(d*x+c)^(3/2)-72/143*d*( -b*d^2*x^2+b*c^2)^(3/2)/b/e^2/(e*x)^(11/2)/(d*x+c)^(3/2)-862/1287*d^2*(-b* d^2*x^2+b*c^2)^(3/2)/b/c/e^3/(e*x)^(9/2)/(d*x+c)^(3/2)-1724/3003*d^3*(-b*d ^2*x^2+b*c^2)^(3/2)/b/c^2/e^4/(e*x)^(7/2)/(d*x+c)^(3/2)-6896/15015*d^4*(-b *d^2*x^2+b*c^2)^(3/2)/b/c^3/e^5/(e*x)^(5/2)/(d*x+c)^(3/2)-13792/45045*d^5* (-b*d^2*x^2+b*c^2)^(3/2)/b/c^4/e^6/(e*x)^(3/2)/(d*x+c)^(3/2)
Time = 6.78 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.37 \[ \int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{15/2}} \, dx=\frac {2 \sqrt {e x} \sqrt {b \left (c^2-d^2 x^2\right )} \left (-3465 c^6-7875 c^5 d x-3745 c^4 d^2 x^2+2155 c^3 d^3 x^3+2586 c^2 d^4 x^4+3448 c d^5 x^5+6896 d^6 x^6\right )}{45045 c^4 e^8 x^7 \sqrt {c+d x}} \] Input:
Integrate[((c + d*x)^(3/2)*Sqrt[b*c^2 - b*d^2*x^2])/(e*x)^(15/2),x]
Output:
(2*Sqrt[e*x]*Sqrt[b*(c^2 - d^2*x^2)]*(-3465*c^6 - 7875*c^5*d*x - 3745*c^4* d^2*x^2 + 2155*c^3*d^3*x^3 + 2586*c^2*d^4*x^4 + 3448*c*d^5*x^5 + 6896*d^6* x^6))/(45045*c^4*e^8*x^7*Sqrt[c + d*x])
Time = 0.53 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {586, 100, 27, 87, 55, 55, 55, 48}
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.
\(\displaystyle \int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{15/2}} \, dx\) |
\(\Big \downarrow \) 586 |
\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \int \frac {(c+d x)^2 \sqrt {b c-b d x}}{(e x)^{15/2}}dx}{\sqrt {c+d x} \sqrt {b c-b d x}}\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {2 \int \frac {b c d e^2 (36 c+13 d x) \sqrt {b c-b d x}}{2 (e x)^{13/2}}dx}{13 b c e^3}-\frac {2 c (b c-b d x)^{3/2}}{13 b e (e x)^{13/2}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {d \int \frac {(36 c+13 d x) \sqrt {b c-b d x}}{(e x)^{13/2}}dx}{13 e}-\frac {2 c (b c-b d x)^{3/2}}{13 b e (e x)^{13/2}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {d \left (\frac {431 d \int \frac {\sqrt {b c-b d x}}{(e x)^{11/2}}dx}{11 e}-\frac {72 (b c-b d x)^{3/2}}{11 b e (e x)^{11/2}}\right )}{13 e}-\frac {2 c (b c-b d x)^{3/2}}{13 b e (e x)^{13/2}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {d \left (\frac {431 d \left (\frac {2 d \int \frac {\sqrt {b c-b d x}}{(e x)^{9/2}}dx}{3 c e}-\frac {2 (b c-b d x)^{3/2}}{9 b c e (e x)^{9/2}}\right )}{11 e}-\frac {72 (b c-b d x)^{3/2}}{11 b e (e x)^{11/2}}\right )}{13 e}-\frac {2 c (b c-b d x)^{3/2}}{13 b e (e x)^{13/2}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {d \left (\frac {431 d \left (\frac {2 d \left (\frac {4 d \int \frac {\sqrt {b c-b d x}}{(e x)^{7/2}}dx}{7 c e}-\frac {2 (b c-b d x)^{3/2}}{7 b c e (e x)^{7/2}}\right )}{3 c e}-\frac {2 (b c-b d x)^{3/2}}{9 b c e (e x)^{9/2}}\right )}{11 e}-\frac {72 (b c-b d x)^{3/2}}{11 b e (e x)^{11/2}}\right )}{13 e}-\frac {2 c (b c-b d x)^{3/2}}{13 b e (e x)^{13/2}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {d \left (\frac {431 d \left (\frac {2 d \left (\frac {4 d \left (\frac {2 d \int \frac {\sqrt {b c-b d x}}{(e x)^{5/2}}dx}{5 c e}-\frac {2 (b c-b d x)^{3/2}}{5 b c e (e x)^{5/2}}\right )}{7 c e}-\frac {2 (b c-b d x)^{3/2}}{7 b c e (e x)^{7/2}}\right )}{3 c e}-\frac {2 (b c-b d x)^{3/2}}{9 b c e (e x)^{9/2}}\right )}{11 e}-\frac {72 (b c-b d x)^{3/2}}{11 b e (e x)^{11/2}}\right )}{13 e}-\frac {2 c (b c-b d x)^{3/2}}{13 b e (e x)^{13/2}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {d \left (\frac {431 d \left (\frac {2 d \left (\frac {4 d \left (-\frac {4 d (b c-b d x)^{3/2}}{15 b c^2 e^2 (e x)^{3/2}}-\frac {2 (b c-b d x)^{3/2}}{5 b c e (e x)^{5/2}}\right )}{7 c e}-\frac {2 (b c-b d x)^{3/2}}{7 b c e (e x)^{7/2}}\right )}{3 c e}-\frac {2 (b c-b d x)^{3/2}}{9 b c e (e x)^{9/2}}\right )}{11 e}-\frac {72 (b c-b d x)^{3/2}}{11 b e (e x)^{11/2}}\right )}{13 e}-\frac {2 c (b c-b d x)^{3/2}}{13 b e (e x)^{13/2}}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\) |
Input:
Int[((c + d*x)^(3/2)*Sqrt[b*c^2 - b*d^2*x^2])/(e*x)^(15/2),x]
Output:
(Sqrt[b*c^2 - b*d^2*x^2]*((-2*c*(b*c - b*d*x)^(3/2))/(13*b*e*(e*x)^(13/2)) + (d*((-72*(b*c - b*d*x)^(3/2))/(11*b*e*(e*x)^(11/2)) + (431*d*((-2*(b*c - b*d*x)^(3/2))/(9*b*c*e*(e*x)^(9/2)) + (2*d*((-2*(b*c - b*d*x)^(3/2))/(7* b*c*e*(e*x)^(7/2)) + (4*d*((-2*(b*c - b*d*x)^(3/2))/(5*b*c*e*(e*x)^(5/2)) - (4*d*(b*c - b*d*x)^(3/2))/(15*b*c^2*e^2*(e*x)^(3/2))))/(7*c*e)))/(3*c*e) ))/(11*e)))/(13*e)))/(Sqrt[c + d*x]*Sqrt[b*c - b*d*x])
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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_) , x_Symbol] :> Simp[(a + b*x^2)^FracPart[p]/((c + d*x)^FracPart[p]*(a/c + ( b*x)/d)^FracPart[p]) Int[(e*x)^m*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0]
Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.32
method | result | size |
gosper | \(-\frac {2 \left (-d x +c \right ) x \left (6896 x^{5} d^{5}+10344 x^{4} c \,d^{4}+12930 c^{2} d^{3} x^{3}+15085 c^{3} d^{2} x^{2}+11340 c^{4} d x +3465 c^{5}\right ) \sqrt {-b \,x^{2} d^{2}+b \,c^{2}}}{45045 c^{4} \sqrt {d x +c}\, \left (e x \right )^{\frac {15}{2}}}\) | \(94\) |
orering | \(-\frac {2 \left (-d x +c \right ) x \left (6896 x^{5} d^{5}+10344 x^{4} c \,d^{4}+12930 c^{2} d^{3} x^{3}+15085 c^{3} d^{2} x^{2}+11340 c^{4} d x +3465 c^{5}\right ) \sqrt {-b \,x^{2} d^{2}+b \,c^{2}}}{45045 c^{4} \sqrt {d x +c}\, \left (e x \right )^{\frac {15}{2}}}\) | \(94\) |
default | \(-\frac {2 \sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, \left (-d x +c \right ) \left (6896 x^{5} d^{5}+10344 x^{4} c \,d^{4}+12930 c^{2} d^{3} x^{3}+15085 c^{3} d^{2} x^{2}+11340 c^{4} d x +3465 c^{5}\right )}{45045 \sqrt {d x +c}\, e^{7} x^{6} \sqrt {e x}\, c^{4}}\) | \(98\) |
Input:
int((d*x+c)^(3/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(15/2),x,method=_RETURNVE RBOSE)
Output:
-2/45045*(-d*x+c)*x*(6896*d^5*x^5+10344*c*d^4*x^4+12930*c^2*d^3*x^3+15085* c^3*d^2*x^2+11340*c^4*d*x+3465*c^5)*(-b*d^2*x^2+b*c^2)^(1/2)/c^4/(d*x+c)^( 1/2)/(e*x)^(15/2)
Time = 0.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.40 \[ \int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{15/2}} \, dx=\frac {2 \, {\left (6896 \, d^{6} x^{6} + 3448 \, c d^{5} x^{5} + 2586 \, c^{2} d^{4} x^{4} + 2155 \, c^{3} d^{3} x^{3} - 3745 \, c^{4} d^{2} x^{2} - 7875 \, c^{5} d x - 3465 \, c^{6}\right )} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} \sqrt {e x}}{45045 \, {\left (c^{4} d e^{8} x^{8} + c^{5} e^{8} x^{7}\right )}} \] Input:
integrate((d*x+c)^(3/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(15/2),x, algorithm ="fricas")
Output:
2/45045*(6896*d^6*x^6 + 3448*c*d^5*x^5 + 2586*c^2*d^4*x^4 + 2155*c^3*d^3*x ^3 - 3745*c^4*d^2*x^2 - 7875*c^5*d*x - 3465*c^6)*sqrt(-b*d^2*x^2 + b*c^2)* sqrt(d*x + c)*sqrt(e*x)/(c^4*d*e^8*x^8 + c^5*e^8*x^7)
Timed out. \[ \int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{15/2}} \, dx=\text {Timed out} \] Input:
integrate((d*x+c)**(3/2)*(-b*d**2*x**2+b*c**2)**(1/2)/(e*x)**(15/2),x)
Output:
Timed out
\[ \int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{15/2}} \, dx=\int { \frac {\sqrt {-b d^{2} x^{2} + b c^{2}} {\left (d x + c\right )}^{\frac {3}{2}}}{\left (e x\right )^{\frac {15}{2}}} \,d x } \] Input:
integrate((d*x+c)^(3/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(15/2),x, algorithm ="maxima")
Output:
integrate(sqrt(-b*d^2*x^2 + b*c^2)*(d*x + c)^(3/2)/(e*x)^(15/2), x)
Time = 0.18 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.64 \[ \int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{15/2}} \, dx=\frac {2 \, {\left (60060 \, b^{11} c d^{13} e^{6} + {\left (156156 \, b^{10} d^{13} e^{6} + 431 \, {\left (\frac {429 \, b^{9} d^{13} e^{6}}{c} + 2 \, {\left (\frac {143 \, b^{8} d^{13} e^{6}}{c^{2}} + 4 \, {\left (\frac {13 \, b^{7} d^{13} e^{6}}{c^{3}} + \frac {2 \, {\left (b d x - b c\right )} b^{6} d^{13} e^{6}}{c^{4}}\right )} {\left (b d x - b c\right )}\right )} {\left (b d x - b c\right )}\right )} {\left (b d x - b c\right )}\right )} {\left (b d x - b c\right )}\right )} {\left (b d x - b c\right )} \sqrt {-b d x + b c} {\left | b \right |} {\left | d \right |}}{45045 \, {\left (b^{2} c d e + {\left (b d x - b c\right )} b d e\right )}^{\frac {13}{2}} d e^{7}} \] Input:
integrate((d*x+c)^(3/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(15/2),x, algorithm ="giac")
Output:
2/45045*(60060*b^11*c*d^13*e^6 + (156156*b^10*d^13*e^6 + 431*(429*b^9*d^13 *e^6/c + 2*(143*b^8*d^13*e^6/c^2 + 4*(13*b^7*d^13*e^6/c^3 + 2*(b*d*x - b*c )*b^6*d^13*e^6/c^4)*(b*d*x - b*c))*(b*d*x - b*c))*(b*d*x - b*c))*(b*d*x - b*c))*(b*d*x - b*c)*sqrt(-b*d*x + b*c)*abs(b)*abs(d)/((b^2*c*d*e + (b*d*x - b*c)*b*d*e)^(13/2)*d*e^7)
Time = 7.23 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.59 \[ \int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{15/2}} \, dx=\frac {\sqrt {b\,c^2-b\,d^2\,x^2}\,\left (\frac {862\,d^2\,x^3\,\sqrt {c+d\,x}}{9009\,c\,e^7}-\frac {50\,c\,x\,\sqrt {c+d\,x}}{143\,e^7}-\frac {214\,d\,x^2\,\sqrt {c+d\,x}}{1287\,e^7}-\frac {2\,c^2\,\sqrt {c+d\,x}}{13\,d\,e^7}+\frac {1724\,d^3\,x^4\,\sqrt {c+d\,x}}{15015\,c^2\,e^7}+\frac {6896\,d^4\,x^5\,\sqrt {c+d\,x}}{45045\,c^3\,e^7}+\frac {13792\,d^5\,x^6\,\sqrt {c+d\,x}}{45045\,c^4\,e^7}\right )}{x^7\,\sqrt {e\,x}+\frac {c\,x^6\,\sqrt {e\,x}}{d}} \] Input:
int(((b*c^2 - b*d^2*x^2)^(1/2)*(c + d*x)^(3/2))/(e*x)^(15/2),x)
Output:
((b*c^2 - b*d^2*x^2)^(1/2)*((862*d^2*x^3*(c + d*x)^(1/2))/(9009*c*e^7) - ( 50*c*x*(c + d*x)^(1/2))/(143*e^7) - (214*d*x^2*(c + d*x)^(1/2))/(1287*e^7) - (2*c^2*(c + d*x)^(1/2))/(13*d*e^7) + (1724*d^3*x^4*(c + d*x)^(1/2))/(15 015*c^2*e^7) + (6896*d^4*x^5*(c + d*x)^(1/2))/(45045*c^3*e^7) + (13792*d^5 *x^6*(c + d*x)^(1/2))/(45045*c^4*e^7)))/(x^7*(e*x)^(1/2) + (c*x^6*(e*x)^(1 /2))/d)
Time = 0.25 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.51 \[ \int \frac {(c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{(e x)^{15/2}} \, dx=\frac {2 \sqrt {e}\, \sqrt {b}\, \left (-3465 \sqrt {x}\, \sqrt {-d x +c}\, c^{6}-7875 \sqrt {x}\, \sqrt {-d x +c}\, c^{5} d x -3745 \sqrt {x}\, \sqrt {-d x +c}\, c^{4} d^{2} x^{2}+2155 \sqrt {x}\, \sqrt {-d x +c}\, c^{3} d^{3} x^{3}+2586 \sqrt {x}\, \sqrt {-d x +c}\, c^{2} d^{4} x^{4}+3448 \sqrt {x}\, \sqrt {-d x +c}\, c \,d^{5} x^{5}+6896 \sqrt {x}\, \sqrt {-d x +c}\, d^{6} x^{6}-6896 \sqrt {d}\, d^{6} i \,x^{7}\right )}{45045 c^{4} e^{8} x^{7}} \] Input:
int((d*x+c)^(3/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(15/2),x)
Output:
(2*sqrt(e)*sqrt(b)*( - 3465*sqrt(x)*sqrt(c - d*x)*c**6 - 7875*sqrt(x)*sqrt (c - d*x)*c**5*d*x - 3745*sqrt(x)*sqrt(c - d*x)*c**4*d**2*x**2 + 2155*sqrt (x)*sqrt(c - d*x)*c**3*d**3*x**3 + 2586*sqrt(x)*sqrt(c - d*x)*c**2*d**4*x* *4 + 3448*sqrt(x)*sqrt(c - d*x)*c*d**5*x**5 + 6896*sqrt(x)*sqrt(c - d*x)*d **6*x**6 - 6896*sqrt(d)*d**6*i*x**7))/(45045*c**4*e**8*x**7)