\(\int \frac {(a+c x^2)^3}{(d+e x)^5} \, dx\) [482]

Optimal result

Integrand size = 17, antiderivative size = 171 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^5} \, dx=-\frac {5 c^3 d x}{e^6}+\frac {c^3 x^2}{2 e^5}-\frac {\left (c d^2+a e^2\right )^3}{4 e^7 (d+e x)^4}+\frac {2 c d \left (c d^2+a e^2\right )^2}{e^7 (d+e x)^3}-\frac {3 c \left (c d^2+a e^2\right ) \left (5 c d^2+a e^2\right )}{2 e^7 (d+e x)^2}+\frac {4 c^2 d \left (5 c d^2+3 a e^2\right )}{e^7 (d+e x)}+\frac {3 c^2 \left (5 c d^2+a e^2\right ) \log (d+e x)}{e^7} \]

[Out]

-5*c^3*d*x/e^6+1/2*c^3*x^2/e^5-1/4*(a*e^2+c*d^2)^3/e^7/(e*x+d)^4+2*c*d*(a*e^2+c*d^2)^2/e^7/(e*x+d)^3-3/2*c*(a*
e^2+c*d^2)*(a*e^2+5*c*d^2)/e^7/(e*x+d)^2+4*c^2*d*(3*a*e^2+5*c*d^2)/e^7/(e*x+d)+3*c^2*(a*e^2+5*c*d^2)*ln(e*x+d)
/e^7

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {711} \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^5} \, dx=\frac {4 c^2 d \left (3 a e^2+5 c d^2\right )}{e^7 (d+e x)}+\frac {3 c^2 \left (a e^2+5 c d^2\right ) \log (d+e x)}{e^7}-\frac {3 c \left (a e^2+c d^2\right ) \left (a e^2+5 c d^2\right )}{2 e^7 (d+e x)^2}+\frac {2 c d \left (a e^2+c d^2\right )^2}{e^7 (d+e x)^3}-\frac {\left (a e^2+c d^2\right )^3}{4 e^7 (d+e x)^4}-\frac {5 c^3 d x}{e^6}+\frac {c^3 x^2}{2 e^5} \]

[In]

Int[(a + c*x^2)^3/(d + e*x)^5,x]

[Out]

(-5*c^3*d*x)/e^6 + (c^3*x^2)/(2*e^5) - (c*d^2 + a*e^2)^3/(4*e^7*(d + e*x)^4) + (2*c*d*(c*d^2 + a*e^2)^2)/(e^7*
(d + e*x)^3) - (3*c*(c*d^2 + a*e^2)*(5*c*d^2 + a*e^2))/(2*e^7*(d + e*x)^2) + (4*c^2*d*(5*c*d^2 + 3*a*e^2))/(e^
7*(d + e*x)) + (3*c^2*(5*c*d^2 + a*e^2)*Log[d + e*x])/e^7

Rule 711

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {5 c^3 d}{e^6}+\frac {c^3 x}{e^5}+\frac {\left (c d^2+a e^2\right )^3}{e^6 (d+e x)^5}-\frac {6 c d \left (c d^2+a e^2\right )^2}{e^6 (d+e x)^4}+\frac {3 c \left (c d^2+a e^2\right ) \left (5 c d^2+a e^2\right )}{e^6 (d+e x)^3}-\frac {4 c^2 d \left (5 c d^2+3 a e^2\right )}{e^6 (d+e x)^2}+\frac {3 c^2 \left (5 c d^2+a e^2\right )}{e^6 (d+e x)}\right ) \, dx \\ & = -\frac {5 c^3 d x}{e^6}+\frac {c^3 x^2}{2 e^5}-\frac {\left (c d^2+a e^2\right )^3}{4 e^7 (d+e x)^4}+\frac {2 c d \left (c d^2+a e^2\right )^2}{e^7 (d+e x)^3}-\frac {3 c \left (c d^2+a e^2\right ) \left (5 c d^2+a e^2\right )}{2 e^7 (d+e x)^2}+\frac {4 c^2 d \left (5 c d^2+3 a e^2\right )}{e^7 (d+e x)}+\frac {3 c^2 \left (5 c d^2+a e^2\right ) \log (d+e x)}{e^7} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^5} \, dx=\frac {-a^3 e^6-a^2 c e^4 \left (d^2+4 d e x+6 e^2 x^2\right )+a c^2 d e^2 \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )+c^3 \left (57 d^6+168 d^5 e x+132 d^4 e^2 x^2-32 d^3 e^3 x^3-68 d^2 e^4 x^4-12 d e^5 x^5+2 e^6 x^6\right )+12 c^2 \left (5 c d^2+a e^2\right ) (d+e x)^4 \log (d+e x)}{4 e^7 (d+e x)^4} \]

[In]

Integrate[(a + c*x^2)^3/(d + e*x)^5,x]

[Out]

(-(a^3*e^6) - a^2*c*e^4*(d^2 + 4*d*e*x + 6*e^2*x^2) + a*c^2*d*e^2*(25*d^3 + 88*d^2*e*x + 108*d*e^2*x^2 + 48*e^
3*x^3) + c^3*(57*d^6 + 168*d^5*e*x + 132*d^4*e^2*x^2 - 32*d^3*e^3*x^3 - 68*d^2*e^4*x^4 - 12*d*e^5*x^5 + 2*e^6*
x^6) + 12*c^2*(5*c*d^2 + a*e^2)*(d + e*x)^4*Log[d + e*x])/(4*e^7*(d + e*x)^4)

Maple [A] (verified)

Time = 2.45 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.19

method	result	size
risch	\(\frac {c^{3} x^{2}}{2 e^{5}}-\frac {5 c^{3} d x}{e^{6}}+\frac {\left (12 d \,e^{4} c^{2} a +20 d^{3} e^{2} c^{3}\right ) x^{3}-\frac {3 e c \left (a^{2} e^{4}-18 a c \,d^{2} e^{2}-35 c^{2} d^{4}\right ) x^{2}}{2}+\left (-d \,e^{4} a^{2} c +22 d^{3} e^{2} c^{2} a +47 d^{5} c^{3}\right ) x -\frac {e^{6} a^{3}+d^{2} e^{4} a^{2} c -25 d^{4} e^{2} c^{2} a -57 c^{3} d^{6}}{4 e}}{e^{6} \left (e x +d \right )^{4}}+\frac {3 c^{2} \ln \left (e x +d \right ) a}{e^{5}}+\frac {15 c^{3} \ln \left (e x +d \right ) d^{2}}{e^{7}}\)	\(203\)
norman	\(\frac {-\frac {e^{6} a^{3}+d^{2} e^{4} a^{2} c -25 d^{4} e^{2} c^{2} a -125 c^{3} d^{6}}{4 e^{7}}+\frac {c^{3} x^{6}}{2 e}-\frac {3 \left (e^{4} a^{2} c -18 d^{2} e^{2} c^{2} a -90 d^{4} c^{3}\right ) x^{2}}{2 e^{5}}-\frac {3 c^{3} d \,x^{5}}{e^{2}}+\frac {4 d \left (3 e^{2} c^{2} a +15 c^{3} d^{2}\right ) x^{3}}{e^{4}}-\frac {d \left (e^{4} a^{2} c -22 d^{2} e^{2} c^{2} a -110 d^{4} c^{3}\right ) x}{e^{6}}}{\left (e x +d \right )^{4}}+\frac {3 c^{2} \left (e^{2} a +5 c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{7}}\)	\(204\)
default	\(-\frac {c^{3} \left (-\frac {1}{2} e \,x^{2}+5 d x \right )}{e^{6}}+\frac {4 c^{2} d \left (3 e^{2} a +5 c \,d^{2}\right )}{e^{7} \left (e x +d \right )}+\frac {2 c d \left (a^{2} e^{4}+2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}{e^{7} \left (e x +d \right )^{3}}-\frac {e^{6} a^{3}+3 d^{2} e^{4} a^{2} c +3 d^{4} e^{2} c^{2} a +c^{3} d^{6}}{4 e^{7} \left (e x +d \right )^{4}}-\frac {3 c \left (a^{2} e^{4}+6 a c \,d^{2} e^{2}+5 c^{2} d^{4}\right )}{2 e^{7} \left (e x +d \right )^{2}}+\frac {3 c^{2} \left (e^{2} a +5 c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{7}}\)	\(205\)
parallelrisch	\(\frac {48 \ln \left (e x +d \right ) x a \,c^{2} d^{3} e^{3}+25 d^{4} e^{2} c^{2} a -d^{2} e^{4} a^{2} c +240 x^{3} c^{3} d^{3} e^{3}+540 x^{2} c^{3} d^{4} e^{2}+440 x \,c^{3} d^{5} e +125 c^{3} d^{6}+48 \ln \left (e x +d \right ) x^{3} a \,c^{2} d \,e^{5}+12 \ln \left (e x +d \right ) x^{4} a \,c^{2} e^{6}+60 \ln \left (e x +d \right ) x^{4} c^{3} d^{2} e^{4}-4 x \,a^{2} c d \,e^{5}+88 x a \,c^{2} d^{3} e^{3}+240 \ln \left (e x +d \right ) x \,c^{3} d^{5} e +2 x^{6} c^{3} e^{6}+72 \ln \left (e x +d \right ) x^{2} a \,c^{2} d^{2} e^{4}-e^{6} a^{3}+12 \ln \left (e x +d \right ) a \,c^{2} d^{4} e^{2}+360 \ln \left (e x +d \right ) x^{2} c^{3} d^{4} e^{2}+48 x^{3} a \,c^{2} d \,e^{5}+108 x^{2} a \,c^{2} d^{2} e^{4}+240 \ln \left (e x +d \right ) x^{3} c^{3} d^{3} e^{3}-12 x^{5} c^{3} d \,e^{5}-6 x^{2} a^{2} c \,e^{6}+60 \ln \left (e x +d \right ) c^{3} d^{6}}{4 e^{7} \left (e x +d \right )^{4}}\)	\(364\)

[In]

int((c*x^2+a)^3/(e*x+d)^5,x,method=_RETURNVERBOSE)

[Out]

1/2*c^3*x^2/e^5-5*c^3*d*x/e^6+((12*a*c^2*d*e^4+20*c^3*d^3*e^2)*x^3-3/2*e*c*(a^2*e^4-18*a*c*d^2*e^2-35*c^2*d^4)
*x^2+(-a^2*c*d*e^4+22*a*c^2*d^3*e^2+47*c^3*d^5)*x-1/4*(a^3*e^6+a^2*c*d^2*e^4-25*a*c^2*d^4*e^2-57*c^3*d^6)/e)/e
^6/(e*x+d)^4+3*c^2*ln(e*x+d)/e^5*a+15*c^3*ln(e*x+d)/e^7*d^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 356 vs. \(2 (165) = 330\).

Time = 0.28 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.08 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^5} \, dx=\frac {2 \, c^{3} e^{6} x^{6} - 12 \, c^{3} d e^{5} x^{5} - 68 \, c^{3} d^{2} e^{4} x^{4} + 57 \, c^{3} d^{6} + 25 \, a c^{2} d^{4} e^{2} - a^{2} c d^{2} e^{4} - a^{3} e^{6} - 16 \, {\left (2 \, c^{3} d^{3} e^{3} - 3 \, a c^{2} d e^{5}\right )} x^{3} + 6 \, {\left (22 \, c^{3} d^{4} e^{2} + 18 \, a c^{2} d^{2} e^{4} - a^{2} c e^{6}\right )} x^{2} + 4 \, {\left (42 \, c^{3} d^{5} e + 22 \, a c^{2} d^{3} e^{3} - a^{2} c d e^{5}\right )} x + 12 \, {\left (5 \, c^{3} d^{6} + a c^{2} d^{4} e^{2} + {\left (5 \, c^{3} d^{2} e^{4} + a c^{2} e^{6}\right )} x^{4} + 4 \, {\left (5 \, c^{3} d^{3} e^{3} + a c^{2} d e^{5}\right )} x^{3} + 6 \, {\left (5 \, c^{3} d^{4} e^{2} + a c^{2} d^{2} e^{4}\right )} x^{2} + 4 \, {\left (5 \, c^{3} d^{5} e + a c^{2} d^{3} e^{3}\right )} x\right )} \log \left (e x + d\right )}{4 \, {\left (e^{11} x^{4} + 4 \, d e^{10} x^{3} + 6 \, d^{2} e^{9} x^{2} + 4 \, d^{3} e^{8} x + d^{4} e^{7}\right )}} \]

[In]

integrate((c*x^2+a)^3/(e*x+d)^5,x, algorithm="fricas")

[Out]

1/4*(2*c^3*e^6*x^6 - 12*c^3*d*e^5*x^5 - 68*c^3*d^2*e^4*x^4 + 57*c^3*d^6 + 25*a*c^2*d^4*e^2 - a^2*c*d^2*e^4 - a
^3*e^6 - 16*(2*c^3*d^3*e^3 - 3*a*c^2*d*e^5)*x^3 + 6*(22*c^3*d^4*e^2 + 18*a*c^2*d^2*e^4 - a^2*c*e^6)*x^2 + 4*(4
2*c^3*d^5*e + 22*a*c^2*d^3*e^3 - a^2*c*d*e^5)*x + 12*(5*c^3*d^6 + a*c^2*d^4*e^2 + (5*c^3*d^2*e^4 + a*c^2*e^6)*
x^4 + 4*(5*c^3*d^3*e^3 + a*c^2*d*e^5)*x^3 + 6*(5*c^3*d^4*e^2 + a*c^2*d^2*e^4)*x^2 + 4*(5*c^3*d^5*e + a*c^2*d^3
*e^3)*x)*log(e*x + d))/(e^11*x^4 + 4*d*e^10*x^3 + 6*d^2*e^9*x^2 + 4*d^3*e^8*x + d^4*e^7)

Sympy [A] (verification not implemented)

Time = 1.89 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.42 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^5} \, dx=- \frac {5 c^{3} d x}{e^{6}} + \frac {c^{3} x^{2}}{2 e^{5}} + \frac {3 c^{2} \left (a e^{2} + 5 c d^{2}\right ) \log {\left (d + e x \right )}}{e^{7}} + \frac {- a^{3} e^{6} - a^{2} c d^{2} e^{4} + 25 a c^{2} d^{4} e^{2} + 57 c^{3} d^{6} + x^{3} \cdot \left (48 a c^{2} d e^{5} + 80 c^{3} d^{3} e^{3}\right ) + x^{2} \left (- 6 a^{2} c e^{6} + 108 a c^{2} d^{2} e^{4} + 210 c^{3} d^{4} e^{2}\right ) + x \left (- 4 a^{2} c d e^{5} + 88 a c^{2} d^{3} e^{3} + 188 c^{3} d^{5} e\right )}{4 d^{4} e^{7} + 16 d^{3} e^{8} x + 24 d^{2} e^{9} x^{2} + 16 d e^{10} x^{3} + 4 e^{11} x^{4}} \]

[In]

integrate((c*x**2+a)**3/(e*x+d)**5,x)

[Out]

-5*c**3*d*x/e**6 + c**3*x**2/(2*e**5) + 3*c**2*(a*e**2 + 5*c*d**2)*log(d + e*x)/e**7 + (-a**3*e**6 - a**2*c*d*
*2*e**4 + 25*a*c**2*d**4*e**2 + 57*c**3*d**6 + x**3*(48*a*c**2*d*e**5 + 80*c**3*d**3*e**3) + x**2*(-6*a**2*c*e
**6 + 108*a*c**2*d**2*e**4 + 210*c**3*d**4*e**2) + x*(-4*a**2*c*d*e**5 + 88*a*c**2*d**3*e**3 + 188*c**3*d**5*e
))/(4*d**4*e**7 + 16*d**3*e**8*x + 24*d**2*e**9*x**2 + 16*d*e**10*x**3 + 4*e**11*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.40 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^5} \, dx=\frac {57 \, c^{3} d^{6} + 25 \, a c^{2} d^{4} e^{2} - a^{2} c d^{2} e^{4} - a^{3} e^{6} + 16 \, {\left (5 \, c^{3} d^{3} e^{3} + 3 \, a c^{2} d e^{5}\right )} x^{3} + 6 \, {\left (35 \, c^{3} d^{4} e^{2} + 18 \, a c^{2} d^{2} e^{4} - a^{2} c e^{6}\right )} x^{2} + 4 \, {\left (47 \, c^{3} d^{5} e + 22 \, a c^{2} d^{3} e^{3} - a^{2} c d e^{5}\right )} x}{4 \, {\left (e^{11} x^{4} + 4 \, d e^{10} x^{3} + 6 \, d^{2} e^{9} x^{2} + 4 \, d^{3} e^{8} x + d^{4} e^{7}\right )}} + \frac {c^{3} e x^{2} - 10 \, c^{3} d x}{2 \, e^{6}} + \frac {3 \, {\left (5 \, c^{3} d^{2} + a c^{2} e^{2}\right )} \log \left (e x + d\right )}{e^{7}} \]

[In]

integrate((c*x^2+a)^3/(e*x+d)^5,x, algorithm="maxima")

[Out]

1/4*(57*c^3*d^6 + 25*a*c^2*d^4*e^2 - a^2*c*d^2*e^4 - a^3*e^6 + 16*(5*c^3*d^3*e^3 + 3*a*c^2*d*e^5)*x^3 + 6*(35*
c^3*d^4*e^2 + 18*a*c^2*d^2*e^4 - a^2*c*e^6)*x^2 + 4*(47*c^3*d^5*e + 22*a*c^2*d^3*e^3 - a^2*c*d*e^5)*x)/(e^11*x
^4 + 4*d*e^10*x^3 + 6*d^2*e^9*x^2 + 4*d^3*e^8*x + d^4*e^7) + 1/2*(c^3*e*x^2 - 10*c^3*d*x)/e^6 + 3*(5*c^3*d^2 +
a*c^2*e^2)*log(e*x + d)/e^7

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.70 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^5} \, dx=\frac {{\left (c^{3} - \frac {12 \, c^{3} d}{e x + d}\right )} {\left (e x + d\right )}^{2}}{2 \, e^{7}} - \frac {3 \, {\left (5 \, c^{3} d^{2} + a c^{2} e^{2}\right )} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{7}} + \frac {\frac {80 \, c^{3} d^{3} e^{29}}{e x + d} - \frac {30 \, c^{3} d^{4} e^{29}}{{\left (e x + d\right )}^{2}} + \frac {8 \, c^{3} d^{5} e^{29}}{{\left (e x + d\right )}^{3}} - \frac {c^{3} d^{6} e^{29}}{{\left (e x + d\right )}^{4}} + \frac {48 \, a c^{2} d e^{31}}{e x + d} - \frac {36 \, a c^{2} d^{2} e^{31}}{{\left (e x + d\right )}^{2}} + \frac {16 \, a c^{2} d^{3} e^{31}}{{\left (e x + d\right )}^{3}} - \frac {3 \, a c^{2} d^{4} e^{31}}{{\left (e x + d\right )}^{4}} - \frac {6 \, a^{2} c e^{33}}{{\left (e x + d\right )}^{2}} + \frac {8 \, a^{2} c d e^{33}}{{\left (e x + d\right )}^{3}} - \frac {3 \, a^{2} c d^{2} e^{33}}{{\left (e x + d\right )}^{4}} - \frac {a^{3} e^{35}}{{\left (e x + d\right )}^{4}}}{4 \, e^{36}} \]

[In]

integrate((c*x^2+a)^3/(e*x+d)^5,x, algorithm="giac")

[Out]

1/2*(c^3 - 12*c^3*d/(e*x + d))*(e*x + d)^2/e^7 - 3*(5*c^3*d^2 + a*c^2*e^2)*log(abs(e*x + d)/((e*x + d)^2*abs(e
)))/e^7 + 1/4*(80*c^3*d^3*e^29/(e*x + d) - 30*c^3*d^4*e^29/(e*x + d)^2 + 8*c^3*d^5*e^29/(e*x + d)^3 - c^3*d^6*
e^29/(e*x + d)^4 + 48*a*c^2*d*e^31/(e*x + d) - 36*a*c^2*d^2*e^31/(e*x + d)^2 + 16*a*c^2*d^3*e^31/(e*x + d)^3 -
3*a*c^2*d^4*e^31/(e*x + d)^4 - 6*a^2*c*e^33/(e*x + d)^2 + 8*a^2*c*d*e^33/(e*x + d)^3 - 3*a^2*c*d^2*e^33/(e*x
+ d)^4 - a^3*e^35/(e*x + d)^4)/e^36

Mupad [B] (verification not implemented)

Time = 9.48 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.38 \[ \int \frac {\left (a+c x^2\right )^3}{(d+e x)^5} \, dx=\frac {x^2\,\left (-\frac {3\,a^2\,c\,e^5}{2}+27\,a\,c^2\,d^2\,e^3+\frac {105\,c^3\,d^4\,e}{2}\right )-\frac {a^3\,e^6+a^2\,c\,d^2\,e^4-25\,a\,c^2\,d^4\,e^2-57\,c^3\,d^6}{4\,e}+x\,\left (-a^2\,c\,d\,e^4+22\,a\,c^2\,d^3\,e^2+47\,c^3\,d^5\right )+x^3\,\left (20\,c^3\,d^3\,e^2+12\,a\,c^2\,d\,e^4\right )}{d^4\,e^6+4\,d^3\,e^7\,x+6\,d^2\,e^8\,x^2+4\,d\,e^9\,x^3+e^{10}\,x^4}+\frac {\ln \left (d+e\,x\right )\,\left (15\,c^3\,d^2+3\,a\,c^2\,e^2\right )}{e^7}+\frac {c^3\,x^2}{2\,e^5}-\frac {5\,c^3\,d\,x}{e^6} \]

[In]

int((a + c*x^2)^3/(d + e*x)^5,x)

[Out]

(x^2*((105*c^3*d^4*e)/2 - (3*a^2*c*e^5)/2 + 27*a*c^2*d^2*e^3) - (a^3*e^6 - 57*c^3*d^6 - 25*a*c^2*d^4*e^2 + a^2
*c*d^2*e^4)/(4*e) + x*(47*c^3*d^5 + 22*a*c^2*d^3*e^2 - a^2*c*d*e^4) + x^3*(20*c^3*d^3*e^2 + 12*a*c^2*d*e^4))/(
d^4*e^6 + e^10*x^4 + 4*d^3*e^7*x + 4*d*e^9*x^3 + 6*d^2*e^8*x^2) + (log(d + e*x)*(15*c^3*d^2 + 3*a*c^2*e^2))/e^
7 + (c^3*x^2)/(2*e^5) - (5*c^3*d*x)/e^6