\(\int \frac {(c d^2-c e^2 x^2)^{3/2}}{(d+e x)^{13/2}} \, dx\) [181]

Optimal result

Integrand size = 29, antiderivative size = 217 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{13/2}} \, dx=\frac {c \sqrt {c d^2-c e^2 x^2}}{8 e (d+e x)^{7/2}}-\frac {c \sqrt {c d^2-c e^2 x^2}}{64 d e (d+e x)^{5/2}}-\frac {3 c \sqrt {c d^2-c e^2 x^2}}{256 d^2 e (d+e x)^{3/2}}-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}}-\frac {3 c^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-c e^2 x^2}}\right )}{256 \sqrt {2} d^{5/2} e} \] Output:

1/8*c*(-c*e^2*x^2+c*d^2)^(1/2)/e/(e*x+d)^(7/2)-1/64*c*(-c*e^2*x^2+c*d^2)^( 
1/2)/d/e/(e*x+d)^(5/2)-3/256*c*(-c*e^2*x^2+c*d^2)^(1/2)/d^2/e/(e*x+d)^(3/2 
)-1/4*(-c*e^2*x^2+c*d^2)^(3/2)/e/(e*x+d)^(11/2)-3/512*c^(3/2)*arctanh(2^(1 
/2)*c^(1/2)*d^(1/2)*(e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(1/2))*2^(1/2)/d^(5/2 
)/e
 

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.71 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{13/2}} \, dx=\frac {\left (c \left (d^2-e^2 x^2\right )\right )^{3/2} \left (-\frac {2 \sqrt {d} \sqrt {d^2-e^2 x^2} \left (39 d^3-79 d^2 e x+13 d e^2 x^2+3 e^3 x^3\right )}{(d+e x)^{9/2}}-3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )\right )}{512 d^{5/2} e \left (d^2-e^2 x^2\right )^{3/2}} \] Input:

Integrate[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(13/2),x]
 

Output:

((c*(d^2 - e^2*x^2))^(3/2)*((-2*Sqrt[d]*Sqrt[d^2 - e^2*x^2]*(39*d^3 - 79*d 
^2*e*x + 13*d*e^2*x^2 + 3*e^3*x^3))/(d + e*x)^(9/2) - 3*Sqrt[2]*ArcTanh[(S 
qrt[2]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[d^2 - e^2*x^2]]))/(512*d^(5/2)*e*(d^2 - 
 e^2*x^2)^(3/2))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {465, 465, 470, 470, 471, 221}

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.

\(\displaystyle \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{13/2}} \, dx\)

\(\Big \downarrow \) 465

\(\displaystyle -\frac {3}{8} c \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{9/2}}dx-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle -\frac {3}{8} c \left (-\frac {1}{6} c \int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}}dx-\frac {\sqrt {c d^2-c e^2 x^2}}{3 e (d+e x)^{7/2}}\right )-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle -\frac {3}{8} c \left (-\frac {1}{6} c \left (\frac {3 \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}dx}{8 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}}\right )-\frac {\sqrt {c d^2-c e^2 x^2}}{3 e (d+e x)^{7/2}}\right )-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle -\frac {3}{8} c \left (-\frac {1}{6} c \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}dx}{4 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}\right )}{8 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}}\right )-\frac {\sqrt {c d^2-c e^2 x^2}}{3 e (d+e x)^{7/2}}\right )-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle -\frac {3}{8} c \left (-\frac {1}{6} c \left (\frac {3 \left (\frac {e \int \frac {1}{\frac {e^2 \left (c d^2-c e^2 x^2\right )}{d+e x}-2 c d e^2}d\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {d+e x}}}{2 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}\right )}{8 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}}\right )-\frac {\sqrt {c d^2-c e^2 x^2}}{3 e (d+e x)^{7/2}}\right )-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {3}{8} c \left (-\frac {1}{6} c \left (\frac {3 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{2 \sqrt {2} \sqrt {c} d^{3/2} e}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}\right )}{8 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}}\right )-\frac {\sqrt {c d^2-c e^2 x^2}}{3 e (d+e x)^{7/2}}\right )-\frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{4 e (d+e x)^{11/2}}\)

Input:

Int[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(13/2),x]
 

Output:

-1/4*(c*d^2 - c*e^2*x^2)^(3/2)/(e*(d + e*x)^(11/2)) - (3*c*(-1/3*Sqrt[c*d^ 
2 - c*e^2*x^2]/(e*(d + e*x)^(7/2)) - (c*(-1/4*Sqrt[c*d^2 - c*e^2*x^2]/(c*d 
*e*(d + e*x)^(5/2)) + (3*(-1/2*Sqrt[c*d^2 - c*e^2*x^2]/(c*d*e*(d + e*x)^(3 
/2)) - ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e 
*x])]/(2*Sqrt[2]*Sqrt[c]*d^(3/2)*e)))/(8*d)))/6))/8
 

Defintions of rubi rules used

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 465

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + p + 1))), x] - Simp[b*(p/(d^2*(n + 
 p + 1)))   Int[(c + d*x)^(n + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, 
b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LtQ[n, -2] || EqQ[n 
+ 2*p + 1, 0]) && NeQ[n + p + 1, 0] && IntegerQ[2*p]
 

rule 470

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 
2*p + 2)/(2*c*(n + p + 1))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; 
 FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + 
 p + 1, 0] && IntegerQ[2*p]
 

rule 471

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim 
p[2*d   Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] 
], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
 

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.40

method	result	size
default	\(-\frac {\sqrt {c \left (-e^{2} x^{2}+d^{2}\right )}\, c \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,e^{4} x^{4}+12 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c d \,e^{3} x^{3}+18 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,d^{2} e^{2} x^{2}+12 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,d^{3} e x +6 e^{3} x^{3} \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}+3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,d^{4}+26 d \,e^{2} x^{2} \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}-158 d^{2} e x \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}+78 \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}\, d^{3}\right )}{512 \left (e x +d \right )^{\frac {9}{2}} \sqrt {c \left (-e x +d \right )}\, e \,d^{2} \sqrt {c d}}\)	\(303\)

Input:

int((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(13/2),x,method=_RETURNVERBOSE)
 

Output:

-1/512*(c*(-e^2*x^2+d^2))^(1/2)*c*(3*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2 
)*2^(1/2)/(c*d)^(1/2))*c*e^4*x^4+12*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2) 
*2^(1/2)/(c*d)^(1/2))*c*d*e^3*x^3+18*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2 
)*2^(1/2)/(c*d)^(1/2))*c*d^2*e^2*x^2+12*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^( 
1/2)*2^(1/2)/(c*d)^(1/2))*c*d^3*e*x+6*e^3*x^3*(c*(-e*x+d))^(1/2)*(c*d)^(1/ 
2)+3*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^(1/2)/(c*d)^(1/2))*c*d^4+26* 
d*e^2*x^2*(c*(-e*x+d))^(1/2)*(c*d)^(1/2)-158*d^2*e*x*(c*(-e*x+d))^(1/2)*(c 
*d)^(1/2)+78*(c*(-e*x+d))^(1/2)*(c*d)^(1/2)*d^3)/(e*x+d)^(9/2)/(c*(-e*x+d) 
)^(1/2)/e/d^2/(c*d)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 518, normalized size of antiderivative = 2.39 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{13/2}} \, dx=\left [\frac {3 \, \sqrt {\frac {1}{2}} {\left (c e^{5} x^{5} + 5 \, c d e^{4} x^{4} + 10 \, c d^{2} e^{3} x^{3} + 10 \, c d^{3} e^{2} x^{2} + 5 \, c d^{4} e x + c d^{5}\right )} \sqrt {\frac {c}{d}} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 4 \, \sqrt {\frac {1}{2}} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d \sqrt {\frac {c}{d}}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 2 \, {\left (3 \, c e^{3} x^{3} + 13 \, c d e^{2} x^{2} - 79 \, c d^{2} e x + 39 \, c d^{3}\right )} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}}{512 \, {\left (d^{2} e^{6} x^{5} + 5 \, d^{3} e^{5} x^{4} + 10 \, d^{4} e^{4} x^{3} + 10 \, d^{5} e^{3} x^{2} + 5 \, d^{6} e^{2} x + d^{7} e\right )}}, -\frac {3 \, \sqrt {\frac {1}{2}} {\left (c e^{5} x^{5} + 5 \, c d e^{4} x^{4} + 10 \, c d^{2} e^{3} x^{3} + 10 \, c d^{3} e^{2} x^{2} + 5 \, c d^{4} e x + c d^{5}\right )} \sqrt {-\frac {c}{d}} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d \sqrt {-\frac {c}{d}}}{c e^{2} x^{2} - c d^{2}}\right ) + {\left (3 \, c e^{3} x^{3} + 13 \, c d e^{2} x^{2} - 79 \, c d^{2} e x + 39 \, c d^{3}\right )} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}}{256 \, {\left (d^{2} e^{6} x^{5} + 5 \, d^{3} e^{5} x^{4} + 10 \, d^{4} e^{4} x^{3} + 10 \, d^{5} e^{3} x^{2} + 5 \, d^{6} e^{2} x + d^{7} e\right )}}\right ] \] Input:

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(13/2),x, algorithm="fricas")
 

Output:

[1/512*(3*sqrt(1/2)*(c*e^5*x^5 + 5*c*d*e^4*x^4 + 10*c*d^2*e^3*x^3 + 10*c*d 
^3*e^2*x^2 + 5*c*d^4*e*x + c*d^5)*sqrt(c/d)*log(-(c*e^2*x^2 - 2*c*d*e*x - 
3*c*d^2 + 4*sqrt(1/2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)*d*sqrt(c/d))/ 
(e^2*x^2 + 2*d*e*x + d^2)) - 2*(3*c*e^3*x^3 + 13*c*d*e^2*x^2 - 79*c*d^2*e* 
x + 39*c*d^3)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d))/(d^2*e^6*x^5 + 5*d^3 
*e^5*x^4 + 10*d^4*e^4*x^3 + 10*d^5*e^3*x^2 + 5*d^6*e^2*x + d^7*e), -1/256* 
(3*sqrt(1/2)*(c*e^5*x^5 + 5*c*d*e^4*x^4 + 10*c*d^2*e^3*x^3 + 10*c*d^3*e^2* 
x^2 + 5*c*d^4*e*x + c*d^5)*sqrt(-c/d)*arctan(2*sqrt(1/2)*sqrt(-c*e^2*x^2 + 
 c*d^2)*sqrt(e*x + d)*d*sqrt(-c/d)/(c*e^2*x^2 - c*d^2)) + (3*c*e^3*x^3 + 1 
3*c*d*e^2*x^2 - 79*c*d^2*e*x + 39*c*d^3)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x 
 + d))/(d^2*e^6*x^5 + 5*d^3*e^5*x^4 + 10*d^4*e^4*x^3 + 10*d^5*e^3*x^2 + 5* 
d^6*e^2*x + d^7*e)]
 

Sympy [F]

\[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{13/2}} \, dx=\int \frac {\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {13}{2}}}\, dx \] Input:

integrate((-c*e**2*x**2+c*d**2)**(3/2)/(e*x+d)**(13/2),x)
 

Output:

Integral((-c*(-d + e*x)*(d + e*x))**(3/2)/(d + e*x)**(13/2), x)
 

Maxima [F]

\[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{13/2}} \, dx=\int { \frac {{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac {3}{2}}}{{\left (e x + d\right )}^{\frac {13}{2}}} \,d x } \] Input:

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(13/2),x, algorithm="maxima")
 

Output:

integrate((-c*e^2*x^2 + c*d^2)^(3/2)/(e*x + d)^(13/2), x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.83 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{13/2}} \, dx=\frac {\frac {3 \, \sqrt {2} c^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{\sqrt {-c d} d^{2}} + \frac {2 \, {\left (24 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c^{5} d^{3} - 44 \, {\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}} c^{4} d^{2} - 22 \, {\left ({\left (e x + d\right )} c - 2 \, c d\right )}^{2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c^{3} d - 3 \, {\left ({\left (e x + d\right )} c - 2 \, c d\right )}^{3} \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c^{2}\right )}}{{\left (e x + d\right )}^{4} c^{4} d^{2}}}{512 \, e} \] Input:

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(13/2),x, algorithm="giac")
 

Output:

1/512*(3*sqrt(2)*c^2*arctan(1/2*sqrt(2)*sqrt(-(e*x + d)*c + 2*c*d)/sqrt(-c 
*d))/(sqrt(-c*d)*d^2) + 2*(24*sqrt(-(e*x + d)*c + 2*c*d)*c^5*d^3 - 44*(-(e 
*x + d)*c + 2*c*d)^(3/2)*c^4*d^2 - 22*((e*x + d)*c - 2*c*d)^2*sqrt(-(e*x + 
 d)*c + 2*c*d)*c^3*d - 3*((e*x + d)*c - 2*c*d)^3*sqrt(-(e*x + d)*c + 2*c*d 
)*c^2)/((e*x + d)^4*c^4*d^2))/e
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{13/2}} \, dx=\int \frac {{\left (c\,d^2-c\,e^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^{13/2}} \,d x \] Input:

int((c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(13/2),x)
 

Output:

int((c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(13/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.68 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{13/2}} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, \sqrt {2}\, c \left (48 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +d}}{\sqrt {d}\, \sqrt {2}}\right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +d}}{\sqrt {d}\, \sqrt {2}}\right )}{2}\right )^{8}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +d}}{\sqrt {d}\, \sqrt {2}}\right )}{2}\right )^{16}-8 \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +d}}{\sqrt {d}\, \sqrt {2}}\right )}{2}\right )^{12}+8 \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +d}}{\sqrt {d}\, \sqrt {2}}\right )}{2}\right )^{4}-1\right )}{8192 \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +d}}{\sqrt {d}\, \sqrt {2}}\right )}{2}\right )^{8} d^{3} e} \] Input:

int((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(13/2),x)
 

Output:

(sqrt(d)*sqrt(c)*sqrt(2)*c*(48*log(tan(asin(sqrt(d + e*x)/(sqrt(d)*sqrt(2) 
))/2))*tan(asin(sqrt(d + e*x)/(sqrt(d)*sqrt(2)))/2)**8 + tan(asin(sqrt(d + 
 e*x)/(sqrt(d)*sqrt(2)))/2)**16 - 8*tan(asin(sqrt(d + e*x)/(sqrt(d)*sqrt(2 
)))/2)**12 + 8*tan(asin(sqrt(d + e*x)/(sqrt(d)*sqrt(2)))/2)**4 - 1))/(8192 
*tan(asin(sqrt(d + e*x)/(sqrt(d)*sqrt(2)))/2)**8*d**3*e)