\(\int x^3 (a+b \sec (c+d \sqrt {x}))^2 \, dx\) [36]

Optimal result

Integrand size = 20, antiderivative size = 749 \[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx =\text {Too large to display} \] Output:

315/2*b^2*polylog(7,-exp(2*I*(c+d*x^(1/2))))/d^8-315*b^2*x*polylog(5,-exp( 
2*I*(c+d*x^(1/2))))/d^6+105*b^2*x^2*polylog(3,-exp(2*I*(c+d*x^(1/2))))/d^4 
+14*b^2*x^3*ln(1+exp(2*I*(c+d*x^(1/2))))/d^2+2*b^2*x^(7/2)*tan(c+d*x^(1/2) 
)/d-2*I*b^2*x^(7/2)/d-10080*I*a*b*x*polylog(6,I*exp(I*(c+d*x^(1/2))))/d^6- 
840*I*a*b*x^2*polylog(4,-I*exp(I*(c+d*x^(1/2))))/d^4-28*I*a*b*x^3*polylog( 
2,I*exp(I*(c+d*x^(1/2))))/d^2-8*I*a*b*x^(7/2)*arctan(exp(I*(c+d*x^(1/2)))) 
/d+10080*I*a*b*x*polylog(6,-I*exp(I*(c+d*x^(1/2))))/d^6+840*I*a*b*x^2*poly 
log(4,I*exp(I*(c+d*x^(1/2))))/d^4+28*I*a*b*x^3*polylog(2,-I*exp(I*(c+d*x^( 
1/2))))/d^2+20160*I*a*b*polylog(8,I*exp(I*(c+d*x^(1/2))))/d^8+210*I*b^2*x^ 
(3/2)*polylog(4,-exp(2*I*(c+d*x^(1/2))))/d^5+20160*a*b*x^(1/2)*polylog(7,I 
*exp(I*(c+d*x^(1/2))))/d^7-20160*a*b*x^(1/2)*polylog(7,-I*exp(I*(c+d*x^(1/ 
2))))/d^7-3360*a*b*x^(3/2)*polylog(5,I*exp(I*(c+d*x^(1/2))))/d^5+3360*a*b* 
x^(3/2)*polylog(5,-I*exp(I*(c+d*x^(1/2))))/d^5+168*a*b*x^(5/2)*polylog(3,I 
*exp(I*(c+d*x^(1/2))))/d^3-168*a*b*x^(5/2)*polylog(3,-I*exp(I*(c+d*x^(1/2) 
)))/d^3-42*I*b^2*x^(5/2)*polylog(2,-exp(2*I*(c+d*x^(1/2))))/d^3-315*I*b^2* 
x^(1/2)*polylog(6,-exp(2*I*(c+d*x^(1/2))))/d^7-20160*I*a*b*polylog(8,-I*ex 
p(I*(c+d*x^(1/2))))/d^8+1/4*a^2*x^4
 

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 739, normalized size of antiderivative = 0.99 \[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx =\text {Too large to display} \] Input:

Integrate[x^3*(a + b*Sec[c + d*Sqrt[x]])^2,x]
 

Output:

((-8*I)*b^2*d^7*x^(7/2) + a^2*d^8*x^4 - (32*I)*a*b*d^7*x^(7/2)*ArcTan[E^(I 
*(c + d*Sqrt[x]))] + 56*b^2*d^6*x^3*Log[1 + E^((2*I)*(c + d*Sqrt[x]))] + ( 
112*I)*a*b*d^6*x^3*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))] - (112*I)*a*b*d^ 
6*x^3*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))] - (168*I)*b^2*d^5*x^(5/2)*PolyLo 
g[2, -E^((2*I)*(c + d*Sqrt[x]))] - 672*a*b*d^5*x^(5/2)*PolyLog[3, (-I)*E^( 
I*(c + d*Sqrt[x]))] + 672*a*b*d^5*x^(5/2)*PolyLog[3, I*E^(I*(c + d*Sqrt[x] 
))] + 420*b^2*d^4*x^2*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))] - (3360*I)*a* 
b*d^4*x^2*PolyLog[4, (-I)*E^(I*(c + d*Sqrt[x]))] + (3360*I)*a*b*d^4*x^2*Po 
lyLog[4, I*E^(I*(c + d*Sqrt[x]))] + (840*I)*b^2*d^3*x^(3/2)*PolyLog[4, -E^ 
((2*I)*(c + d*Sqrt[x]))] + 13440*a*b*d^3*x^(3/2)*PolyLog[5, (-I)*E^(I*(c + 
 d*Sqrt[x]))] - 13440*a*b*d^3*x^(3/2)*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))] 
- 1260*b^2*d^2*x*PolyLog[5, -E^((2*I)*(c + d*Sqrt[x]))] + (40320*I)*a*b*d^ 
2*x*PolyLog[6, (-I)*E^(I*(c + d*Sqrt[x]))] - (40320*I)*a*b*d^2*x*PolyLog[6 
, I*E^(I*(c + d*Sqrt[x]))] - (1260*I)*b^2*d*Sqrt[x]*PolyLog[6, -E^((2*I)*( 
c + d*Sqrt[x]))] - 80640*a*b*d*Sqrt[x]*PolyLog[7, (-I)*E^(I*(c + d*Sqrt[x] 
))] + 80640*a*b*d*Sqrt[x]*PolyLog[7, I*E^(I*(c + d*Sqrt[x]))] + 630*b^2*Po 
lyLog[7, -E^((2*I)*(c + d*Sqrt[x]))] - (80640*I)*a*b*PolyLog[8, (-I)*E^(I* 
(c + d*Sqrt[x]))] + (80640*I)*a*b*PolyLog[8, I*E^(I*(c + d*Sqrt[x]))] + 8* 
b^2*d^7*x^(7/2)*Tan[c + d*Sqrt[x]])/(4*d^8)
 

Rubi [A] (verified)

Time = 1.13 (sec) , antiderivative size = 756, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4692, 3042, 4678, 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.

\(\displaystyle \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx\)

\(\Big \downarrow \) 4692

\(\displaystyle 2 \int x^{7/2} \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2d\sqrt {x}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int x^{7/2} \left (a+b \csc \left (c+d \sqrt {x}+\frac {\pi }{2}\right )\right )^2d\sqrt {x}\)

\(\Big \downarrow \) 4678

\(\displaystyle 2 \int \left (a^2 x^{7/2}+b^2 \sec ^2\left (c+d \sqrt {x}\right ) x^{7/2}+2 a b \sec \left (c+d \sqrt {x}\right ) x^{7/2}\right )d\sqrt {x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {a^2 x^4}{8}-\frac {4 i a b x^{7/2} \arctan \left (e^{i \left (c+d \sqrt {x}\right )}\right )}{d}-\frac {10080 i a b \operatorname {PolyLog}\left (8,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}+\frac {10080 i a b \operatorname {PolyLog}\left (8,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^8}-\frac {10080 a b \sqrt {x} \operatorname {PolyLog}\left (7,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {10080 a b \sqrt {x} \operatorname {PolyLog}\left (7,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^7}+\frac {5040 i a b x \operatorname {PolyLog}\left (6,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}-\frac {5040 i a b x \operatorname {PolyLog}\left (6,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^6}+\frac {1680 a b x^{3/2} \operatorname {PolyLog}\left (5,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {1680 a b x^{3/2} \operatorname {PolyLog}\left (5,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^5}-\frac {420 i a b x^2 \operatorname {PolyLog}\left (4,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}+\frac {420 i a b x^2 \operatorname {PolyLog}\left (4,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^4}-\frac {84 a b x^{5/2} \operatorname {PolyLog}\left (3,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {84 a b x^{5/2} \operatorname {PolyLog}\left (3,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {14 i a b x^3 \operatorname {PolyLog}\left (2,-i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}-\frac {14 i a b x^3 \operatorname {PolyLog}\left (2,i e^{i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {315 b^2 \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{4 d^8}-\frac {315 i b^2 \sqrt {x} \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^7}-\frac {315 b^2 x \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^6}+\frac {105 i b^2 x^{3/2} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^5}+\frac {105 b^2 x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{2 d^4}-\frac {21 i b^2 x^{5/2} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^3}+\frac {7 b^2 x^3 \log \left (1+e^{2 i \left (c+d \sqrt {x}\right )}\right )}{d^2}+\frac {b^2 x^{7/2} \tan \left (c+d \sqrt {x}\right )}{d}-\frac {i b^2 x^{7/2}}{d}\right )\)

Input:

Int[x^3*(a + b*Sec[c + d*Sqrt[x]])^2,x]
 

Output:

2*(((-I)*b^2*x^(7/2))/d + (a^2*x^4)/8 - ((4*I)*a*b*x^(7/2)*ArcTan[E^(I*(c 
+ d*Sqrt[x]))])/d + (7*b^2*x^3*Log[1 + E^((2*I)*(c + d*Sqrt[x]))])/d^2 + ( 
(14*I)*a*b*x^3*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))])/d^2 - ((14*I)*a*b*x 
^3*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - ((21*I)*b^2*x^(5/2)*PolyLog[ 
2, -E^((2*I)*(c + d*Sqrt[x]))])/d^3 - (84*a*b*x^(5/2)*PolyLog[3, (-I)*E^(I 
*(c + d*Sqrt[x]))])/d^3 + (84*a*b*x^(5/2)*PolyLog[3, I*E^(I*(c + d*Sqrt[x] 
))])/d^3 + (105*b^2*x^2*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/(2*d^4) - 
((420*I)*a*b*x^2*PolyLog[4, (-I)*E^(I*(c + d*Sqrt[x]))])/d^4 + ((420*I)*a* 
b*x^2*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + ((105*I)*b^2*x^(3/2)*Poly 
Log[4, -E^((2*I)*(c + d*Sqrt[x]))])/d^5 + (1680*a*b*x^(3/2)*PolyLog[5, (-I 
)*E^(I*(c + d*Sqrt[x]))])/d^5 - (1680*a*b*x^(3/2)*PolyLog[5, I*E^(I*(c + d 
*Sqrt[x]))])/d^5 - (315*b^2*x*PolyLog[5, -E^((2*I)*(c + d*Sqrt[x]))])/(2*d 
^6) + ((5040*I)*a*b*x*PolyLog[6, (-I)*E^(I*(c + d*Sqrt[x]))])/d^6 - ((5040 
*I)*a*b*x*PolyLog[6, I*E^(I*(c + d*Sqrt[x]))])/d^6 - (((315*I)/2)*b^2*Sqrt 
[x]*PolyLog[6, -E^((2*I)*(c + d*Sqrt[x]))])/d^7 - (10080*a*b*Sqrt[x]*PolyL 
og[7, (-I)*E^(I*(c + d*Sqrt[x]))])/d^7 + (10080*a*b*Sqrt[x]*PolyLog[7, I*E 
^(I*(c + d*Sqrt[x]))])/d^7 + (315*b^2*PolyLog[7, -E^((2*I)*(c + d*Sqrt[x]) 
)])/(4*d^8) - ((10080*I)*a*b*PolyLog[8, (-I)*E^(I*(c + d*Sqrt[x]))])/d^8 + 
 ((10080*I)*a*b*PolyLog[8, I*E^(I*(c + d*Sqrt[x]))])/d^8 + (b^2*x^(7/2)*Ta 
n[c + d*Sqrt[x]])/d)
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4678

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Csc[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]
 

rule 4692

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sec[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 
 1)/n], 0] && IntegerQ[p]
 

Maple [F]

\[\int x^{3} \left (a +b \sec \left (c +d \sqrt {x}\right )\right )^{2}d x\]

Input:

int(x^3*(a+b*sec(c+d*x^(1/2)))^2,x)
 

Output:

int(x^3*(a+b*sec(c+d*x^(1/2)))^2,x)
 

Fricas [F]

\[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )}^{2} x^{3} \,d x } \] Input:

integrate(x^3*(a+b*sec(c+d*x^(1/2)))^2,x, algorithm="fricas")
 

Output:

integral(b^2*x^3*sec(d*sqrt(x) + c)^2 + 2*a*b*x^3*sec(d*sqrt(x) + c) + a^2 
*x^3, x)
 

Sympy [F]

\[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int x^{3} \left (a + b \sec {\left (c + d \sqrt {x} \right )}\right )^{2}\, dx \] Input:

integrate(x**3*(a+b*sec(c+d*x**(1/2)))**2,x)
 

Output:

Integral(x**3*(a + b*sec(c + d*sqrt(x)))**2, x)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 6347 vs. \(2 (574) = 1148\).

Time = 0.45 (sec) , antiderivative size = 6347, normalized size of antiderivative = 8.47 \[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\text {Too large to display} \] Input:

integrate(x^3*(a+b*sec(c+d*x^(1/2)))^2,x, algorithm="maxima")
 

Output:

1/4*((d*sqrt(x) + c)^8*a^2 - 8*(d*sqrt(x) + c)^7*a^2*c + 28*(d*sqrt(x) + c 
)^6*a^2*c^2 - 56*(d*sqrt(x) + c)^5*a^2*c^3 + 70*(d*sqrt(x) + c)^4*a^2*c^4 
- 56*(d*sqrt(x) + c)^3*a^2*c^5 + 28*(d*sqrt(x) + c)^2*a^2*c^6 - 8*(d*sqrt( 
x) + c)*a^2*c^7 - 16*a*b*c^7*log(sec(d*sqrt(x) + c) + tan(d*sqrt(x) + c)) 
- 8*(60*b^2*c^7 + 60*((d*sqrt(x) + c)^7*a*b - 7*(d*sqrt(x) + c)^6*a*b*c + 
21*(d*sqrt(x) + c)^5*a*b*c^2 - 35*(d*sqrt(x) + c)^4*a*b*c^3 + 35*(d*sqrt(x 
) + c)^3*a*b*c^4 - 21*(d*sqrt(x) + c)^2*a*b*c^5 + 7*(d*sqrt(x) + c)*a*b*c^ 
6 + ((d*sqrt(x) + c)^7*a*b - 7*(d*sqrt(x) + c)^6*a*b*c + 21*(d*sqrt(x) + c 
)^5*a*b*c^2 - 35*(d*sqrt(x) + c)^4*a*b*c^3 + 35*(d*sqrt(x) + c)^3*a*b*c^4 
- 21*(d*sqrt(x) + c)^2*a*b*c^5 + 7*(d*sqrt(x) + c)*a*b*c^6)*cos(2*d*sqrt(x 
) + 2*c) + (I*(d*sqrt(x) + c)^7*a*b - 7*I*(d*sqrt(x) + c)^6*a*b*c + 21*I*( 
d*sqrt(x) + c)^5*a*b*c^2 - 35*I*(d*sqrt(x) + c)^4*a*b*c^3 + 35*I*(d*sqrt(x 
) + c)^3*a*b*c^4 - 21*I*(d*sqrt(x) + c)^2*a*b*c^5 + 7*I*(d*sqrt(x) + c)*a* 
b*c^6)*sin(2*d*sqrt(x) + 2*c))*arctan2(cos(d*sqrt(x) + c), sin(d*sqrt(x) + 
 c) + 1) + 60*((d*sqrt(x) + c)^7*a*b - 7*(d*sqrt(x) + c)^6*a*b*c + 21*(d*s 
qrt(x) + c)^5*a*b*c^2 - 35*(d*sqrt(x) + c)^4*a*b*c^3 + 35*(d*sqrt(x) + c)^ 
3*a*b*c^4 - 21*(d*sqrt(x) + c)^2*a*b*c^5 + 7*(d*sqrt(x) + c)*a*b*c^6 + ((d 
*sqrt(x) + c)^7*a*b - 7*(d*sqrt(x) + c)^6*a*b*c + 21*(d*sqrt(x) + c)^5*a*b 
*c^2 - 35*(d*sqrt(x) + c)^4*a*b*c^3 + 35*(d*sqrt(x) + c)^3*a*b*c^4 - 21*(d 
*sqrt(x) + c)^2*a*b*c^5 + 7*(d*sqrt(x) + c)*a*b*c^6)*cos(2*d*sqrt(x) + ...
 

Giac [F]

\[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int { {\left (b \sec \left (d \sqrt {x} + c\right ) + a\right )}^{2} x^{3} \,d x } \] Input:

integrate(x^3*(a+b*sec(c+d*x^(1/2)))^2,x, algorithm="giac")
 

Output:

integrate((b*sec(d*sqrt(x) + c) + a)^2*x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\int x^3\,{\left (a+\frac {b}{\cos \left (c+d\,\sqrt {x}\right )}\right )}^2 \,d x \] Input:

int(x^3*(a + b/cos(c + d*x^(1/2)))^2,x)
 

Output:

int(x^3*(a + b/cos(c + d*x^(1/2)))^2, x)
 

Reduce [F]

\[ \int x^3 \left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2 \, dx=\text {too large to display} \] Input:

int(x^3*(a+b*sec(c+d*x^(1/2)))^2,x)
 

Output:

(16*sqrt(x)*cos(sqrt(x)*d + c)*tan((sqrt(x)*d + c)/2)*a*b*d**7*x**3 + 6720 
*sqrt(x)*cos(sqrt(x)*d + c)*tan((sqrt(x)*d + c)/2)*a*b*d**5*x**2 - 215040* 
sqrt(x)*cos(sqrt(x)*d + c)*tan((sqrt(x)*d + c)/2)*a*b*d**3*x - 645120*sqrt 
(x)*cos(sqrt(x)*d + c)*tan((sqrt(x)*d + c)/2)*a*b*d - 16*sqrt(x)*cos(sqrt( 
x)*d + c)*tan((sqrt(x)*d + c)/2)*b**2*d**7*x**3 - 6720*sqrt(x)*cos(sqrt(x) 
*d + c)*tan((sqrt(x)*d + c)/2)*b**2*d**5*x**2 + 215040*sqrt(x)*cos(sqrt(x) 
*d + c)*tan((sqrt(x)*d + c)/2)*b**2*d**3*x + 645120*sqrt(x)*cos(sqrt(x)*d 
+ c)*tan((sqrt(x)*d + c)/2)*b**2*d + 16*cos(sqrt(x)*d + c)*int(x**3/(tan(( 
sqrt(x)*d + c)/2)**4 - 2*tan((sqrt(x)*d + c)/2)**2 + 1),x)*tan((sqrt(x)*d 
+ c)/2)**2*a*b*d**8 - 16*cos(sqrt(x)*d + c)*int(x**3/(tan((sqrt(x)*d + c)/ 
2)**4 - 2*tan((sqrt(x)*d + c)/2)**2 + 1),x)*a*b*d**8 + 3360*cos(sqrt(x)*d 
+ c)*int(x**2/(tan((sqrt(x)*d + c)/2)**4 - 2*tan((sqrt(x)*d + c)/2)**2 + 1 
),x)*tan((sqrt(x)*d + c)/2)**2*a*b*d**6 - 3360*cos(sqrt(x)*d + c)*int(x**2 
/(tan((sqrt(x)*d + c)/2)**4 - 2*tan((sqrt(x)*d + c)/2)**2 + 1),x)*tan((sqr 
t(x)*d + c)/2)**2*b**2*d**6 - 3360*cos(sqrt(x)*d + c)*int(x**2/(tan((sqrt( 
x)*d + c)/2)**4 - 2*tan((sqrt(x)*d + c)/2)**2 + 1),x)*a*b*d**6 + 3360*cos( 
sqrt(x)*d + c)*int(x**2/(tan((sqrt(x)*d + c)/2)**4 - 2*tan((sqrt(x)*d + c) 
/2)**2 + 1),x)*b**2*d**6 + 2688*cos(sqrt(x)*d + c)*int((tan((sqrt(x)*d + c 
)/2)**4*x**2)/(tan((sqrt(x)*d + c)/2)**4 - 2*tan((sqrt(x)*d + c)/2)**2 + 1 
),x)*tan((sqrt(x)*d + c)/2)**2*a*b*d**6 - 2688*cos(sqrt(x)*d + c)*int((...