\(\int \frac {x}{(a+b \sec (c+d \sqrt {x}))^2} \, dx\) [48]

Optimal result

Integrand size = 18, antiderivative size = 1523 \[ \int \frac {x}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2} \, dx =\text {Too large to display} \] Output:

6*b^2*x*ln(1+a*exp(I*(c+d*x^(1/2)))/(b+I*(a^2-b^2)^(1/2)))/a^2/(a^2-b^2)/d 
^2+6*b^2*x*ln(1+a*exp(I*(c+d*x^(1/2)))/(b-I*(a^2-b^2)^(1/2)))/a^2/(a^2-b^2 
)/d^2-12*b*x*polylog(2,-a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a^2/( 
-a^2+b^2)^(1/2)/d^2+12*b*x*polylog(2,-a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2) 
^(1/2)))/a^2/(-a^2+b^2)^(1/2)/d^2+6*b^3*x*polylog(2,-a*exp(I*(c+d*x^(1/2)) 
)/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d^2-6*b^3*x*polylog(2,-a*exp( 
I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d^2-2*I*b^2*x^ 
(3/2)/a^2/(a^2-b^2)/d+1/2*x^2/a^2+12*b^2*polylog(3,-a*exp(I*(c+d*x^(1/2))) 
/(b-I*(a^2-b^2)^(1/2)))/a^2/(a^2-b^2)/d^4-12*b^3*polylog(4,-a*exp(I*(c+d*x 
^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d^4+12*b^3*polylog(4,- 
a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d^4+24*b 
*polylog(4,-a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1 
/2)/d^4-24*b*polylog(4,-a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a^2/( 
-a^2+b^2)^(1/2)/d^4+12*b^2*polylog(3,-a*exp(I*(c+d*x^(1/2)))/(b+I*(a^2-b^2 
)^(1/2)))/a^2/(a^2-b^2)/d^4+4*I*b*x^(3/2)*ln(1+a*exp(I*(c+d*x^(1/2)))/(b-( 
-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/d+12*I*b^3*x^(1/2)*polylog(3,-a*exp 
(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d^3+24*I*b*x^ 
(1/2)*polylog(3,-a*exp(I*(c+d*x^(1/2)))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^ 
2)^(1/2)/d^3+2*I*b^3*x^(3/2)*ln(1+a*exp(I*(c+d*x^(1/2)))/(b+(-a^2+b^2)^(1/ 
2)))/a^2/(-a^2+b^2)^(3/2)/d+2*b^2*x^(3/2)*sin(c+d*x^(1/2))/a/(a^2-b^2)/...
 

Mathematica [A] (verified)

Time = 12.45 (sec) , antiderivative size = 1695, normalized size of antiderivative = 1.11 \[ \int \frac {x}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((b + a*Cos[c + d*Sqrt[x]])*Sec[c + d*Sqrt[x]]^2*(x^2*(b + a*Cos[c + d*Sqr 
t[x]]) + (4*b*(b + a*Cos[c + d*Sqrt[x]])*(((-2*I)*b*d^3*E^((2*I)*c)*x^(3/2 
))/(1 + E^((2*I)*c)) + (3*b*d^2*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]*x*Log[1 + ( 
a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 
 (2*I)*a^2*d^3*E^(I*c)*x^(3/2)*Log[1 + (a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I 
*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - I*b^2*d^3*E^(I*c)*x^(3/2)*Log[1 + 
 (a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] 
 + 3*b*d^2*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]*x*Log[1 + (a*E^(I*(2*c + d*Sqrt[ 
x])))/(b*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - (2*I)*a^2*d^3*E^(I*c 
)*x^(3/2)*Log[1 + (a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) + Sqrt[(-a^2 + b^ 
2)*E^((2*I)*c)])] + I*b^2*d^3*E^(I*c)*x^(3/2)*Log[1 + (a*E^(I*(2*c + d*Sqr 
t[x])))/(b*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 3*d*((2*I)*b*Sqrt[ 
(-a^2 + b^2)*E^((2*I)*c)] - 2*a^2*d*E^(I*c)*Sqrt[x] + b^2*d*E^(I*c)*Sqrt[x 
])*Sqrt[x]*PolyLog[2, -((a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^ 
2 + b^2)*E^((2*I)*c)]))] + 3*d*((-2*I)*b*Sqrt[(-a^2 + b^2)*E^((2*I)*c)] - 
2*a^2*d*E^(I*c)*Sqrt[x] + b^2*d*E^(I*c)*Sqrt[x])*Sqrt[x]*PolyLog[2, -((a*E 
^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] + 6 
*b*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]*PolyLog[3, -((a*E^(I*(2*c + d*Sqrt[x]))) 
/(b*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] + (12*I)*a^2*d*E^(I*c)*Sqr 
t[x]*PolyLog[3, -((a*E^(I*(2*c + d*Sqrt[x])))/(b*E^(I*c) - Sqrt[(-a^2 +...
 

Rubi [A] (verified)

Time = 2.78 (sec) , antiderivative size = 1524, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4692, 3042, 4679, 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.

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

\(\Big \downarrow \) 4692

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

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int \frac {x^{3/2}}{\left (a+b \csc \left (c+d \sqrt {x}+\frac {\pi }{2}\right )\right )^2}d\sqrt {x}\)

\(\Big \downarrow \) 4679

\(\displaystyle 2 \int \left (\frac {x^{3/2} b^2}{a^2 \left (b+a \cos \left (c+d \sqrt {x}\right )\right )^2}-\frac {2 x^{3/2} b}{a^2 \left (b+a \cos \left (c+d \sqrt {x}\right )\right )}+\frac {x^{3/2}}{a^2}\right )d\sqrt {x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (-\frac {i x^{3/2} \log \left (\frac {e^{i \left (c+d \sqrt {x}\right )} a}{b-\sqrt {b^2-a^2}}+1\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d}+\frac {i x^{3/2} \log \left (\frac {e^{i \left (c+d \sqrt {x}\right )} a}{b+\sqrt {b^2-a^2}}+1\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d}-\frac {3 x \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^2}+\frac {3 x \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^2}-\frac {6 i \sqrt {x} \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^3}+\frac {6 i \sqrt {x} \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^3}+\frac {6 \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^4}-\frac {6 \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right ) b^3}{a^2 \left (b^2-a^2\right )^{3/2} d^4}-\frac {i x^{3/2} b^2}{a^2 \left (a^2-b^2\right ) d}+\frac {3 x \log \left (\frac {e^{i \left (c+d \sqrt {x}\right )} a}{b-i \sqrt {a^2-b^2}}+1\right ) b^2}{a^2 \left (a^2-b^2\right ) d^2}+\frac {3 x \log \left (\frac {e^{i \left (c+d \sqrt {x}\right )} a}{b+i \sqrt {a^2-b^2}}+1\right ) b^2}{a^2 \left (a^2-b^2\right ) d^2}-\frac {6 i \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-i \sqrt {a^2-b^2}}\right ) b^2}{a^2 \left (a^2-b^2\right ) d^3}-\frac {6 i \sqrt {x} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+i \sqrt {a^2-b^2}}\right ) b^2}{a^2 \left (a^2-b^2\right ) d^3}+\frac {6 \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-i \sqrt {a^2-b^2}}\right ) b^2}{a^2 \left (a^2-b^2\right ) d^4}+\frac {6 \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+i \sqrt {a^2-b^2}}\right ) b^2}{a^2 \left (a^2-b^2\right ) d^4}+\frac {x^{3/2} \sin \left (c+d \sqrt {x}\right ) b^2}{a \left (a^2-b^2\right ) d \left (b+a \cos \left (c+d \sqrt {x}\right )\right )}+\frac {2 i x^{3/2} \log \left (\frac {e^{i \left (c+d \sqrt {x}\right )} a}{b-\sqrt {b^2-a^2}}+1\right ) b}{a^2 \sqrt {b^2-a^2} d}-\frac {2 i x^{3/2} \log \left (\frac {e^{i \left (c+d \sqrt {x}\right )} a}{b+\sqrt {b^2-a^2}}+1\right ) b}{a^2 \sqrt {b^2-a^2} d}+\frac {6 x \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^2}-\frac {6 x \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^2}+\frac {12 i \sqrt {x} \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^3}-\frac {12 i \sqrt {x} \operatorname {PolyLog}\left (3,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^3}-\frac {12 \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b-\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^4}+\frac {12 \operatorname {PolyLog}\left (4,-\frac {a e^{i \left (c+d \sqrt {x}\right )}}{b+\sqrt {b^2-a^2}}\right ) b}{a^2 \sqrt {b^2-a^2} d^4}+\frac {x^2}{4 a^2}\right )\)

Input:

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

Output:

2*(((-I)*b^2*x^(3/2))/(a^2*(a^2 - b^2)*d) + x^2/(4*a^2) + (3*b^2*x*Log[1 + 
 (a*E^(I*(c + d*Sqrt[x])))/(b - I*Sqrt[a^2 - b^2])])/(a^2*(a^2 - b^2)*d^2) 
 + (3*b^2*x*Log[1 + (a*E^(I*(c + d*Sqrt[x])))/(b + I*Sqrt[a^2 - b^2])])/(a 
^2*(a^2 - b^2)*d^2) - (I*b^3*x^(3/2)*Log[1 + (a*E^(I*(c + d*Sqrt[x])))/(b 
- Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*d) + ((2*I)*b*x^(3/2)*Log[1 
+ (a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2] 
*d) + (I*b^3*x^(3/2)*Log[1 + (a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^ 
2])])/(a^2*(-a^2 + b^2)^(3/2)*d) - ((2*I)*b*x^(3/2)*Log[1 + (a*E^(I*(c + d 
*Sqrt[x])))/(b + Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*d) - ((6*I)*b^2 
*Sqrt[x]*PolyLog[2, -((a*E^(I*(c + d*Sqrt[x])))/(b - I*Sqrt[a^2 - b^2]))]) 
/(a^2*(a^2 - b^2)*d^3) - ((6*I)*b^2*Sqrt[x]*PolyLog[2, -((a*E^(I*(c + d*Sq 
rt[x])))/(b + I*Sqrt[a^2 - b^2]))])/(a^2*(a^2 - b^2)*d^3) - (3*b^3*x*PolyL 
og[2, -((a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 + b 
^2)^(3/2)*d^2) + (6*b*x*PolyLog[2, -((a*E^(I*(c + d*Sqrt[x])))/(b - Sqrt[- 
a^2 + b^2]))])/(a^2*Sqrt[-a^2 + b^2]*d^2) + (3*b^3*x*PolyLog[2, -((a*E^(I* 
(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 + b^2)^(3/2)*d^2) - 
 (6*b*x*PolyLog[2, -((a*E^(I*(c + d*Sqrt[x])))/(b + Sqrt[-a^2 + b^2]))])/( 
a^2*Sqrt[-a^2 + b^2]*d^2) + (6*b^2*PolyLog[3, -((a*E^(I*(c + d*Sqrt[x])))/ 
(b - I*Sqrt[a^2 - b^2]))])/(a^2*(a^2 - b^2)*d^4) + (6*b^2*PolyLog[3, -((a* 
E^(I*(c + d*Sqrt[x])))/(b + I*Sqrt[a^2 - b^2]))])/(a^2*(a^2 - b^2)*d^4)...
 

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 4679

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si 
n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt 
Q[m, 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{\left (a+b \sec \left (c+d \sqrt {x}\right )\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 384*cos(sqrt(x)*d + c)*int((sqrt(x)*tan((sqrt(x)*d + c)/2))/(4*tan((sq 
rt(x)*d + c)/2)**4*a**6 - 12*tan((sqrt(x)*d + c)/2)**4*a**5*b + 9*tan((sqr 
t(x)*d + c)/2)**4*a**4*b**2 + 4*tan((sqrt(x)*d + c)/2)**4*a**3*b**3 - 6*ta 
n((sqrt(x)*d + c)/2)**4*a**2*b**4 + tan((sqrt(x)*d + c)/2)**4*b**6 - 8*tan 
((sqrt(x)*d + c)/2)**2*a**6 + 8*tan((sqrt(x)*d + c)/2)**2*a**5*b + 14*tan( 
(sqrt(x)*d + c)/2)**2*a**4*b**2 - 12*tan((sqrt(x)*d + c)/2)**2*a**3*b**3 - 
 8*tan((sqrt(x)*d + c)/2)**2*a**2*b**4 + 4*tan((sqrt(x)*d + c)/2)**2*a*b** 
5 + 2*tan((sqrt(x)*d + c)/2)**2*b**6 + 4*a**6 + 4*a**5*b - 7*a**4*b**2 - 8 
*a**3*b**3 + 2*a**2*b**4 + 4*a*b**5 + b**6),x)*a**9*b*d**3 + 576*cos(sqrt( 
x)*d + c)*int((sqrt(x)*tan((sqrt(x)*d + c)/2))/(4*tan((sqrt(x)*d + c)/2)** 
4*a**6 - 12*tan((sqrt(x)*d + c)/2)**4*a**5*b + 9*tan((sqrt(x)*d + c)/2)**4 
*a**4*b**2 + 4*tan((sqrt(x)*d + c)/2)**4*a**3*b**3 - 6*tan((sqrt(x)*d + c) 
/2)**4*a**2*b**4 + tan((sqrt(x)*d + c)/2)**4*b**6 - 8*tan((sqrt(x)*d + c)/ 
2)**2*a**6 + 8*tan((sqrt(x)*d + c)/2)**2*a**5*b + 14*tan((sqrt(x)*d + c)/2 
)**2*a**4*b**2 - 12*tan((sqrt(x)*d + c)/2)**2*a**3*b**3 - 8*tan((sqrt(x)*d 
 + c)/2)**2*a**2*b**4 + 4*tan((sqrt(x)*d + c)/2)**2*a*b**5 + 2*tan((sqrt(x 
)*d + c)/2)**2*b**6 + 4*a**6 + 4*a**5*b - 7*a**4*b**2 - 8*a**3*b**3 + 2*a* 
*2*b**4 + 4*a*b**5 + b**6),x)*a**8*b**2*d**3 + 288*cos(sqrt(x)*d + c)*int( 
(sqrt(x)*tan((sqrt(x)*d + c)/2))/(4*tan((sqrt(x)*d + c)/2)**4*a**6 - 12*ta 
n((sqrt(x)*d + c)/2)**4*a**5*b + 9*tan((sqrt(x)*d + c)/2)**4*a**4*b**2 ...