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

Optimal result
Mathematica [A] (warning: unable to verify)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F]
Maxima [F(-2)]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 18, antiderivative size = 1565 \[ \int \frac {x}{\left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2} \, dx =\text {Too large to display} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 11.10 (sec) , antiderivative size = 1729, normalized size of antiderivative = 1.10 \[ \int \frac {x}{\left (a+b \csc \left (c+d \sqrt {x}\right )\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 2.84 (sec) , antiderivative size = 1567, 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 = {4693, 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.

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

\(\Big \downarrow \) 4693

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

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

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (-\frac {i x^{3/2} \log \left (1-\frac {i 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}+\frac {i x^{3/2} \log \left (1-\frac {i 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}-\frac {3 x \operatorname {PolyLog}\left (2,\frac {i 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 {i 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 {i 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 {i 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 {i 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 {i 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}{i b-\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}{i b+\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 )}}{i b-\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 )}}{i b+\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 )}}{i b-\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 )}}{i b+\sqrt {a^2-b^2}}\right ) b^2}{a^2 \left (a^2-b^2\right ) d^4}-\frac {x^{3/2} \cos \left (c+d \sqrt {x}\right ) b^2}{a \left (a^2-b^2\right ) d \left (b+a \sin \left (c+d \sqrt {x}\right )\right )}+\frac {2 i x^{3/2} \log \left (1-\frac {i 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}-\frac {2 i x^{3/2} \log \left (1-\frac {i 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}+\frac {6 x \operatorname {PolyLog}\left (2,\frac {i 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 {i 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 {i 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 {i 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 {i 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 {i 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*Csc[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])))/(I*b - 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])))/(I*b + Sqrt[a^2 - b^2])])/(a 
^2*(a^2 - b^2)*d^2) - (I*b^3*x^(3/2)*Log[1 - (I*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 - (I*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 - (I*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 - (I*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])))/(I*b - 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*Sqrt[x])))/(I*b + Sqrt[a^2 - b^2]))])/(a^2*(a^2 - b^2)*d^3) - (3*b^3 
*x*PolyLog[2, (I*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, (I*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, (I*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, (I*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] 
)))/(I*b - 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])))/(I*b + Sqrt[a^2 - b^2]))])/(a^2*(a^2 - b^2)*...
 

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 4693
Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Csc[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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x/(a+b*csc(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(864*sqrt( - a**2 + b**2)*atan((tan((sqrt(x)*d + c)/2)*b + a)/sqrt( - a**2 
 + b**2))*sin(sqrt(x)*d + c)*a**5 - 672*sqrt( - a**2 + b**2)*atan((tan((sq 
rt(x)*d + c)/2)*b + a)/sqrt( - a**2 + b**2))*sin(sqrt(x)*d + c)*a**3*b**2 
+ 48*sqrt( - a**2 + b**2)*atan((tan((sqrt(x)*d + c)/2)*b + a)/sqrt( - a**2 
 + b**2))*sin(sqrt(x)*d + c)*a*b**4 + 864*sqrt( - a**2 + b**2)*atan((tan(( 
sqrt(x)*d + c)/2)*b + a)/sqrt( - a**2 + b**2))*a**4*b - 672*sqrt( - a**2 + 
 b**2)*atan((tan((sqrt(x)*d + c)/2)*b + a)/sqrt( - a**2 + b**2))*a**2*b**3 
 + 48*sqrt( - a**2 + b**2)*atan((tan((sqrt(x)*d + c)/2)*b + a)/sqrt( - a** 
2 + b**2))*b**5 + 432*sqrt(x)*cos(sqrt(x)*d + c)*a**6*d + 8*sqrt(x)*cos(sq 
rt(x)*d + c)*a**4*b**2*d**3*x - 552*sqrt(x)*cos(sqrt(x)*d + c)*a**4*b**2*d 
 - 8*sqrt(x)*cos(sqrt(x)*d + c)*a**2*b**4*d**3*x + 120*sqrt(x)*cos(sqrt(x) 
*d + c)*a**2*b**4*d - 72*cos(sqrt(x)*d + c)*a**5*b*d**2*x + 84*cos(sqrt(x) 
*d + c)*a**3*b**3*d**2*x - 12*cos(sqrt(x)*d + c)*a*b**5*d**2*x - 12*sqrt(x 
)*sin(sqrt(x)*d + c)*a**5*b*d**3*x + 16*sqrt(x)*sin(sqrt(x)*d + c)*a**3*b* 
*3*d**3*x - 4*sqrt(x)*sin(sqrt(x)*d + c)*a*b**5*d**3*x + 432*sqrt(x)*a**6* 
d - 4*sqrt(x)*a**4*b**2*d**3*x - 768*sqrt(x)*a**4*b**2*d + 4*sqrt(x)*a**2* 
b**4*d**3*x + 360*sqrt(x)*a**2*b**4*d - 24*sqrt(x)*b**6*d - 72*int(sqrt(x) 
/(tan((sqrt(x)*d + c)/2)**4*b**2 + 4*tan((sqrt(x)*d + c)/2)**3*a*b + 4*tan 
((sqrt(x)*d + c)/2)**2*a**2 + 2*tan((sqrt(x)*d + c)/2)**2*b**2 + 4*tan((sq 
rt(x)*d + c)/2)*a*b + b**2),x)*sin(sqrt(x)*d + c)*a**7*b*d**3 + 96*int(...