\(\int \frac {\sin ^3(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [42]

Optimal result

Integrand size = 23, antiderivative size = 114 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\sqrt {b} (3 a+5 b) \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{2 a^{7/2} f}-\frac {(a+2 b) \cos (e+f x)}{a^3 f}+\frac {\cos ^3(e+f x)}{3 a^2 f}-\frac {b (a+b) \cos (e+f x)}{2 a^3 f \left (b+a \cos ^2(e+f x)\right )} \] Output:

1/2*b^(1/2)*(3*a+5*b)*arctan(a^(1/2)*cos(f*x+e)/b^(1/2))/a^(7/2)/f-(a+2*b) 
*cos(f*x+e)/a^3/f+1/3*cos(f*x+e)^3/a^2/f-1/2*b*(a+b)*cos(f*x+e)/a^3/f/(b+a 
*cos(f*x+e)^2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.91 (sec) , antiderivative size = 403, normalized size of antiderivative = 3.54 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {3 \left (3 a^3+192 a b^2+320 b^3\right ) \arctan \left (\frac {\left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )}{b^{3/2}}+\frac {3 \left (3 a^3+192 a b^2+320 b^3\right ) \arctan \left (\frac {\left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )}{b^{3/2}}-\frac {9 a^3 \arctan \left (\frac {\sqrt {a}-\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{b^{3/2}}-\frac {9 a^3 \arctan \left (\frac {\sqrt {a}+\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{b^{3/2}}-\frac {32 \sqrt {a} \cos (e+f x) \left (9 a^2+56 a b+60 b^2+4 a (2 a+5 b) \cos (2 (e+f x))-a^2 \cos (4 (e+f x))\right )}{a+2 b+a \cos (2 (e+f x))}}{384 a^{7/2} f} \] Input:

Integrate[Sin[e + f*x]^3/(a + b*Sec[e + f*x]^2)^2,x]
 

Output:

((3*(3*a^3 + 192*a*b^2 + 320*b^3)*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[( 
Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] - Sqrt[a + b] 
*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]])/b^(3/2) + (3*(3*a^3 
+ 192*a*b^2 + 320*b^3)*ArcTan[((-Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cos[e] - I* 
Sin[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] + Sqrt[a + b]*Sqrt[(Cos[ 
e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]])/b^(3/2) - (9*a^3*ArcTan[(Sqrt[a 
] - Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]])/b^(3/2) - (9*a^3*ArcTan[(Sqrt[ 
a] + Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]])/b^(3/2) - (32*Sqrt[a]*Cos[e + 
 f*x]*(9*a^2 + 56*a*b + 60*b^2 + 4*a*(2*a + 5*b)*Cos[2*(e + f*x)] - a^2*Co 
s[4*(e + f*x)]))/(a + 2*b + a*Cos[2*(e + f*x)]))/(384*a^(7/2)*f)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4621, 360, 1467, 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.

\(\displaystyle \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)^3}{\left (a+b \sec (e+f x)^2\right )^2}dx\)

\(\Big \downarrow \) 4621

\(\displaystyle -\frac {\int \frac {\cos ^4(e+f x) \left (1-\cos ^2(e+f x)\right )}{\left (a \cos ^2(e+f x)+b\right )^2}d\cos (e+f x)}{f}\)

\(\Big \downarrow \) 360

\(\displaystyle -\frac {\frac {b (a+b) \cos (e+f x)}{2 a^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {\int \frac {2 a^2 \cos ^4(e+f x)-2 a (a+b) \cos ^2(e+f x)+b (a+b)}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{2 a^3}}{f}\)

\(\Big \downarrow \) 1467

\(\displaystyle -\frac {\frac {b (a+b) \cos (e+f x)}{2 a^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {\int \left (2 a \cos ^2(e+f x)-2 (a+2 b)+\frac {5 b^2+3 a b}{a \cos ^2(e+f x)+b}\right )d\cos (e+f x)}{2 a^3}}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {b (a+b) \cos (e+f x)}{2 a^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {\frac {\sqrt {b} (3 a+5 b) \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{\sqrt {a}}-2 (a+2 b) \cos (e+f x)+\frac {2}{3} a \cos ^3(e+f x)}{2 a^3}}{f}\)

Input:

Int[Sin[e + f*x]^3/(a + b*Sec[e + f*x]^2)^2,x]
 

Output:

-(((b*(a + b)*Cos[e + f*x])/(2*a^3*(b + a*Cos[e + f*x]^2)) - ((Sqrt[b]*(3* 
a + 5*b)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/Sqrt[a] - 2*(a + 2*b)*Cos 
[e + f*x] + (2*a*Cos[e + f*x]^3)/3)/(2*a^3))/f)
 

Defintions of rubi rules used

rule 360

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

rule 1467

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
 x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], 
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e 
 + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

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 4621

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ 
)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f 
   Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff*x)^n)^p/(ff*x)^(n*p)), 
x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/ 
2] && IntegerQ[n] && IntegerQ[p]
 

Maple [A] (verified)

Time = 1.92 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {\frac {\frac {a \cos \left (f x +e \right )^{3}}{3}-\cos \left (f x +e \right ) a -2 \cos \left (f x +e \right ) b}{a^{3}}+\frac {b \left (\frac {\left (-\frac {a}{2}-\frac {b}{2}\right ) \cos \left (f x +e \right )}{b +a \cos \left (f x +e \right )^{2}}+\frac {\left (3 a +5 b \right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}}{f}\)	\(102\)
default	\(\frac {\frac {\frac {a \cos \left (f x +e \right )^{3}}{3}-\cos \left (f x +e \right ) a -2 \cos \left (f x +e \right ) b}{a^{3}}+\frac {b \left (\frac {\left (-\frac {a}{2}-\frac {b}{2}\right ) \cos \left (f x +e \right )}{b +a \cos \left (f x +e \right )^{2}}+\frac {\left (3 a +5 b \right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}}{f}\)	\(102\)
risch	\(\frac {{\mathrm e}^{3 i \left (f x +e \right )}}{24 a^{2} f}-\frac {3 \,{\mathrm e}^{i \left (f x +e \right )}}{8 f \,a^{2}}-\frac {{\mathrm e}^{i \left (f x +e \right )} b}{f \,a^{3}}-\frac {3 \,{\mathrm e}^{-i \left (f x +e \right )}}{8 f \,a^{2}}-\frac {{\mathrm e}^{-i \left (f x +e \right )} b}{f \,a^{3}}+\frac {{\mathrm e}^{-3 i \left (f x +e \right )}}{24 a^{2} f}-\frac {\left (a +b \right ) b \left ({\mathrm e}^{3 i \left (f x +e \right )}+{\mathrm e}^{i \left (f x +e \right )}\right )}{a^{3} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {3 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{4 a^{3} f}+\frac {5 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b}{4 a^{4} f}-\frac {3 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{4 a^{3} f}-\frac {5 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b}{4 a^{4} f}\)	\(362\)

Input:

int(sin(f*x+e)^3/(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
 

Output:

1/f*(1/a^3*(1/3*a*cos(f*x+e)^3-cos(f*x+e)*a-2*cos(f*x+e)*b)+b/a^3*((-1/2*a 
-1/2*b)*cos(f*x+e)/(b+a*cos(f*x+e)^2)+1/2*(3*a+5*b)/(a*b)^(1/2)*arctan(a*c 
os(f*x+e)/(a*b)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.61 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [\frac {4 \, a^{2} \cos \left (f x + e\right )^{5} - 4 \, {\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left ({\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 5 \, b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) - 6 \, {\left (3 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )}{12 \, {\left (a^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}}, \frac {2 \, a^{2} \cos \left (f x + e\right )^{5} - 2 \, {\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left ({\left (3 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 5 \, b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right ) - 3 \, {\left (3 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )}{6 \, {\left (a^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}}\right ] \] Input:

integrate(sin(f*x+e)^3/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")
 

Output:

[1/12*(4*a^2*cos(f*x + e)^5 - 4*(3*a^2 + 5*a*b)*cos(f*x + e)^3 + 3*((3*a^2 
 + 5*a*b)*cos(f*x + e)^2 + 3*a*b + 5*b^2)*sqrt(-b/a)*log(-(a*cos(f*x + e)^ 
2 + 2*a*sqrt(-b/a)*cos(f*x + e) - b)/(a*cos(f*x + e)^2 + b)) - 6*(3*a*b + 
5*b^2)*cos(f*x + e))/(a^4*f*cos(f*x + e)^2 + a^3*b*f), 1/6*(2*a^2*cos(f*x 
+ e)^5 - 2*(3*a^2 + 5*a*b)*cos(f*x + e)^3 + 3*((3*a^2 + 5*a*b)*cos(f*x + e 
)^2 + 3*a*b + 5*b^2)*sqrt(b/a)*arctan(a*sqrt(b/a)*cos(f*x + e)/b) - 3*(3*a 
*b + 5*b^2)*cos(f*x + e))/(a^4*f*cos(f*x + e)^2 + a^3*b*f)]
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Timed out} \] Input:

integrate(sin(f*x+e)**3/(a+b*sec(f*x+e)**2)**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.91 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {3 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )}{a^{4} \cos \left (f x + e\right )^{2} + a^{3} b} - \frac {3 \, {\left (3 \, a b + 5 \, b^{2}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (a \cos \left (f x + e\right )^{3} - 3 \, {\left (a + 2 \, b\right )} \cos \left (f x + e\right )\right )}}{a^{3}}}{6 \, f} \] Input:

integrate(sin(f*x+e)^3/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")
 

Output:

-1/6*(3*(a*b + b^2)*cos(f*x + e)/(a^4*cos(f*x + e)^2 + a^3*b) - 3*(3*a*b + 
 5*b^2)*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^3) - 2*(a*cos(f*x + 
e)^3 - 3*(a + 2*b)*cos(f*x + e))/a^3)/f
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.17 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {{\left (3 \, a b + 5 \, b^{2}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3} f} - \frac {a b \cos \left (f x + e\right ) + b^{2} \cos \left (f x + e\right )}{2 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )} a^{3} f} + \frac {a^{4} f^{2} \cos \left (f x + e\right )^{3} - 3 \, a^{4} f^{2} \cos \left (f x + e\right ) - 6 \, a^{3} b f^{2} \cos \left (f x + e\right )}{3 \, a^{6} f^{3}} \] Input:

integrate(sin(f*x+e)^3/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")
 

Output:

1/2*(3*a*b + 5*b^2)*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^3*f) - 1 
/2*(a*b*cos(f*x + e) + b^2*cos(f*x + e))/((a*cos(f*x + e)^2 + b)*a^3*f) + 
1/3*(a^4*f^2*cos(f*x + e)^3 - 3*a^4*f^2*cos(f*x + e) - 6*a^3*b*f^2*cos(f*x 
 + e))/(a^6*f^3)
 

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.14 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {{\cos \left (e+f\,x\right )}^3}{3\,a^2\,f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {2\,b}{a^3}+\frac {1}{a^2}\right )}{f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {b^2}{2}+\frac {a\,b}{2}\right )}{f\,\left (a^4\,{\cos \left (e+f\,x\right )}^2+b\,a^3\right )}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\cos \left (e+f\,x\right )\,\left (3\,a+5\,b\right )}{5\,b^2+3\,a\,b}\right )\,\left (3\,a+5\,b\right )}{2\,a^{7/2}\,f} \] Input:

int(sin(e + f*x)^3/(a + b/cos(e + f*x)^2)^2,x)
 

Output:

cos(e + f*x)^3/(3*a^2*f) - (cos(e + f*x)*((2*b)/a^3 + 1/a^2))/f - (cos(e + 
 f*x)*((a*b)/2 + b^2/2))/(f*(a^3*b + a^4*cos(e + f*x)^2)) + (b^(1/2)*atan( 
(a^(1/2)*b^(1/2)*cos(e + f*x)*(3*a + 5*b))/(3*a*b + 5*b^2))*(3*a + 5*b))/( 
2*a^(7/2)*f)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 523, normalized size of antiderivative = 4.59 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:

int(sin(f*x+e)^3/(a+b*sec(f*x+e)^2)^2,x)
 

Output:

( - 9*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b 
))*sin(e + f*x)**2*a**2 - 15*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f* 
x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a*b + 9*sqrt(b)*sqrt(a)*atan((sq 
rt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**2 + 24*sqrt(b)*sqrt(a)*a 
tan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a*b + 15*sqrt(b)*sqr 
t(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*b**2 + 9*sqrt( 
b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + 
f*x)**2*a**2 + 15*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqr 
t(a))/sqrt(b))*sin(e + f*x)**2*a*b - 9*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*t 
an((e + f*x)/2) + sqrt(a))/sqrt(b))*a**2 - 24*sqrt(b)*sqrt(a)*atan((sqrt(a 
 + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*a*b - 15*sqrt(b)*sqrt(a)*atan(( 
sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*b**2 - 2*cos(e + f*x)*sin 
(e + f*x)**4*a**3 - 2*cos(e + f*x)*sin(e + f*x)**2*a**3 - 10*cos(e + f*x)* 
sin(e + f*x)**2*a**2*b + 4*cos(e + f*x)*a**3 + 19*cos(e + f*x)*a**2*b + 15 
*cos(e + f*x)*a*b**2 + 4*sin(e + f*x)**2*a**3 + 15*sin(e + f*x)**2*a**2*b 
- 4*a**3 - 19*a**2*b - 15*a*b**2)/(6*a**4*f*(sin(e + f*x)**2*a - a - b))