\(\int \frac {\sec ^6(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [199]

Optimal result

Integrand size = 23, antiderivative size = 100 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {a (3 a+4 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 b^{5/2} (a+b)^{3/2} f}+\frac {\tan (e+f x)}{b^2 f}+\frac {a^2 \tan (e+f x)}{2 b^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 3.02 (sec) , antiderivative size = 248, normalized size of antiderivative = 2.48 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^4(e+f x) \left (\frac {a (3 a+4 b) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (a+2 b+a \cos (2 (e+f x))) (\cos (2 e)-i \sin (2 e))}{(a+b)^{3/2} \sqrt {b (\cos (e)-i \sin (e))^4}}+2 (a+2 b+a \cos (2 (e+f x))) \sec (e) \sec (e+f x) \sin (f x)+\frac {a (-((a+2 b) \sin (2 e))+a \sin (2 f x))}{(a+b) (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}\right )}{8 b^2 f \left (a+b \sec ^2(e+f x)\right )^2} \] Input:

Integrate[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^2,x]
 

Output:

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^4*((a*(3*a + 4*b)*ArcTan[(Sec 
[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/ 
(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(a + 2*b + a*Cos[2*(e + f*x 
)])*(Cos[2*e] - I*Sin[2*e]))/((a + b)^(3/2)*Sqrt[b*(Cos[e] - I*Sin[e])^4]) 
 + 2*(a + 2*b + a*Cos[2*(e + f*x)])*Sec[e]*Sec[e + f*x]*Sin[f*x] + (a*(-(( 
a + 2*b)*Sin[2*e]) + a*Sin[2*f*x]))/((a + b)*(Cos[e] - Sin[e])*(Cos[e] + S 
in[e]))))/(8*b^2*f*(a + b*Sec[e + f*x]^2)^2)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4634, 300, 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.

\(\displaystyle \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (e+f x)^6}{\left (a+b \sec (e+f x)^2\right )^2}dx\)

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 300

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {a^2 \tan (e+f x)}{2 b^2 (a+b) \left (a+b \tan ^2(e+f x)+b\right )}-\frac {a (3 a+4 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 b^{5/2} (a+b)^{3/2}}+\frac {\tan (e+f x)}{b^2}}{f}\)

Input:

Int[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2)^2,x]
 

Output:

(-1/2*(a*(3*a + 4*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(b^(5/2)* 
(a + b)^(3/2)) + Tan[e + f*x]/b^2 + (a^2*Tan[e + f*x])/(2*b^2*(a + b)*(a + 
 b + b*Tan[e + f*x]^2)))/f
 

Defintions of rubi rules used

rule 300

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

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 4634

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

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {-\frac {a \left (-\frac {a \tan \left (f x +e \right )}{2 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (3 a +4 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{b^{2}}+\frac {\tan \left (f x +e \right )}{b^{2}}}{f}\)	\(89\)
default	\(\frac {-\frac {a \left (-\frac {a \tan \left (f x +e \right )}{2 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (3 a +4 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{b^{2}}+\frac {\tan \left (f x +e \right )}{b^{2}}}{f}\)	\(89\)
risch	\(\frac {i \left (3 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+4 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+6 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+14 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+8 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+3 a^{2}+2 a b \right )}{\left (a +b \right ) b^{2} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i b a +2 i b^{2}-a \sqrt {-a b -b^{2}}-2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right ) f \,b^{2}}-\frac {a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i b a +2 i b^{2}-a \sqrt {-a b -b^{2}}-2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{\sqrt {-a b -b^{2}}\, \left (a +b \right ) f b}+\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right ) f \,b^{2}}+\frac {a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{\sqrt {-a b -b^{2}}\, \left (a +b \right ) f b}\)	\(529\)

Input:

int(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
 

Output:

1/f*(-1/b^2*a*(-1/2*a/(a+b)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)+1/2*(3*a+4*b)/ 
(a+b)/((a+b)*b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2)))+tan(f*x+e)/b^2 
)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (88) = 176\).

Time = 0.12 (sec) , antiderivative size = 516, normalized size of antiderivative = 5.16 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [-\frac {{\left ({\left (3 \, a^{3} + 4 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) - 4 \, {\left (2 \, a^{2} b^{2} + 4 \, a b^{3} + 2 \, b^{4} + {\left (3 \, a^{3} b + 5 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{8 \, {\left ({\left (a^{3} b^{3} + 2 \, a^{2} b^{4} + a b^{5}\right )} f \cos \left (f x + e\right )^{3} + {\left (a^{2} b^{4} + 2 \, a b^{5} + b^{6}\right )} f \cos \left (f x + e\right )\right )}}, \frac {{\left ({\left (3 \, a^{3} + 4 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} + {\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a b + b^{2}} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + 2 \, {\left (2 \, a^{2} b^{2} + 4 \, a b^{3} + 2 \, b^{4} + {\left (3 \, a^{3} b + 5 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{4 \, {\left ({\left (a^{3} b^{3} + 2 \, a^{2} b^{4} + a b^{5}\right )} f \cos \left (f x + e\right )^{3} + {\left (a^{2} b^{4} + 2 \, a b^{5} + b^{6}\right )} f \cos \left (f x + e\right )\right )}}\right ] \] Input:

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")
 

Output:

[-1/8*(((3*a^3 + 4*a^2*b)*cos(f*x + e)^3 + (3*a^2*b + 4*a*b^2)*cos(f*x + e 
))*sqrt(-a*b - b^2)*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 
 4*b^2)*cos(f*x + e)^2 - 4*((a + 2*b)*cos(f*x + e)^3 - b*cos(f*x + e))*sqr 
t(-a*b - b^2)*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e) 
^2 + b^2)) - 4*(2*a^2*b^2 + 4*a*b^3 + 2*b^4 + (3*a^3*b + 5*a^2*b^2 + 2*a*b 
^3)*cos(f*x + e)^2)*sin(f*x + e))/((a^3*b^3 + 2*a^2*b^4 + a*b^5)*f*cos(f*x 
 + e)^3 + (a^2*b^4 + 2*a*b^5 + b^6)*f*cos(f*x + e)), 1/4*(((3*a^3 + 4*a^2* 
b)*cos(f*x + e)^3 + (3*a^2*b + 4*a*b^2)*cos(f*x + e))*sqrt(a*b + b^2)*arct 
an(1/2*((a + 2*b)*cos(f*x + e)^2 - b)/(sqrt(a*b + b^2)*cos(f*x + e)*sin(f* 
x + e))) + 2*(2*a^2*b^2 + 4*a*b^3 + 2*b^4 + (3*a^3*b + 5*a^2*b^2 + 2*a*b^3 
)*cos(f*x + e)^2)*sin(f*x + e))/((a^3*b^3 + 2*a^2*b^4 + a*b^5)*f*cos(f*x + 
 e)^3 + (a^2*b^4 + 2*a*b^5 + b^6)*f*cos(f*x + e))]
 

Sympy [F]

\[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\int \frac {\sec ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \] Input:

integrate(sec(f*x+e)**6/(a+b*sec(f*x+e)**2)**2,x)
 

Output:

Integral(sec(e + f*x)**6/(a + b*sec(e + f*x)**2)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.10 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {a^{2} \tan \left (f x + e\right )}{a^{2} b^{2} + 2 \, a b^{3} + b^{4} + {\left (a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{2}} - \frac {{\left (3 \, a^{2} + 4 \, a b\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {2 \, \tan \left (f x + e\right )}{b^{2}}}{2 \, f} \] Input:

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.20 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {a^{2} \tan \left (f x + e\right )}{{\left (a b^{2} + b^{3}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} - \frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (3 \, a^{2} + 4 \, a b\right )}}{{\left (a b^{2} + b^{3}\right )} \sqrt {a b + b^{2}}} + \frac {2 \, \tan \left (f x + e\right )}{b^{2}}}{2 \, f} \] Input:

integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")
 

Output:

1/2*(a^2*tan(f*x + e)/((a*b^2 + b^3)*(b*tan(f*x + e)^2 + a + b)) - (pi*flo 
or((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*(3 
*a^2 + 4*a*b)/((a*b^2 + b^3)*sqrt(a*b + b^2)) + 2*tan(f*x + e)/b^2)/f
 

Mupad [B] (verification not implemented)

Time = 15.69 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )}{b^2\,f}+\frac {a^2\,\mathrm {tan}\left (e+f\,x\right )}{2\,f\,\left (a+b\right )\,\left (b^3\,{\mathrm {tan}\left (e+f\,x\right )}^2+b^3+a\,b^2\right )}-\frac {a\,\mathrm {atan}\left (\frac {a\,\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (3\,a+4\,b\right )}{\sqrt {a+b}\,\left (3\,a^2+4\,b\,a\right )}\right )\,\left (3\,a+4\,b\right )}{2\,b^{5/2}\,f\,{\left (a+b\right )}^{3/2}} \] Input:

int(1/(cos(e + f*x)^6*(a + b/cos(e + f*x)^2)^2),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 622, normalized size of antiderivative = 6.22 \[ \int \frac {\sec ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:

int(sec(f*x+e)^6/(a+b*sec(f*x+e)^2)^2,x)
 

Output:

( - 3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sq 
rt(b))*cos(e + f*x)*sin(e + f*x)**2*a**3 - 4*sqrt(b)*sqrt(a + b)*atan((sqr 
t(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*cos(e + f*x)*sin(e + f*x)**2 
*a**2*b + 3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt( 
a))/sqrt(b))*cos(e + f*x)*a**3 + 7*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*t 
an((e + f*x)/2) - sqrt(a))/sqrt(b))*cos(e + f*x)*a**2*b + 4*sqrt(b)*sqrt(a 
 + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*cos(e + f*x)* 
a*b**2 - 3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a 
))/sqrt(b))*cos(e + f*x)*sin(e + f*x)**2*a**3 - 4*sqrt(b)*sqrt(a + b)*atan 
((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*cos(e + f*x)*sin(e + f* 
x)**2*a**2*b + 3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + 
sqrt(a))/sqrt(b))*cos(e + f*x)*a**3 + 7*sqrt(b)*sqrt(a + b)*atan((sqrt(a + 
 b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*cos(e + f*x)*a**2*b + 4*sqrt(b)*s 
qrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*cos(e + 
f*x)*a*b**2 + 3*sin(e + f*x)**3*a**3*b + 5*sin(e + f*x)**3*a**2*b**2 + 2*s 
in(e + f*x)**3*a*b**3 - 3*sin(e + f*x)*a**3*b - 7*sin(e + f*x)*a**2*b**2 - 
 6*sin(e + f*x)*a*b**3 - 2*sin(e + f*x)*b**4)/(2*cos(e + f*x)*b**3*f*(sin( 
e + f*x)**2*a**3 + 2*sin(e + f*x)**2*a**2*b + sin(e + f*x)**2*a*b**2 - a** 
3 - 3*a**2*b - 3*a*b**2 - b**3))