Integrand size = 23, antiderivative size = 77 \[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {a^2 \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{b^{5/2} \sqrt {a+b} f}-\frac {(a-b) \tan (e+f x)}{b^2 f}+\frac {\tan ^3(e+f x)}{3 b f} \] Output:
a^2*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/b^(5/2)/(a+b)^(1/2)/f-(a-b)*tan (f*x+e)/b^2/f+1/3*tan(f*x+e)^3/b/f
Result contains complex when optimal does not.
Time = 2.98 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.91 \[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (-3 a^2 \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))+\sqrt {a+b} \sec (e+f x) \sqrt {b (i \cos (e)+\sin (e))^4} \left (\sec (e) \left (-3 a+2 b+b \sec ^2(e+f x)\right ) \sin (f x)+b \sec (e+f x) \tan (e)\right )\right )}{6 b^2 \sqrt {a+b} f \left (a+b \sec ^2(e+f x)\right ) \sqrt {b (\cos (e)-i \sin (e))^4}} \] Input:
Integrate[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2),x]
Output:
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(-3*a^2*ArcTan[(Sec[f*x]*(C os[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])]*(Cos[2*e] - I*Sin[2*e]) + Sqrt[a + b]*Sec[e + f*x]*Sqrt[b*(I*Cos[e] + Sin[e])^4]*(Sec[e]*(-3*a + 2*b + b*Sec[ e + f*x]^2)*Sin[f*x] + b*Sec[e + f*x]*Tan[e])))/(6*b^2*Sqrt[a + b]*f*(a + b*Sec[e + f*x]^2)*Sqrt[b*(Cos[e] - I*Sin[e])^4])
Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94, 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.
\(\displaystyle \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (e+f x)^6}{a+b \sec (e+f x)^2}dx\) |
\(\Big \downarrow \) 4634 |
\(\displaystyle \frac {\int \frac {\left (\tan ^2(e+f x)+1\right )^2}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {\int \left (\frac {a^2}{b^2 \left (b \tan ^2(e+f x)+a+b\right )}+\frac {\tan ^2(e+f x)}{b}-\frac {a-b}{b^2}\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^2 \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{b^{5/2} \sqrt {a+b}}-\frac {(a-b) \tan (e+f x)}{b^2}+\frac {\tan ^3(e+f x)}{3 b}}{f}\) |
Input:
Int[Sec[e + f*x]^6/(a + b*Sec[e + f*x]^2),x]
Output:
((a^2*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(b^(5/2)*Sqrt[a + b]) - ((a - b)*Tan[e + f*x])/b^2 + Tan[e + f*x]^3/(3*b))/f
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]
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]
Time = 0.81 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(\frac {-\frac {-\frac {b \tan \left (f x +e \right )^{3}}{3}+a \tan \left (f x +e \right )-b \tan \left (f x +e \right )}{b^{2}}+\frac {a^{2} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{b^{2} \sqrt {\left (a +b \right ) b}}}{f}\) | \(70\) |
default | \(\frac {-\frac {-\frac {b \tan \left (f x +e \right )^{3}}{3}+a \tan \left (f x +e \right )-b \tan \left (f x +e \right )}{b^{2}}+\frac {a^{2} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{b^{2} \sqrt {\left (a +b \right ) b}}}{f}\) | \(70\) |
risch | \(-\frac {2 i \left (3 a \,{\mathrm e}^{4 i \left (f x +e \right )}+6 a \,{\mathrm e}^{2 i \left (f x +e \right )}-6 b \,{\mathrm e}^{2 i \left (f x +e \right )}+3 a -2 b \right )}{3 f \,b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{2 \sqrt {-a b -b^{2}}\, f \,b^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {-2 i b a -2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{2 \sqrt {-a b -b^{2}}\, f \,b^{2}}\) | \(249\) |
Input:
int(sec(f*x+e)^6/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(-1/b^2*(-1/3*b*tan(f*x+e)^3+a*tan(f*x+e)-b*tan(f*x+e))+a^2/b^2/((a+b) *b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (67) = 134\).
Time = 0.10 (sec) , antiderivative size = 354, normalized size of antiderivative = 4.60 \[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [-\frac {3 \, \sqrt {-a b - b^{2}} a^{2} \cos \left (f x + e\right )^{3} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) - 4 \, {\left (a b^{2} + b^{3} - {\left (3 \, a^{2} b + a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{12 \, {\left (a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{3}}, -\frac {3 \, \sqrt {a b + b^{2}} a^{2} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{3} - 2 \, {\left (a b^{2} + b^{3} - {\left (3 \, a^{2} b + a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{6 \, {\left (a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{3}}\right ] \] Input:
integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
[-1/12*(3*sqrt(-a*b - b^2)*a^2*cos(f*x + e)^3*log(((a^2 + 8*a*b + 8*b^2)*c os(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))*sqrt(-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*(a*b^2 + b^3 - (3*a^2*b + a*b^2 - 2*b^3)*cos(f*x + e)^2)*sin(f*x + e))/((a*b^3 + b^4)*f*cos(f*x + e)^3), -1 /6*(3*sqrt(a*b + b^2)*a^2*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)/(sqrt( a*b + b^2)*cos(f*x + e)*sin(f*x + e)))*cos(f*x + e)^3 - 2*(a*b^2 + b^3 - ( 3*a^2*b + a*b^2 - 2*b^3)*cos(f*x + e)^2)*sin(f*x + e))/((a*b^3 + b^4)*f*co s(f*x + e)^3)]
\[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\sec ^{6}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \] Input:
integrate(sec(f*x+e)**6/(a+b*sec(f*x+e)**2),x)
Output:
Integral(sec(e + f*x)**6/(a + b*sec(e + f*x)**2), x)
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84 \[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {3 \, a^{2} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} b^{2}} + \frac {b \tan \left (f x + e\right )^{3} - 3 \, {\left (a - b\right )} \tan \left (f x + e\right )}{b^{2}}}{3 \, f} \] Input:
integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/3*(3*a^2*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)*b^2) + (b*tan(f*x + e)^3 - 3*(a - b)*tan(f*x + e))/b^2)/f
Time = 0.14 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} a^{2}}{\sqrt {a b + b^{2}} b^{2}} + \frac {b^{2} \tan \left (f x + e\right )^{3} - 3 \, a b \tan \left (f x + e\right ) + 3 \, b^{2} \tan \left (f x + e\right )}{b^{3}}}{3 \, f} \] Input:
integrate(sec(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/3*(3*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a *b + b^2)))*a^2/(sqrt(a*b + b^2)*b^2) + (b^2*tan(f*x + e)^3 - 3*a*b*tan(f* x + e) + 3*b^2*tan(f*x + e))/b^3)/f
Time = 17.17 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94 \[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {{\mathrm {tan}\left (e+f\,x\right )}^3}{3\,b\,f}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {a+b}{b^2}-\frac {2}{b}\right )}{f}+\frac {a^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a+b}}\right )}{b^{5/2}\,f\,\sqrt {a+b}} \] Input:
int(1/(cos(e + f*x)^6*(a + b/cos(e + f*x)^2)),x)
Output:
tan(e + f*x)^3/(3*b*f) - (tan(e + f*x)*((a + b)/b^2 - 2/b))/f + (a^2*atan( (b^(1/2)*tan(e + f*x))/(a + b)^(1/2)))/(b^(5/2)*f*(a + b)^(1/2))
Time = 0.19 (sec) , antiderivative size = 290, normalized size of antiderivative = 3.77 \[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {3 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a^{2}-3 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) \cos \left (f x +e \right ) a^{2}+3 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a^{2}-3 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) \cos \left (f x +e \right ) a^{2}-3 \sin \left (f x +e \right )^{3} a^{2} b -\sin \left (f x +e \right )^{3} a \,b^{2}+2 \sin \left (f x +e \right )^{3} b^{3}+3 \sin \left (f x +e \right ) a^{2} b -3 \sin \left (f x +e \right ) b^{3}}{3 \cos \left (f x +e \right ) b^{3} f \left (\sin \left (f x +e \right )^{2} a +\sin \left (f x +e \right )^{2} b -a -b \right )} \] Input:
int(sec(f*x+e)^6/(a+b*sec(f*x+e)^2),x)
Output:
(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**2 - 3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*cos(e + f*x)*a**2 + 3*sqrt(b)*s qrt(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 - 3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*cos(e + f*x)*a**2 - 3*sin(e + f*x)**3*a**2* b - sin(e + f*x)**3*a*b**2 + 2*sin(e + f*x)**3*b**3 + 3*sin(e + f*x)*a**2* b - 3*sin(e + f*x)*b**3)/(3*cos(e + f*x)*b**3*f*(sin(e + f*x)**2*a + sin(e + f*x)**2*b - a - b))