\(\int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) [186]

Optimal result

Integrand size = 23, antiderivative size = 77 \[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {a^2 \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{b^{5/2} \sqrt {a+b} f}-\frac {(a-b) \tan (e+f x)}{b^2 f}+\frac {\tan ^3(e+f x)}{3 b f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4231, 398, 211} \[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {a^2 \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{b^{5/2} f \sqrt {a+b}}-\frac {(a-b) \tan (e+f x)}{b^2 f}+\frac {\tan ^3(e+f x)}{3 b f} \]

[In]

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

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 398

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

Rule 4231

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {a-b}{b^2}+\frac {x^2}{b}+\frac {a^2}{b^2 \left (a+b+b x^2\right )}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {(a-b) \tan (e+f x)}{b^2 f}+\frac {\tan ^3(e+f x)}{3 b f}+\frac {a^2 \text {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{b^2 f} \\ & = \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} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.79 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.91 \[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (-3 a^2 \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))+\sqrt {a+b} \sec (e+f x) \sqrt {b (i \cos (e)+\sin (e))^4} \left (\sec (e) \left (-3 a+2 b+b \sec ^2(e+f x)\right ) \sin (f x)+b \sec (e+f x) \tan (e)\right )\right )}{6 b^2 \sqrt {a+b} f \left (a+b \sec ^2(e+f x)\right ) \sqrt {b (\cos (e)-i \sin (e))^4}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.97 (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	\(-\frac {2 i \left (3 a \,{\mathrm e}^{4 i \left (f x +e \right )}+6 a \,{\mathrm e}^{2 i \left (f x +e \right )}-6 b \,{\mathrm e}^{2 i \left (f x +e \right )}+3 a -2 b \right )}{3 f \,b^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}-\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{2 \sqrt {-a b -b^{2}}\, f \,b^{2}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i b a +2 i b^{2}-a \sqrt {-a b -b^{2}}-2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{2 \sqrt {-a b -b^{2}}\, f \,b^{2}}\)	\(251\)

[In]

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

[Out]

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)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (67) = 134\).

Time = 0.28 (sec) , antiderivative size = 354, normalized size of antiderivative = 4.60 \[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [-\frac {3 \, \sqrt {-a b - b^{2}} a^{2} \cos \left (f x + e\right )^{3} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) - 4 \, {\left (a b^{2} + b^{3} - {\left (3 \, a^{2} b + a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{12 \, {\left (a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{3}}, -\frac {3 \, \sqrt {a b + b^{2}} a^{2} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{3} - 2 \, {\left (a b^{2} + b^{3} - {\left (3 \, a^{2} b + a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{6 \, {\left (a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{3}}\right ] \]

[In]

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

[Out]

[-1/12*(3*sqrt(-a*b - b^2)*a^2*cos(f*x + e)^3*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*co
s(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)/(s
qrt(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*cos(f*x + e)^3)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84 \[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {3 \, a^{2} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} b^{2}} + \frac {b \tan \left (f x + e\right )^{3} - 3 \, {\left (a - b\right )} \tan \left (f x + e\right )}{b^{2}}}{3 \, f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} a^{2}}{\sqrt {a b + b^{2}} b^{2}} + \frac {b^{2} \tan \left (f x + e\right )^{3} - 3 \, a b \tan \left (f x + e\right ) + 3 \, b^{2} \tan \left (f x + e\right )}{b^{3}}}{3 \, f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 18.55 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94 \[ \int \frac {\sec ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {{\mathrm {tan}\left (e+f\,x\right )}^3}{3\,b\,f}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (\frac {a+b}{b^2}-\frac {2}{b}\right )}{f}+\frac {a^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a+b}}\right )}{b^{5/2}\,f\,\sqrt {a+b}} \]

[In]

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

[Out]

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))