\(\int \frac {\sec ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) [187]

Optimal result

Integrand size = 23, antiderivative size = 52 \[ \int \frac {\sec ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {a \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{b^{3/2} \sqrt {a+b} f}+\frac {\tan (e+f x)}{b f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4231, 396, 211} \[ \int \frac {\sec ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\tan (e+f x)}{b f}-\frac {a \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{b^{3/2} f \sqrt {a+b}} \]

[In]

Int[Sec[e + f*x]^4/(a + b*Sec[e + f*x]^2),x]

[Out]

-((a*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(b^(3/2)*Sqrt[a + b]*f)) + Tan[e + f*x]/(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 396

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(
p + 1) + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.65 (sec) , antiderivative size = 192, normalized size of antiderivative = 3.69 \[ \int \frac {\sec ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (a \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))+\sqrt {a+b} \sec (e) \sec (e+f x) \sqrt {b (i \cos (e)+\sin (e))^4} \sin (f x)\right )}{2 b \sqrt {a+b} f \left (a+b \sec ^2(e+f x)\right ) \sqrt {b (\cos (e)-i \sin (e))^4}} \]

[In]

Integrate[Sec[e + f*x]^4/(a + b*Sec[e + f*x]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (f x +e \right )}{b}-\frac {a \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{b \sqrt {\left (a +b \right ) b}}}{f}\)	\(45\)
default	\(\frac {\frac {\tan \left (f x +e \right )}{b}-\frac {a \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{b \sqrt {\left (a +b \right ) b}}}{f}\)	\(45\)
risch	\(\frac {2 i}{f b \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {-2 i b a -2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{2 \sqrt {-a b -b^{2}}\, f b}+\frac {a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{2 \sqrt {-a b -b^{2}}\, f b}\)	\(202\)

[In]

int(sec(f*x+e)^4/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

1/f*(tan(f*x+e)/b-a/b/((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. 103 vs. \(2 (44) = 88\).

Time = 0.28 (sec) , antiderivative size = 286, normalized size of antiderivative = 5.50 \[ \int \frac {\sec ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [-\frac {\sqrt {-a b - b^{2}} a \cos \left (f x + e\right ) \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) - 4 \, {\left (a b + b^{2}\right )} \sin \left (f x + e\right )}{4 \, {\left (a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )}, \frac {\sqrt {a b + b^{2}} a \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right ) + 2 \, {\left (a b + b^{2}\right )} \sin \left (f x + e\right )}{2 \, {\left (a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )}\right ] \]

[In]

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

[-1/4*(sqrt(-a*b - b^2)*a*cos(f*x + e)*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))*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 + b^2)*sin(f*x + e))/((a*b^2 + b^3)*f*cos(f*x + e)), 1/2*(sqrt(a*b +
b^2)*a*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) + 2
*(a*b + b^2)*sin(f*x + e))/((a*b^2 + b^3)*f*cos(f*x + e))]

Sympy [F]

\[ \int \frac {\sec ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\sec ^{4}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]

[In]

integrate(sec(f*x+e)**4/(a+b*sec(f*x+e)**2),x)

[Out]

Integral(sec(e + f*x)**4/(a + b*sec(e + f*x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87 \[ \int \frac {\sec ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {a \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} b} - \frac {\tan \left (f x + e\right )}{b}}{f} \]

[In]

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

-(a*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)*b) - tan(f*x + e)/b)/f

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.27 \[ \int \frac {\sec ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} a}{\sqrt {a b + b^{2}} b} - \frac {\tan \left (f x + e\right )}{b}}{f} \]

[In]

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

-((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*a/(sqrt(a*b + b^2)*b) - tan(f
*x + e)/b)/f

Mupad [B] (verification not implemented)

Time = 18.46 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^4(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )}{b\,f}-\frac {a\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a+b}}\right )}{b^{3/2}\,f\,\sqrt {a+b}} \]

[In]

int(1/(cos(e + f*x)^4*(a + b/cos(e + f*x)^2)),x)

[Out]

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