\(\int \frac {\sec ^4(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [200]

Optimal result

Integrand size = 23, antiderivative size = 82 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 b^{3/2} (a+b)^{3/2} f}-\frac {a \tan (e+f x)}{2 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 82, 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, 393, 211} \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 b^{3/2} f (a+b)^{3/2}}-\frac {a \tan (e+f x)}{2 b f (a+b) \left (a+b \tan ^2(e+f x)+b\right )} \]

[In]

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

[Out]

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

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 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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}{\left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {a \tan (e+f x)}{2 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {(a+2 b) \text {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 b (a+b) f} \\ & = \frac {(a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 b^{3/2} (a+b)^{3/2} f}-\frac {a \tan (e+f x)}{2 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {(a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {a \sqrt {b} \sin (2 (e+f x))}{(a+b) (a+2 b+a \cos (2 (e+f x)))}}{2 b^{3/2} f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {-\frac {a \tan \left (f x +e \right )}{2 \left (a +b \right ) b \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (a +2 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) b \sqrt {\left (a +b \right ) b}}}{f}\)	\(76\)
default	\(\frac {-\frac {a \tan \left (f x +e \right )}{2 \left (a +b \right ) b \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (a +2 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) b \sqrt {\left (a +b \right ) b}}}{f}\)	\(76\)
risch	\(-\frac {i \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}{b f \left (a +b \right ) \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}-\frac {a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right ) f b}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{2 \sqrt {-a b -b^{2}}\, \left (a +b \right ) f}+\frac {a \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i b a +2 i b^{2}-a \sqrt {-a b -b^{2}}-2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right ) f b}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i b a +2 i b^{2}-a \sqrt {-a b -b^{2}}-2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{2 \sqrt {-a b -b^{2}}\, \left (a +b \right ) f}\)	\(452\)

[In]

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

[Out]

1/f*(-1/2*a/(a+b)/b*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)+1/2*(a+2*b)/(a+b)/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. 163 vs. \(2 (70) = 140\).

Time = 0.28 (sec) , antiderivative size = 406, normalized size of antiderivative = 4.95 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [-\frac {4 \, {\left (a^{2} b + a b^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt {-a b - b^{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{8 \, {\left ({\left (a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} f\right )}}, -\frac {2 \, {\left (a^{2} b + a b^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt {a b + b^{2}} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{4 \, {\left ({\left (a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} f\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {a \tan \left (f x + e\right )}{a^{2} b + 2 \, a b^{2} + b^{3} + {\left (a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{2}} - \frac {{\left (a + 2 \, b\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} {\left (a b + b^{2}\right )}}}{2 \, f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (a + 2 \, b\right )}}{{\left (a b + b^{2}\right )}^{\frac {3}{2}}} - \frac {a \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )} {\left (a b + b^{2}\right )}}}{2 \, f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 19.61 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a+b}}\right )\,\left (a+2\,b\right )}{2\,b^{3/2}\,f\,{\left (a+b\right )}^{3/2}}-\frac {a\,\mathrm {tan}\left (e+f\,x\right )}{2\,b\,f\,\left (a+b\right )\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )} \]

[In]

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

[Out]

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