\(\int \frac {\sec (e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [195]

Optimal result

Integrand size = 21, antiderivative size = 83 \[ \int \frac {\sec (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(2 a+b) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{2 a^{3/2} (a+b)^{3/2} f}-\frac {b \sin (e+f x)}{2 a (a+b) f \left (a+b-a \sin ^2(e+f x)\right )} \]

[Out]

1/2*(2*a+b)*arctanh(sin(f*x+e)*a^(1/2)/(a+b)^(1/2))/a^(3/2)/(a+b)^(3/2)/f-1/2*b*sin(f*x+e)/a/(a+b)/f/(a+b-a*si
n(f*x+e)^2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4232, 393, 214} \[ \int \frac {\sec (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(2 a+b) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{2 a^{3/2} f (a+b)^{3/2}}-\frac {b \sin (e+f x)}{2 a f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )} \]

[In]

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

[Out]

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

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[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 4232

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

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int \frac {\sec (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {(2 a+b) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {2 \sqrt {a} b \sin (e+f x)}{(a+b) (a+2 b+a \cos (2 (e+f x)))}}{2 a^{3/2} f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {\frac {b \sin \left (f x +e \right )}{2 a \left (a +b \right ) \left (a \sin \left (f x +e \right )^{2}-a -b \right )}+\frac {\left (2 a +b \right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 a \left (a +b \right ) \sqrt {a \left (a +b \right )}}}{f}\)	\(80\)
default	\(\frac {\frac {b \sin \left (f x +e \right )}{2 a \left (a +b \right ) \left (a \sin \left (f x +e \right )^{2}-a -b \right )}+\frac {\left (2 a +b \right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{2 a \left (a +b \right ) \sqrt {a \left (a +b \right )}}}{f}\)	\(80\)
risch	\(\frac {i b \left ({\mathrm e}^{3 i \left (f x +e \right )}-{\mathrm e}^{i \left (f x +e \right )}\right )}{a f \left (a +b \right ) \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{2 \sqrt {a^{2}+a b}\, \left (a +b \right ) f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right ) b}{4 \sqrt {a^{2}+a b}\, \left (a +b \right ) f a}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{2 \sqrt {a^{2}+a b}\, \left (a +b \right ) f}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right ) b}{4 \sqrt {a^{2}+a b}\, \left (a +b \right ) f a}\)	\(305\)

[In]

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

[Out]

1/f*(1/2*b/a/(a+b)*sin(f*x+e)/(a*sin(f*x+e)^2-a-b)+1/2*(2*a+b)/a/(a+b)/(a*(a+b))^(1/2)*arctanh(a*sin(f*x+e)/(a
*(a+b))^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 301, normalized size of antiderivative = 3.63 \[ \int \frac {\sec (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [\frac {{\left ({\left (2 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + 2 \, a b + b^{2}\right )} \sqrt {a^{2} + a b} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a^{2} + a b} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) - 2 \, {\left (a^{2} b + a b^{2}\right )} \sin \left (f x + e\right )}{4 \, {\left ({\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f\right )}}, -\frac {{\left ({\left (2 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + 2 \, a b + b^{2}\right )} \sqrt {-a^{2} - a b} \arctan \left (\frac {\sqrt {-a^{2} - a b} \sin \left (f x + e\right )}{a + b}\right ) + {\left (a^{2} b + a b^{2}\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.34 \[ \int \frac {\sec (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {2 \, b \sin \left (f x + e\right )}{a^{3} + 2 \, a^{2} b + a b^{2} - {\left (a^{3} + a^{2} b\right )} \sin \left (f x + e\right )^{2}} + \frac {{\left (2 \, a + b\right )} \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} {\left (a^{2} + a b\right )}}}{4 \, f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.10 \[ \int \frac {\sec (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {\frac {{\left (2 \, a + b\right )} \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{{\left (a^{2} + a b\right )} \sqrt {-a^{2} - a b}} - \frac {b \sin \left (f x + e\right )}{{\left (a \sin \left (f x + e\right )^{2} - a - b\right )} {\left (a^{2} + a b\right )}}}{2 \, f} \]

[In]

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

[Out]

-1/2*((2*a + b)*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/((a^2 + a*b)*sqrt(-a^2 - a*b)) - b*sin(f*x + e)/((a*si
n(f*x + e)^2 - a - b)*(a^2 + a*b)))/f

Mupad [B] (verification not implemented)

Time = 19.74 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.86 \[ \int \frac {\sec (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )}{\sqrt {a+b}}\right )\,\left (2\,a+b\right )}{2\,a^{3/2}\,f\,{\left (a+b\right )}^{3/2}}-\frac {b\,\sin \left (e+f\,x\right )}{2\,a\,f\,\left (a+b\right )\,\left (-a\,{\sin \left (e+f\,x\right )}^2+a+b\right )} \]

[In]

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

[Out]

(atanh((a^(1/2)*sin(e + f*x))/(a + b)^(1/2))*(2*a + b))/(2*a^(3/2)*f*(a + b)^(3/2)) - (b*sin(e + f*x))/(2*a*f*
(a + b)*(a + b - a*sin(e + f*x)^2))