Integrand size = 21, antiderivative size = 52 \[ \int \frac {\cos (e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {b \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{a^{3/2} \sqrt {a+b} f}+\frac {\sin (e+f x)}{a f} \] Output:
-b*arctanh(a^(1/2)*sin(f*x+e)/(a+b)^(1/2))/a^(3/2)/(a+b)^(1/2)/f+sin(f*x+e )/a/f
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {-\frac {b \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {a+b}}+\sqrt {a} \sin (e+f x)}{a^{3/2} f} \] Input:
Integrate[Cos[e + f*x]/(a + b*Sec[e + f*x]^2),x]
Output:
(-((b*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/Sqrt[a + b]) + Sqrt[a]* Sin[e + f*x])/(a^(3/2)*f)
Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 4635, 299, 221}
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.
\(\displaystyle \int \frac {\cos (e+f x)}{a+b \sec ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sec (e+f x) \left (a+b \sec (e+f x)^2\right )}dx\) |
\(\Big \downarrow \) 4635 |
\(\displaystyle \frac {\int \frac {1-\sin ^2(e+f x)}{-a \sin ^2(e+f x)+a+b}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\frac {\sin (e+f x)}{a}-\frac {b \int \frac {1}{-a \sin ^2(e+f x)+a+b}d\sin (e+f x)}{a}}{f}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\sin (e+f x)}{a}-\frac {b \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{a^{3/2} \sqrt {a+b}}}{f}\) |
Input:
Int[Cos[e + f*x]/(a + b*Sec[e + f*x]^2),x]
Output:
(-((b*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(a^(3/2)*Sqrt[a + b])) + Sin[e + f*x]/a)/f
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]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_ ))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[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] && In tegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Time = 0.58 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87
method | result | size |
derivativedivides | \(\frac {\frac {\sin \left (f x +e \right )}{a}-\frac {b \,\operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{a \sqrt {a \left (a +b \right )}}}{f}\) | \(45\) |
default | \(\frac {\frac {\sin \left (f x +e \right )}{a}-\frac {b \,\operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{a \sqrt {a \left (a +b \right )}}}{f}\) | \(45\) |
risch | \(-\frac {i {\mathrm e}^{i \left (f x +e \right )}}{2 a f}+\frac {i {\mathrm e}^{-i \left (f x +e \right )}}{2 a f}+\frac {b \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{2 \sqrt {a^{2}+a b}\, f a}-\frac {b \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{2 \sqrt {a^{2}+a b}\, f a}\) | \(146\) |
Input:
int(cos(f*x+e)/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/f*(sin(f*x+e)/a-b/a/(a*(a+b))^(1/2)*arctanh(a*sin(f*x+e)/(a*(a+b))^(1/2) ))
Time = 0.09 (sec) , antiderivative size = 164, normalized size of antiderivative = 3.15 \[ \int \frac {\cos (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [\frac {\sqrt {a^{2} + a b} b \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {a^{2} + a b} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 2 \, {\left (a^{2} + a b\right )} \sin \left (f x + e\right )}{2 \, {\left (a^{3} + a^{2} b\right )} f}, \frac {\sqrt {-a^{2} - a b} b \arctan \left (\frac {\sqrt {-a^{2} - a b} \sin \left (f x + e\right )}{a + b}\right ) + {\left (a^{2} + a b\right )} \sin \left (f x + e\right )}{{\left (a^{3} + a^{2} b\right )} f}\right ] \] Input:
integrate(cos(f*x+e)/(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
[1/2*(sqrt(a^2 + a*b)*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 + a*b)*sin(f*x + e))/((a ^3 + a^2*b)*f), (sqrt(-a^2 - a*b)*b*arctan(sqrt(-a^2 - a*b)*sin(f*x + e)/( a + b)) + (a^2 + a*b)*sin(f*x + e))/((a^3 + a^2*b)*f)]
\[ \int \frac {\cos (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\cos {\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \] Input:
integrate(cos(f*x+e)/(a+b*sec(f*x+e)**2),x)
Output:
Integral(cos(e + f*x)/(a + b*sec(e + f*x)**2), x)
Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.29 \[ \int \frac {\cos (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {b \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} a} + \frac {2 \, \sin \left (f x + e\right )}{a}}{2 \, f} \] Input:
integrate(cos(f*x+e)/(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/2*(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*sin(f*x + e)/a)/f
Time = 0.14 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02 \[ \int \frac {\cos (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {b \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{\sqrt {-a^{2} - a b} a} + \frac {\sin \left (f x + e\right )}{a}}{f} \] Input:
integrate(cos(f*x+e)/(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
(b*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/(sqrt(-a^2 - a*b)*a) + sin(f*x + e)/a)/f
Time = 16.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85 \[ \int \frac {\cos (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\sin \left (e+f\,x\right )}{a\,f}-\frac {b\,\mathrm {atanh}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )}{\sqrt {a+b}}\right )}{a^{3/2}\,f\,\sqrt {a+b}} \] Input:
int(cos(e + f*x)/(a + b/cos(e + f*x)^2),x)
Output:
sin(e + f*x)/(a*f) - (b*atanh((a^(1/2)*sin(e + f*x))/(a + b)^(1/2)))/(a^(3 /2)*f*(a + b)^(1/2))
Time = 0.16 (sec) , antiderivative size = 122, normalized size of antiderivative = 2.35 \[ \int \frac {\cos (e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\sqrt {a}\, \sqrt {a +b}\, \mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sqrt {a +b}-2 \sqrt {a}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b -\sqrt {a}\, \sqrt {a +b}\, \mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sqrt {a +b}+2 \sqrt {a}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b +2 \sin \left (f x +e \right ) a^{2}+2 \sin \left (f x +e \right ) a b}{2 a^{2} f \left (a +b \right )} \] Input:
int(cos(f*x+e)/(a+b*sec(f*x+e)^2),x)
Output:
(sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2 *sqrt(a)*tan((e + f*x)/2))*b - sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*x)/2))*b + 2*sin(e + f*x )*a**2 + 2*sin(e + f*x)*a*b)/(2*a**2*f*(a + b))