Integrand size = 13, antiderivative size = 41 \[ \int \frac {\sec (x)}{a+b \cos ^2(x)} \, dx=\frac {\text {arctanh}(\sin (x))}{a}-\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}} \]
Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \frac {\sec (x)}{a+b \cos ^2(x)} \, dx=\frac {\text {arctanh}(\sin (x))-\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{\sqrt {a+b}}}{a} \]
Time = 0.22 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 3665, 303, 219, 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.
\(\displaystyle \int \frac {\sec (x)}{a+b \cos ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin \left (x+\frac {\pi }{2}\right ) \left (a+b \sin \left (x+\frac {\pi }{2}\right )^2\right )}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle \int \frac {1}{\left (1-\sin ^2(x)\right ) \left (a-b \sin ^2(x)+b\right )}d\sin (x)\) |
\(\Big \downarrow \) 303 |
\(\displaystyle \frac {\int \frac {1}{1-\sin ^2(x)}d\sin (x)}{a}-\frac {b \int \frac {1}{-b \sin ^2(x)+a+b}d\sin (x)}{a}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\text {arctanh}(\sin (x))}{a}-\frac {b \int \frac {1}{-b \sin ^2(x)+a+b}d\sin (x)}{a}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\text {arctanh}(\sin (x))}{a}-\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sin (x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}\) |
3.1.32.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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[1/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Simp[b/(b *c - a*d) Int[1/(a + b*x^2), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x ^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.39 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.15
method | result | size |
default | \(-\frac {\ln \left (\sin \left (x \right )-1\right )}{2 a}+\frac {\ln \left (\sin \left (x \right )+1\right )}{2 a}-\frac {b \,\operatorname {arctanh}\left (\frac {b \sin \left (x \right )}{\sqrt {\left (a +b \right ) b}}\right )}{a \sqrt {\left (a +b \right ) b}}\) | \(47\) |
risch | \(\frac {\ln \left ({\mathrm e}^{i x}+i\right )}{a}-\frac {\ln \left ({\mathrm e}^{i x}-i\right )}{a}+\frac {\sqrt {\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{i x}}{b}-1\right )}{2 \left (a +b \right ) a}-\frac {\sqrt {\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {\left (a +b \right ) b}\, {\mathrm e}^{i x}}{b}-1\right )}{2 \left (a +b \right ) a}\) | \(115\) |
-1/2/a*ln(sin(x)-1)+1/2/a*ln(sin(x)+1)-b/a/((a+b)*b)^(1/2)*arctanh(b*sin(x )/((a+b)*b)^(1/2))
Time = 0.25 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.90 \[ \int \frac {\sec (x)}{a+b \cos ^2(x)} \, dx=\left [\frac {\sqrt {\frac {b}{a + b}} \log \left (-\frac {b \cos \left (x\right )^{2} + 2 \, {\left (a + b\right )} \sqrt {\frac {b}{a + b}} \sin \left (x\right ) - a - 2 \, b}{b \cos \left (x\right )^{2} + a}\right ) + \log \left (\sin \left (x\right ) + 1\right ) - \log \left (-\sin \left (x\right ) + 1\right )}{2 \, a}, \frac {2 \, \sqrt {-\frac {b}{a + b}} \arctan \left (\sqrt {-\frac {b}{a + b}} \sin \left (x\right )\right ) + \log \left (\sin \left (x\right ) + 1\right ) - \log \left (-\sin \left (x\right ) + 1\right )}{2 \, a}\right ] \]
[1/2*(sqrt(b/(a + b))*log(-(b*cos(x)^2 + 2*(a + b)*sqrt(b/(a + b))*sin(x) - a - 2*b)/(b*cos(x)^2 + a)) + log(sin(x) + 1) - log(-sin(x) + 1))/a, 1/2* (2*sqrt(-b/(a + b))*arctan(sqrt(-b/(a + b))*sin(x)) + log(sin(x) + 1) - lo g(-sin(x) + 1))/a]
\[ \int \frac {\sec (x)}{a+b \cos ^2(x)} \, dx=\int \frac {\sec {\left (x \right )}}{a + b \cos ^{2}{\left (x \right )}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.56 \[ \int \frac {\sec (x)}{a+b \cos ^2(x)} \, dx=\frac {b \log \left (\frac {b \sin \left (x\right ) - \sqrt {{\left (a + b\right )} b}}{b \sin \left (x\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{2 \, \sqrt {{\left (a + b\right )} b} a} + \frac {\log \left (\sin \left (x\right ) + 1\right )}{2 \, a} - \frac {\log \left (\sin \left (x\right ) - 1\right )}{2 \, a} \]
1/2*b*log((b*sin(x) - sqrt((a + b)*b))/(b*sin(x) + sqrt((a + b)*b)))/(sqrt ((a + b)*b)*a) + 1/2*log(sin(x) + 1)/a - 1/2*log(sin(x) - 1)/a
Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.39 \[ \int \frac {\sec (x)}{a+b \cos ^2(x)} \, dx=\frac {b \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {-a b - b^{2}}}\right )}{\sqrt {-a b - b^{2}} a} + \frac {\log \left (\sin \left (x\right ) + 1\right )}{2 \, a} - \frac {\log \left (-\sin \left (x\right ) + 1\right )}{2 \, a} \]
b*arctan(b*sin(x)/sqrt(-a*b - b^2))/(sqrt(-a*b - b^2)*a) + 1/2*log(sin(x) + 1)/a - 1/2*log(-sin(x) + 1)/a
Time = 2.69 (sec) , antiderivative size = 414, normalized size of antiderivative = 10.10 \[ \int \frac {\sec (x)}{a+b \cos ^2(x)} \, dx=\frac {\mathrm {atanh}\left (\sin \left (x\right )\right )}{a}+\frac {\mathrm {atan}\left (\frac {\frac {\left (2\,b^3\,\sin \left (x\right )+\frac {\left (2\,a^2\,b^2-\frac {\sin \left (x\right )\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}\,1{}\mathrm {i}}{a^2+b\,a}+\frac {\left (2\,b^3\,\sin \left (x\right )-\frac {\left (2\,a^2\,b^2+\frac {\sin \left (x\right )\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}\,1{}\mathrm {i}}{a^2+b\,a}}{\frac {\left (2\,b^3\,\sin \left (x\right )+\frac {\left (2\,a^2\,b^2-\frac {\sin \left (x\right )\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}}{a^2+b\,a}-\frac {\left (2\,b^3\,\sin \left (x\right )-\frac {\left (2\,a^2\,b^2+\frac {\sin \left (x\right )\,\left (8\,a^3\,b^2+16\,a^2\,b^3\right )\,\sqrt {b\,\left (a+b\right )}}{4\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {b\,\left (a+b\right )}}{a^2+b\,a}}\right )\,\sqrt {b\,\left (a+b\right )}\,1{}\mathrm {i}}{a^2+b\,a} \]
atanh(sin(x))/a + (atan((((2*b^3*sin(x) + ((2*a^2*b^2 - (sin(x)*(16*a^2*b^ 3 + 8*a^3*b^2)*(b*(a + b))^(1/2))/(4*(a*b + a^2)))*(b*(a + b))^(1/2))/(2*( a*b + a^2)))*(b*(a + b))^(1/2)*1i)/(a*b + a^2) + ((2*b^3*sin(x) - ((2*a^2* b^2 + (sin(x)*(16*a^2*b^3 + 8*a^3*b^2)*(b*(a + b))^(1/2))/(4*(a*b + a^2))) *(b*(a + b))^(1/2))/(2*(a*b + a^2)))*(b*(a + b))^(1/2)*1i)/(a*b + a^2))/(( (2*b^3*sin(x) + ((2*a^2*b^2 - (sin(x)*(16*a^2*b^3 + 8*a^3*b^2)*(b*(a + b)) ^(1/2))/(4*(a*b + a^2)))*(b*(a + b))^(1/2))/(2*(a*b + a^2)))*(b*(a + b))^( 1/2))/(a*b + a^2) - ((2*b^3*sin(x) - ((2*a^2*b^2 + (sin(x)*(16*a^2*b^3 + 8 *a^3*b^2)*(b*(a + b))^(1/2))/(4*(a*b + a^2)))*(b*(a + b))^(1/2))/(2*(a*b + a^2)))*(b*(a + b))^(1/2))/(a*b + a^2)))*(b*(a + b))^(1/2)*1i)/(a*b + a^2)