Integrand size = 13, antiderivative size = 40 \[ \int \frac {\sec (x)}{a+b \sin ^2(x)} \, dx=\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}+\frac {\text {arctanh}(\sin (x))}{a+b} \]
Leaf count is larger than twice the leaf count of optimal. \(96\) vs. \(2(40)=80\).
Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 2.40 \[ \int \frac {\sec (x)}{a+b \sin ^2(x)} \, dx=\frac {-\sqrt {b} \arctan \left (\frac {\sqrt {a} \csc (x)}{\sqrt {b}}\right )+\sqrt {b} \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )+2 \sqrt {a} \left (-\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )\right )}{2 \sqrt {a} (a+b)} \]
(-(Sqrt[b]*ArcTan[(Sqrt[a]*Csc[x])/Sqrt[b]]) + Sqrt[b]*ArcTan[(Sqrt[b]*Sin [x])/Sqrt[a]] + 2*Sqrt[a]*(-Log[Cos[x/2] - Sin[x/2]] + Log[Cos[x/2] + Sin[ x/2]]))/(2*Sqrt[a]*(a + b))
Time = 0.22 (sec) , antiderivative size = 40, 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, 3669, 303, 218, 219}
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.
\(\displaystyle \int \frac {\sec (x)}{a+b \sin ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (x) \left (a+b \sin (x)^2\right )}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \int \frac {1}{\left (1-\sin ^2(x)\right ) \left (a+b \sin ^2(x)\right )}d\sin (x)\) |
\(\Big \downarrow \) 303 |
\(\displaystyle \frac {\int \frac {1}{1-\sin ^2(x)}d\sin (x)}{a+b}+\frac {b \int \frac {1}{b \sin ^2(x)+a}d\sin (x)}{a+b}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\int \frac {1}{1-\sin ^2(x)}d\sin (x)}{a+b}+\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}+\frac {\text {arctanh}(\sin (x))}{a+b}\) |
3.4.8.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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[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[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f S ubst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x] /ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Time = 0.46 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.38
method | result | size |
default | \(-\frac {\ln \left (\sin \left (x \right )-1\right )}{2 a +2 b}+\frac {b \arctan \left (\frac {b \sin \left (x \right )}{\sqrt {a b}}\right )}{\left (a +b \right ) \sqrt {a b}}+\frac {\ln \left (1+\sin \left (x \right )\right )}{2 a +2 b}\) | \(55\) |
risch | \(\frac {\ln \left ({\mathrm e}^{i x}+i\right )}{a +b}-\frac {\ln \left ({\mathrm e}^{i x}-i\right )}{a +b}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-a b}\, {\mathrm e}^{i x}}{b}-1\right )}{2 a \left (a +b \right )}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-a b}\, {\mathrm e}^{i x}}{b}-1\right )}{2 a \left (a +b \right )}\) | \(115\) |
-1/(2*a+2*b)*ln(sin(x)-1)+b/(a+b)/(a*b)^(1/2)*arctan(b*sin(x)/(a*b)^(1/2)) +1/(2*a+2*b)*ln(1+sin(x))
Time = 0.33 (sec) , antiderivative size = 116, normalized size of antiderivative = 2.90 \[ \int \frac {\sec (x)}{a+b \sin ^2(x)} \, dx=\left [\frac {\sqrt {-\frac {b}{a}} \log \left (-\frac {b \cos \left (x\right )^{2} - 2 \, a \sqrt {-\frac {b}{a}} \sin \left (x\right ) + a - b}{b \cos \left (x\right )^{2} - a - b}\right ) + \log \left (\sin \left (x\right ) + 1\right ) - \log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a + b\right )}}, \frac {2 \, \sqrt {\frac {b}{a}} \arctan \left (\sqrt {\frac {b}{a}} \sin \left (x\right )\right ) + \log \left (\sin \left (x\right ) + 1\right ) - \log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a + b\right )}}\right ] \]
[1/2*(sqrt(-b/a)*log(-(b*cos(x)^2 - 2*a*sqrt(-b/a)*sin(x) + a - b)/(b*cos( x)^2 - a - b)) + log(sin(x) + 1) - log(-sin(x) + 1))/(a + b), 1/2*(2*sqrt( b/a)*arctan(sqrt(b/a)*sin(x)) + log(sin(x) + 1) - log(-sin(x) + 1))/(a + b )]
\[ \int \frac {\sec (x)}{a+b \sin ^2(x)} \, dx=\int \frac {\sec {\left (x \right )}}{a + b \sin ^{2}{\left (x \right )}}\, dx \]
Time = 0.36 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.18 \[ \int \frac {\sec (x)}{a+b \sin ^2(x)} \, dx=\frac {b \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{\sqrt {a b} {\left (a + b\right )}} + \frac {\log \left (\sin \left (x\right ) + 1\right )}{2 \, {\left (a + b\right )}} - \frac {\log \left (\sin \left (x\right ) - 1\right )}{2 \, {\left (a + b\right )}} \]
b*arctan(b*sin(x)/sqrt(a*b))/(sqrt(a*b)*(a + b)) + 1/2*log(sin(x) + 1)/(a + b) - 1/2*log(sin(x) - 1)/(a + b)
Time = 0.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.22 \[ \int \frac {\sec (x)}{a+b \sin ^2(x)} \, dx=\frac {b \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{\sqrt {a b} {\left (a + b\right )}} + \frac {\log \left (\sin \left (x\right ) + 1\right )}{2 \, {\left (a + b\right )}} - \frac {\log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a + b\right )}} \]
b*arctan(b*sin(x)/sqrt(a*b))/(sqrt(a*b)*(a + b)) + 1/2*log(sin(x) + 1)/(a + b) - 1/2*log(-sin(x) + 1)/(a + b)
Time = 15.20 (sec) , antiderivative size = 856, normalized size of antiderivative = 21.40 \[ \int \frac {\sec (x)}{a+b \sin ^2(x)} \, dx=-\frac {\mathrm {atan}\left (\frac {\frac {\left (4\,b^3\,\sin \left (x\right )+\frac {8\,a\,b^3+4\,b^4+4\,a^2\,b^2-\frac {\sin \left (x\right )\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{2\,\left (a+b\right )}}{2\,\left (a+b\right )}\right )\,1{}\mathrm {i}}{2\,\left (a+b\right )}+\frac {\left (4\,b^3\,\sin \left (x\right )-\frac {8\,a\,b^3+4\,b^4+4\,a^2\,b^2+\frac {\sin \left (x\right )\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{2\,\left (a+b\right )}}{2\,\left (a+b\right )}\right )\,1{}\mathrm {i}}{2\,\left (a+b\right )}}{\frac {4\,b^3\,\sin \left (x\right )+\frac {8\,a\,b^3+4\,b^4+4\,a^2\,b^2-\frac {\sin \left (x\right )\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{2\,\left (a+b\right )}}{2\,\left (a+b\right )}}{2\,\left (a+b\right )}-\frac {4\,b^3\,\sin \left (x\right )-\frac {8\,a\,b^3+4\,b^4+4\,a^2\,b^2+\frac {\sin \left (x\right )\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{2\,\left (a+b\right )}}{2\,\left (a+b\right )}}{2\,\left (a+b\right )}}\right )\,1{}\mathrm {i}}{a+b}-\frac {\mathrm {atan}\left (\frac {\frac {\left (2\,b^3\,\sin \left (x\right )+\frac {\sqrt {-a\,b}\,\left (4\,a\,b^3+2\,b^4+2\,a^2\,b^2-\frac {\sin \left (x\right )\,\sqrt {-a\,b}\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{4\,\left (a^2+b\,a\right )}\right )}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-a\,b}\,1{}\mathrm {i}}{a^2+b\,a}+\frac {\left (2\,b^3\,\sin \left (x\right )-\frac {\sqrt {-a\,b}\,\left (4\,a\,b^3+2\,b^4+2\,a^2\,b^2+\frac {\sin \left (x\right )\,\sqrt {-a\,b}\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{4\,\left (a^2+b\,a\right )}\right )}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-a\,b}\,1{}\mathrm {i}}{a^2+b\,a}}{\frac {\left (2\,b^3\,\sin \left (x\right )+\frac {\sqrt {-a\,b}\,\left (4\,a\,b^3+2\,b^4+2\,a^2\,b^2-\frac {\sin \left (x\right )\,\sqrt {-a\,b}\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{4\,\left (a^2+b\,a\right )}\right )}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-a\,b}}{a^2+b\,a}-\frac {\left (2\,b^3\,\sin \left (x\right )-\frac {\sqrt {-a\,b}\,\left (4\,a\,b^3+2\,b^4+2\,a^2\,b^2+\frac {\sin \left (x\right )\,\sqrt {-a\,b}\,\left (-8\,a^3\,b^2-8\,a^2\,b^3+8\,a\,b^4+8\,b^5\right )}{4\,\left (a^2+b\,a\right )}\right )}{2\,\left (a^2+b\,a\right )}\right )\,\sqrt {-a\,b}}{a^2+b\,a}}\right )\,\sqrt {-a\,b}\,1{}\mathrm {i}}{a\,\left (a+b\right )} \]
- (atan((((4*b^3*sin(x) + (8*a*b^3 + 4*b^4 + 4*a^2*b^2 - (sin(x)*(8*a*b^4 + 8*b^5 - 8*a^2*b^3 - 8*a^3*b^2))/(2*(a + b)))/(2*(a + b)))*1i)/(2*(a + b) ) + ((4*b^3*sin(x) - (8*a*b^3 + 4*b^4 + 4*a^2*b^2 + (sin(x)*(8*a*b^4 + 8*b ^5 - 8*a^2*b^3 - 8*a^3*b^2))/(2*(a + b)))/(2*(a + b)))*1i)/(2*(a + b)))/(( 4*b^3*sin(x) + (8*a*b^3 + 4*b^4 + 4*a^2*b^2 - (sin(x)*(8*a*b^4 + 8*b^5 - 8 *a^2*b^3 - 8*a^3*b^2))/(2*(a + b)))/(2*(a + b)))/(2*(a + b)) - (4*b^3*sin( x) - (8*a*b^3 + 4*b^4 + 4*a^2*b^2 + (sin(x)*(8*a*b^4 + 8*b^5 - 8*a^2*b^3 - 8*a^3*b^2))/(2*(a + b)))/(2*(a + b)))/(2*(a + b))))*1i)/(a + b) - (atan(( ((2*b^3*sin(x) + ((-a*b)^(1/2)*(4*a*b^3 + 2*b^4 + 2*a^2*b^2 - (sin(x)*(-a* b)^(1/2)*(8*a*b^4 + 8*b^5 - 8*a^2*b^3 - 8*a^3*b^2))/(4*(a*b + a^2))))/(2*( a*b + a^2)))*(-a*b)^(1/2)*1i)/(a*b + a^2) + ((2*b^3*sin(x) - ((-a*b)^(1/2) *(4*a*b^3 + 2*b^4 + 2*a^2*b^2 + (sin(x)*(-a*b)^(1/2)*(8*a*b^4 + 8*b^5 - 8* a^2*b^3 - 8*a^3*b^2))/(4*(a*b + a^2))))/(2*(a*b + a^2)))*(-a*b)^(1/2)*1i)/ (a*b + a^2))/(((2*b^3*sin(x) + ((-a*b)^(1/2)*(4*a*b^3 + 2*b^4 + 2*a^2*b^2 - (sin(x)*(-a*b)^(1/2)*(8*a*b^4 + 8*b^5 - 8*a^2*b^3 - 8*a^3*b^2))/(4*(a*b + a^2))))/(2*(a*b + a^2)))*(-a*b)^(1/2))/(a*b + a^2) - ((2*b^3*sin(x) - (( -a*b)^(1/2)*(4*a*b^3 + 2*b^4 + 2*a^2*b^2 + (sin(x)*(-a*b)^(1/2)*(8*a*b^4 + 8*b^5 - 8*a^2*b^3 - 8*a^3*b^2))/(4*(a*b + a^2))))/(2*(a*b + a^2)))*(-a*b) ^(1/2))/(a*b + a^2)))*(-a*b)^(1/2)*1i)/(a*(a + b))