Integrand size = 13, antiderivative size = 73 \[ \int \frac {\sec (x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\frac {\sqrt {b} (3 a+b) \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{2 a^{3/2} (a+b)^2}+\frac {\text {arctanh}(\sin (x))}{(a+b)^2}+\frac {b \sin (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )} \]
arctanh(sin(x))/(a+b)^2+1/2*b*sin(x)/a/(a+b)/(a+b*sin(x)^2)+1/2*(3*a+b)*ar ctan(sin(x)*b^(1/2)/a^(1/2))*b^(1/2)/a^(3/2)/(a+b)^2
Time = 0.56 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.78 \[ \int \frac {\sec (x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\frac {-\frac {\sqrt {b} (3 a+b) \arctan \left (\frac {\sqrt {a} \csc (x)}{\sqrt {b}}\right )}{a^{3/2}}+\frac {\sqrt {b} (3 a+b) \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{a^{3/2}}+4 \left (-\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )+\frac {b (a+b) \sin (x)}{a (2 a+b-b \cos (2 x))}\right )}{4 (a+b)^2} \]
(-((Sqrt[b]*(3*a + b)*ArcTan[(Sqrt[a]*Csc[x])/Sqrt[b]])/a^(3/2)) + (Sqrt[b ]*(3*a + b)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/a^(3/2) + 4*(-Log[Cos[x/2] - Sin[x/2]] + Log[Cos[x/2] + Sin[x/2]] + (b*(a + b)*Sin[x])/(a*(2*a + b - b *Cos[2*x]))))/(4*(a + b)^2)
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.18, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3042, 3669, 316, 397, 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.
| \(\displaystyle \int \frac {\sec (x)}{\left (a+b \sin ^2(x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (x) \left (a+b \sin (x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 3669 |
| \(\displaystyle \int \frac {1}{\left (1-\sin ^2(x)\right ) \left (a+b \sin ^2(x)\right )^2}d\sin (x)\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {b \sin (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )}-\frac {\int \frac {b \sin ^2(x)+b-2 (a+b)}{\left (1-\sin ^2(x)\right ) \left (b \sin ^2(x)+a\right )}d\sin (x)}{2 a (a+b)}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {b \sin (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )}-\frac {-\frac {2 a \int \frac {1}{1-\sin ^2(x)}d\sin (x)}{a+b}-\frac {b (3 a+b) \int \frac {1}{b \sin ^2(x)+a}d\sin (x)}{a+b}}{2 a (a+b)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {b \sin (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )}-\frac {-\frac {2 a \int \frac {1}{1-\sin ^2(x)}d\sin (x)}{a+b}-\frac {\sqrt {b} (3 a+b) \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}}{2 a (a+b)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {b \sin (x)}{2 a (a+b) \left (a+b \sin ^2(x)\right )}-\frac {-\frac {\sqrt {b} (3 a+b) \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} (a+b)}-\frac {2 a \text {arctanh}(\sin (x))}{a+b}}{2 a (a+b)}\) |
-1/2*(-((Sqrt[b]*(3*a + b)*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(Sqrt[a]*(a + b))) - (2*a*ArcTanh[Sin[x]])/(a + b))/(a*(a + b)) + (b*Sin[x])/(2*a*(a + b)*(a + b*Sin[x]^2))
3.4.20.3.1 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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
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 = 0.84 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.08
| method | result | size |
| default | \(\frac {\ln \left (1+\sin \left (x \right )\right )}{2 \left (a +b \right )^{2}}+\frac {b \left (\frac {\left (a +b \right ) \sin \left (x \right )}{2 a \left (a +b \left (\sin ^{2}\left (x \right )\right )\right )}+\frac {\left (3 a +b \right ) \arctan \left (\frac {b \sin \left (x \right )}{\sqrt {a b}}\right )}{2 a \sqrt {a b}}\right )}{\left (a +b \right )^{2}}-\frac {\ln \left (\sin \left (x \right )-1\right )}{2 \left (a +b \right )^{2}}\) | \(79\) |
| risch | \(\frac {i b \left ({\mathrm e}^{3 i x}-{\mathrm e}^{i x}\right )}{a \left (a +b \right ) \left (b \,{\mathrm e}^{4 i x}-4 a \,{\mathrm e}^{2 i x}-2 b \,{\mathrm e}^{2 i x}+b \right )}+\frac {\ln \left ({\mathrm e}^{i x}+i\right )}{a^{2}+2 a b +b^{2}}-\frac {\ln \left ({\mathrm e}^{i x}-i\right )}{a^{2}+2 a b +b^{2}}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-a b}\, {\mathrm e}^{i x}}{b}-1\right )}{4 a \left (a +b \right )^{2}}+\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-a b}\, {\mathrm e}^{i x}}{b}-1\right ) b}{4 a^{2} \left (a +b \right )^{2}}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-a b}\, {\mathrm e}^{i x}}{b}-1\right )}{4 a \left (a +b \right )^{2}}-\frac {\sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-a b}\, {\mathrm e}^{i x}}{b}-1\right ) b}{4 a^{2} \left (a +b \right )^{2}}\) | \(267\) |
1/2/(a+b)^2*ln(1+sin(x))+b/(a+b)^2*(1/2*(a+b)/a*sin(x)/(a+b*sin(x)^2)+1/2* (3*a+b)/a/(a*b)^(1/2)*arctan(b*sin(x)/(a*b)^(1/2)))-1/2/(a+b)^2*ln(sin(x)- 1)
Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (61) = 122\).
Time = 0.35 (sec) , antiderivative size = 354, normalized size of antiderivative = 4.85 \[ \int \frac {\sec (x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\left [-\frac {{\left ({\left (3 \, a b + b^{2}\right )} \cos \left (x\right )^{2} - 3 \, a^{2} - 4 \, a b - b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {b \cos \left (x\right )^{2} - 2 \, a \sqrt {-\frac {b}{a}} \sin \left (x\right ) + a - b}{b \cos \left (x\right )^{2} - a - b}\right ) + 2 \, {\left (a b \cos \left (x\right )^{2} - a^{2} - a b\right )} \log \left (\sin \left (x\right ) + 1\right ) - 2 \, {\left (a b \cos \left (x\right )^{2} - a^{2} - a b\right )} \log \left (-\sin \left (x\right ) + 1\right ) - 2 \, {\left (a b + b^{2}\right )} \sin \left (x\right )}{4 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3} - {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (x\right )^{2}\right )}}, -\frac {{\left ({\left (3 \, a b + b^{2}\right )} \cos \left (x\right )^{2} - 3 \, a^{2} - 4 \, a b - b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\sqrt {\frac {b}{a}} \sin \left (x\right )\right ) + {\left (a b \cos \left (x\right )^{2} - a^{2} - a b\right )} \log \left (\sin \left (x\right ) + 1\right ) - {\left (a b \cos \left (x\right )^{2} - a^{2} - a b\right )} \log \left (-\sin \left (x\right ) + 1\right ) - {\left (a b + b^{2}\right )} \sin \left (x\right )}{2 \, {\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3} - {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} \cos \left (x\right )^{2}\right )}}\right ] \]
[-1/4*(((3*a*b + b^2)*cos(x)^2 - 3*a^2 - 4*a*b - b^2)*sqrt(-b/a)*log(-(b*c os(x)^2 - 2*a*sqrt(-b/a)*sin(x) + a - b)/(b*cos(x)^2 - a - b)) + 2*(a*b*co s(x)^2 - a^2 - a*b)*log(sin(x) + 1) - 2*(a*b*cos(x)^2 - a^2 - a*b)*log(-si n(x) + 1) - 2*(a*b + b^2)*sin(x))/(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3 - (a^ 3*b + 2*a^2*b^2 + a*b^3)*cos(x)^2), -1/2*(((3*a*b + b^2)*cos(x)^2 - 3*a^2 - 4*a*b - b^2)*sqrt(b/a)*arctan(sqrt(b/a)*sin(x)) + (a*b*cos(x)^2 - a^2 - a*b)*log(sin(x) + 1) - (a*b*cos(x)^2 - a^2 - a*b)*log(-sin(x) + 1) - (a*b + b^2)*sin(x))/(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3 - (a^3*b + 2*a^2*b^2 + a *b^3)*cos(x)^2)]
\[ \int \frac {\sec (x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\int \frac {\sec {\left (x \right )}}{\left (a + b \sin ^{2}{\left (x \right )}\right )^{2}}\, dx \]
Time = 0.38 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.58 \[ \int \frac {\sec (x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\frac {b \sin \left (x\right )}{2 \, {\left (a^{3} + a^{2} b + {\left (a^{2} b + a b^{2}\right )} \sin \left (x\right )^{2}\right )}} + \frac {{\left (3 \, a b + b^{2}\right )} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {a b}} + \frac {\log \left (\sin \left (x\right ) + 1\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {\log \left (\sin \left (x\right ) - 1\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} \]
1/2*b*sin(x)/(a^3 + a^2*b + (a^2*b + a*b^2)*sin(x)^2) + 1/2*(3*a*b + b^2)* arctan(b*sin(x)/sqrt(a*b))/((a^3 + 2*a^2*b + a*b^2)*sqrt(a*b)) + 1/2*log(s in(x) + 1)/(a^2 + 2*a*b + b^2) - 1/2*log(sin(x) - 1)/(a^2 + 2*a*b + b^2)
Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.49 \[ \int \frac {\sec (x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\frac {{\left (3 \, a b + b^{2}\right )} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {a b}} + \frac {\log \left (\sin \left (x\right ) + 1\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {\log \left (-\sin \left (x\right ) + 1\right )}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {b \sin \left (x\right )}{2 \, {\left (b \sin \left (x\right )^{2} + a\right )} {\left (a^{2} + a b\right )}} \]
1/2*(3*a*b + b^2)*arctan(b*sin(x)/sqrt(a*b))/((a^3 + 2*a^2*b + a*b^2)*sqrt (a*b)) + 1/2*log(sin(x) + 1)/(a^2 + 2*a*b + b^2) - 1/2*log(-sin(x) + 1)/(a ^2 + 2*a*b + b^2) + 1/2*b*sin(x)/((b*sin(x)^2 + a)*(a^2 + a*b))
Time = 14.92 (sec) , antiderivative size = 2213, normalized size of antiderivative = 30.32 \[ \int \frac {\sec (x)}{\left (a+b \sin ^2(x)\right )^2} \, dx=\text {Too large to display} \]
(b*sin(x))/(2*a*(a + b)*(a + b*sin(x)^2)) - (atan((((3*a + b)*(-a^3*b)^(1/ 2)*((sin(x)*(6*a*b^4 + b^5 + 13*a^2*b^3))/(2*(2*a^3*b + a^4 + a^2*b^2)) + ((3*a + b)*(-a^3*b)^(1/2)*((2*a*b^7 + 12*a^2*b^6 + 28*a^3*b^5 + 32*a^4*b^4 + 18*a^5*b^3 + 4*a^6*b^2)/(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2) - (sin(x) *(3*a + b)*(-a^3*b)^(1/2)*(16*a^2*b^7 + 48*a^3*b^6 + 32*a^4*b^5 - 32*a^5*b ^4 - 48*a^6*b^3 - 16*a^7*b^2))/(8*(2*a^3*b + a^4 + a^2*b^2)*(2*a^4*b + a^5 + a^3*b^2))))/(4*(2*a^4*b + a^5 + a^3*b^2)))*1i)/(4*(2*a^4*b + a^5 + a^3* b^2)) + ((3*a + b)*(-a^3*b)^(1/2)*((sin(x)*(6*a*b^4 + b^5 + 13*a^2*b^3))/( 2*(2*a^3*b + a^4 + a^2*b^2)) - ((3*a + b)*(-a^3*b)^(1/2)*((2*a*b^7 + 12*a^ 2*b^6 + 28*a^3*b^5 + 32*a^4*b^4 + 18*a^5*b^3 + 4*a^6*b^2)/(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2) + (sin(x)*(3*a + b)*(-a^3*b)^(1/2)*(16*a^2*b^7 + 48* a^3*b^6 + 32*a^4*b^5 - 32*a^5*b^4 - 48*a^6*b^3 - 16*a^7*b^2))/(8*(2*a^3*b + a^4 + a^2*b^2)*(2*a^4*b + a^5 + a^3*b^2))))/(4*(2*a^4*b + a^5 + a^3*b^2) ))*1i)/(4*(2*a^4*b + a^5 + a^3*b^2)))/(((3*a*b^3)/2 + b^4/2)/(3*a^4*b + a^ 5 + a^2*b^3 + 3*a^3*b^2) + ((3*a + b)*(-a^3*b)^(1/2)*((sin(x)*(6*a*b^4 + b ^5 + 13*a^2*b^3))/(2*(2*a^3*b + a^4 + a^2*b^2)) + ((3*a + b)*(-a^3*b)^(1/2 )*((2*a*b^7 + 12*a^2*b^6 + 28*a^3*b^5 + 32*a^4*b^4 + 18*a^5*b^3 + 4*a^6*b^ 2)/(3*a^4*b + a^5 + a^2*b^3 + 3*a^3*b^2) - (sin(x)*(3*a + b)*(-a^3*b)^(1/2 )*(16*a^2*b^7 + 48*a^3*b^6 + 32*a^4*b^5 - 32*a^5*b^4 - 48*a^6*b^3 - 16*a^7 *b^2))/(8*(2*a^3*b + a^4 + a^2*b^2)*(2*a^4*b + a^5 + a^3*b^2))))/(4*(2*...