Integrand size = 15, antiderivative size = 59 \[ \int \frac {\cos ^4(x)}{a+b \sin ^2(x)} \, dx=-\frac {(2 a+3 b) x}{2 b^2}+\frac {(a+b)^{3/2} \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} b^2}-\frac {\cos (x) \sin (x)}{2 b} \]
-1/2*(2*a+3*b)*x/b^2-1/2*cos(x)*sin(x)/b+(a+b)^(3/2)*arctan((a+b)^(1/2)*ta n(x)/a^(1/2))/b^2/a^(1/2)
Time = 0.19 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^4(x)}{a+b \sin ^2(x)} \, dx=\frac {\frac {4 (a+b)^{3/2} \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a}}-2 (2 a x+3 b x+b \cos (x) \sin (x))}{4 b^2} \]
((4*(a + b)^(3/2)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/Sqrt[a] - 2*(2*a*x + 3*b*x + b*Cos[x]*Sin[x]))/(4*b^2)
Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.25, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3670, 316, 397, 216, 218}
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.
\(\displaystyle \int \frac {\cos ^4(x)}{a+b \sin ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (x)^4}{a+b \sin (x)^2}dx\) |
\(\Big \downarrow \) 3670 |
\(\displaystyle \int \frac {1}{\left (\tan ^2(x)+1\right )^2 \left ((a+b) \tan ^2(x)+a\right )}d\tan (x)\) |
\(\Big \downarrow \) 316 |
\(\displaystyle \frac {\int \frac {-\left ((a+b) \tan ^2(x)\right )+a+2 b}{\left (\tan ^2(x)+1\right ) \left ((a+b) \tan ^2(x)+a\right )}d\tan (x)}{2 b}-\frac {\tan (x)}{2 b \left (\tan ^2(x)+1\right )}\) |
\(\Big \downarrow \) 397 |
\(\displaystyle \frac {\frac {2 (a+b)^2 \int \frac {1}{(a+b) \tan ^2(x)+a}d\tan (x)}{b}-\frac {(2 a+3 b) \int \frac {1}{\tan ^2(x)+1}d\tan (x)}{b}}{2 b}-\frac {\tan (x)}{2 b \left (\tan ^2(x)+1\right )}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {2 (a+b)^2 \int \frac {1}{(a+b) \tan ^2(x)+a}d\tan (x)}{b}-\frac {(2 a+3 b) \arctan (\tan (x))}{b}}{2 b}-\frac {\tan (x)}{2 b \left (\tan ^2(x)+1\right )}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {2 (a+b)^{3/2} \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} b}-\frac {(2 a+3 b) \arctan (\tan (x))}{b}}{2 b}-\frac {\tan (x)}{2 b \left (\tan ^2(x)+1\right )}\) |
(-(((2*a + 3*b)*ArcTan[Tan[x]])/b) + (2*(a + b)^(3/2)*ArcTan[(Sqrt[a + b]* Tan[x])/Sqrt[a]])/(Sqrt[a]*b))/(2*b) - Tan[x]/(2*b*(1 + Tan[x]^2))
3.4.4.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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)^(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[Tan[e + f*x], x]}, Simp[ff/f Su bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Time = 0.72 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.07
method | result | size |
default | \(-\frac {\frac {b \tan \left (x \right )}{2+2 \left (\tan ^{2}\left (x \right )\right )}+\frac {\left (2 a +3 b \right ) \arctan \left (\tan \left (x \right )\right )}{2}}{b^{2}}+\frac {\left (a +b \right )^{2} \arctan \left (\frac {\left (a +b \right ) \tan \left (x \right )}{\sqrt {a \left (a +b \right )}}\right )}{b^{2} \sqrt {a \left (a +b \right )}}\) | \(63\) |
risch | \(-\frac {a x}{b^{2}}-\frac {3 x}{2 b}+\frac {i {\mathrm e}^{2 i x}}{8 b}-\frac {i {\mathrm e}^{-2 i x}}{8 b}+\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 b^{2}}+\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 a b}-\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 b^{2}}-\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 a b}\) | \(209\) |
-1/b^2*(1/2*b*tan(x)/(1+tan(x)^2)+1/2*(2*a+3*b)*arctan(tan(x)))+(a+b)^2/b^ 2/(a*(a+b))^(1/2)*arctan((a+b)*tan(x)/(a*(a+b))^(1/2))
Time = 0.34 (sec) , antiderivative size = 239, normalized size of antiderivative = 4.05 \[ \int \frac {\cos ^4(x)}{a+b \sin ^2(x)} \, dx=\left [-\frac {2 \, b \cos \left (x\right ) \sin \left (x\right ) - {\left (a + b\right )} \sqrt {-\frac {a + b}{a}} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (x\right )^{2} - 4 \, {\left ({\left (2 \, a^{2} + a b\right )} \cos \left (x\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (x\right )\right )} \sqrt {-\frac {a + b}{a}} \sin \left (x\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (x\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (x\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) + 2 \, {\left (2 \, a + 3 \, b\right )} x}{4 \, b^{2}}, -\frac {b \cos \left (x\right ) \sin \left (x\right ) + {\left (a + b\right )} \sqrt {\frac {a + b}{a}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a - b\right )} \sqrt {\frac {a + b}{a}}}{2 \, {\left (a + b\right )} \cos \left (x\right ) \sin \left (x\right )}\right ) + {\left (2 \, a + 3 \, b\right )} x}{2 \, b^{2}}\right ] \]
[-1/4*(2*b*cos(x)*sin(x) - (a + b)*sqrt(-(a + b)/a)*log(((8*a^2 + 8*a*b + b^2)*cos(x)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(x)^2 - 4*((2*a^2 + a*b)*cos(x) ^3 - (a^2 + a*b)*cos(x))*sqrt(-(a + b)/a)*sin(x) + a^2 + 2*a*b + b^2)/(b^2 *cos(x)^4 - 2*(a*b + b^2)*cos(x)^2 + a^2 + 2*a*b + b^2)) + 2*(2*a + 3*b)*x )/b^2, -1/2*(b*cos(x)*sin(x) + (a + b)*sqrt((a + b)/a)*arctan(1/2*((2*a + b)*cos(x)^2 - a - b)*sqrt((a + b)/a)/((a + b)*cos(x)*sin(x))) + (2*a + 3*b )*x)/b^2]
Timed out. \[ \int \frac {\cos ^4(x)}{a+b \sin ^2(x)} \, dx=\text {Timed out} \]
Time = 0.35 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^4(x)}{a+b \sin ^2(x)} \, dx=-\frac {{\left (2 \, a + 3 \, b\right )} x}{2 \, b^{2}} - \frac {\tan \left (x\right )}{2 \, {\left (b \tan \left (x\right )^{2} + b\right )}} + \frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} b^{2}} \]
-1/2*(2*a + 3*b)*x/b^2 - 1/2*tan(x)/(b*tan(x)^2 + b) + (a^2 + 2*a*b + b^2) *arctan((a + b)*tan(x)/sqrt((a + b)*a))/(sqrt((a + b)*a)*b^2)
Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.56 \[ \int \frac {\cos ^4(x)}{a+b \sin ^2(x)} \, dx=-\frac {{\left (2 \, a + 3 \, b\right )} x}{2 \, b^{2}} + \frac {{\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (x\right ) + b \tan \left (x\right )}{\sqrt {a^{2} + a b}}\right )\right )} {\left (a^{2} + 2 \, a b + b^{2}\right )}}{\sqrt {a^{2} + a b} b^{2}} - \frac {\tan \left (x\right )}{2 \, {\left (\tan \left (x\right )^{2} + 1\right )} b} \]
-1/2*(2*a + 3*b)*x/b^2 + (pi*floor(x/pi + 1/2)*sgn(2*a + 2*b) + arctan((a* tan(x) + b*tan(x))/sqrt(a^2 + a*b)))*(a^2 + 2*a*b + b^2)/(sqrt(a^2 + a*b)* b^2) - 1/2*tan(x)/((tan(x)^2 + 1)*b)
Time = 13.65 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.02 \[ \int \frac {\cos ^4(x)}{a+b \sin ^2(x)} \, dx=-\frac {3\,\mathrm {atan}\left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right )}{2\,b}-\frac {a\,\mathrm {atan}\left (\frac {\sin \left (x\right )}{\cos \left (x\right )}\right )}{b^2}-\frac {\cos \left (x\right )\,\sin \left (x\right )}{2\,b}-\frac {\mathrm {atanh}\left (\frac {\sin \left (x\right )\,\sqrt {-a^4-3\,a^3\,b-3\,a^2\,b^2-a\,b^3}}{\cos \left (x\right )\,a^2+b\,\cos \left (x\right )\,a}\right )\,\sqrt {-a^4-3\,a^3\,b-3\,a^2\,b^2-a\,b^3}}{a\,b^2} \]