Integrand size = 15, antiderivative size = 87 \[ \int \frac {\cos ^6(x)}{a+b \sin ^2(x)} \, dx=-\frac {\left (8 a^2+20 a b+15 b^2\right ) x}{8 b^3}+\frac {(a+b)^{5/2} \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} b^3}-\frac {(4 a+7 b) \cos (x) \sin (x)}{8 b^2}-\frac {\cos ^3(x) \sin (x)}{4 b} \]
-1/8*(8*a^2+20*a*b+15*b^2)*x/b^3-1/8*(4*a+7*b)*cos(x)*sin(x)/b^2-1/4*cos(x )^3*sin(x)/b+(a+b)^(5/2)*arctan((a+b)^(1/2)*tan(x)/a^(1/2))/b^3/a^(1/2)
Time = 0.23 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^6(x)}{a+b \sin ^2(x)} \, dx=\frac {(a+b)^{5/2} \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} b^3}-\frac {4 \left (8 a^2+20 a b+15 b^2\right ) x+8 b (a+2 b) \sin (2 x)+b^2 \sin (4 x)}{32 b^3} \]
((a + b)^(5/2)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(Sqrt[a]*b^3) - (4*(8 *a^2 + 20*a*b + 15*b^2)*x + 8*b*(a + 2*b)*Sin[2*x] + b^2*Sin[4*x])/(32*b^3 )
Time = 0.34 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.31, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 3670, 316, 402, 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.
| \(\displaystyle \int \frac {\cos ^6(x)}{a+b \sin ^2(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (x)^6}{a+b \sin (x)^2}dx\) |
| \(\Big \downarrow \) 3670 |
| \(\displaystyle \int \frac {1}{\left (\tan ^2(x)+1\right )^3 \left ((a+b) \tan ^2(x)+a\right )}d\tan (x)\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\int \frac {-3 (a+b) \tan ^2(x)+a+4 b}{\left (\tan ^2(x)+1\right )^2 \left ((a+b) \tan ^2(x)+a\right )}d\tan (x)}{4 b}-\frac {\tan (x)}{4 b \left (\tan ^2(x)+1\right )^2}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\int \frac {4 a^2+9 b a+8 b^2-(a+b) (4 a+7 b) \tan ^2(x)}{\left (\tan ^2(x)+1\right ) \left ((a+b) \tan ^2(x)+a\right )}d\tan (x)}{2 b}-\frac {(4 a+7 b) \tan (x)}{2 b \left (\tan ^2(x)+1\right )}}{4 b}-\frac {\tan (x)}{4 b \left (\tan ^2(x)+1\right )^2}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {8 (a+b)^3 \int \frac {1}{(a+b) \tan ^2(x)+a}d\tan (x)}{b}-\frac {\left (8 a^2+20 a b+15 b^2\right ) \int \frac {1}{\tan ^2(x)+1}d\tan (x)}{b}}{2 b}-\frac {(4 a+7 b) \tan (x)}{2 b \left (\tan ^2(x)+1\right )}}{4 b}-\frac {\tan (x)}{4 b \left (\tan ^2(x)+1\right )^2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {8 (a+b)^3 \int \frac {1}{(a+b) \tan ^2(x)+a}d\tan (x)}{b}-\frac {\left (8 a^2+20 a b+15 b^2\right ) \arctan (\tan (x))}{b}}{2 b}-\frac {(4 a+7 b) \tan (x)}{2 b \left (\tan ^2(x)+1\right )}}{4 b}-\frac {\tan (x)}{4 b \left (\tan ^2(x)+1\right )^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {8 (a+b)^{5/2} \arctan \left (\frac {\sqrt {a+b} \tan (x)}{\sqrt {a}}\right )}{\sqrt {a} b}-\frac {\left (8 a^2+20 a b+15 b^2\right ) \arctan (\tan (x))}{b}}{2 b}-\frac {(4 a+7 b) \tan (x)}{2 b \left (\tan ^2(x)+1\right )}}{4 b}-\frac {\tan (x)}{4 b \left (\tan ^2(x)+1\right )^2}\) |
-1/4*Tan[x]/(b*(1 + Tan[x]^2)^2) + ((-(((8*a^2 + 20*a*b + 15*b^2)*ArcTan[T an[x]])/b) + (8*(a + b)^(5/2)*ArcTan[(Sqrt[a + b]*Tan[x])/Sqrt[a]])/(Sqrt[ a]*b))/(2*b) - ((4*a + 7*b)*Tan[x])/(2*b*(1 + Tan[x]^2)))/(4*b)
3.4.2.3.1 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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
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 = 0.97 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.10
| method | result | size |
| default | \(\frac {\left (a +b \right )^{3} \arctan \left (\frac {\left (a +b \right ) \tan \left (x \right )}{\sqrt {a \left (a +b \right )}}\right )}{b^{3} \sqrt {a \left (a +b \right )}}-\frac {\frac {\left (\frac {1}{2} a b +\frac {7}{8} b^{2}\right ) \left (\tan ^{3}\left (x \right )\right )+\left (\frac {1}{2} a b +\frac {9}{8} b^{2}\right ) \tan \left (x \right )}{\left (1+\tan ^{2}\left (x \right )\right )^{2}}+\frac {\left (8 a^{2}+20 a b +15 b^{2}\right ) \arctan \left (\tan \left (x \right )\right )}{8}}{b^{3}}\) | \(96\) |
| risch | \(-\frac {x \,a^{2}}{b^{3}}-\frac {5 a x}{2 b^{2}}-\frac {15 x}{8 b}+\frac {i {\mathrm e}^{2 i x} a}{8 b^{2}}+\frac {i {\mathrm e}^{2 i x}}{4 b}-\frac {i {\mathrm e}^{-2 i x} a}{8 b^{2}}-\frac {i {\mathrm e}^{-2 i x}}{4 b}+\frac {a \sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 b^{3}}+\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{b^{2}}+\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 a b}-\frac {a \sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 b^{3}}-\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{b^{2}}-\frac {\sqrt {-a \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 i x}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 a b}-\frac {\sin \left (4 x \right )}{32 b}\) | \(335\) |
(a+b)^3/b^3/(a*(a+b))^(1/2)*arctan((a+b)*tan(x)/(a*(a+b))^(1/2))-1/b^3*((( 1/2*a*b+7/8*b^2)*tan(x)^3+(1/2*a*b+9/8*b^2)*tan(x))/(1+tan(x)^2)^2+1/8*(8* a^2+20*a*b+15*b^2)*arctan(tan(x)))
Time = 0.32 (sec) , antiderivative size = 312, normalized size of antiderivative = 3.59 \[ \int \frac {\cos ^6(x)}{a+b \sin ^2(x)} \, dx=\left [\frac {2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-\frac {a + b}{a}} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (x\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (x\right )^{2} - 4 \, {\left ({\left (2 \, a^{2} + a b\right )} \cos \left (x\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (x\right )\right )} \sqrt {-\frac {a + b}{a}} \sin \left (x\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (x\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (x\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - {\left (8 \, a^{2} + 20 \, a b + 15 \, b^{2}\right )} x - {\left (2 \, b^{2} \cos \left (x\right )^{3} + {\left (4 \, a b + 7 \, b^{2}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{8 \, b^{3}}, -\frac {4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {\frac {a + b}{a}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (x\right )^{2} - a - b\right )} \sqrt {\frac {a + b}{a}}}{2 \, {\left (a + b\right )} \cos \left (x\right ) \sin \left (x\right )}\right ) + {\left (8 \, a^{2} + 20 \, a b + 15 \, b^{2}\right )} x + {\left (2 \, b^{2} \cos \left (x\right )^{3} + {\left (4 \, a b + 7 \, b^{2}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{8 \, b^{3}}\right ] \]
[1/8*(2*(a^2 + 2*a*b + b^2)*sqrt(-(a + b)/a)*log(((8*a^2 + 8*a*b + b^2)*co s(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)) - (8*a^2 + 20*a*b + 15*b ^2)*x - (2*b^2*cos(x)^3 + (4*a*b + 7*b^2)*cos(x))*sin(x))/b^3, -1/8*(4*(a^ 2 + 2*a*b + b^2)*sqrt((a + b)/a)*arctan(1/2*((2*a + b)*cos(x)^2 - a - b)*s qrt((a + b)/a)/((a + b)*cos(x)*sin(x))) + (8*a^2 + 20*a*b + 15*b^2)*x + (2 *b^2*cos(x)^3 + (4*a*b + 7*b^2)*cos(x))*sin(x))/b^3]
Timed out. \[ \int \frac {\cos ^6(x)}{a+b \sin ^2(x)} \, dx=\text {Timed out} \]
Time = 0.34 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.31 \[ \int \frac {\cos ^6(x)}{a+b \sin ^2(x)} \, dx=-\frac {{\left (4 \, a + 7 \, b\right )} \tan \left (x\right )^{3} + {\left (4 \, a + 9 \, b\right )} \tan \left (x\right )}{8 \, {\left (b^{2} \tan \left (x\right )^{4} + 2 \, b^{2} \tan \left (x\right )^{2} + b^{2}\right )}} - \frac {{\left (8 \, a^{2} + 20 \, a b + 15 \, b^{2}\right )} x}{8 \, b^{3}} + \frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (x\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} b^{3}} \]
-1/8*((4*a + 7*b)*tan(x)^3 + (4*a + 9*b)*tan(x))/(b^2*tan(x)^4 + 2*b^2*tan (x)^2 + b^2) - 1/8*(8*a^2 + 20*a*b + 15*b^2)*x/b^3 + (a^3 + 3*a^2*b + 3*a* b^2 + b^3)*arctan((a + b)*tan(x)/sqrt((a + b)*a))/(sqrt((a + b)*a)*b^3)
Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.51 \[ \int \frac {\cos ^6(x)}{a+b \sin ^2(x)} \, dx=-\frac {{\left (8 \, a^{2} + 20 \, a b + 15 \, b^{2}\right )} x}{8 \, b^{3}} + \frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (x\right ) + b \tan \left (x\right )}{\sqrt {a^{2} + a b}}\right )\right )}}{\sqrt {a^{2} + a b} b^{3}} - \frac {4 \, a \tan \left (x\right )^{3} + 7 \, b \tan \left (x\right )^{3} + 4 \, a \tan \left (x\right ) + 9 \, b \tan \left (x\right )}{8 \, {\left (\tan \left (x\right )^{2} + 1\right )}^{2} b^{2}} \]
-1/8*(8*a^2 + 20*a*b + 15*b^2)*x/b^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(pi *floor(x/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(x) + b*tan(x))/sqrt(a^2 + a*b)))/(sqrt(a^2 + a*b)*b^3) - 1/8*(4*a*tan(x)^3 + 7*b*tan(x)^3 + 4*a*ta n(x) + 9*b*tan(x))/((tan(x)^2 + 1)^2*b^2)
Time = 14.52 (sec) , antiderivative size = 1804, normalized size of antiderivative = 20.74 \[ \int \frac {\cos ^6(x)}{a+b \sin ^2(x)} \, dx=\text {Too large to display} \]
(atan(((((tan(x)*(1723*a*b^6 + 960*a^6*b + 128*a^7 + 289*b^7 + 4459*a^2*b^ 5 + 6505*a^3*b^4 + 5784*a^4*b^3 + 3136*a^5*b^2))/(32*b^4) - ((((25*a*b^9)/ 2 + 4*b^10 + 15*a^2*b^8 + (17*a^3*b^7)/2 + 2*a^4*b^6)/b^6 - (tan(x)*(a*b*2 0i + a^2*8i + b^2*15i)*(1024*a*b^8 + 256*b^9 + 1280*a^2*b^7 + 512*a^3*b^6) )/(512*b^7))*(a*b*20i + a^2*8i + b^2*15i))/(16*b^3))*(a*b*20i + a^2*8i + b ^2*15i)*1i)/(16*b^3) + (((tan(x)*(1723*a*b^6 + 960*a^6*b + 128*a^7 + 289*b ^7 + 4459*a^2*b^5 + 6505*a^3*b^4 + 5784*a^4*b^3 + 3136*a^5*b^2))/(32*b^4) + ((((25*a*b^9)/2 + 4*b^10 + 15*a^2*b^8 + (17*a^3*b^7)/2 + 2*a^4*b^6)/b^6 + (tan(x)*(a*b*20i + a^2*8i + b^2*15i)*(1024*a*b^8 + 256*b^9 + 1280*a^2*b^ 7 + 512*a^3*b^6))/(512*b^7))*(a*b*20i + a^2*8i + b^2*15i))/(16*b^3))*(a*b* 20i + a^2*8i + b^2*15i)*1i)/(16*b^3))/(((725*a*b^7)/32 + (37*a^7*b)/4 + a^ 8 + (105*b^8)/32 + (1093*a^2*b^6)/16 + (1881*a^3*b^5)/16 + (4045*a^4*b^4)/ 32 + (2785*a^5*b^3)/32 + (75*a^6*b^2)/2)/b^6 - (((tan(x)*(1723*a*b^6 + 960 *a^6*b + 128*a^7 + 289*b^7 + 4459*a^2*b^5 + 6505*a^3*b^4 + 5784*a^4*b^3 + 3136*a^5*b^2))/(32*b^4) - ((((25*a*b^9)/2 + 4*b^10 + 15*a^2*b^8 + (17*a^3* b^7)/2 + 2*a^4*b^6)/b^6 - (tan(x)*(a*b*20i + a^2*8i + b^2*15i)*(1024*a*b^8 + 256*b^9 + 1280*a^2*b^7 + 512*a^3*b^6))/(512*b^7))*(a*b*20i + a^2*8i + b ^2*15i))/(16*b^3))*(a*b*20i + a^2*8i + b^2*15i))/(16*b^3) + (((tan(x)*(172 3*a*b^6 + 960*a^6*b + 128*a^7 + 289*b^7 + 4459*a^2*b^5 + 6505*a^3*b^4 + 57 84*a^4*b^3 + 3136*a^5*b^2))/(32*b^4) + ((((25*a*b^9)/2 + 4*b^10 + 15*a^...