Integrand size = 15, antiderivative size = 78 \[ \int \frac {\cos ^7(x)}{a+b \sin ^2(x)} \, dx=\frac {(a+b)^3 \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2}}-\frac {\left (a^2+3 a b+3 b^2\right ) \sin (x)}{b^3}+\frac {(a+3 b) \sin ^3(x)}{3 b^2}-\frac {\sin ^5(x)}{5 b} \]
-(a^2+3*a*b+3*b^2)*sin(x)/b^3+1/3*(a+3*b)*sin(x)^3/b^2-1/5*sin(x)^5/b+(a+b )^3*arctan(sin(x)*b^(1/2)/a^(1/2))/b^(7/2)/a^(1/2)
Time = 0.31 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.40 \[ \int \frac {\cos ^7(x)}{a+b \sin ^2(x)} \, dx=\frac {-120 (a+b)^3 \arctan \left (\frac {\sqrt {a} \csc (x)}{\sqrt {b}}\right )+120 (a+b)^3 \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )-2 \sqrt {a} \sqrt {b} \left (120 a^2+340 a b+309 b^2+4 b (5 a+12 b) \cos (2 x)+3 b^2 \cos (4 x)\right ) \sin (x)}{240 \sqrt {a} b^{7/2}} \]
(-120*(a + b)^3*ArcTan[(Sqrt[a]*Csc[x])/Sqrt[b]] + 120*(a + b)^3*ArcTan[(S qrt[b]*Sin[x])/Sqrt[a]] - 2*Sqrt[a]*Sqrt[b]*(120*a^2 + 340*a*b + 309*b^2 + 4*b*(5*a + 12*b)*Cos[2*x] + 3*b^2*Cos[4*x])*Sin[x])/(240*Sqrt[a]*b^(7/2))
Time = 0.28 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 3669, 300, 2009}
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.
\(\displaystyle \int \frac {\cos ^7(x)}{a+b \sin ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (x)^7}{a+b \sin (x)^2}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle \int \frac {\left (1-\sin ^2(x)\right )^3}{a+b \sin ^2(x)}d\sin (x)\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \int \left (-\frac {a^2+3 a b+3 b^2}{b^3}+\frac {a^3+3 a^2 b+3 a b^2+b^3}{b^3 \left (a+b \sin ^2(x)\right )}+\frac {(a+3 b) \sin ^2(x)}{b^2}-\frac {\sin ^4(x)}{b}\right )d\sin (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (a^2+3 a b+3 b^2\right ) \sin (x)}{b^3}+\frac {(a+b)^3 \arctan \left (\frac {\sqrt {b} \sin (x)}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2}}+\frac {(a+3 b) \sin ^3(x)}{3 b^2}-\frac {\sin ^5(x)}{5 b}\) |
((a + b)^3*ArcTan[(Sqrt[b]*Sin[x])/Sqrt[a]])/(Sqrt[a]*b^(7/2)) - ((a^2 + 3 *a*b + 3*b^2)*Sin[x])/b^3 + ((a + 3*b)*Sin[x]^3)/(3*b^2) - Sin[x]^5/(5*b)
3.4.1.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
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 = 1.31 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.23
method | result | size |
derivativedivides | \(-\frac {\frac {\left (\sin ^{5}\left (x \right )\right ) b^{2}}{5}-\frac {a b \left (\sin ^{3}\left (x \right )\right )}{3}-b^{2} \left (\sin ^{3}\left (x \right )\right )+\sin \left (x \right ) a^{2}+3 a b \sin \left (x \right )+3 b^{2} \sin \left (x \right )}{b^{3}}-\frac {\left (-a^{3}-3 a^{2} b -3 a \,b^{2}-b^{3}\right ) \arctan \left (\frac {b \sin \left (x \right )}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\) | \(96\) |
default | \(-\frac {\frac {\left (\sin ^{5}\left (x \right )\right ) b^{2}}{5}-\frac {a b \left (\sin ^{3}\left (x \right )\right )}{3}-b^{2} \left (\sin ^{3}\left (x \right )\right )+\sin \left (x \right ) a^{2}+3 a b \sin \left (x \right )+3 b^{2} \sin \left (x \right )}{b^{3}}-\frac {\left (-a^{3}-3 a^{2} b -3 a \,b^{2}-b^{3}\right ) \arctan \left (\frac {b \sin \left (x \right )}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\) | \(96\) |
risch | \(-\frac {i {\mathrm e}^{-i x} a^{2}}{2 b^{3}}-\frac {11 i {\mathrm e}^{-i x} a}{8 b^{2}}-\frac {19 i {\mathrm e}^{-i x}}{16 b}+\frac {11 i {\mathrm e}^{i x} a}{8 b^{2}}+\frac {19 i {\mathrm e}^{i x}}{16 b}+\frac {i {\mathrm e}^{i x} a^{2}}{2 b^{3}}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right ) a^{3}}{2 \sqrt {-a b}\, b^{3}}-\frac {3 \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right ) a^{2}}{2 \sqrt {-a b}\, b^{2}}-\frac {3 \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right ) a}{2 \sqrt {-a b}\, b}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{2 \sqrt {-a b}}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right ) a^{3}}{2 \sqrt {-a b}\, b^{3}}+\frac {3 \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right ) a^{2}}{2 \sqrt {-a b}\, b^{2}}+\frac {3 \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right ) a}{2 \sqrt {-a b}\, b}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {-a b}}-1\right )}{2 \sqrt {-a b}}-\frac {\sin \left (5 x \right )}{80 b}-\frac {\sin \left (3 x \right ) a}{12 b^{2}}-\frac {3 \sin \left (3 x \right )}{16 b}\) | \(384\) |
-1/b^3*(1/5*sin(x)^5*b^2-1/3*a*b*sin(x)^3-b^2*sin(x)^3+sin(x)*a^2+3*a*b*si n(x)+3*b^2*sin(x))-(-a^3-3*a^2*b-3*a*b^2-b^3)/b^3/(a*b)^(1/2)*arctan(b*sin (x)/(a*b)^(1/2))
Time = 0.32 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.99 \[ \int \frac {\cos ^7(x)}{a+b \sin ^2(x)} \, dx=\left [-\frac {15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {-a b} \log \left (-\frac {b \cos \left (x\right )^{2} + 2 \, \sqrt {-a b} \sin \left (x\right ) + a - b}{b \cos \left (x\right )^{2} - a - b}\right ) + 2 \, {\left (3 \, a b^{3} \cos \left (x\right )^{4} + 15 \, a^{3} b + 40 \, a^{2} b^{2} + 33 \, a b^{3} + {\left (5 \, a^{2} b^{2} + 9 \, a b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{30 \, a b^{4}}, \frac {15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} \sin \left (x\right )}{a}\right ) - {\left (3 \, a b^{3} \cos \left (x\right )^{4} + 15 \, a^{3} b + 40 \, a^{2} b^{2} + 33 \, a b^{3} + {\left (5 \, a^{2} b^{2} + 9 \, a b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{15 \, a b^{4}}\right ] \]
[-1/30*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(-a*b)*log(-(b*cos(x)^2 + 2 *sqrt(-a*b)*sin(x) + a - b)/(b*cos(x)^2 - a - b)) + 2*(3*a*b^3*cos(x)^4 + 15*a^3*b + 40*a^2*b^2 + 33*a*b^3 + (5*a^2*b^2 + 9*a*b^3)*cos(x)^2)*sin(x)) /(a*b^4), 1/15*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b)*arctan(sqrt(a *b)*sin(x)/a) - (3*a*b^3*cos(x)^4 + 15*a^3*b + 40*a^2*b^2 + 33*a*b^3 + (5* a^2*b^2 + 9*a*b^3)*cos(x)^2)*sin(x))/(a*b^4)]
Timed out. \[ \int \frac {\cos ^7(x)}{a+b \sin ^2(x)} \, dx=\text {Timed out} \]
Time = 0.37 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^7(x)}{a+b \sin ^2(x)} \, dx=\frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {3 \, b^{2} \sin \left (x\right )^{5} - 5 \, {\left (a b + 3 \, b^{2}\right )} \sin \left (x\right )^{3} + 15 \, {\left (a^{2} + 3 \, a b + 3 \, b^{2}\right )} \sin \left (x\right )}{15 \, b^{3}} \]
(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*arctan(b*sin(x)/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/15*(3*b^2*sin(x)^5 - 5*(a*b + 3*b^2)*sin(x)^3 + 15*(a^2 + 3*a*b + 3*b ^2)*sin(x))/b^3
Time = 0.29 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^7(x)}{a+b \sin ^2(x)} \, dx=\frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \sin \left (x\right )}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} - \frac {3 \, b^{4} \sin \left (x\right )^{5} - 5 \, a b^{3} \sin \left (x\right )^{3} - 15 \, b^{4} \sin \left (x\right )^{3} + 15 \, a^{2} b^{2} \sin \left (x\right ) + 45 \, a b^{3} \sin \left (x\right ) + 45 \, b^{4} \sin \left (x\right )}{15 \, b^{5}} \]
(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*arctan(b*sin(x)/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/15*(3*b^4*sin(x)^5 - 5*a*b^3*sin(x)^3 - 15*b^4*sin(x)^3 + 15*a^2*b^2* sin(x) + 45*a*b^3*sin(x) + 45*b^4*sin(x))/b^5
Time = 0.11 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.27 \[ \int \frac {\cos ^7(x)}{a+b \sin ^2(x)} \, dx={\sin \left (x\right )}^3\,\left (\frac {a}{3\,b^2}+\frac {1}{b}\right )-\sin \left (x\right )\,\left (\frac {3}{b}+\frac {a\,\left (\frac {a}{b^2}+\frac {3}{b}\right )}{b}\right )-\frac {{\sin \left (x\right )}^5}{5\,b}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sin \left (x\right )\,{\left (a+b\right )}^3}{\sqrt {a}\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}\right )\,{\left (a+b\right )}^3}{\sqrt {a}\,b^{7/2}} \]