Integrand size = 15, antiderivative size = 78 \[ \int \frac {\sin ^7(x)}{a+b \cos ^2(x)} \, dx=-\frac {(a+b)^3 \arctan \left (\frac {\sqrt {b} \cos (x)}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2}}+\frac {\left (a^2+3 a b+3 b^2\right ) \cos (x)}{b^3}-\frac {(a+3 b) \cos ^3(x)}{3 b^2}+\frac {\cos ^5(x)}{5 b} \]
(a^2+3*a*b+3*b^2)*cos(x)/b^3-1/3*(a+3*b)*cos(x)^3/b^2+1/5*cos(x)^5/b-(a+b) ^3*arctan(cos(x)*b^(1/2)/a^(1/2))/b^(7/2)/a^(1/2)
Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.83 \[ \int \frac {\sin ^7(x)}{a+b \cos ^2(x)} \, dx=-\frac {(a+b)^3 \arctan \left (\frac {\sqrt {b}-\sqrt {a+b} \tan \left (\frac {x}{2}\right )}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2}}-\frac {(a+b)^3 \arctan \left (\frac {\sqrt {b}+\sqrt {a+b} \tan \left (\frac {x}{2}\right )}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2}}+\frac {\left (8 a^2+22 a b+19 b^2\right ) \cos (x)}{8 b^3}-\frac {(4 a+9 b) \cos (3 x)}{48 b^2}+\frac {\cos (5 x)}{80 b} \]
-(((a + b)^3*ArcTan[(Sqrt[b] - Sqrt[a + b]*Tan[x/2])/Sqrt[a]])/(Sqrt[a]*b^ (7/2))) - ((a + b)^3*ArcTan[(Sqrt[b] + Sqrt[a + b]*Tan[x/2])/Sqrt[a]])/(Sq rt[a]*b^(7/2)) + ((8*a^2 + 22*a*b + 19*b^2)*Cos[x])/(8*b^3) - ((4*a + 9*b) *Cos[3*x])/(48*b^2) + Cos[5*x]/(80*b)
Time = 0.31 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 25, 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.
\(\displaystyle \int \frac {\sin ^7(x)}{a+b \cos ^2(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cos \left (x+\frac {\pi }{2}\right )^7}{a+b \sin \left (x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cos \left (x+\frac {\pi }{2}\right )^7}{b \sin \left (x+\frac {\pi }{2}\right )^2+a}dx\) |
\(\Big \downarrow \) 3669 |
\(\displaystyle -\int \frac {\left (1-\cos ^2(x)\right )^3}{b \cos ^2(x)+a}d\cos (x)\) |
\(\Big \downarrow \) 300 |
\(\displaystyle -\int \left (-\frac {\cos ^4(x)}{b}+\frac {(a+3 b) \cos ^2(x)}{b^2}-\frac {a^2+3 b a+3 b^2}{b^3}+\frac {a^3+3 b a^2+3 b^2 a+b^3}{b^3 \left (b \cos ^2(x)+a\right )}\right )d\cos (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a^2+3 a b+3 b^2\right ) \cos (x)}{b^3}-\frac {(a+b)^3 \arctan \left (\frac {\sqrt {b} \cos (x)}{\sqrt {a}}\right )}{\sqrt {a} b^{7/2}}-\frac {(a+3 b) \cos ^3(x)}{3 b^2}+\frac {\cos ^5(x)}{5 b}\) |
-(((a + b)^3*ArcTan[(Sqrt[b]*Cos[x])/Sqrt[a]])/(Sqrt[a]*b^(7/2))) + ((a^2 + 3*a*b + 3*b^2)*Cos[x])/b^3 - ((a + 3*b)*Cos[x]^3)/(3*b^2) + Cos[x]^5/(5* b)
3.1.10.3.1 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 = 1.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.21
method | result | size |
derivativedivides | \(\frac {\frac {\left (\cos ^{5}\left (x \right )\right ) b^{2}}{5}-\frac {b \left (\cos ^{3}\left (x \right )\right ) a}{3}-b^{2} \left (\cos ^{3}\left (x \right )\right )+a^{2} \cos \left (x \right )+3 b \cos \left (x \right ) a +3 b^{2} \cos \left (x \right )}{b^{3}}+\frac {\left (-a^{3}-3 a^{2} b -3 a \,b^{2}-b^{3}\right ) \arctan \left (\frac {b \cos \left (x \right )}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\) | \(94\) |
default | \(\frac {\frac {\left (\cos ^{5}\left (x \right )\right ) b^{2}}{5}-\frac {b \left (\cos ^{3}\left (x \right )\right ) a}{3}-b^{2} \left (\cos ^{3}\left (x \right )\right )+a^{2} \cos \left (x \right )+3 b \cos \left (x \right ) a +3 b^{2} \cos \left (x \right )}{b^{3}}+\frac {\left (-a^{3}-3 a^{2} b -3 a \,b^{2}-b^{3}\right ) \arctan \left (\frac {b \cos \left (x \right )}{\sqrt {a b}}\right )}{b^{3} \sqrt {a b}}\) | \(94\) |
risch | \(\frac {{\mathrm e}^{i x} a^{2}}{2 b^{3}}+\frac {11 \,{\mathrm e}^{i x} a}{8 b^{2}}+\frac {19 \,{\mathrm e}^{i x}}{16 b}+\frac {{\mathrm e}^{-i x} a^{2}}{2 b^{3}}+\frac {11 \,{\mathrm e}^{-i x} a}{8 b^{2}}+\frac {19 \,{\mathrm e}^{-i x}}{16 b}+\frac {3 i \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a}{2 \sqrt {a b}\, b}+\frac {3 i \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a^{2}}{2 \sqrt {a b}\, b^{2}}-\frac {i \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right )}{2 \sqrt {a b}}+\frac {i \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a^{3}}{2 \sqrt {a b}\, b^{3}}-\frac {3 i \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a}{2 \sqrt {a b}\, b}-\frac {3 i \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a^{2}}{2 \sqrt {a b}\, b^{2}}-\frac {i \ln \left ({\mathrm e}^{2 i x}+\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right ) a^{3}}{2 \sqrt {a b}\, b^{3}}+\frac {i \ln \left ({\mathrm e}^{2 i x}-\frac {2 i a \,{\mathrm e}^{i x}}{\sqrt {a b}}+1\right )}{2 \sqrt {a b}}+\frac {\cos \left (5 x \right )}{80 b}-\frac {3 \cos \left (3 x \right )}{16 b}-\frac {\cos \left (3 x \right ) a}{12 b^{2}}\) | \(370\) |
1/b^3*(1/5*cos(x)^5*b^2-1/3*b*cos(x)^3*a-b^2*cos(x)^3+a^2*cos(x)+3*b*cos(x )*a+3*b^2*cos(x))+(-a^3-3*a^2*b-3*a*b^2-b^3)/b^3/(a*b)^(1/2)*arctan(b*cos( x)/(a*b)^(1/2))
Time = 0.27 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.88 \[ \int \frac {\sin ^7(x)}{a+b \cos ^2(x)} \, dx=\left [\frac {6 \, a b^{3} \cos \left (x\right )^{5} - 10 \, {\left (a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (x\right )^{3} - 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {-a b} \log \left (-\frac {b \cos \left (x\right )^{2} + 2 \, \sqrt {-a b} \cos \left (x\right ) - a}{b \cos \left (x\right )^{2} + a}\right ) + 30 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (x\right )}{30 \, a b^{4}}, \frac {3 \, a b^{3} \cos \left (x\right )^{5} - 5 \, {\left (a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (x\right )^{3} - 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} \cos \left (x\right )}{a}\right ) + 15 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (x\right )}{15 \, a b^{4}}\right ] \]
[1/30*(6*a*b^3*cos(x)^5 - 10*(a^2*b^2 + 3*a*b^3)*cos(x)^3 - 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)*cos(x) - a )/(b*cos(x)^2 + a)) + 30*(a^3*b + 3*a^2*b^2 + 3*a*b^3)*cos(x))/(a*b^4), 1/ 15*(3*a*b^3*cos(x)^5 - 5*(a^2*b^2 + 3*a*b^3)*cos(x)^3 - 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b)*arctan(sqrt(a*b)*cos(x)/a) + 15*(a^3*b + 3*a^2* b^2 + 3*a*b^3)*cos(x))/(a*b^4)]
Timed out. \[ \int \frac {\sin ^7(x)}{a+b \cos ^2(x)} \, dx=\text {Timed out} \]
Time = 0.30 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.12 \[ \int \frac {\sin ^7(x)}{a+b \cos ^2(x)} \, dx=-\frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \cos \left (x\right )}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {3 \, b^{2} \cos \left (x\right )^{5} - 5 \, {\left (a b + 3 \, b^{2}\right )} \cos \left (x\right )^{3} + 15 \, {\left (a^{2} + 3 \, a b + 3 \, b^{2}\right )} \cos \left (x\right )}{15 \, b^{3}} \]
-(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*arctan(b*cos(x)/sqrt(a*b))/(sqrt(a*b)*b^3 ) + 1/15*(3*b^2*cos(x)^5 - 5*(a*b + 3*b^2)*cos(x)^3 + 15*(a^2 + 3*a*b + 3* b^2)*cos(x))/b^3
Time = 0.34 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.27 \[ \int \frac {\sin ^7(x)}{a+b \cos ^2(x)} \, dx=-\frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \cos \left (x\right )}{\sqrt {a b}}\right )}{\sqrt {a b} b^{3}} + \frac {3 \, b^{4} \cos \left (x\right )^{5} - 5 \, a b^{3} \cos \left (x\right )^{3} - 15 \, b^{4} \cos \left (x\right )^{3} + 15 \, a^{2} b^{2} \cos \left (x\right ) + 45 \, a b^{3} \cos \left (x\right ) + 45 \, b^{4} \cos \left (x\right )}{15 \, b^{5}} \]
-(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*arctan(b*cos(x)/sqrt(a*b))/(sqrt(a*b)*b^3 ) + 1/15*(3*b^4*cos(x)^5 - 5*a*b^3*cos(x)^3 - 15*b^4*cos(x)^3 + 15*a^2*b^2 *cos(x) + 45*a*b^3*cos(x) + 45*b^4*cos(x))/b^5
Time = 2.41 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.28 \[ \int \frac {\sin ^7(x)}{a+b \cos ^2(x)} \, dx=\cos \left (x\right )\,\left (\frac {3}{b}+\frac {a\,\left (\frac {a}{b^2}+\frac {3}{b}\right )}{b}\right )-{\cos \left (x\right )}^3\,\left (\frac {a}{3\,b^2}+\frac {1}{b}\right )+\frac {{\cos \left (x\right )}^5}{5\,b}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\cos \left (x\right )\,{\left (a+b\right )}^3}{\sqrt {a}\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}\right )\,{\left (a+b\right )}^3}{\sqrt {a}\,b^{7/2}} \]