Integrand size = 25, antiderivative size = 130 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{b^{5/2} f}+\frac {(a+b) \cos ^2(e+f x) \sin (e+f x)}{3 a b f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {(3 a-2 b) (a+b) \sin (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sin ^2(e+f x)}} \]
arctanh(sin(f*x+e)*b^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/b^(5/2)/f+1/3*(a+b)*c os(f*x+e)^2*sin(f*x+e)/a/b/f/(a+b*sin(f*x+e)^2)^(3/2)-1/3*(3*a-2*b)*(a+b)* sin(f*x+e)/a^2/b^2/f/(a+b*sin(f*x+e)^2)^(1/2)
Time = 0.87 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\frac {3 \arctan \left (\frac {\sqrt {2} \sqrt {-b} \sin (e+f x)}{\sqrt {2 a+b-b \cos (2 (e+f x))}}\right )}{\sqrt {-b}}+\frac {2 \sqrt {2} (a+b) \left (-3 a^2+a b+b^2+(2 a-b) b \cos (2 (e+f x))\right ) \sin (e+f x)}{a^2 (2 a+b-b \cos (2 (e+f x)))^{3/2}}}{3 b^2 f} \]
((3*ArcTan[(Sqrt[2]*Sqrt[-b]*Sin[e + f*x])/Sqrt[2*a + b - b*Cos[2*(e + f*x )]]])/Sqrt[-b] + (2*Sqrt[2]*(a + b)*(-3*a^2 + a*b + b^2 + (2*a - b)*b*Cos[ 2*(e + f*x)])*Sin[e + f*x])/(a^2*(2*a + b - b*Cos[2*(e + f*x)])^(3/2)))/(3 *b^2*f)
Time = 0.32 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3669, 315, 25, 298, 224, 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.
| \(\displaystyle \int \frac {\cos ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (e+f x)^5}{\left (a+b \sin (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 3669 |
| \(\displaystyle \frac {\int \frac {\left (1-\sin ^2(e+f x)\right )^2}{\left (b \sin ^2(e+f x)+a\right )^{5/2}}d\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 315 |
| \(\displaystyle \frac {\frac {\int -\frac {-3 a \sin ^2(e+f x)+a-2 b}{\left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin (e+f x)}{3 a b}+\frac {(a+b) \sin (e+f x) \left (1-\sin ^2(e+f x)\right )}{3 a b \left (a+b \sin ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {(a+b) \sin (e+f x) \left (1-\sin ^2(e+f x)\right )}{3 a b \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\int \frac {-3 a \sin ^2(e+f x)+a-2 b}{\left (b \sin ^2(e+f x)+a\right )^{3/2}}d\sin (e+f x)}{3 a b}}{f}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle \frac {\frac {(a+b) \sin (e+f x) \left (1-\sin ^2(e+f x)\right )}{3 a b \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\frac {\left (\frac {3 a}{b}-\frac {2 b}{a}+1\right ) \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}-\frac {3 a \int \frac {1}{\sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{b}}{3 a b}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {(a+b) \sin (e+f x) \left (1-\sin ^2(e+f x)\right )}{3 a b \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\frac {\left (\frac {3 a}{b}-\frac {2 b}{a}+1\right ) \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}-\frac {3 a \int \frac {1}{1-\frac {b \sin ^2(e+f x)}{b \sin ^2(e+f x)+a}}d\frac {\sin (e+f x)}{\sqrt {b \sin ^2(e+f x)+a}}}{b}}{3 a b}}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {(a+b) \sin (e+f x) \left (1-\sin ^2(e+f x)\right )}{3 a b \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\frac {\left (\frac {3 a}{b}-\frac {2 b}{a}+1\right ) \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}-\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{b^{3/2}}}{3 a b}}{f}\) |
(((a + b)*Sin[e + f*x]*(1 - Sin[e + f*x]^2))/(3*a*b*(a + b*Sin[e + f*x]^2) ^(3/2)) - ((-3*a*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2] ])/b^(3/2) + ((1 + (3*a)/b - (2*b)/a)*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x ]^2])/(3*a*b))/f
3.4.63.3.1 Defintions of rubi rules used
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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, 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 = 0.41 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.60
| method | result | size |
| derivativedivides | \(\frac {\sin \left (f x +e \right )}{3 a f {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}}}+\frac {2 \sin \left (f x +e \right )}{3 a^{2} f \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}-\frac {\sin ^{3}\left (f x +e \right )}{3 f b {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}}}-\frac {\sin \left (f x +e \right )}{f \,b^{2} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}+\frac {\ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{f \,b^{\frac {5}{2}}}+\frac {2 \sin \left (f x +e \right )}{3 f b {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}}}-\frac {2 \sin \left (f x +e \right )}{3 f b a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}\) | \(208\) |
| default | \(\frac {\sin \left (f x +e \right )}{3 a f {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}}}+\frac {2 \sin \left (f x +e \right )}{3 a^{2} f \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}-\frac {\sin ^{3}\left (f x +e \right )}{3 f b {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}}}-\frac {\sin \left (f x +e \right )}{f \,b^{2} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}+\frac {\ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{f \,b^{\frac {5}{2}}}+\frac {2 \sin \left (f x +e \right )}{3 f b {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{\frac {3}{2}}}-\frac {2 \sin \left (f x +e \right )}{3 f b a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}\) | \(208\) |
1/3*sin(f*x+e)/a/f/(a+b*sin(f*x+e)^2)^(3/2)+2/3*sin(f*x+e)/a^2/f/(a+b*sin( f*x+e)^2)^(1/2)-1/3/f*sin(f*x+e)^3/b/(a+b*sin(f*x+e)^2)^(3/2)-1/f/b^2*sin( f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)+1/f/b^(5/2)*ln(b^(1/2)*sin(f*x+e)+(a+b*sin (f*x+e)^2)^(1/2))+2/3/f/b*sin(f*x+e)/(a+b*sin(f*x+e)^2)^(3/2)-2/3/f/b/a*si n(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 345 vs. \(2 (116) = 232\).
Time = 1.11 (sec) , antiderivative size = 799, normalized size of antiderivative = 6.15 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a^{2} b^{2} \cos \left (f x + e\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} - 2 \, {\left (a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{2} b^{2} + 24 \, a b^{3} + 24 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 32 \, a^{3} b + 160 \, a^{2} b^{2} + 256 \, a b^{3} + 128 \, b^{4} - 32 \, {\left (a^{3} b + 10 \, a^{2} b^{2} + 24 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )^{2} - 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{6} - 24 \, {\left (a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} - 10 \, a^{2} b - 24 \, a b^{2} - 16 \, b^{3} + 2 \, {\left (5 \, a^{2} b + 24 \, a b^{2} + 24 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b} \sin \left (f x + e\right )\right ) - 8 \, {\left (3 \, a^{3} b + 4 \, a^{2} b^{2} - a b^{3} - 2 \, b^{4} - 2 \, {\left (2 \, a^{2} b^{2} + a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{24 \, {\left (a^{2} b^{5} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b^{3} + 2 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}, -\frac {3 \, {\left (a^{2} b^{2} \cos \left (f x + e\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} - 2 \, {\left (a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{4} + a^{2} b + 3 \, a b^{2} + 2 \, b^{3} - {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, {\left (3 \, a^{3} b + 4 \, a^{2} b^{2} - a b^{3} - 2 \, b^{4} - 2 \, {\left (2 \, a^{2} b^{2} + a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{12 \, {\left (a^{2} b^{5} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b^{3} + 2 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}\right ] \]
[1/24*(3*(a^2*b^2*cos(f*x + e)^4 + a^4 + 2*a^3*b + a^2*b^2 - 2*(a^3*b + a^ 2*b^2)*cos(f*x + e)^2)*sqrt(b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + 2 *b^4)*cos(f*x + e)^6 + 32*(5*a^2*b^2 + 24*a*b^3 + 24*b^4)*cos(f*x + e)^4 + a^4 + 32*a^3*b + 160*a^2*b^2 + 256*a*b^3 + 128*b^4 - 32*(a^3*b + 10*a^2*b ^2 + 24*a*b^3 + 16*b^4)*cos(f*x + e)^2 - 8*(16*b^3*cos(f*x + e)^6 - 24*(a* b^2 + 2*b^3)*cos(f*x + e)^4 - a^3 - 10*a^2*b - 24*a*b^2 - 16*b^3 + 2*(5*a^ 2*b + 24*a*b^2 + 24*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*s qrt(b)*sin(f*x + e)) - 8*(3*a^3*b + 4*a^2*b^2 - a*b^3 - 2*b^4 - 2*(2*a^2*b ^2 + a*b^3 - b^4)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/(a^2*b^5*f*cos(f*x + e)^4 - 2*(a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^2 + (a^4*b^3 + 2*a^3*b^4 + a^2*b^5)*f), -1/12*(3*(a^2*b^2*cos(f*x + e)^4 + a^ 4 + 2*a^3*b + a^2*b^2 - 2*(a^3*b + a^2*b^2)*cos(f*x + e)^2)*sqrt(-b)*arcta n(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + 2*b^2)*cos(f*x + e)^2 + a^2 + 8*a*b + 8*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b)/((2*b^3*cos(f*x + e)^4 + a^2*b + 3*a*b^2 + 2*b^3 - (3*a*b^2 + 4*b^3)*cos(f*x + e)^2)*sin(f*x + e) )) + 4*(3*a^3*b + 4*a^2*b^2 - a*b^3 - 2*b^4 - 2*(2*a^2*b^2 + a*b^3 - b^4)* cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/(a^2*b^5*f*c os(f*x + e)^4 - 2*(a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^2 + (a^4*b^3 + 2*a^3* b^4 + a^2*b^5)*f)]
Timed out. \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.59 \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {{\left (\frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} \sin \left (f x + e\right ) - \frac {3 \, \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} - \frac {2 \, \sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {\sin \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a} + \frac {\sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} b^{2}} - \frac {2 \, \sin \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, \sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a b}}{3 \, f} \]
-1/3*((3*sin(f*x + e)^2/((b*sin(f*x + e)^2 + a)^(3/2)*b) + 2*a/((b*sin(f*x + e)^2 + a)^(3/2)*b^2))*sin(f*x + e) - 3*arcsinh(b*sin(f*x + e)/sqrt(a*b) )/b^(5/2) - 2*sin(f*x + e)/(sqrt(b*sin(f*x + e)^2 + a)*a^2) - sin(f*x + e) /((b*sin(f*x + e)^2 + a)^(3/2)*a) + sin(f*x + e)/(sqrt(b*sin(f*x + e)^2 + a)*b^2) - 2*sin(f*x + e)/((b*sin(f*x + e)^2 + a)^(3/2)*b) + 2*sin(f*x + e) /(sqrt(b*sin(f*x + e)^2 + a)*a*b))/f
\[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{5}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^5}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]