Integrand size = 23, antiderivative size = 53 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {(a+b)^2 \sin (e+f x)}{f}-\frac {2 a (a+b) \sin ^3(e+f x)}{3 f}+\frac {a^2 \sin ^5(e+f x)}{5 f} \] Output:
(a+b)^2*sin(f*x+e)/f-2/3*a*(a+b)*sin(f*x+e)^3/f+1/5*a^2*sin(f*x+e)^5/f
Time = 0.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 2.00 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {b^2 \cos (f x) \sin (e)}{f}+\frac {b^2 \cos (e) \sin (f x)}{f}+\frac {a^2 \sin (e+f x)}{f}+\frac {2 a b \sin (e+f x)}{f}-\frac {2 a^2 \sin ^3(e+f x)}{3 f}-\frac {2 a b \sin ^3(e+f x)}{3 f}+\frac {a^2 \sin ^5(e+f x)}{5 f} \] Input:
Integrate[Cos[e + f*x]^5*(a + b*Sec[e + f*x]^2)^2,x]
Output:
(b^2*Cos[f*x]*Sin[e])/f + (b^2*Cos[e]*Sin[f*x])/f + (a^2*Sin[e + f*x])/f + (2*a*b*Sin[e + f*x])/f - (2*a^2*Sin[e + f*x]^3)/(3*f) - (2*a*b*Sin[e + f* x]^3)/(3*f) + (a^2*Sin[e + f*x]^5)/(5*f)
Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4635, 210, 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.
\(\displaystyle \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^2}{\sec (e+f x)^5}dx\) |
\(\Big \downarrow \) 4635 |
\(\displaystyle \frac {\int \left (-a \sin ^2(e+f x)+a+b\right )^2d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 210 |
\(\displaystyle \frac {\int \left (a^2 \sin ^4(e+f x)-2 a^2 \left (\frac {b}{a}+1\right ) \sin ^2(e+f x)+a^2 \left (\frac {b (2 a+b)}{a^2}+1\right )\right )d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{5} a^2 \sin ^5(e+f x)-\frac {2}{3} a (a+b) \sin ^3(e+f x)+(a+b)^2 \sin (e+f x)}{f}\) |
Input:
Int[Cos[e + f*x]^5*(a + b*Sec[e + f*x]^2)^2,x]
Output:
((a + b)^2*Sin[e + f*x] - (2*a*(a + b)*Sin[e + f*x]^3)/3 + (a^2*Sin[e + f* x]^5)/5)/f
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^2 )^p, x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_ ))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && In tegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Time = 1.72 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.26
method | result | size |
derivativedivides | \(\frac {\frac {a^{2} \left (\frac {8}{3}+\cos \left (f x +e \right )^{4}+\frac {4 \cos \left (f x +e \right )^{2}}{3}\right ) \sin \left (f x +e \right )}{5}+\frac {2 a b \left (2+\cos \left (f x +e \right )^{2}\right ) \sin \left (f x +e \right )}{3}+\sin \left (f x +e \right ) b^{2}}{f}\) | \(67\) |
default | \(\frac {\frac {a^{2} \left (\frac {8}{3}+\cos \left (f x +e \right )^{4}+\frac {4 \cos \left (f x +e \right )^{2}}{3}\right ) \sin \left (f x +e \right )}{5}+\frac {2 a b \left (2+\cos \left (f x +e \right )^{2}\right ) \sin \left (f x +e \right )}{3}+\sin \left (f x +e \right ) b^{2}}{f}\) | \(67\) |
parallelrisch | \(\frac {150 \sin \left (f x +e \right ) a^{2}+360 \sin \left (f x +e \right ) a b +240 \sin \left (f x +e \right ) b^{2}+3 \sin \left (5 f x +5 e \right ) a^{2}+25 \sin \left (3 f x +3 e \right ) a^{2}+40 \sin \left (3 f x +3 e \right ) a b}{240 f}\) | \(80\) |
risch | \(\frac {5 a^{2} \sin \left (f x +e \right )}{8 f}+\frac {3 \sin \left (f x +e \right ) a b}{2 f}+\frac {\sin \left (f x +e \right ) b^{2}}{f}+\frac {a^{2} \sin \left (5 f x +5 e \right )}{80 f}+\frac {5 a^{2} \sin \left (3 f x +3 e \right )}{48 f}+\frac {\sin \left (3 f x +3 e \right ) a b}{6 f}\) | \(92\) |
norman | \(\frac {-\frac {2 \left (a^{2}+2 a b +b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {2 \left (a^{2}+2 a b +b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{15}}{f}+\frac {2 \left (5 a^{2}+2 a b -3 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 f}-\frac {2 \left (5 a^{2}+2 a b -3 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{13}}{3 f}-\frac {2 \left (43 a^{2}-50 a b -45 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{15 f}+\frac {2 \left (43 a^{2}-50 a b -45 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{15 f}+\frac {2 \left (109 a^{2}+10 a b +45 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{15 f}-\frac {2 \left (109 a^{2}+10 a b +45 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{15 f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3}}\) | \(271\) |
Input:
int(cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(1/5*a^2*(8/3+cos(f*x+e)^4+4/3*cos(f*x+e)^2)*sin(f*x+e)+2/3*a*b*(2+cos (f*x+e)^2)*sin(f*x+e)+sin(f*x+e)*b^2)
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.11 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {{\left (3 \, a^{2} \cos \left (f x + e\right )^{4} + 2 \, {\left (2 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} + 20 \, a b + 15 \, b^{2}\right )} \sin \left (f x + e\right )}{15 \, f} \] Input:
integrate(cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")
Output:
1/15*(3*a^2*cos(f*x + e)^4 + 2*(2*a^2 + 5*a*b)*cos(f*x + e)^2 + 8*a^2 + 20 *a*b + 15*b^2)*sin(f*x + e)/f
Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\text {Timed out} \] Input:
integrate(cos(f*x+e)**5*(a+b*sec(f*x+e)**2)**2,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, a^{2} \sin \left (f x + e\right )^{5} - 10 \, {\left (a^{2} + a b\right )} \sin \left (f x + e\right )^{3} + 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \left (f x + e\right )}{15 \, f} \] Input:
integrate(cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")
Output:
1/15*(3*a^2*sin(f*x + e)^5 - 10*(a^2 + a*b)*sin(f*x + e)^3 + 15*(a^2 + 2*a *b + b^2)*sin(f*x + e))/f
Time = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.43 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, a^{2} \sin \left (f x + e\right )^{5} - 10 \, a^{2} \sin \left (f x + e\right )^{3} - 10 \, a b \sin \left (f x + e\right )^{3} + 15 \, a^{2} \sin \left (f x + e\right ) + 30 \, a b \sin \left (f x + e\right ) + 15 \, b^{2} \sin \left (f x + e\right )}{15 \, f} \] Input:
integrate(cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")
Output:
1/15*(3*a^2*sin(f*x + e)^5 - 10*a^2*sin(f*x + e)^3 - 10*a*b*sin(f*x + e)^3 + 15*a^2*sin(f*x + e) + 30*a*b*sin(f*x + e) + 15*b^2*sin(f*x + e))/f
Time = 14.99 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {\sin \left (e+f\,x\right )\,{\left (a+b\right )}^2+\frac {a^2\,{\sin \left (e+f\,x\right )}^5}{5}-\frac {2\,a\,{\sin \left (e+f\,x\right )}^3\,\left (a+b\right )}{3}}{f} \] Input:
int(cos(e + f*x)^5*(a + b/cos(e + f*x)^2)^2,x)
Output:
(sin(e + f*x)*(a + b)^2 + (a^2*sin(e + f*x)^5)/5 - (2*a*sin(e + f*x)^3*(a + b))/3)/f
Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.21 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {\sin \left (f x +e \right ) \left (3 \sin \left (f x +e \right )^{4} a^{2}-10 \sin \left (f x +e \right )^{2} a^{2}-10 \sin \left (f x +e \right )^{2} a b +15 a^{2}+30 a b +15 b^{2}\right )}{15 f} \] Input:
int(cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x)
Output:
(sin(e + f*x)*(3*sin(e + f*x)**4*a**2 - 10*sin(e + f*x)**2*a**2 - 10*sin(e + f*x)**2*a*b + 15*a**2 + 30*a*b + 15*b**2))/(15*f)