Integrand size = 21, antiderivative size = 66 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^5(e+f x) \, dx=-\frac {(a-2 b) \cos (e+f x)}{f}+\frac {(2 a-b) \cos ^3(e+f x)}{3 f}-\frac {a \cos ^5(e+f x)}{5 f}+\frac {b \sec (e+f x)}{f} \] Output:
-(a-2*b)*cos(f*x+e)/f+1/3*(2*a-b)*cos(f*x+e)^3/f-1/5*a*cos(f*x+e)^5/f+b*se c(f*x+e)/f
Time = 0.17 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.33 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^5(e+f x) \, dx=-\frac {5 a \cos (e+f x)}{8 f}+\frac {7 b \cos (e+f x)}{4 f}+\frac {5 a \cos (3 (e+f x))}{48 f}-\frac {b \cos (3 (e+f x))}{12 f}-\frac {a \cos (5 (e+f x))}{80 f}+\frac {b \sec (e+f x)}{f} \] Input:
Integrate[(a + b*Sec[e + f*x]^2)*Sin[e + f*x]^5,x]
Output:
(-5*a*Cos[e + f*x])/(8*f) + (7*b*Cos[e + f*x])/(4*f) + (5*a*Cos[3*(e + f*x )])/(48*f) - (b*Cos[3*(e + f*x)])/(12*f) - (a*Cos[5*(e + f*x)])/(80*f) + ( b*Sec[e + f*x])/f
Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 4621, 355, 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.
\(\displaystyle \int \sin ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (e+f x)^5 \left (a+b \sec (e+f x)^2\right )dx\) |
\(\Big \downarrow \) 4621 |
\(\displaystyle -\frac {\int \left (1-\cos ^2(e+f x)\right )^2 \left (a \cos ^2(e+f x)+b\right ) \sec ^2(e+f x)d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 355 |
\(\displaystyle -\frac {\int \left (a \cos ^4(e+f x)-(2 a-b) \cos ^2(e+f x)+b \sec ^2(e+f x)+a \left (1-\frac {2 b}{a}\right )\right )d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{3} (2 a-b) \cos ^3(e+f x)+(a-2 b) \cos (e+f x)+\frac {1}{5} a \cos ^5(e+f x)-b \sec (e+f x)}{f}\) |
Input:
Int[(a + b*Sec[e + f*x]^2)*Sin[e + f*x]^5,x]
Output:
-(((a - 2*b)*Cos[e + f*x] - ((2*a - b)*Cos[e + f*x]^3)/3 + (a*Cos[e + f*x] ^5)/5 - b*Sec[e + f*x])/f)
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] & & IGtQ[q, 0]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff*x)^n)^p/(ff*x)^(n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/ 2] && IntegerQ[n] && IntegerQ[p]
Time = 1.05 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.23
method | result | size |
parallelrisch | \(\frac {\left (-125 a +400 b \right ) \cos \left (2 f x +2 e \right )+\left (22 a -20 b \right ) \cos \left (4 f x +4 e \right )-3 \cos \left (6 f x +6 e \right ) a +\left (-256 a +1280 b \right ) \cos \left (f x +e \right )-150 a +900 b}{480 f \cos \left (f x +e \right )}\) | \(81\) |
derivativedivides | \(\frac {-\frac {a \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5}+b \left (\frac {\sin \left (f x +e \right )^{6}}{\cos \left (f x +e \right )}+\left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )\right )}{f}\) | \(82\) |
default | \(\frac {-\frac {a \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5}+b \left (\frac {\sin \left (f x +e \right )^{6}}{\cos \left (f x +e \right )}+\left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )\right )}{f}\) | \(82\) |
parts | \(-\frac {a \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5 f}+\frac {b \left (\frac {\sin \left (f x +e \right )^{6}}{\cos \left (f x +e \right )}+\left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )\right )}{f}\) | \(84\) |
norman | \(\frac {\frac {16 a -80 b}{15 f}-\frac {32 \left (a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 f}+\frac {4 \left (16 a -80 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{15 f}+\frac {\left (16 a -80 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{3 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right ) \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{5}}\) | \(110\) |
risch | \(-\frac {5 \,{\mathrm e}^{i \left (f x +e \right )} a}{16 f}+\frac {7 \,{\mathrm e}^{i \left (f x +e \right )} b}{8 f}-\frac {5 \,{\mathrm e}^{-i \left (f x +e \right )} a}{16 f}+\frac {7 \,{\mathrm e}^{-i \left (f x +e \right )} b}{8 f}+\frac {2 b \,{\mathrm e}^{i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {\cos \left (5 f x +5 e \right ) a}{80 f}+\frac {5 \cos \left (3 f x +3 e \right ) a}{48 f}-\frac {\cos \left (3 f x +3 e \right ) b}{12 f}\) | \(135\) |
Input:
int((a+b*sec(f*x+e)^2)*sin(f*x+e)^5,x,method=_RETURNVERBOSE)
Output:
1/480*((-125*a+400*b)*cos(2*f*x+2*e)+(22*a-20*b)*cos(4*f*x+4*e)-3*cos(6*f* x+6*e)*a+(-256*a+1280*b)*cos(f*x+e)-150*a+900*b)/f/cos(f*x+e)
Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.91 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^5(e+f x) \, dx=-\frac {3 \, a \cos \left (f x + e\right )^{6} - 5 \, {\left (2 \, a - b\right )} \cos \left (f x + e\right )^{4} + 15 \, {\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{2} - 15 \, b}{15 \, f \cos \left (f x + e\right )} \] Input:
integrate((a+b*sec(f*x+e)^2)*sin(f*x+e)^5,x, algorithm="fricas")
Output:
-1/15*(3*a*cos(f*x + e)^6 - 5*(2*a - b)*cos(f*x + e)^4 + 15*(a - 2*b)*cos( f*x + e)^2 - 15*b)/(f*cos(f*x + e))
\[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^5(e+f x) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \sin ^{5}{\left (e + f x \right )}\, dx \] Input:
integrate((a+b*sec(f*x+e)**2)*sin(f*x+e)**5,x)
Output:
Integral((a + b*sec(e + f*x)**2)*sin(e + f*x)**5, x)
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.88 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^5(e+f x) \, dx=-\frac {3 \, a \cos \left (f x + e\right )^{5} - 5 \, {\left (2 \, a - b\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (a - 2 \, b\right )} \cos \left (f x + e\right ) - \frac {15 \, b}{\cos \left (f x + e\right )}}{15 \, f} \] Input:
integrate((a+b*sec(f*x+e)^2)*sin(f*x+e)^5,x, algorithm="maxima")
Output:
-1/15*(3*a*cos(f*x + e)^5 - 5*(2*a - b)*cos(f*x + e)^3 + 15*(a - 2*b)*cos( f*x + e) - 15*b/cos(f*x + e))/f
Time = 0.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.03 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^5(e+f x) \, dx=-\frac {3 \, a \cos \left (f x + e\right )^{5} - 10 \, a \cos \left (f x + e\right )^{3} + 5 \, b \cos \left (f x + e\right )^{3} + 15 \, a \cos \left (f x + e\right ) - 30 \, b \cos \left (f x + e\right ) - \frac {15 \, b}{\cos \left (f x + e\right )}}{15 \, f} \] Input:
integrate((a+b*sec(f*x+e)^2)*sin(f*x+e)^5,x, algorithm="giac")
Output:
-1/15*(3*a*cos(f*x + e)^5 - 10*a*cos(f*x + e)^3 + 5*b*cos(f*x + e)^3 + 15* a*cos(f*x + e) - 30*b*cos(f*x + e) - 15*b/cos(f*x + e))/f
Time = 12.51 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.83 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^5(e+f x) \, dx=\frac {{\cos \left (e+f\,x\right )}^3\,\left (\frac {2\,a}{3}-\frac {b}{3}\right )-\cos \left (e+f\,x\right )\,\left (a-2\,b\right )-\frac {a\,{\cos \left (e+f\,x\right )}^5}{5}+\frac {b}{\cos \left (e+f\,x\right )}}{f} \] Input:
int(sin(e + f*x)^5*(a + b/cos(e + f*x)^2),x)
Output:
(cos(e + f*x)^3*((2*a)/3 - b/3) - cos(e + f*x)*(a - 2*b) - (a*cos(e + f*x) ^5)/5 + b/cos(e + f*x))/f
Time = 0.15 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.61 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^5(e+f x) \, dx=\frac {-3 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{4} a -4 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2} a -8 \cos \left (f x +e \right )^{2} a +8 \cos \left (f x +e \right ) a -40 \cos \left (f x +e \right ) b -5 \sin \left (f x +e \right )^{4} b -20 \sin \left (f x +e \right )^{2} b +40 b}{15 \cos \left (f x +e \right ) f} \] Input:
int((a+b*sec(f*x+e)^2)*sin(f*x+e)^5,x)
Output:
( - 3*cos(e + f*x)**2*sin(e + f*x)**4*a - 4*cos(e + f*x)**2*sin(e + f*x)** 2*a - 8*cos(e + f*x)**2*a + 8*cos(e + f*x)*a - 40*cos(e + f*x)*b - 5*sin(e + f*x)**4*b - 20*sin(e + f*x)**2*b + 40*b)/(15*cos(e + f*x)*f)