Integrand size = 21, antiderivative size = 44 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^3(e+f x) \, dx=-\frac {(a-b) \cos (e+f x)}{f}+\frac {a \cos ^3(e+f x)}{3 f}+\frac {b \sec (e+f x)}{f} \] Output:
-(a-b)*cos(f*x+e)/f+1/3*a*cos(f*x+e)^3/f+b*sec(f*x+e)/f
Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.20 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^3(e+f x) \, dx=-\frac {3 a \cos (e+f x)}{4 f}+\frac {b \cos (e+f x)}{f}+\frac {a \cos (3 (e+f x))}{12 f}+\frac {b \sec (e+f x)}{f} \] Input:
Integrate[(a + b*Sec[e + f*x]^2)*Sin[e + f*x]^3,x]
Output:
(-3*a*Cos[e + f*x])/(4*f) + (b*Cos[e + f*x])/f + (a*Cos[3*(e + f*x)])/(12* f) + (b*Sec[e + f*x])/f
Time = 0.23 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, 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.
\(\displaystyle \int \sin ^3(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (e+f x)^3 \left (a+b \sec (e+f x)^2\right )dx\) |
\(\Big \downarrow \) 4621 |
\(\displaystyle -\frac {\int \left (1-\cos ^2(e+f x)\right ) \left (a \cos ^2(e+f x)+b\right ) \sec ^2(e+f x)d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 355 |
\(\displaystyle -\frac {\int \left (-a \cos ^2(e+f x)+b \sec ^2(e+f x)+a \left (1-\frac {b}{a}\right )\right )d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {(a-b) \cos (e+f x)-\frac {1}{3} a \cos ^3(e+f x)-b \sec (e+f x)}{f}\) |
Input:
Int[(a + b*Sec[e + f*x]^2)*Sin[e + f*x]^3,x]
Output:
-(((a - b)*Cos[e + f*x] - (a*Cos[e + f*x]^3)/3 - 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 = 0.91 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.41
method | result | size |
derivativedivides | \(\frac {-\frac {a \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}+b \left (\frac {\sin \left (f x +e \right )^{4}}{\cos \left (f x +e \right )}+\left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )\right )}{f}\) | \(62\) |
default | \(\frac {-\frac {a \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}+b \left (\frac {\sin \left (f x +e \right )^{4}}{\cos \left (f x +e \right )}+\left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )\right )}{f}\) | \(62\) |
parallelrisch | \(\frac {\left (-8 a +12 b \right ) \cos \left (2 f x +2 e \right )+\cos \left (4 f x +4 e \right ) a +\left (-16 a +48 b \right ) \cos \left (f x +e \right )-9 a +36 b}{24 f \cos \left (f x +e \right )}\) | \(63\) |
parts | \(-\frac {a \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3 f}+\frac {b \left (\frac {\sin \left (f x +e \right )^{4}}{\cos \left (f x +e \right )}+\left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )\right )}{f}\) | \(64\) |
norman | \(\frac {\frac {4 a -12 b}{3 f}-\frac {4 \left (a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{f}+\frac {2 \left (4 a -12 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right ) \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3}}\) | \(87\) |
risch | \(-\frac {3 \,{\mathrm e}^{i \left (f x +e \right )} a}{8 f}+\frac {{\mathrm e}^{i \left (f x +e \right )} b}{2 f}-\frac {3 \,{\mathrm e}^{-i \left (f x +e \right )} a}{8 f}+\frac {{\mathrm e}^{-i \left (f x +e \right )} b}{2 f}+\frac {2 b \,{\mathrm e}^{i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {\cos \left (3 f x +3 e \right ) a}{12 f}\) | \(105\) |
Input:
int((a+b*sec(f*x+e)^2)*sin(f*x+e)^3,x,method=_RETURNVERBOSE)
Output:
1/f*(-1/3*a*(2+sin(f*x+e)^2)*cos(f*x+e)+b*(sin(f*x+e)^4/cos(f*x+e)+(2+sin( f*x+e)^2)*cos(f*x+e)))
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^3(e+f x) \, dx=\frac {a \cos \left (f x + e\right )^{4} - 3 \, {\left (a - b\right )} \cos \left (f x + e\right )^{2} + 3 \, b}{3 \, f \cos \left (f x + e\right )} \] Input:
integrate((a+b*sec(f*x+e)^2)*sin(f*x+e)^3,x, algorithm="fricas")
Output:
1/3*(a*cos(f*x + e)^4 - 3*(a - b)*cos(f*x + e)^2 + 3*b)/(f*cos(f*x + e))
\[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^3(e+f x) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \sin ^{3}{\left (e + f x \right )}\, dx \] Input:
integrate((a+b*sec(f*x+e)**2)*sin(f*x+e)**3,x)
Output:
Integral((a + b*sec(e + f*x)**2)*sin(e + f*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.91 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^3(e+f x) \, dx=\frac {a \cos \left (f x + e\right )^{3} - 3 \, {\left (a - b\right )} \cos \left (f x + e\right ) + \frac {3 \, b}{\cos \left (f x + e\right )}}{3 \, f} \] Input:
integrate((a+b*sec(f*x+e)^2)*sin(f*x+e)^3,x, algorithm="maxima")
Output:
1/3*(a*cos(f*x + e)^3 - 3*(a - b)*cos(f*x + e) + 3*b/cos(f*x + e))/f
Time = 0.13 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.02 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^3(e+f x) \, dx=\frac {a \cos \left (f x + e\right )^{3} - 3 \, a \cos \left (f x + e\right ) + 3 \, b \cos \left (f x + e\right ) + \frac {3 \, b}{\cos \left (f x + e\right )}}{3 \, f} \] Input:
integrate((a+b*sec(f*x+e)^2)*sin(f*x+e)^3,x, algorithm="giac")
Output:
1/3*(a*cos(f*x + e)^3 - 3*a*cos(f*x + e) + 3*b*cos(f*x + e) + 3*b/cos(f*x + e))/f
Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.89 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^3(e+f x) \, dx=\frac {\frac {a\,{\cos \left (e+f\,x\right )}^3}{3}-\cos \left (e+f\,x\right )\,\left (a-b\right )+\frac {b}{\cos \left (e+f\,x\right )}}{f} \] Input:
int(sin(e + f*x)^3*(a + b/cos(e + f*x)^2),x)
Output:
((a*cos(e + f*x)^3)/3 - cos(e + f*x)*(a - b) + b/cos(e + f*x))/f
Time = 0.15 (sec) , antiderivative size = 125, normalized size of antiderivative = 2.84 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^3(e+f x) \, dx=\frac {-\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} a +\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a -2 \cos \left (f x +e \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} a +2 \cos \left (f x +e \right ) a +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} a -12 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} b -2 a}{3 f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-1\right )} \] Input:
int((a+b*sec(f*x+e)^2)*sin(f*x+e)^3,x)
Output:
( - cos(e + f*x)*sin(e + f*x)**2*tan((e + f*x)/2)**4*a + cos(e + f*x)*sin( e + f*x)**2*a - 2*cos(e + f*x)*tan((e + f*x)/2)**4*a + 2*cos(e + f*x)*a + 2*tan((e + f*x)/2)**4*a - 12*tan((e + f*x)/2)**4*b - 2*a)/(3*f*(tan((e + f *x)/2)**4 - 1))