Integrand size = 21, antiderivative size = 42 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^2(e+f x) \, dx=\frac {1}{2} (a-2 b) x-\frac {a \cos (e+f x) \sin (e+f x)}{2 f}+\frac {b \tan (e+f x)}{f} \] Output:
1/2*(a-2*b)*x-1/2*a*cos(f*x+e)*sin(f*x+e)/f+b*tan(f*x+e)/f
Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.29 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^2(e+f x) \, dx=\frac {a (e+f x)}{2 f}-\frac {b \arctan (\tan (e+f x))}{f}-\frac {a \sin (2 (e+f x))}{4 f}+\frac {b \tan (e+f x)}{f} \] Input:
Integrate[(a + b*Sec[e + f*x]^2)*Sin[e + f*x]^2,x]
Output:
(a*(e + f*x))/(2*f) - (b*ArcTan[Tan[e + f*x]])/f - (a*Sin[2*(e + f*x)])/(4 *f) + (b*Tan[e + f*x])/f
Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.31, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4620, 360, 25, 299, 216}
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.
\(\displaystyle \int \sin ^2(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (e+f x)^2 \left (a+b \sec (e+f x)^2\right )dx\) |
\(\Big \downarrow \) 4620 |
\(\displaystyle \frac {\int \frac {\tan ^2(e+f x) \left (b \tan ^2(e+f x)+a+b\right )}{\left (\tan ^2(e+f x)+1\right )^2}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 360 |
\(\displaystyle \frac {-\frac {1}{2} \int -\frac {2 b \tan ^2(e+f x)+a}{\tan ^2(e+f x)+1}d\tan (e+f x)-\frac {a \tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {2 b \tan ^2(e+f x)+a}{\tan ^2(e+f x)+1}d\tan (e+f x)-\frac {a \tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}}{f}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\frac {1}{2} \left ((a-2 b) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)+2 b \tan (e+f x)\right )-\frac {a \tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {1}{2} ((a-2 b) \arctan (\tan (e+f x))+2 b \tan (e+f x))-\frac {a \tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}}{f}\) |
Input:
Int[(a + b*Sec[e + f*x]^2)*Sin[e + f*x]^2,x]
Output:
(((a - 2*b)*ArcTan[Tan[e + f*x]] + 2*b*Tan[e + f*x])/2 - (a*Tan[e + f*x])/ (2*(1 + Tan[e + f*x]^2)))/f
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1)/f Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]
Time = 0.45 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.10
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+b \left (\tan \left (f x +e \right )-f x -e \right )}{f}\) | \(46\) |
default | \(\frac {a \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+b \left (\tan \left (f x +e \right )-f x -e \right )}{f}\) | \(46\) |
parts | \(\frac {a \left (-\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {b \left (\tan \left (f x +e \right )-f x -e \right )}{f}\) | \(48\) |
parallelrisch | \(\frac {-\sin \left (3 f x +3 e \right ) a +4 f x \left (a -2 b \right ) \cos \left (f x +e \right )-\sin \left (f x +e \right ) \left (a -8 b \right )}{8 \cos \left (f x +e \right ) f}\) | \(55\) |
risch | \(\frac {a x}{2}-x b +\frac {i {\mathrm e}^{2 i \left (f x +e \right )} a}{8 f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} a}{8 f}+\frac {2 i b}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\) | \(62\) |
norman | \(\frac {\left (-\frac {a}{2}+b \right ) x +\frac {\left (a -2 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {\left (a -2 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{f}+\left (-\frac {a}{2}+b \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (\frac {a}{2}-b \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (\frac {a}{2}-b \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}-\frac {2 \left (a +2 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right ) \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}\) | \(157\) |
Input:
int((a+b*sec(f*x+e)^2)*sin(f*x+e)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(a*(-1/2*cos(f*x+e)*sin(f*x+e)+1/2*f*x+1/2*e)+b*(tan(f*x+e)-f*x-e))
Time = 0.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.19 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^2(e+f x) \, dx=\frac {{\left (a - 2 \, b\right )} f x \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right )^{2} - 2 \, b\right )} \sin \left (f x + e\right )}{2 \, f \cos \left (f x + e\right )} \] Input:
integrate((a+b*sec(f*x+e)^2)*sin(f*x+e)^2,x, algorithm="fricas")
Output:
1/2*((a - 2*b)*f*x*cos(f*x + e) - (a*cos(f*x + e)^2 - 2*b)*sin(f*x + e))/( f*cos(f*x + e))
\[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^2(e+f x) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \sin ^{2}{\left (e + f x \right )}\, dx \] Input:
integrate((a+b*sec(f*x+e)**2)*sin(f*x+e)**2,x)
Output:
Integral((a + b*sec(e + f*x)**2)*sin(e + f*x)**2, x)
Time = 0.11 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.12 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^2(e+f x) \, dx=\frac {{\left (f x + e\right )} {\left (a - 2 \, b\right )} + 2 \, b \tan \left (f x + e\right ) - \frac {a \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}}{2 \, f} \] Input:
integrate((a+b*sec(f*x+e)^2)*sin(f*x+e)^2,x, algorithm="maxima")
Output:
1/2*((f*x + e)*(a - 2*b) + 2*b*tan(f*x + e) - a*tan(f*x + e)/(tan(f*x + e) ^2 + 1))/f
Time = 0.15 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.12 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^2(e+f x) \, dx=\frac {{\left (f x + e\right )} {\left (a - 2 \, b\right )} + 2 \, b \tan \left (f x + e\right ) - \frac {a \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}}{2 \, f} \] Input:
integrate((a+b*sec(f*x+e)^2)*sin(f*x+e)^2,x, algorithm="giac")
Output:
1/2*((f*x + e)*(a - 2*b) + 2*b*tan(f*x + e) - a*tan(f*x + e)/(tan(f*x + e) ^2 + 1))/f
Time = 12.54 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.83 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^2(e+f x) \, dx=\frac {b\,\mathrm {tan}\left (e+f\,x\right )-\frac {a\,\sin \left (2\,e+2\,f\,x\right )}{4}+f\,x\,\left (\frac {a}{2}-b\right )}{f} \] Input:
int(sin(e + f*x)^2*(a + b/cos(e + f*x)^2),x)
Output:
(b*tan(e + f*x) - (a*sin(2*e + 2*f*x))/4 + f*x*(a/2 - b))/f
Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.45 \[ \int \left (a+b \sec ^2(e+f x)\right ) \sin ^2(e+f x) \, dx=\frac {-\cos \left (f x +e \right )^{2} \sin \left (f x +e \right ) a +\cos \left (f x +e \right ) a f x -2 \cos \left (f x +e \right ) b f x +2 \sin \left (f x +e \right ) b}{2 \cos \left (f x +e \right ) f} \] Input:
int((a+b*sec(f*x+e)^2)*sin(f*x+e)^2,x)
Output:
( - cos(e + f*x)**2*sin(e + f*x)*a + cos(e + f*x)*a*f*x - 2*cos(e + f*x)*b *f*x + 2*sin(e + f*x)*b)/(2*cos(e + f*x)*f)