Integrand size = 21, antiderivative size = 46 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin (e+f x) \, dx=-\frac {a^2 \cos (e+f x)}{f}+\frac {2 a b \sec (e+f x)}{f}+\frac {b^2 \sec ^3(e+f x)}{3 f} \] Output:
-a^2*cos(f*x+e)/f+2*a*b*sec(f*x+e)/f+1/3*b^2*sec(f*x+e)^3/f
Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.63 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin (e+f x) \, dx=\frac {4 \left (b+a \cos ^2(e+f x)\right )^2 \left (b^2+6 a b \cos ^2(e+f x)-3 a^2 \cos ^4(e+f x)\right ) \sec ^3(e+f x)}{3 f (a+2 b+a \cos (2 (e+f x)))^2} \] Input:
Integrate[(a + b*Sec[e + f*x]^2)^2*Sin[e + f*x],x]
Output:
(4*(b + a*Cos[e + f*x]^2)^2*(b^2 + 6*a*b*Cos[e + f*x]^2 - 3*a^2*Cos[e + f* x]^4)*Sec[e + f*x]^3)/(3*f*(a + 2*b + a*Cos[2*(e + f*x)])^2)
Time = 0.23 (sec) , antiderivative size = 41, 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, 244, 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.
\(\displaystyle \int \sin (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (e+f x) \left (a+b \sec (e+f x)^2\right )^2dx\) |
\(\Big \downarrow \) 4621 |
\(\displaystyle -\frac {\int \left (a \cos ^2(e+f x)+b\right )^2 \sec ^4(e+f x)d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {\int \left (b^2 \sec ^4(e+f x)+2 a b \sec ^2(e+f x)+a^2\right )d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 \cos (e+f x)-2 a b \sec (e+f x)-\frac {1}{3} b^2 \sec ^3(e+f x)}{f}\) |
Input:
Int[(a + b*Sec[e + f*x]^2)^2*Sin[e + f*x],x]
Output:
-((a^2*Cos[e + f*x] - 2*a*b*Sec[e + f*x] - (b^2*Sec[e + f*x]^3)/3)/f)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 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 = 0.47 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(\frac {\frac {b^{2} \sec \left (f x +e \right )^{3}}{3}+2 \sec \left (f x +e \right ) a b -\frac {a^{2}}{\sec \left (f x +e \right )}}{f}\) | \(42\) |
default | \(\frac {\frac {b^{2} \sec \left (f x +e \right )^{3}}{3}+2 \sec \left (f x +e \right ) a b -\frac {a^{2}}{\sec \left (f x +e \right )}}{f}\) | \(42\) |
parts | \(-\frac {a^{2} \cos \left (f x +e \right )}{f}+\frac {2 a b \sec \left (f x +e \right )}{f}+\frac {b^{2} \sec \left (f x +e \right )^{3}}{3 f}\) | \(45\) |
risch | \(-\frac {{\mathrm e}^{i \left (f x +e \right )} a^{2}}{2 f}-\frac {{\mathrm e}^{-i \left (f x +e \right )} a^{2}}{2 f}+\frac {4 b \left (3 a \,{\mathrm e}^{5 i \left (f x +e \right )}+6 a \,{\mathrm e}^{3 i \left (f x +e \right )}+2 b \,{\mathrm e}^{3 i \left (f x +e \right )}+3 a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}\) | \(104\) |
parallelrisch | \(\frac {2 \left (-3 a^{2}+6 a b +b^{2}\right ) \cos \left (3 f x +3 e \right )-12 a \left (a -2 b \right ) \cos \left (2 f x +2 e \right )-3 a^{2} \cos \left (4 f x +4 e \right )+6 \left (-3 a^{2}+6 a b +b^{2}\right ) \cos \left (f x +e \right )-9 a^{2}+24 a b +8 b^{2}}{6 f \left (\cos \left (3 f x +3 e \right )+3 \cos \left (f x +e \right )\right )}\) | \(117\) |
norman | \(\frac {\frac {6 a^{2}-12 a b -2 b^{2}}{3 f}+\frac {2 \left (3 a^{2}+2 a b -b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{f}-\frac {2 \left (a^{2}+2 a b +b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{f}-\frac {2 \left (9 a^{2}-6 a b +b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3} \left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )}\) | \(140\) |
Input:
int((a+b*sec(f*x+e)^2)^2*sin(f*x+e),x,method=_RETURNVERBOSE)
Output:
1/f*(1/3*b^2*sec(f*x+e)^3+2*sec(f*x+e)*a*b-a^2/sec(f*x+e))
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin (e+f x) \, dx=-\frac {3 \, a^{2} \cos \left (f x + e\right )^{4} - 6 \, a b \cos \left (f x + e\right )^{2} - b^{2}}{3 \, f \cos \left (f x + e\right )^{3}} \] Input:
integrate((a+b*sec(f*x+e)^2)^2*sin(f*x+e),x, algorithm="fricas")
Output:
-1/3*(3*a^2*cos(f*x + e)^4 - 6*a*b*cos(f*x + e)^2 - b^2)/(f*cos(f*x + e)^3 )
\[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin (e+f x) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \sin {\left (e + f x \right )}\, dx \] Input:
integrate((a+b*sec(f*x+e)**2)**2*sin(f*x+e),x)
Output:
Integral((a + b*sec(e + f*x)**2)**2*sin(e + f*x), x)
Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.91 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin (e+f x) \, dx=-\frac {3 \, a^{2} \cos \left (f x + e\right ) - \frac {6 \, a b}{\cos \left (f x + e\right )} - \frac {b^{2}}{\cos \left (f x + e\right )^{3}}}{3 \, f} \] Input:
integrate((a+b*sec(f*x+e)^2)^2*sin(f*x+e),x, algorithm="maxima")
Output:
-1/3*(3*a^2*cos(f*x + e) - 6*a*b/cos(f*x + e) - b^2/cos(f*x + e)^3)/f
Time = 0.13 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.93 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin (e+f x) \, dx=-\frac {3 \, a^{2} \cos \left (f x + e\right ) - \frac {6 \, a b \cos \left (f x + e\right )^{2} + b^{2}}{\cos \left (f x + e\right )^{3}}}{3 \, f} \] Input:
integrate((a+b*sec(f*x+e)^2)^2*sin(f*x+e),x, algorithm="giac")
Output:
-1/3*(3*a^2*cos(f*x + e) - (6*a*b*cos(f*x + e)^2 + b^2)/cos(f*x + e)^3)/f
Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.98 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin (e+f x) \, dx=\frac {\frac {b^2}{3}+2\,a\,b\,{\cos \left (e+f\,x\right )}^2}{f\,{\cos \left (e+f\,x\right )}^3}-\frac {a^2\,\cos \left (e+f\,x\right )}{f} \] Input:
int(sin(e + f*x)*(a + b/cos(e + f*x)^2)^2,x)
Output:
(b^2/3 + 2*a*b*cos(e + f*x)^2)/(f*cos(e + f*x)^3) - (a^2*cos(e + f*x))/f
Time = 0.16 (sec) , antiderivative size = 138, normalized size of antiderivative = 3.00 \[ \int \left (a+b \sec ^2(e+f x)\right )^2 \sin (e+f x) \, dx=\frac {-3 \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2} a^{2}+3 \cos \left (f x +e \right )^{2} a^{2}-6 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a b -\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b^{2}+6 \cos \left (f x +e \right ) a b +\cos \left (f x +e \right ) b^{2}+6 \sin \left (f x +e \right )^{2} a b -6 a b -b^{2}}{3 \cos \left (f x +e \right ) f \left (\sin \left (f x +e \right )^{2}-1\right )} \] Input:
int((a+b*sec(f*x+e)^2)^2*sin(f*x+e),x)
Output:
( - 3*cos(e + f*x)**2*sin(e + f*x)**2*a**2 + 3*cos(e + f*x)**2*a**2 - 6*co s(e + f*x)*sin(e + f*x)**2*a*b - cos(e + f*x)*sin(e + f*x)**2*b**2 + 6*cos (e + f*x)*a*b + cos(e + f*x)*b**2 + 6*sin(e + f*x)**2*a*b - 6*a*b - b**2)/ (3*cos(e + f*x)*f*(sin(e + f*x)**2 - 1))