Integrand size = 21, antiderivative size = 61 \[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {1}{8} (3 a+4 b) x+\frac {(3 a+4 b) \cos (e+f x) \sin (e+f x)}{8 f}+\frac {a \cos ^3(e+f x) \sin (e+f x)}{4 f} \] Output:
1/8*(3*a+4*b)*x+1/8*(3*a+4*b)*cos(f*x+e)*sin(f*x+e)/f+1/4*a*cos(f*x+e)^3*s in(f*x+e)/f
Time = 0.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74 \[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {4 (3 a+4 b) (e+f x)+8 (a+b) \sin (2 (e+f x))+a \sin (4 (e+f x))}{32 f} \] Input:
Integrate[Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2),x]
Output:
(4*(3*a + 4*b)*(e + f*x) + 8*(a + b)*Sin[2*(e + f*x)] + a*Sin[4*(e + f*x)] )/(32*f)
Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4533, 3042, 3115, 24}
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.
\(\displaystyle \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \csc \left (e+f x+\frac {\pi }{2}\right )^2}{\csc \left (e+f x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 4533 |
\(\displaystyle \frac {1}{4} (3 a+4 b) \int \cos ^2(e+f x)dx+\frac {a \sin (e+f x) \cos ^3(e+f x)}{4 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} (3 a+4 b) \int \sin \left (e+f x+\frac {\pi }{2}\right )^2dx+\frac {a \sin (e+f x) \cos ^3(e+f x)}{4 f}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{4} (3 a+4 b) \left (\frac {\int 1dx}{2}+\frac {\sin (e+f x) \cos (e+f x)}{2 f}\right )+\frac {a \sin (e+f x) \cos ^3(e+f x)}{4 f}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{4} (3 a+4 b) \left (\frac {\sin (e+f x) \cos (e+f x)}{2 f}+\frac {x}{2}\right )+\frac {a \sin (e+f x) \cos ^3(e+f x)}{4 f}\) |
Input:
Int[Cos[e + f*x]^4*(a + b*Sec[e + f*x]^2),x]
Output:
(a*Cos[e + f*x]^3*Sin[e + f*x])/(4*f) + ((3*a + 4*b)*(x/2 + (Cos[e + f*x]* Sin[e + f*x])/(2*f)))/4
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
Time = 0.94 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72
method | result | size |
parallelrisch | \(\frac {\left (8 a +8 b \right ) \sin \left (2 f x +2 e \right )+a \sin \left (4 f x +4 e \right )+12 x \left (a +\frac {4 b}{3}\right ) f}{32 f}\) | \(44\) |
risch | \(\frac {3 a x}{8}+\frac {x b}{2}+\frac {a \sin \left (4 f x +4 e \right )}{32 f}+\frac {a \sin \left (2 f x +2 e \right )}{4 f}+\frac {\sin \left (2 f x +2 e \right ) b}{4 f}\) | \(55\) |
derivativedivides | \(\frac {a \left (\frac {\left (\cos \left (f x +e \right )^{3}+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+b \left (\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\) | \(65\) |
default | \(\frac {a \left (\frac {\left (\cos \left (f x +e \right )^{3}+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+b \left (\frac {\cos \left (f x +e \right ) \sin \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\) | \(65\) |
norman | \(\frac {\left (-\frac {3 a}{8}-\frac {b}{2}\right ) x +\left (-\frac {9 a}{8}-\frac {3 b}{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (-\frac {3 a}{4}-b \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (\frac {3 a}{4}+b \right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+\left (\frac {3 a}{8}+\frac {b}{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}+\left (\frac {9 a}{8}+\frac {3 b}{2}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}+\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f}+\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{f}-\frac {\left (3 a -4 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{2 f}-\frac {\left (5 a +4 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}-\frac {\left (5 a +4 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{4 f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )}\) | \(241\) |
Input:
int(cos(f*x+e)^4*(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
1/32*((8*a+8*b)*sin(2*f*x+2*e)+a*sin(4*f*x+4*e)+12*x*(a+4/3*b)*f)/f
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.80 \[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {{\left (3 \, a + 4 \, b\right )} f x + {\left (2 \, a \cos \left (f x + e\right )^{3} + {\left (3 \, a + 4 \, b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, f} \] Input:
integrate(cos(f*x+e)^4*(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
1/8*((3*a + 4*b)*f*x + (2*a*cos(f*x + e)^3 + (3*a + 4*b)*cos(f*x + e))*sin (f*x + e))/f
\[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \cos ^{4}{\left (e + f x \right )}\, dx \] Input:
integrate(cos(f*x+e)**4*(a+b*sec(f*x+e)**2),x)
Output:
Integral((a + b*sec(e + f*x)**2)*cos(e + f*x)**4, x)
Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.20 \[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {{\left (f x + e\right )} {\left (3 \, a + 4 \, b\right )} + \frac {{\left (3 \, a + 4 \, b\right )} \tan \left (f x + e\right )^{3} + {\left (5 \, a + 4 \, b\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}}{8 \, f} \] Input:
integrate(cos(f*x+e)^4*(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
1/8*((f*x + e)*(3*a + 4*b) + ((3*a + 4*b)*tan(f*x + e)^3 + (5*a + 4*b)*tan (f*x + e))/(tan(f*x + e)^4 + 2*tan(f*x + e)^2 + 1))/f
Time = 0.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.20 \[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {{\left (f x + e\right )} {\left (3 \, a + 4 \, b\right )} + \frac {3 \, a \tan \left (f x + e\right )^{3} + 4 \, b \tan \left (f x + e\right )^{3} + 5 \, a \tan \left (f x + e\right ) + 4 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2}}}{8 \, f} \] Input:
integrate(cos(f*x+e)^4*(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
1/8*((f*x + e)*(3*a + 4*b) + (3*a*tan(f*x + e)^3 + 4*b*tan(f*x + e)^3 + 5* a*tan(f*x + e) + 4*b*tan(f*x + e))/(tan(f*x + e)^2 + 1)^2)/f
Time = 15.97 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.10 \[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=x\,\left (\frac {3\,a}{8}+\frac {b}{2}\right )+\frac {\left (\frac {3\,a}{8}+\frac {b}{2}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (\frac {5\,a}{8}+\frac {b}{2}\right )\,\mathrm {tan}\left (e+f\,x\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^4+2\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )} \] Input:
int(cos(e + f*x)^4*(a + b/cos(e + f*x)^2),x)
Output:
x*((3*a)/8 + b/2) + (tan(e + f*x)^3*((3*a)/8 + b/2) + tan(e + f*x)*((5*a)/ 8 + b/2))/(f*(2*tan(e + f*x)^2 + tan(e + f*x)^4 + 1))
Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03 \[ \int \cos ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {-2 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a +5 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a +4 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b +3 a f x +4 b f x}{8 f} \] Input:
int(cos(f*x+e)^4*(a+b*sec(f*x+e)^2),x)
Output:
( - 2*cos(e + f*x)*sin(e + f*x)**3*a + 5*cos(e + f*x)*sin(e + f*x)*a + 4*c os(e + f*x)*sin(e + f*x)*b + 3*a*f*x + 4*b*f*x)/(8*f)