Integrand size = 21, antiderivative size = 61 \[ \int \cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} (3 A+4 C) x+\frac {(3 A+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) \sin (c+d x)}{4 d} \] Output:
1/8*(3*A+4*C)*x+1/8*(3*A+4*C)*cos(d*x+c)*sin(d*x+c)/d+1/4*A*cos(d*x+c)^3*s in(d*x+c)/d
Time = 0.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.74 \[ \int \cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 (3 A+4 C) (c+d x)+8 (A+C) \sin (2 (c+d x))+A \sin (4 (c+d x))}{32 d} \] Input:
Integrate[Cos[c + d*x]^4*(A + C*Sec[c + d*x]^2),x]
Output:
(4*(3*A + 4*C)*(c + d*x) + 8*(A + C)*Sin[2*(c + d*x)] + A*Sin[4*(c + d*x)] )/(32*d)
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.
\(\displaystyle \int \cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 4533 |
\(\displaystyle \frac {1}{4} (3 A+4 C) \int \cos ^2(c+d x)dx+\frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{4} (3 A+4 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {1}{4} (3 A+4 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{4} (3 A+4 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d}\) |
Input:
Int[Cos[c + d*x]^4*(A + C*Sec[c + d*x]^2),x]
Output:
(A*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + ((3*A + 4*C)*(x/2 + (Cos[c + d*x]* Sin[c + d*x])/(2*d)))/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 = 0.33 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.72
method | result | size |
parallelrisch | \(\frac {\left (8 A +8 C \right ) \sin \left (2 d x +2 c \right )+A \sin \left (4 d x +4 c \right )+12 x d \left (A +\frac {4 C}{3}\right )}{32 d}\) | \(44\) |
risch | \(\frac {3 A x}{8}+\frac {C x}{2}+\frac {A \sin \left (4 d x +4 c \right )}{32 d}+\frac {A \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C}{4 d}\) | \(55\) |
derivativedivides | \(\frac {A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(65\) |
default | \(\frac {A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(65\) |
norman | \(\frac {\left (-\frac {3 A}{8}-\frac {C}{2}\right ) x +\left (-\frac {9 A}{8}-\frac {3 C}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {3 A}{4}-C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {3 A}{4}+C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (\frac {3 A}{8}+\frac {C}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {9 A}{8}+\frac {3 C}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}+\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {\left (3 A -4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}-\frac {\left (5 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (5 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\) | \(241\) |
Input:
int(cos(d*x+c)^4*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/32*((8*A+8*C)*sin(2*d*x+2*c)+A*sin(4*d*x+4*c)+12*x*d*(A+4/3*C))/d
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.80 \[ \int \cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (3 \, A + 4 \, C\right )} d x + {\left (2 \, A \cos \left (d x + c\right )^{3} + {\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \] Input:
integrate(cos(d*x+c)^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
Output:
1/8*((3*A + 4*C)*d*x + (2*A*cos(d*x + c)^3 + (3*A + 4*C)*cos(d*x + c))*sin (d*x + c))/d
\[ \int \cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)**4*(A+C*sec(d*x+c)**2),x)
Output:
Integral((A + C*sec(c + d*x)**2)*cos(c + d*x)**4, x)
Time = 0.11 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.20 \[ \int \cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (d x + c\right )} {\left (3 \, A + 4 \, C\right )} + \frac {{\left (3 \, A + 4 \, C\right )} \tan \left (d x + c\right )^{3} + {\left (5 \, A + 4 \, C\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{8 \, d} \] Input:
integrate(cos(d*x+c)^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")
Output:
1/8*((d*x + c)*(3*A + 4*C) + ((3*A + 4*C)*tan(d*x + c)^3 + (5*A + 4*C)*tan (d*x + c))/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1))/d
Time = 0.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.20 \[ \int \cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (d x + c\right )} {\left (3 \, A + 4 \, C\right )} + \frac {3 \, A \tan \left (d x + c\right )^{3} + 4 \, C \tan \left (d x + c\right )^{3} + 5 \, A \tan \left (d x + c\right ) + 4 \, C \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{2}}}{8 \, d} \] Input:
integrate(cos(d*x+c)^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")
Output:
1/8*((d*x + c)*(3*A + 4*C) + (3*A*tan(d*x + c)^3 + 4*C*tan(d*x + c)^3 + 5* A*tan(d*x + c) + 4*C*tan(d*x + c))/(tan(d*x + c)^2 + 1)^2)/d
Time = 11.98 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.10 \[ \int \cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=x\,\left (\frac {3\,A}{8}+\frac {C}{2}\right )+\frac {\left (\frac {3\,A}{8}+\frac {C}{2}\right )\,{\mathrm {tan}\left (c+d\,x\right )}^3+\left (\frac {5\,A}{8}+\frac {C}{2}\right )\,\mathrm {tan}\left (c+d\,x\right )}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4+2\,{\mathrm {tan}\left (c+d\,x\right )}^2+1\right )} \] Input:
int(cos(c + d*x)^4*(A + C/cos(c + d*x)^2),x)
Output:
x*((3*A)/8 + C/2) + (tan(c + d*x)*((5*A)/8 + C/2) + tan(c + d*x)^3*((3*A)/ 8 + C/2))/(d*(2*tan(c + d*x)^2 + tan(c + d*x)^4 + 1))
Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03 \[ \int \cos ^4(c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +5 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +4 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c +3 a d x +4 c d x}{8 d} \] Input:
int(cos(d*x+c)^4*(A+C*sec(d*x+c)^2),x)
Output:
( - 2*cos(c + d*x)*sin(c + d*x)**3*a + 5*cos(c + d*x)*sin(c + d*x)*a + 4*c os(c + d*x)*sin(c + d*x)*c + 3*a*d*x + 4*c*d*x)/(8*d)