Integrand size = 19, antiderivative size = 65 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {3 a x}{8}-\frac {b \cos ^4(c+d x)}{4 d}+\frac {3 a \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d} \]
Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {3 a (c+d x)}{8 d}-\frac {b \cos ^4(c+d x)}{4 d}+\frac {a \sin (2 (c+d x))}{4 d}+\frac {a \sin (4 (c+d x))}{32 d} \]
(3*a*(c + d*x))/(8*d) - (b*Cos[c + d*x]^4)/(4*d) + (a*Sin[2*(c + d*x)])/(4 *d) + (a*Sin[4*(c + d*x)])/(32*d)
Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 3967, 3042, 3115, 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.
\(\displaystyle \int \cos ^4(c+d x) (a+b \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \tan (c+d x)}{\sec (c+d x)^4}dx\) |
\(\Big \downarrow \) 3967 |
\(\displaystyle a \int \cos ^4(c+d x)dx-\frac {b \cos ^4(c+d x)}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \sin \left (c+d x+\frac {\pi }{2}\right )^4dx-\frac {b \cos ^4(c+d x)}{4 d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle a \left (\frac {3}{4} \int \cos ^2(c+d x)dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {b \cos ^4(c+d x)}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {3}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {b \cos ^4(c+d x)}{4 d}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle a \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {b \cos ^4(c+d x)}{4 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle a \left (\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )-\frac {b \cos ^4(c+d x)}{4 d}\) |
-1/4*(b*Cos[c + d*x]^4)/d + a*((Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (3*(x /2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4)
3.6.16.3.1 Defintions of rubi rules used
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[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a Int[(d *Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] || NeQ[a^2 + b^2, 0])
Time = 4.86 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.80
method | result | size |
derivativedivides | \(\frac {-\frac {b \left (\cos ^{4}\left (d x +c \right )\right )}{4}+a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(52\) |
default | \(\frac {-\frac {b \left (\cos ^{4}\left (d x +c \right )\right )}{4}+a \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(52\) |
risch | \(\frac {3 a x}{8}-\frac {b \cos \left (4 d x +4 c \right )}{32 d}+\frac {a \sin \left (4 d x +4 c \right )}{32 d}-\frac {b \cos \left (2 d x +2 c \right )}{8 d}+\frac {a \sin \left (2 d x +2 c \right )}{4 d}\) | \(66\) |
Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {2 \, b \cos \left (d x + c\right )^{4} - 3 \, a d x - {\left (2 \, a \cos \left (d x + c\right )^{3} + 3 \, a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]
-1/8*(2*b*cos(d*x + c)^4 - 3*a*d*x - (2*a*cos(d*x + c)^3 + 3*a*cos(d*x + c ))*sin(d*x + c))/d
\[ \int \cos ^4(c+d x) (a+b \tan (c+d x)) \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )}\, dx \]
Time = 0.52 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {3 \, {\left (d x + c\right )} a + \frac {3 \, a \tan \left (d x + c\right )^{3} + 5 \, a \tan \left (d x + c\right ) - 2 \, b}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1}}{8 \, d} \]
1/8*(3*(d*x + c)*a + (3*a*tan(d*x + c)^3 + 5*a*tan(d*x + c) - 2*b)/(tan(d* x + c)^4 + 2*tan(d*x + c)^2 + 1))/d
Leaf count of result is larger than twice the leaf count of optimal. 426 vs. \(2 (57) = 114\).
Time = 0.52 (sec) , antiderivative size = 426, normalized size of antiderivative = 6.55 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {12 \, a d x \tan \left (d x\right )^{4} \tan \left (c\right )^{4} + 24 \, a d x \tan \left (d x\right )^{4} \tan \left (c\right )^{2} + 24 \, a d x \tan \left (d x\right )^{2} \tan \left (c\right )^{4} - 5 \, b \tan \left (d x\right )^{4} \tan \left (c\right )^{4} - 20 \, a \tan \left (d x\right )^{4} \tan \left (c\right )^{3} - 20 \, a \tan \left (d x\right )^{3} \tan \left (c\right )^{4} + 12 \, a d x \tan \left (d x\right )^{4} + 48 \, a d x \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + 6 \, b \tan \left (d x\right )^{4} \tan \left (c\right )^{2} + 32 \, b \tan \left (d x\right )^{3} \tan \left (c\right )^{3} + 12 \, a d x \tan \left (c\right )^{4} + 6 \, b \tan \left (d x\right )^{2} \tan \left (c\right )^{4} - 12 \, a \tan \left (d x\right )^{4} \tan \left (c\right ) + 24 \, a \tan \left (d x\right )^{3} \tan \left (c\right )^{2} + 24 \, a \tan \left (d x\right )^{2} \tan \left (c\right )^{3} - 12 \, a \tan \left (d x\right ) \tan \left (c\right )^{4} + 24 \, a d x \tan \left (d x\right )^{2} + 3 \, b \tan \left (d x\right )^{4} + 24 \, a d x \tan \left (c\right )^{2} - 36 \, b \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + 3 \, b \tan \left (c\right )^{4} + 12 \, a \tan \left (d x\right )^{3} - 24 \, a \tan \left (d x\right )^{2} \tan \left (c\right ) - 24 \, a \tan \left (d x\right ) \tan \left (c\right )^{2} + 12 \, a \tan \left (c\right )^{3} + 12 \, a d x + 6 \, b \tan \left (d x\right )^{2} + 32 \, b \tan \left (d x\right ) \tan \left (c\right ) + 6 \, b \tan \left (c\right )^{2} + 20 \, a \tan \left (d x\right ) + 20 \, a \tan \left (c\right ) - 5 \, b}{32 \, {\left (d \tan \left (d x\right )^{4} \tan \left (c\right )^{4} + 2 \, d \tan \left (d x\right )^{4} \tan \left (c\right )^{2} + 2 \, d \tan \left (d x\right )^{2} \tan \left (c\right )^{4} + d \tan \left (d x\right )^{4} + 4 \, d \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + d \tan \left (c\right )^{4} + 2 \, d \tan \left (d x\right )^{2} + 2 \, d \tan \left (c\right )^{2} + d\right )}} \]
1/32*(12*a*d*x*tan(d*x)^4*tan(c)^4 + 24*a*d*x*tan(d*x)^4*tan(c)^2 + 24*a*d *x*tan(d*x)^2*tan(c)^4 - 5*b*tan(d*x)^4*tan(c)^4 - 20*a*tan(d*x)^4*tan(c)^ 3 - 20*a*tan(d*x)^3*tan(c)^4 + 12*a*d*x*tan(d*x)^4 + 48*a*d*x*tan(d*x)^2*t an(c)^2 + 6*b*tan(d*x)^4*tan(c)^2 + 32*b*tan(d*x)^3*tan(c)^3 + 12*a*d*x*ta n(c)^4 + 6*b*tan(d*x)^2*tan(c)^4 - 12*a*tan(d*x)^4*tan(c) + 24*a*tan(d*x)^ 3*tan(c)^2 + 24*a*tan(d*x)^2*tan(c)^3 - 12*a*tan(d*x)*tan(c)^4 + 24*a*d*x* tan(d*x)^2 + 3*b*tan(d*x)^4 + 24*a*d*x*tan(c)^2 - 36*b*tan(d*x)^2*tan(c)^2 + 3*b*tan(c)^4 + 12*a*tan(d*x)^3 - 24*a*tan(d*x)^2*tan(c) - 24*a*tan(d*x) *tan(c)^2 + 12*a*tan(c)^3 + 12*a*d*x + 6*b*tan(d*x)^2 + 32*b*tan(d*x)*tan( c) + 6*b*tan(c)^2 + 20*a*tan(d*x) + 20*a*tan(c) - 5*b)/(d*tan(d*x)^4*tan(c )^4 + 2*d*tan(d*x)^4*tan(c)^2 + 2*d*tan(d*x)^2*tan(c)^4 + d*tan(d*x)^4 + 4 *d*tan(d*x)^2*tan(c)^2 + d*tan(c)^4 + 2*d*tan(d*x)^2 + 2*d*tan(c)^2 + d)
Time = 4.06 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.63 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {3\,a\,x}{8}+\frac {{\cos \left (c+d\,x\right )}^4\,\left (\frac {3\,a\,{\mathrm {tan}\left (c+d\,x\right )}^3}{8}+\frac {5\,a\,\mathrm {tan}\left (c+d\,x\right )}{8}-\frac {b}{4}\right )}{d} \]