3.3.47 \(\int \cos (a+b x) \tan ^3(c+b x) \, dx\) [247]

3.3.47.1 Optimal result

Integrand size = 15, antiderivative size = 72 \[ \int \cos (a+b x) \tan ^3(c+b x) \, dx=\frac {\cos (a+b x)}{b}+\frac {\cos (a-c) \sec (c+b x)}{b}+\frac {3 \text {arctanh}(\sin (c+b x)) \sin (a-c)}{2 b}-\frac {\sec (c+b x) \sin (a-c) \tan (c+b x)}{2 b} \]

output

cos(b*x+a)/b+cos(a-c)*sec(b*x+c)/b+3/2*arctanh(sin(b*x+c))*sin(a-c)/b-1/2* 
sec(b*x+c)*sin(a-c)*tan(b*x+c)/b
 

3.3.47.2 Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97 \[ \int \cos (a+b x) \tan ^3(c+b x) \, dx=\frac {(2 \cos (a-2 c-b x)+5 \cos (a+b x)+\cos (a+2 c+3 b x)) \sec ^2(c+b x)+12 \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {b x}{2}\right )\right ) \sin (a-c)}{4 b} \]

input

Integrate[Cos[a + b*x]*Tan[c + b*x]^3,x]
 

output

((2*Cos[a - 2*c - b*x] + 5*Cos[a + b*x] + Cos[a + 2*c + 3*b*x])*Sec[c + b* 
x]^2 + 12*ArcTanh[Sin[c] + Cos[c]*Tan[(b*x)/2]]*Sin[a - c])/(4*b)
 

3.3.47.3 Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.19, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.867, Rules used = {5090, 3042, 3091, 3042, 4257, 5087, 3042, 3086, 24, 5090, 3042, 3118, 4257}

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.

\(\displaystyle \int \cos (a+b x) \tan ^3(b x+c) \, dx\)

\(\Big \downarrow \) 5090

\(\displaystyle \int \sin (a+b x) \tan ^2(c+b x)dx-\sin (a-c) \int \sec (c+b x) \tan ^2(c+b x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (a+b x) \tan ^2(c+b x)dx-\sin (a-c) \int \sec (c+b x) \tan (c+b x)^2dx\)

\(\Big \downarrow \) 3091

\(\displaystyle \int \sin (a+b x) \tan ^2(c+b x)dx-\sin (a-c) \left (\frac {\tan (b x+c) \sec (b x+c)}{2 b}-\frac {1}{2} \int \sec (c+b x)dx\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (a+b x) \tan ^2(c+b x)dx-\sin (a-c) \left (\frac {\tan (b x+c) \sec (b x+c)}{2 b}-\frac {1}{2} \int \csc \left (c+b x+\frac {\pi }{2}\right )dx\right )\)

\(\Big \downarrow \) 4257

\(\displaystyle \int \sin (a+b x) \tan ^2(c+b x)dx-\sin (a-c) \left (\frac {\tan (b x+c) \sec (b x+c)}{2 b}-\frac {\text {arctanh}(\sin (b x+c))}{2 b}\right )\)

\(\Big \downarrow \) 5087

\(\displaystyle -\int \cos (a+b x) \tan (c+b x)dx+\cos (a-c) \int \sec (c+b x) \tan (c+b x)dx-\sin (a-c) \left (\frac {\tan (b x+c) \sec (b x+c)}{2 b}-\frac {\text {arctanh}(\sin (b x+c))}{2 b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -\int \cos (a+b x) \tan (c+b x)dx+\cos (a-c) \int \sec (c+b x) \tan (c+b x)dx-\sin (a-c) \left (\frac {\tan (b x+c) \sec (b x+c)}{2 b}-\frac {\text {arctanh}(\sin (b x+c))}{2 b}\right )\)

\(\Big \downarrow \) 3086

\(\displaystyle -\int \cos (a+b x) \tan (c+b x)dx+\frac {\cos (a-c) \int 1d\sec (c+b x)}{b}-\sin (a-c) \left (\frac {\tan (b x+c) \sec (b x+c)}{2 b}-\frac {\text {arctanh}(\sin (b x+c))}{2 b}\right )\)

\(\Big \downarrow \) 24

\(\displaystyle -\int \cos (a+b x) \tan (c+b x)dx-\sin (a-c) \left (\frac {\tan (b x+c) \sec (b x+c)}{2 b}-\frac {\text {arctanh}(\sin (b x+c))}{2 b}\right )+\frac {\cos (a-c) \sec (b x+c)}{b}\)

\(\Big \downarrow \) 5090

\(\displaystyle \sin (a-c) \int \sec (c+b x)dx-\int \sin (a+b x)dx-\sin (a-c) \left (\frac {\tan (b x+c) \sec (b x+c)}{2 b}-\frac {\text {arctanh}(\sin (b x+c))}{2 b}\right )+\frac {\cos (a-c) \sec (b x+c)}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \sin (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )dx-\int \sin (a+b x)dx-\sin (a-c) \left (\frac {\tan (b x+c) \sec (b x+c)}{2 b}-\frac {\text {arctanh}(\sin (b x+c))}{2 b}\right )+\frac {\cos (a-c) \sec (b x+c)}{b}\)

\(\Big \downarrow \) 3118

\(\displaystyle \sin (a-c) \int \csc \left (c+b x+\frac {\pi }{2}\right )dx-\sin (a-c) \left (\frac {\tan (b x+c) \sec (b x+c)}{2 b}-\frac {\text {arctanh}(\sin (b x+c))}{2 b}\right )+\frac {\cos (a-c) \sec (b x+c)}{b}+\frac {\cos (a+b x)}{b}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\sin (a-c) \text {arctanh}(\sin (b x+c))}{b}-\sin (a-c) \left (\frac {\tan (b x+c) \sec (b x+c)}{2 b}-\frac {\text {arctanh}(\sin (b x+c))}{2 b}\right )+\frac {\cos (a-c) \sec (b x+c)}{b}+\frac {\cos (a+b x)}{b}\)

input

Int[Cos[a + b*x]*Tan[c + b*x]^3,x]
 

output

Cos[a + b*x]/b + (Cos[a - c]*Sec[c + b*x])/b + (ArcTanh[Sin[c + b*x]]*Sin[ 
a - c])/b - Sin[a - c]*(-1/2*ArcTanh[Sin[c + b*x]]/b + (Sec[c + b*x]*Tan[c 
 + b*x])/(2*b))
 

3.3.47.3.1 Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3086

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_.), x_Symbol] :> Simp[a/f   Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 
), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 
] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])
 

rule 3091

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m 
 + n - 1))), x] - Simp[b^2*((n - 1)/(m + n - 1))   Int[(a*Sec[e + f*x])^m*( 
b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] & 
& NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
 

rule 3118

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ 
[{c, d}, x]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 5087

Int[Sin[v_]*Tan[w_]^(n_.), x_Symbol] :> -Int[Cos[v]*Tan[w]^(n - 1), x] + Si 
mp[Cos[v - w]   Int[Sec[w]*Tan[w]^(n - 1), x], x] /; GtQ[n, 0] && FreeQ[v - 
 w, x] && NeQ[w, v]
 

rule 5090

Int[Cos[v_]*Tan[w_]^(n_.), x_Symbol] :> Int[Sin[v]*Tan[w]^(n - 1), x] - Sim 
p[Sin[v - w]   Int[Sec[w]*Tan[w]^(n - 1), x], x] /; GtQ[n, 0] && FreeQ[v - 
w, x] && NeQ[w, v]
 

3.3.47.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.51

method	result	size
risch	\(\frac {{\mathrm e}^{i \left (x b +a \right )}}{2 b}+\frac {{\mathrm e}^{-i \left (x b +a \right )}}{2 b}+\frac {3 \,{\mathrm e}^{i \left (3 x b +5 a +2 c \right )}+{\mathrm e}^{i \left (3 x b +3 a +4 c \right )}+{\mathrm e}^{i \left (x b +5 a \right )}+3 \,{\mathrm e}^{i \left (x b +3 a +2 c \right )}}{2 b \left ({\mathrm e}^{2 i \left (x b +a +c \right )}+{\mathrm e}^{2 i a}\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (x b +a \right )}-i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{2 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (x b +a \right )}+i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{2 b}\)	\(181\)

input

int(cos(b*x+a)*tan(b*x+c)^3,x,method=_RETURNVERBOSE)
 

output

1/2*exp(I*(b*x+a))/b+1/2/b*exp(-I*(b*x+a))+1/2/b/(exp(2*I*(b*x+a+c))+exp(2 
*I*a))^2*(3*exp(I*(3*b*x+5*a+2*c))+exp(I*(3*b*x+3*a+4*c))+exp(I*(b*x+5*a)) 
+3*exp(I*(b*x+3*a+2*c)))-3/2*ln(exp(I*(b*x+a))-I*exp(I*(a-c)))/b*sin(a-c)+ 
3/2*ln(exp(I*(b*x+a))+I*exp(I*(a-c)))/b*sin(a-c)
 

3.3.47.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (68) = 136\).

Time = 0.28 (sec) , antiderivative size = 366, normalized size of antiderivative = 5.08 \[ \int \cos (a+b x) \tan ^3(c+b x) \, dx=\frac {16 \, \cos \left (b x + a\right )^{3} \cos \left (-2 \, a + 2 \, c\right ) - 4 \, {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - 4 \, {\left (\cos \left (-2 \, a + 2 \, c\right ) - 5\right )} \cos \left (b x + a\right ) + \frac {3 \, \sqrt {2} {\left (2 \, {\left (\cos \left (-2 \, a + 2 \, c\right )^{2} - 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + {\left (2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (-2 \, a + 2 \, c\right )\right )} \log \left (-\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + \frac {2 \, \sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (b x + a\right ) + \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) - 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1}\right )}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}}}{8 \, {\left (2 \, b \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, b \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - b \cos \left (-2 \, a + 2 \, c\right ) + b\right )}} \]

input

integrate(cos(b*x+a)*tan(b*x+c)^3,x, algorithm="fricas")
 

output

1/8*(16*cos(b*x + a)^3*cos(-2*a + 2*c) - 4*(4*cos(b*x + a)^2 + 1)*sin(b*x 
+ a)*sin(-2*a + 2*c) - 4*(cos(-2*a + 2*c) - 5)*cos(b*x + a) + 3*sqrt(2)*(2 
*(cos(-2*a + 2*c)^2 - 1)*cos(b*x + a)*sin(b*x + a) + (2*cos(b*x + a)^2*cos 
(-2*a + 2*c) - cos(-2*a + 2*c) + 1)*sin(-2*a + 2*c))*log(-(2*cos(b*x + a)^ 
2*cos(-2*a + 2*c) - 2*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) + 2*sqrt(2 
)*((cos(-2*a + 2*c) + 1)*sin(b*x + a) + cos(b*x + a)*sin(-2*a + 2*c))/sqrt 
(cos(-2*a + 2*c) + 1) - cos(-2*a + 2*c) - 3)/(2*cos(b*x + a)^2*cos(-2*a + 
2*c) - 2*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) - cos(-2*a + 2*c) + 1)) 
/sqrt(cos(-2*a + 2*c) + 1))/(2*b*cos(b*x + a)^2*cos(-2*a + 2*c) - 2*b*cos( 
b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) - b*cos(-2*a + 2*c) + b)
 

3.3.47.6 Sympy [F]

\[ \int \cos (a+b x) \tan ^3(c+b x) \, dx=\int \cos {\left (a + b x \right )} \tan ^{3}{\left (b x + c \right )}\, dx \]

input

integrate(cos(b*x+a)*tan(b*x+c)**3,x)
 

output

Integral(cos(a + b*x)*tan(b*x + c)**3, x)
 

3.3.47.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1027 vs. \(2 (68) = 136\).

Time = 0.39 (sec) , antiderivative size = 1027, normalized size of antiderivative = 14.26 \[ \int \cos (a+b x) \tan ^3(c+b x) \, dx=\text {Too large to display} \]

input

integrate(cos(b*x+a)*tan(b*x+c)^3,x, algorithm="maxima")
 

output

1/4*(2*(cos(5*b*x + a + 4*c) + 2*cos(3*b*x + a + 2*c) + cos(b*x + a))*cos( 
6*b*x + 2*a + 4*c) + 2*(5*cos(4*b*x + 2*a + 2*c) + 2*cos(4*b*x + 4*c) + 2* 
cos(2*b*x + 2*a) + 5*cos(2*b*x + 2*c) + 1)*cos(5*b*x + a + 4*c) + 10*(2*co 
s(3*b*x + a + 2*c) + cos(b*x + a))*cos(4*b*x + 2*a + 2*c) + 4*(2*cos(3*b*x 
 + a + 2*c) + cos(b*x + a))*cos(4*b*x + 4*c) + 4*(2*cos(2*b*x + 2*a) + 5*c 
os(2*b*x + 2*c) + 1)*cos(3*b*x + a + 2*c) + 4*cos(2*b*x + 2*a)*cos(b*x + a 
) + 10*cos(2*b*x + 2*c)*cos(b*x + a) + 3*(cos(5*b*x + a + 4*c)^2*sin(-a + 
c) + 4*cos(3*b*x + a + 2*c)^2*sin(-a + c) + 4*cos(3*b*x + a + 2*c)*cos(b*x 
 + a)*sin(-a + c) + cos(b*x + a)^2*sin(-a + c) + sin(5*b*x + a + 4*c)^2*si 
n(-a + c) + 4*sin(3*b*x + a + 2*c)^2*sin(-a + c) + 4*sin(3*b*x + a + 2*c)* 
sin(b*x + a)*sin(-a + c) + sin(b*x + a)^2*sin(-a + c) + 2*(2*cos(3*b*x + a 
 + 2*c)*sin(-a + c) + cos(b*x + a)*sin(-a + c))*cos(5*b*x + a + 4*c) + 2*( 
2*sin(3*b*x + a + 2*c)*sin(-a + c) + sin(b*x + a)*sin(-a + c))*sin(5*b*x + 
 a + 4*c))*log((cos(b*x + 2*c)^2 + cos(c)^2 - 2*cos(c)*sin(b*x + 2*c) + si 
n(b*x + 2*c)^2 + 2*cos(b*x + 2*c)*sin(c) + sin(c)^2)/(cos(b*x + 2*c)^2 + c 
os(c)^2 + 2*cos(c)*sin(b*x + 2*c) + sin(b*x + 2*c)^2 - 2*cos(b*x + 2*c)*si 
n(c) + sin(c)^2)) + 2*(sin(5*b*x + a + 4*c) + 2*sin(3*b*x + a + 2*c) + sin 
(b*x + a))*sin(6*b*x + 2*a + 4*c) + 2*(5*sin(4*b*x + 2*a + 2*c) + 2*sin(4* 
b*x + 4*c) + 2*sin(2*b*x + 2*a) + 5*sin(2*b*x + 2*c))*sin(5*b*x + a + 4*c) 
 + 10*(2*sin(3*b*x + a + 2*c) + sin(b*x + a))*sin(4*b*x + 2*a + 2*c) + ...
 

3.3.47.8 Giac [F]

\[ \int \cos (a+b x) \tan ^3(c+b x) \, dx=\int { \cos \left (b x + a\right ) \tan \left (b x + c\right )^{3} \,d x } \]

input

integrate(cos(b*x+a)*tan(b*x+c)^3,x, algorithm="giac")
 

output

integrate(cos(b*x + a)*tan(b*x + c)^3, x)
 

3.3.47.9 Mupad [F(-1)]

Timed out. \[ \int \cos (a+b x) \tan ^3(c+b x) \, dx=\text {Hanged} \]

input

int(cos(a + b*x)*tan(c + b*x)^3,x)
 

output

\text{Hanged}