\(\int \cos ^5(c+d x) (a+b \tan (c+d x))^3 \, dx\) [549]

Optimal result

Integrand size = 21, antiderivative size = 111 \[ \int \cos ^5(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {b^3 \cos ^3(c+d x)}{3 d}-\frac {b \left (3 a^2-b^2\right ) \cos ^5(c+d x)}{5 d}+\frac {a^3 \sin (c+d x)}{d}-\frac {a \left (2 a^2-3 b^2\right ) \sin ^3(c+d x)}{3 d}+\frac {a \left (a^2-3 b^2\right ) \sin ^5(c+d x)}{5 d} \] Output:

-1/3*b^3*cos(d*x+c)^3/d-1/5*b*(3*a^2-b^2)*cos(d*x+c)^5/d+a^3*sin(d*x+c)/d- 
1/3*a*(2*a^2-3*b^2)*sin(d*x+c)^3/d+1/5*a*(a^2-3*b^2)*sin(d*x+c)^5/d
 

Mathematica [A] (verified)

Time = 0.82 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.09 \[ \int \cos ^5(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {-9 a^2 b \cos ^5(c+d x)+15 a^3 \sin (c+d x)-5 a \left (2 a^2-3 b^2\right ) \sin ^3(c+d x)+3 a \left (a^2-3 b^2\right ) \sin ^5(c+d x)+b^3 \cos (c+d x) \left (-2+\frac {2}{\sqrt {\cos ^2(c+d x)}}-\sin ^2(c+d x)+3 \sin ^4(c+d x)\right )}{15 d} \] Input:

Integrate[Cos[c + d*x]^5*(a + b*Tan[c + d*x])^3,x]
 

Output:

(-9*a^2*b*Cos[c + d*x]^5 + 15*a^3*Sin[c + d*x] - 5*a*(2*a^2 - 3*b^2)*Sin[c 
 + d*x]^3 + 3*a*(a^2 - 3*b^2)*Sin[c + d*x]^5 + b^3*Cos[c + d*x]*(-2 + 2/Sq 
rt[Cos[c + d*x]^2] - Sin[c + d*x]^2 + 3*Sin[c + d*x]^4))/(15*d)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3991, 3042, 4159, 27, 290, 2009, 4857, 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.

\(\displaystyle \int \cos ^5(c+d x) (a+b \tan (c+d x))^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (c+d x))^3}{\sec (c+d x)^5}dx\)

\(\Big \downarrow \) 3991

\(\displaystyle \int \cos ^5(c+d x) \left (a^3+3 b^2 \tan ^2(c+d x) a\right )dx+\int \cos ^4(c+d x) \sin (c+d x) \left (\tan ^2(c+d x) b^3+3 a^2 b\right )dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {a^3+3 b^2 \tan (c+d x)^2 a}{\sec (c+d x)^5}dx+\int \frac {\sin (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )}{\sec (c+d x)^4}dx\)

\(\Big \downarrow \) 4159

\(\displaystyle \int \frac {\sin (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )}{\sec (c+d x)^4}dx+\frac {\int a \left (1-\sin ^2(c+d x)\right ) \left (a^2-\left (a^2-3 b^2\right ) \sin ^2(c+d x)\right )d\sin (c+d x)}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {\sin (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )}{\sec (c+d x)^4}dx+\frac {a \int \left (1-\sin ^2(c+d x)\right ) \left (a^2-\left (a^2-3 b^2\right ) \sin ^2(c+d x)\right )d\sin (c+d x)}{d}\)

\(\Big \downarrow \) 290

\(\displaystyle \int \frac {\sin (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )}{\sec (c+d x)^4}dx+\frac {a \int \left (\left (a^2-3 b^2\right ) \sin ^4(c+d x)-\left (2 a^2-3 b^2\right ) \sin ^2(c+d x)+a^2\right )d\sin (c+d x)}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \int \frac {\sin (c+d x) \left (\tan (c+d x)^2 b^3+3 a^2 b\right )}{\sec (c+d x)^4}dx+\frac {a \left (\frac {1}{5} \left (a^2-3 b^2\right ) \sin ^5(c+d x)-\frac {1}{3} \left (2 a^2-3 b^2\right ) \sin ^3(c+d x)+a^2 \sin (c+d x)\right )}{d}\)

\(\Big \downarrow \) 4857

\(\displaystyle \frac {a \left (\frac {1}{5} \left (a^2-3 b^2\right ) \sin ^5(c+d x)-\frac {1}{3} \left (2 a^2-3 b^2\right ) \sin ^3(c+d x)+a^2 \sin (c+d x)\right )}{d}-\frac {\int \left (3 a^2 b \cos ^4(c+d x)+b^3 \left (1-\cos ^2(c+d x)\right ) \cos ^2(c+d x)\right )d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a \left (\frac {1}{5} \left (a^2-3 b^2\right ) \sin ^5(c+d x)-\frac {1}{3} \left (2 a^2-3 b^2\right ) \sin ^3(c+d x)+a^2 \sin (c+d x)\right )}{d}-\frac {\frac {3}{5} a^2 b \cos ^5(c+d x)-\frac {1}{5} b^3 \cos ^5(c+d x)+\frac {1}{3} b^3 \cos ^3(c+d x)}{d}\)

Input:

Int[Cos[c + d*x]^5*(a + b*Tan[c + d*x])^3,x]
 

Output:

-(((b^3*Cos[c + d*x]^3)/3 + (3*a^2*b*Cos[c + d*x]^5)/5 - (b^3*Cos[c + d*x] 
^5)/5)/d) + (a*(a^2*Sin[c + d*x] - ((2*a^2 - 3*b^2)*Sin[c + d*x]^3)/3 + (( 
a^2 - 3*b^2)*Sin[c + d*x]^5)/5))/d
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 290

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I 
nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d 
}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3991

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n 
_), x_Symbol] :> Module[{k}, Int[Sec[e + f*x]^m*Sum[Binomial[n, 2*k]*a^(n - 
 2*k)*b^(2*k)*Tan[e + f*x]^(2*k), {k, 0, n/2}], x] + Int[Sec[e + f*x]^m*Tan 
[e + f*x]*Sum[Binomial[n, 2*k + 1]*a^(n - 2*k - 1)*b^(2*k + 1)*Tan[e + f*x] 
^(2*k), {k, 0, (n - 1)/2}], x]] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 + b^2, 
 0] && IntegerQ[(m - 1)/2] && IGtQ[n, 0]
 

rule 4159

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_ 
))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f 
  Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2 
*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f} 
, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
 

rule 4857

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFacto 
rs[Cos[c*(a + b*x)], x]}, Simp[-d/(b*c)   Subst[Int[SubstFor[1, Cos[c*(a + 
b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a + b* 
x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Sin] || EqQ[F, sin])
 

Maple [A] (verified)

Time = 35.88 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.13

method	result	size
derivativedivides	\(\frac {b^{3} \left (-\frac {\cos \left (d x +c \right )^{3} \sin \left (d x +c \right )^{2}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )+3 a \,b^{2} \left (-\frac {\cos \left (d x +c \right )^{4} \sin \left (d x +c \right )}{5}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{15}\right )-\frac {3 a^{2} b \cos \left (d x +c \right )^{5}}{5}+\frac {a^{3} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(125\)
default	\(\frac {b^{3} \left (-\frac {\cos \left (d x +c \right )^{3} \sin \left (d x +c \right )^{2}}{5}-\frac {2 \cos \left (d x +c \right )^{3}}{15}\right )+3 a \,b^{2} \left (-\frac {\cos \left (d x +c \right )^{4} \sin \left (d x +c \right )}{5}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{15}\right )-\frac {3 a^{2} b \cos \left (d x +c \right )^{5}}{5}+\frac {a^{3} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(125\)
risch	\(-\frac {3 b \cos \left (d x +c \right ) a^{2}}{8 d}-\frac {b^{3} \cos \left (d x +c \right )}{8 d}+\frac {5 a^{3} \sin \left (d x +c \right )}{8 d}+\frac {3 a \sin \left (d x +c \right ) b^{2}}{8 d}-\frac {3 b \cos \left (5 d x +5 c \right ) a^{2}}{80 d}+\frac {b^{3} \cos \left (5 d x +5 c \right )}{80 d}+\frac {a^{3} \sin \left (5 d x +5 c \right )}{80 d}-\frac {3 a \sin \left (5 d x +5 c \right ) b^{2}}{80 d}-\frac {3 b \cos \left (3 d x +3 c \right ) a^{2}}{16 d}-\frac {b^{3} \cos \left (3 d x +3 c \right )}{48 d}+\frac {5 a^{3} \sin \left (3 d x +3 c \right )}{48 d}-\frac {a \sin \left (3 d x +3 c \right ) b^{2}}{16 d}\)	\(200\)

Input:

int(cos(d*x+c)^5*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(b^3*(-1/5*cos(d*x+c)^3*sin(d*x+c)^2-2/15*cos(d*x+c)^3)+3*a*b^2*(-1/5* 
cos(d*x+c)^4*sin(d*x+c)+1/15*(2+cos(d*x+c)^2)*sin(d*x+c))-3/5*a^2*b*cos(d* 
x+c)^5+1/5*a^3*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.92 \[ \int \cos ^5(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {5 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{5} - {\left (3 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{3} + 6 \, a b^{2} + {\left (4 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \] Input:

integrate(cos(d*x+c)^5*(a+b*tan(d*x+c))^3,x, algorithm="fricas")
 

Output:

-1/15*(5*b^3*cos(d*x + c)^3 + 3*(3*a^2*b - b^3)*cos(d*x + c)^5 - (3*(a^3 - 
 3*a*b^2)*cos(d*x + c)^4 + 8*a^3 + 6*a*b^2 + (4*a^3 + 3*a*b^2)*cos(d*x + c 
)^2)*sin(d*x + c))/d
 

Sympy [F]

\[ \int \cos ^5(c+d x) (a+b \tan (c+d x))^3 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{3} \cos ^{5}{\left (c + d x \right )}\, dx \] Input:

integrate(cos(d*x+c)**5*(a+b*tan(d*x+c))**3,x)
 

Output:

Integral((a + b*tan(c + d*x))**3*cos(c + d*x)**5, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96 \[ \int \cos ^5(c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {9 \, a^{2} b \cos \left (d x + c\right )^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3} + 3 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a b^{2} - {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} b^{3}}{15 \, d} \] Input:

integrate(cos(d*x+c)^5*(a+b*tan(d*x+c))^3,x, algorithm="maxima")
 

Output:

-1/15*(9*a^2*b*cos(d*x + c)^5 - (3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15 
*sin(d*x + c))*a^3 + 3*(3*sin(d*x + c)^5 - 5*sin(d*x + c)^3)*a*b^2 - (3*co 
s(d*x + c)^5 - 5*cos(d*x + c)^3)*b^3)/d
 

Giac [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) (a+b \tan (c+d x))^3 \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)^5*(a+b*tan(d*x+c))^3,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 0.97 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.32 \[ \int \cos ^5(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {2\,\left (\frac {3\,\sin \left (c+d\,x\right )\,a^3\,{\cos \left (c+d\,x\right )}^4}{2}+2\,\sin \left (c+d\,x\right )\,a^3\,{\cos \left (c+d\,x\right )}^2+4\,\sin \left (c+d\,x\right )\,a^3-\frac {9\,a^2\,b\,{\cos \left (c+d\,x\right )}^5}{2}-\frac {9\,\sin \left (c+d\,x\right )\,a\,b^2\,{\cos \left (c+d\,x\right )}^4}{2}+\frac {3\,\sin \left (c+d\,x\right )\,a\,b^2\,{\cos \left (c+d\,x\right )}^2}{2}+3\,\sin \left (c+d\,x\right )\,a\,b^2+\frac {3\,b^3\,{\cos \left (c+d\,x\right )}^5}{2}-\frac {5\,b^3\,{\cos \left (c+d\,x\right )}^3}{2}\right )}{15\,d} \] Input:

int(cos(c + d*x)^5*(a + b*tan(c + d*x))^3,x)
 

Output:

(2*(4*a^3*sin(c + d*x) - (5*b^3*cos(c + d*x)^3)/2 + (3*b^3*cos(c + d*x)^5) 
/2 - (9*a^2*b*cos(c + d*x)^5)/2 + 2*a^3*cos(c + d*x)^2*sin(c + d*x) + (3*a 
^3*cos(c + d*x)^4*sin(c + d*x))/2 + 3*a*b^2*sin(c + d*x) + (3*a*b^2*cos(c 
+ d*x)^2*sin(c + d*x))/2 - (9*a*b^2*cos(c + d*x)^4*sin(c + d*x))/2))/(15*d 
)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.65 \[ \int \cos ^5(c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {-9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{2} b +3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{3}+18 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2} b -\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{3}-9 \cos \left (d x +c \right ) a^{2} b -2 \cos \left (d x +c \right ) b^{3}+3 \sin \left (d x +c \right )^{5} a^{3}-9 \sin \left (d x +c \right )^{5} a \,b^{2}-10 \sin \left (d x +c \right )^{3} a^{3}+15 \sin \left (d x +c \right )^{3} a \,b^{2}+15 \sin \left (d x +c \right ) a^{3}+9 a^{2} b +2 b^{3}}{15 d} \] Input:

int(cos(d*x+c)^5*(a+b*tan(d*x+c))^3,x)
 

Output:

( - 9*cos(c + d*x)*sin(c + d*x)**4*a**2*b + 3*cos(c + d*x)*sin(c + d*x)**4 
*b**3 + 18*cos(c + d*x)*sin(c + d*x)**2*a**2*b - cos(c + d*x)*sin(c + d*x) 
**2*b**3 - 9*cos(c + d*x)*a**2*b - 2*cos(c + d*x)*b**3 + 3*sin(c + d*x)**5 
*a**3 - 9*sin(c + d*x)**5*a*b**2 - 10*sin(c + d*x)**3*a**3 + 15*sin(c + d* 
x)**3*a*b**2 + 15*sin(c + d*x)*a**3 + 9*a**2*b + 2*b**3)/(15*d)