\(\int \cos (c+d x) (a+b \tan (c+d x))^n \, dx\) [662]

Optimal result

Integrand size = 19, antiderivative size = 161 \[ \int \cos (c+d x) (a+b \tan (c+d x))^n \, dx=\frac {\operatorname {AppellF1}\left (1+n,\frac {3}{2},\frac {3}{2},2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}},\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) \cos ^3(c+d x) (a+b \tan (c+d x))^{1+n} \left (1-\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )^{3/2} \left (1-\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )^{3/2}}{b d (1+n)} \] Output:

AppellF1(1+n,3/2,3/2,2+n,(a+b*tan(d*x+c))/(a-(-b^2)^(1/2)),(a+b*tan(d*x+c) 
)/(a+(-b^2)^(1/2)))*cos(d*x+c)^3*(a+b*tan(d*x+c))^(1+n)*(1-(a+b*tan(d*x+c) 
)/(a-(-b^2)^(1/2)))^(3/2)*(1-(a+b*tan(d*x+c))/(a+(-b^2)^(1/2)))^(3/2)/b/d/ 
(1+n)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 4.08 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.12 \[ \int \cos (c+d x) (a+b \tan (c+d x))^n \, dx=\frac {2 \left (a^2+b^2\right )^2 (2+n) \operatorname {AppellF1}\left (1+n,\frac {3}{2},\frac {3}{2},2+n,\frac {a+b \tan (c+d x)}{a-i b},\frac {a+b \tan (c+d x)}{a+i b}\right ) \cos ^5(c+d x) (-i+\tan (c+d x)) (i+\tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{(a-i b) (a+i b) b d (1+n) \left (2 \left (a^2+b^2\right ) (2+n) \operatorname {AppellF1}\left (1+n,\frac {3}{2},\frac {3}{2},2+n,\frac {a+b \tan (c+d x)}{a-i b},\frac {a+b \tan (c+d x)}{a+i b}\right )+3 \left ((a-i b) \operatorname {AppellF1}\left (2+n,\frac {3}{2},\frac {5}{2},3+n,\frac {a+b \tan (c+d x)}{a-i b},\frac {a+b \tan (c+d x)}{a+i b}\right )+(a+i b) \operatorname {AppellF1}\left (2+n,\frac {5}{2},\frac {3}{2},3+n,\frac {a+b \tan (c+d x)}{a-i b},\frac {a+b \tan (c+d x)}{a+i b}\right )\right ) (a+b \tan (c+d x))\right )} \] Input:

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

Output:

(2*(a^2 + b^2)^2*(2 + n)*AppellF1[1 + n, 3/2, 3/2, 2 + n, (a + b*Tan[c + d 
*x])/(a - I*b), (a + b*Tan[c + d*x])/(a + I*b)]*Cos[c + d*x]^5*(-I + Tan[c 
 + d*x])*(I + Tan[c + d*x])*(a + b*Tan[c + d*x])^(1 + n))/((a - I*b)*(a + 
I*b)*b*d*(1 + n)*(2*(a^2 + b^2)*(2 + n)*AppellF1[1 + n, 3/2, 3/2, 2 + n, ( 
a + b*Tan[c + d*x])/(a - I*b), (a + b*Tan[c + d*x])/(a + I*b)] + 3*((a - I 
*b)*AppellF1[2 + n, 3/2, 5/2, 3 + n, (a + b*Tan[c + d*x])/(a - I*b), (a + 
b*Tan[c + d*x])/(a + I*b)] + (a + I*b)*AppellF1[2 + n, 5/2, 3/2, 3 + n, (a 
 + b*Tan[c + d*x])/(a - I*b), (a + b*Tan[c + d*x])/(a + I*b)])*(a + b*Tan[ 
c + d*x])))
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3992, 514, 150}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3992

\(\displaystyle \frac {\sec (c+d x) \int \frac {(a+b \tan (c+d x))^n}{\left (\tan ^2(c+d x)+1\right )^{3/2}}d(b \tan (c+d x))}{b d \sqrt {\sec ^2(c+d x)}}\)

\(\Big \downarrow \) 514

\(\displaystyle \frac {\sec (c+d x) \left (1-\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )^{3/2} \left (1-\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )^{3/2} \int \frac {(a+b \tan (c+d x))^n}{\left (1-\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )^{3/2} \left (1-\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )^{3/2}}d(a+b \tan (c+d x))}{b d \left (\tan ^2(c+d x)+1\right )^{3/2} \sqrt {\sec ^2(c+d x)}}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {\sec (c+d x) \left (1-\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )^{3/2} \left (1-\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )^{3/2} (a+b \tan (c+d x))^{n+1} \operatorname {AppellF1}\left (n+1,\frac {3}{2},\frac {3}{2},n+2,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}},\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{b d (n+1) \left (\tan ^2(c+d x)+1\right )^{3/2} \sqrt {\sec ^2(c+d x)}}\)

Input:

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

Output:

(AppellF1[1 + n, 3/2, 3/2, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2]), ( 
a + b*Tan[c + d*x])/(a + Sqrt[-b^2])]*Sec[c + d*x]*(a + b*Tan[c + d*x])^(1 
 + n)*(1 - (a + b*Tan[c + d*x])/(a - Sqrt[-b^2]))^(3/2)*(1 - (a + b*Tan[c 
+ d*x])/(a + Sqrt[-b^2]))^(3/2))/(b*d*(1 + n)*Sqrt[Sec[c + d*x]^2]*(1 + Ta 
n[c + d*x]^2)^(3/2))
 

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 514

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Rt[-a/b, 2]}, Simp[(a + b*x^2)^p/(d*(1 - (c + d*x)/(c - d*q))^p*(1 - ( 
c + d*x)/(c + d*q))^p)   Subst[Int[x^n*Simp[1 - x/(c + d*q), x]^p*Simp[1 - 
x/(c - d*q), x]^p, x], x, c + d*x], x]] /; FreeQ[{a, b, c, d, n, p}, x] && 
NeQ[b*c^2 + a*d^2, 0]
 

rule 3042

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

rule 3992

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

Maple [F]

\[\int \cos \left (d x +c \right ) \left (a +b \tan \left (d x +c \right )\right )^{n}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b*tan(d*x + c) + a)^n*cos(d*x + c), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \cos (c+d x) (a+b \tan (c+d x))^n \, dx=\int \cos \left (c+d\,x\right )\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

int((tan(c + d*x)*b + a)**n*cos(c + d*x),x)