\(\int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx\) [708]

Optimal result

Integrand size = 23, antiderivative size = 227 \[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=\frac {2 a b (d \cos (e+f x))^m \operatorname {Hypergeometric2F1}\left (2,-\frac {m}{2},1-\frac {m}{2},\frac {b^2 \sec ^2(e+f x)}{a^2+b^2}\right )}{\left (a^2+b^2\right )^2 f m}+\frac {\operatorname {AppellF1}\left (\frac {1}{2},2,\frac {2+m}{2},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2} \tan (e+f x)}{a^2 f}+\frac {b^2 \operatorname {AppellF1}\left (\frac {3}{2},2,\frac {2+m}{2},\frac {5}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \cos (e+f x))^m \sec ^2(e+f x)^{m/2} \tan ^3(e+f x)}{3 a^4 f} \] Output:

2*a*b*(d*cos(f*x+e))^m*hypergeom([2, -1/2*m],[1-1/2*m],b^2*sec(f*x+e)^2/(a 
^2+b^2))/(a^2+b^2)^2/f/m+AppellF1(1/2,2,1+1/2*m,3/2,b^2*tan(f*x+e)^2/a^2,- 
tan(f*x+e)^2)*(d*cos(f*x+e))^m*(sec(f*x+e)^2)^(1/2*m)*tan(f*x+e)/a^2/f+1/3 
*b^2*AppellF1(3/2,2,1+1/2*m,5/2,b^2*tan(f*x+e)^2/a^2,-tan(f*x+e)^2)*(d*cos 
(f*x+e))^m*(sec(f*x+e)^2)^(1/2*m)*tan(f*x+e)^3/a^4/f
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 15.68 (sec) , antiderivative size = 2502, normalized size of antiderivative = 11.02 \[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=\text {Result too large to show} \] Input:

Integrate[(d*Cos[e + f*x])^m/(a + b*Tan[e + f*x])^2,x]
 

Output:

((d*Cos[e + f*x])^m*((2*a*b*(-1 + (Sec[e + f*x]^2)^(-1/2*m)))/m + a^2*Hype 
rgeometric2F1[1/2, 1 + m/2, 3/2, -Tan[e + f*x]^2]*Tan[e + f*x] - b^2*Hyper 
geometric2F1[1/2, 1 + m/2, 3/2, -Tan[e + f*x]^2]*Tan[e + f*x] - (2*a*b*App 
ellF1[m, m/2, m/2, 1 + m, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b 
*Tan[e + f*x])]*((b*(-I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2)*((b*( 
I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2))/(m*(Sec[e + f*x]^2)^(m/2)) 
 - (b*(a^2 + b^2)*AppellF1[1 + m, m/2, m/2, 2 + m, (a - I*b)/(a + b*Tan[e 
+ f*x]), (a + I*b)/(a + b*Tan[e + f*x])]*((b*(-I + Tan[e + f*x]))/(a + b*T 
an[e + f*x]))^(m/2)*((b*(I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2))/( 
(1 + m)*(Sec[e + f*x]^2)^(m/2)*(a + b*Tan[e + f*x]))))/(f*(a + b*Tan[e + f 
*x])^2*(a^2*Hypergeometric2F1[1/2, 1 + m/2, 3/2, -Tan[e + f*x]^2]*Sec[e + 
f*x]^2 - b^2*Hypergeometric2F1[1/2, 1 + m/2, 3/2, -Tan[e + f*x]^2]*Sec[e + 
 f*x]^2 - (2*a*b*Tan[e + f*x])/(Sec[e + f*x]^2)^(m/2) + (2*a*b*AppellF1[m, 
 m/2, m/2, 1 + m, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + 
 f*x])]*Tan[e + f*x]*((b*(-I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2)* 
((b*(I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2))/(Sec[e + f*x]^2)^(m/2 
) + (b^2*(a^2 + b^2)*AppellF1[1 + m, m/2, m/2, 2 + m, (a - I*b)/(a + b*Tan 
[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])]*(Sec[e + f*x]^2)^(1 - m/2)*((b 
*(-I + Tan[e + f*x]))/(a + b*Tan[e + f*x]))^(m/2)*((b*(I + Tan[e + f*x]))/ 
(a + b*Tan[e + f*x]))^(m/2))/((1 + m)*(a + b*Tan[e + f*x])^2) + (b*(a^2...
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3998, 3042, 3994, 505, 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.

\(\displaystyle \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2}dx\)

\(\Big \downarrow \) 3998

\(\displaystyle (d \cos (e+f x))^m (d \sec (e+f x))^m \int \frac {(d \sec (e+f x))^{-m}}{(a+b \tan (e+f x))^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle (d \cos (e+f x))^m (d \sec (e+f x))^m \int \frac {(d \sec (e+f x))^{-m}}{(a+b \tan (e+f x))^2}dx\)

\(\Big \downarrow \) 3994

\(\displaystyle \frac {\sec ^2(e+f x)^{m/2} (d \cos (e+f x))^m \int \frac {\left (\tan ^2(e+f x)+1\right )^{-\frac {m}{2}-1}}{(a+b \tan (e+f x))^2}d(b \tan (e+f x))}{b f}\)

\(\Big \downarrow \) 505

\(\displaystyle \frac {\sec ^2(e+f x)^{m/2} (d \cos (e+f x))^m \int \left (\frac {a^2 \left (\tan ^2(e+f x)+1\right )^{-\frac {m}{2}-1}}{\left (a^2-b^2 \tan ^2(e+f x)\right )^2}-\frac {2 a b \tan (e+f x) \left (\tan ^2(e+f x)+1\right )^{-\frac {m}{2}-1}}{\left (a^2-b^2 \tan ^2(e+f x)\right )^2}+\frac {b^2 \tan ^2(e+f x) \left (\tan ^2(e+f x)+1\right )^{-\frac {m}{2}-1}}{\left (b^2 \tan ^2(e+f x)-a^2\right )^2}\right )d(b \tan (e+f x))}{b f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sec ^2(e+f x)^{m/2} (d \cos (e+f x))^m \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},2,\frac {m+2}{2},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a^2}+\frac {2 a b^2 \left (\tan ^2(e+f x)+1\right )^{-m/2} \operatorname {Hypergeometric2F1}\left (2,-\frac {m}{2},1-\frac {m}{2},\frac {\tan ^2(e+f x) b^2+b^2}{a^2+b^2}\right )}{m \left (a^2+b^2\right )^2}+\frac {b^3 \tan ^3(e+f x) \operatorname {AppellF1}\left (\frac {3}{2},2,\frac {m+2}{2},\frac {5}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{3 a^4}\right )}{b f}\)

Input:

Int[(d*Cos[e + f*x])^m/(a + b*Tan[e + f*x])^2,x]
 

Output:

((d*Cos[e + f*x])^m*(Sec[e + f*x]^2)^(m/2)*((b*AppellF1[1/2, 2, (2 + m)/2, 
 3/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*Tan[e + f*x])/a^2 + (b^3* 
AppellF1[3/2, 2, (2 + m)/2, 5/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2 
]*Tan[e + f*x]^3)/(3*a^4) + (2*a*b^2*Hypergeometric2F1[2, -1/2*m, 1 - m/2, 
 (b^2 + b^2*Tan[e + f*x]^2)/(a^2 + b^2)])/((a^2 + b^2)^2*m*(1 + Tan[e + f* 
x]^2)^(m/2))))/(b*f)
 

Defintions of rubi rules used

rule 505

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[E 
xpandIntegrand[(a + b*x^2)^p, (c/(c^2 - d^2*x^2) - d*(x/(c^2 - d^2*x^2)))^( 
-n), x], x] /; FreeQ[{a, b, c, d, p}, x] && ILtQ[n, -1] && PosQ[a/b]
 

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 3994

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[d^(2*IntPart[m/2])*((d*Sec[e + f*x])^(2*FracP 
art[m/2])/(b*f*(Sec[e + f*x]^2)^FracPart[m/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, d, e, f, m, 
n}, x] && NeQ[a^2 + b^2, 0] &&  !IntegerQ[m] && IntegerQ[n]
 

rule 3998

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_.), x_Symbol] :> Simp[(d*Cos[e + f*x])^m*(d*Sec[e + f*x])^m   Int[( 
a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m 
, n}, x] &&  !IntegerQ[m]
 

Maple [F]

\[\int \frac {\left (d \cos \left (f x +e \right )\right )^{m}}{\left (a +b \tan \left (f x +e \right )\right )^{2}}d x\]

Input:

int((d*cos(f*x+e))^m/(a+b*tan(f*x+e))^2,x)
 

Output:

int((d*cos(f*x+e))^m/(a+b*tan(f*x+e))^2,x)
 

Fricas [F]

\[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\left (d \cos \left (f x + e\right )\right )^{m}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((d*cos(f*x+e))^m/(a+b*tan(f*x+e))^2,x, algorithm="fricas")
 

Output:

integral((d*cos(f*x + e))^m/(b^2*tan(f*x + e)^2 + 2*a*b*tan(f*x + e) + a^2 
), x)
 

Sympy [F]

\[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=\int \frac {\left (d \cos {\left (e + f x \right )}\right )^{m}}{\left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \] Input:

integrate((d*cos(f*x+e))**m/(a+b*tan(f*x+e))**2,x)
 

Output:

Integral((d*cos(e + f*x))**m/(a + b*tan(e + f*x))**2, x)
 

Maxima [F]

\[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\left (d \cos \left (f x + e\right )\right )^{m}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((d*cos(f*x+e))^m/(a+b*tan(f*x+e))^2,x, algorithm="maxima")
 

Output:

integrate((d*cos(f*x + e))^m/(b*tan(f*x + e) + a)^2, x)
 

Giac [F]

\[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\left (d \cos \left (f x + e\right )\right )^{m}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((d*cos(f*x+e))^m/(a+b*tan(f*x+e))^2,x, algorithm="giac")
 

Output:

integrate((d*cos(f*x + e))^m/(b*tan(f*x + e) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=\int \frac {{\left (d\,\cos \left (e+f\,x\right )\right )}^m}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \] Input:

int((d*cos(e + f*x))^m/(a + b*tan(e + f*x))^2,x)
 

Output:

int((d*cos(e + f*x))^m/(a + b*tan(e + f*x))^2, x)
 

Reduce [F]

\[ \int \frac {(d \cos (e+f x))^m}{(a+b \tan (e+f x))^2} \, dx=d^{m} \left (\int \frac {\cos \left (f x +e \right )^{m}}{\tan \left (f x +e \right )^{2} b^{2}+2 \tan \left (f x +e \right ) a b +a^{2}}d x \right ) \] Input:

int((d*cos(f*x+e))^m/(a+b*tan(f*x+e))^2,x)
 

Output:

d**m*int(cos(e + f*x)**m/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2 
),x)