\(\int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx\) [212]

Optimal result

Integrand size = 23, antiderivative size = 130 \[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx=\frac {e \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1+m),\frac {1+m}{2},-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-1+m}}{a d (1-m)}-\frac {e \cos ^2(c+d x)^{m/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-1+m),\frac {m}{2},\frac {1+m}{2},\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-1+m}}{a d (1-m)} \] Output:

e*hypergeom([1, -1/2+1/2*m],[1/2+1/2*m],-tan(d*x+c)^2)*(e*tan(d*x+c))^(-1+ 
m)/a/d/(1-m)-e*(cos(d*x+c)^2)^(1/2*m)*hypergeom([1/2*m, -1/2+1/2*m],[1/2+1 
/2*m],sin(d*x+c)^2)*sec(d*x+c)*(e*tan(d*x+c))^(-1+m)/a/d/(1-m)
 

Mathematica [F]

\[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx=\int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx \] Input:

Integrate[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x]),x]
 

Output:

Integrate[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x]), x]
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4376, 25, 3042, 4372, 3042, 3097, 3957, 278}

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.

\(\displaystyle \int \frac {(e \tan (c+d x))^m}{a \sec (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^m}{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}dx\)

\(\Big \downarrow \) 4376

\(\displaystyle \frac {e^2 \int -\left ((a-a \sec (c+d x)) (e \tan (c+d x))^{m-2}\right )dx}{a^2}\)

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {e^2 \int \left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{m-2} \left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}\)

\(\Big \downarrow \) 4372

\(\displaystyle -\frac {e^2 \left (a \int (e \tan (c+d x))^{m-2}dx-a \int \sec (c+d x) (e \tan (c+d x))^{m-2}dx\right )}{a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {e^2 \left (a \int (e \tan (c+d x))^{m-2}dx-a \int \sec (c+d x) (e \tan (c+d x))^{m-2}dx\right )}{a^2}\)

\(\Big \downarrow \) 3097

\(\displaystyle -\frac {e^2 \left (a \int (e \tan (c+d x))^{m-2}dx+\frac {a \sec (c+d x) \cos ^2(c+d x)^{m/2} (e \tan (c+d x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {m-1}{2},\frac {m}{2},\frac {m+1}{2},\sin ^2(c+d x)\right )}{d e (1-m)}\right )}{a^2}\)

\(\Big \downarrow \) 3957

\(\displaystyle -\frac {e^2 \left (\frac {a e \int \frac {(e \tan (c+d x))^{m-2}}{\tan ^2(c+d x) e^2+e^2}d(e \tan (c+d x))}{d}+\frac {a \sec (c+d x) \cos ^2(c+d x)^{m/2} (e \tan (c+d x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {m-1}{2},\frac {m}{2},\frac {m+1}{2},\sin ^2(c+d x)\right )}{d e (1-m)}\right )}{a^2}\)

\(\Big \downarrow \) 278

\(\displaystyle -\frac {e^2 \left (\frac {a \sec (c+d x) \cos ^2(c+d x)^{m/2} (e \tan (c+d x))^{m-1} \operatorname {Hypergeometric2F1}\left (\frac {m-1}{2},\frac {m}{2},\frac {m+1}{2},\sin ^2(c+d x)\right )}{d e (1-m)}-\frac {a (e \tan (c+d x))^{m-1} \operatorname {Hypergeometric2F1}\left (1,\frac {m-1}{2},\frac {m+1}{2},-\tan ^2(c+d x)\right )}{d e (1-m)}\right )}{a^2}\)

Input:

Int[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x]),x]
 

Output:

-((e^2*(-((a*Hypergeometric2F1[1, (-1 + m)/2, (1 + m)/2, -Tan[c + d*x]^2]* 
(e*Tan[c + d*x])^(-1 + m))/(d*e*(1 - m))) + (a*(Cos[c + d*x]^2)^(m/2)*Hype 
rgeometric2F1[(-1 + m)/2, m/2, (1 + m)/2, Sin[c + d*x]^2]*Sec[c + d*x]*(e* 
Tan[c + d*x])^(-1 + m))/(d*e*(1 - m))))/a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 3042

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

rule 3097

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( 
n_), x_Symbol] :> Simp[(a*Sec[e + f*x])^m*(b*Tan[e + f*x])^(n + 1)*((Cos[e 
+ f*x]^2)^((m + n + 1)/2)/(b*f*(n + 1)))*Hypergeometric2F1[(n + 1)/2, (m + 
n + 1)/2, (n + 3)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x] && 
  !IntegerQ[(n - 1)/2] &&  !IntegerQ[m/2]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

rule 4372

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(e*Cot[c + d*x])^m, x], x] + Simp[b   Int[ 
(e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]
 

rule 4376

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( 
a_))^(n_), x_Symbol] :> Simp[a^(2*n)/e^(2*n)   Int[(e*Cot[c + d*x])^(m + 2* 
n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[a 
^2 - b^2, 0] && ILtQ[n, 0]
 

Maple [F]

\[\int \frac {\left (e \tan \left (d x +c \right )\right )^{m}}{a +a \sec \left (d x +c \right )}d x\]

Input:

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c)),x)
 

Output:

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c)),x)
 

Fricas [F]

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

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c)),x, algorithm="fricas")
 

Output:

integral((e*tan(d*x + c))^m/(a*sec(d*x + c) + a), x)
 

Sympy [F]

\[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{m}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:

integrate((e*tan(d*x+c))**m/(a+a*sec(d*x+c)),x)
 

Output:

Integral((e*tan(c + d*x))**m/(sec(c + d*x) + 1), x)/a
 

Maxima [F]

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

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c)),x, algorithm="maxima")
 

Output:

integrate((e*tan(d*x + c))^m/(a*sec(d*x + c) + a), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c)),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{-1,[0,1,0]%%%} / %%%{2,[0,0,1]%%%} Error: Bad Argument Val 
ue
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx=\int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \] Input:

int((e*tan(c + d*x))^m/(a + a/cos(c + d*x)),x)
 

Output:

int((cos(c + d*x)*(e*tan(c + d*x))^m)/(a*(cos(c + d*x) + 1)), x)
 

Reduce [F]

\[ \int \frac {(e \tan (c+d x))^m}{a+a \sec (c+d x)} \, dx=\frac {e^{m} \left (\int \frac {\tan \left (d x +c \right )^{m}}{\sec \left (d x +c \right )+1}d x \right )}{a} \] Input:

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c)),x)
 

Output:

(e**m*int(tan(c + d*x)**m/(sec(c + d*x) + 1),x))/a