\(\int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx\) [213]

Optimal result

Integrand size = 23, antiderivative size = 169 \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx=-\frac {e^3 (e \tan (c+d x))^{-3+m}}{a^2 d (3-m)}-\frac {e^3 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-3+m}}{a^2 d (3-m)}+\frac {2 e^3 \cos ^2(c+d x)^{\frac {1}{2} (-2+m)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-3+m),\frac {1}{2} (-2+m),\frac {1}{2} (-1+m),\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-3+m}}{a^2 d (3-m)} \] Output:

-e^3*(e*tan(d*x+c))^(-3+m)/a^2/d/(3-m)-e^3*hypergeom([1, -3/2+1/2*m],[-1/2 
+1/2*m],-tan(d*x+c)^2)*(e*tan(d*x+c))^(-3+m)/a^2/d/(3-m)+2*e^3*(cos(d*x+c) 
^2)^(-1+1/2*m)*hypergeom([-1+1/2*m, -3/2+1/2*m],[-1/2+1/2*m],sin(d*x+c)^2) 
*sec(d*x+c)*(e*tan(d*x+c))^(-3+m)/a^2/d/(3-m)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 9.51 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.95 \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx=\frac {\left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^m \tan \left (\frac {1}{2} (c+d x)\right ) \left (-3 (3+m) \operatorname {Hypergeometric2F1}\left (m,\frac {1+m}{2},\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+(1+m) \operatorname {Hypergeometric2F1}\left (m,\frac {3+m}{2},\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (e \tan (c+d x))^m}{2 a^2 d (1+m) (3+m)}+\frac {i 2^{1-m} \left (-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (2^m \operatorname {Hypergeometric2F1}\left (1,m,1+m,-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )-\left (1+e^{2 i (c+d x)}\right )^m \operatorname {Hypergeometric2F1}\left (m,m,1+m,\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )\right ) \sec ^2(c+d x) \tan ^{-m}(c+d x) (e \tan (c+d x))^m}{d m (a+a \sec (c+d x))^2} \] Input:

Integrate[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x])^2,x]
 

Output:

((Cos[c + d*x]*Sec[(c + d*x)/2]^2)^m*Tan[(c + d*x)/2]*(-3*(3 + m)*Hypergeo 
metric2F1[m, (1 + m)/2, (3 + m)/2, Tan[(c + d*x)/2]^2] + (1 + m)*Hypergeom 
etric2F1[m, (3 + m)/2, (5 + m)/2, Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2)* 
(e*Tan[c + d*x])^m)/(2*a^2*d*(1 + m)*(3 + m)) + (I*2^(1 - m)*(((-I)*(-1 + 
E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x))))^m*Cos[c/2 + (d*x)/2]^4*(2 
^m*Hypergeometric2F1[1, m, 1 + m, -((-1 + E^((2*I)*(c + d*x)))/(1 + E^((2* 
I)*(c + d*x))))] - (1 + E^((2*I)*(c + d*x)))^m*Hypergeometric2F1[m, m, 1 + 
 m, (1 - E^((2*I)*(c + d*x)))/2])*Sec[c + d*x]^2*(e*Tan[c + d*x])^m)/(d*m* 
(a + a*Sec[c + d*x])^2*Tan[c + d*x]^m)
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4376, 3042, 4374, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4376

\(\displaystyle \frac {e^4 \int (a-a \sec (c+d x))^2 (e \tan (c+d x))^{m-4}dx}{a^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {e^4 \int \left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{m-4} \left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2dx}{a^4}\)

\(\Big \downarrow \) 4374

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {e^4 \left (-\frac {a^2 (e \tan (c+d x))^{m-3} \operatorname {Hypergeometric2F1}\left (1,\frac {m-3}{2},\frac {m-1}{2},-\tan ^2(c+d x)\right )}{d e (3-m)}+\frac {2 a^2 \sec (c+d x) \cos ^2(c+d x)^{\frac {m-2}{2}} (e \tan (c+d x))^{m-3} \operatorname {Hypergeometric2F1}\left (\frac {m-3}{2},\frac {m-2}{2},\frac {m-1}{2},\sin ^2(c+d x)\right )}{d e (3-m)}-\frac {a^2 (e \tan (c+d x))^{m-3}}{d e (3-m)}\right )}{a^4}\)

Input:

Int[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x])^2,x]
 

Output:

(e^4*(-((a^2*(e*Tan[c + d*x])^(-3 + m))/(d*e*(3 - m))) - (a^2*Hypergeometr 
ic2F1[1, (-3 + m)/2, (-1 + m)/2, -Tan[c + d*x]^2]*(e*Tan[c + d*x])^(-3 + m 
))/(d*e*(3 - m)) + (2*a^2*(Cos[c + d*x]^2)^((-2 + m)/2)*Hypergeometric2F1[ 
(-3 + m)/2, (-2 + m)/2, (-1 + m)/2, Sin[c + d*x]^2]*Sec[c + d*x]*(e*Tan[c 
+ d*x])^(-3 + m))/(d*e*(3 - m))))/a^4
 

Defintions of rubi rules used

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 4374

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

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]

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

Input:

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^2,x)
 

Output:

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^2,x)
 

Fricas [F]

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

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^2,x, algorithm="fricas")
 

Output:

integral((e*tan(d*x + c))^m/(a^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a^2 
), x)
 

Sympy [F]

\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{m}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:

integrate((e*tan(d*x+c))**m/(a+a*sec(d*x+c))**2,x)
 

Output:

Integral((e*tan(c + d*x))**m/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a* 
*2
 

Maxima [F]

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

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^2,x, algorithm="maxima")
 

Output:

integrate((e*tan(d*x + c))^m/(a*sec(d*x + c) + a)^2, x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^2,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,[0,1,2,0]%%%}+%%%{-3,[0,1,0,0]%%%} / %%%{4,[0,0,0,2]%%%} 
 Error: B
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \] Input:

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

Output:

int((cos(c + d*x)^2*(e*tan(c + d*x))^m)/(a^2*(cos(c + d*x) + 1)^2), x)
 

Reduce [F]

\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx=\frac {e^{m} \left (\int \frac {\tan \left (d x +c \right )^{m}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right )}{a^{2}} \] Input:

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^2,x)
 

Output:

(e**m*int(tan(c + d*x)**m/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1),x))/a**2