\(\int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx\) [214]

Optimal result

Integrand size = 23, antiderivative size = 252 \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\frac {3 e^5 (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}+\frac {e^5 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {3 e^5 \cos ^2(c+d x)^{\frac {1}{2} (-4+m)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-5+m),\frac {1}{2} (-4+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {e^5 \cos ^2(c+d x)^{\frac {1}{2} (-2+m)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-5+m),\frac {1}{2} (-2+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)} \] Output:

3*e^5*(e*tan(d*x+c))^(-5+m)/a^3/d/(5-m)+e^5*hypergeom([1, -5/2+1/2*m],[-3/ 
2+1/2*m],-tan(d*x+c)^2)*(e*tan(d*x+c))^(-5+m)/a^3/d/(5-m)-3*e^5*(cos(d*x+c 
)^2)^(-2+1/2*m)*hypergeom([-2+1/2*m, -5/2+1/2*m],[-3/2+1/2*m],sin(d*x+c)^2 
)*sec(d*x+c)*(e*tan(d*x+c))^(-5+m)/a^3/d/(5-m)-e^5*(cos(d*x+c)^2)^(-1+1/2* 
m)*hypergeom([-1+1/2*m, -5/2+1/2*m],[-3/2+1/2*m],sin(d*x+c)^2)*sec(d*x+c)^ 
3*(e*tan(d*x+c))^(-5+m)/a^3/d/(5-m)
 

Mathematica [F]

\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx \] Input:

Integrate[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x])^3,x]
 

Output:

Integrate[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x])^3, x]
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4376, 25, 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.

\(\displaystyle \int \frac {(e \tan (c+d x))^m}{(a \sec (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^m}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\)

\(\Big \downarrow \) 4376

\(\displaystyle \frac {e^6 \int -(a-a \sec (c+d x))^3 (e \tan (c+d x))^{m-6}dx}{a^6}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {e^6 \int (a-a \sec (c+d x))^3 (e \tan (c+d x))^{m-6}dx}{a^6}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {e^6 \int \left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{m-6} \left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3dx}{a^6}\)

\(\Big \downarrow \) 4374

\(\displaystyle -\frac {e^6 \int \left (a^3 (e \tan (c+d x))^{m-6}-a^3 \sec ^3(c+d x) (e \tan (c+d x))^{m-6}+3 a^3 \sec ^2(c+d x) (e \tan (c+d x))^{m-6}-3 a^3 \sec (c+d x) (e \tan (c+d x))^{m-6}\right )dx}{a^6}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {e^6 \left (-\frac {a^3 (e \tan (c+d x))^{m-5} \operatorname {Hypergeometric2F1}\left (1,\frac {m-5}{2},\frac {m-3}{2},-\tan ^2(c+d x)\right )}{d e (5-m)}+\frac {a^3 \sec ^3(c+d x) \cos ^2(c+d x)^{\frac {m-2}{2}} (e \tan (c+d x))^{m-5} \operatorname {Hypergeometric2F1}\left (\frac {m-5}{2},\frac {m-2}{2},\frac {m-3}{2},\sin ^2(c+d x)\right )}{d e (5-m)}+\frac {3 a^3 \sec (c+d x) \cos ^2(c+d x)^{\frac {m-4}{2}} (e \tan (c+d x))^{m-5} \operatorname {Hypergeometric2F1}\left (\frac {m-5}{2},\frac {m-4}{2},\frac {m-3}{2},\sin ^2(c+d x)\right )}{d e (5-m)}-\frac {3 a^3 (e \tan (c+d x))^{m-5}}{d e (5-m)}\right )}{a^6}\)

Input:

Int[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x])^3,x]
 

Output:

-((e^6*((-3*a^3*(e*Tan[c + d*x])^(-5 + m))/(d*e*(5 - m)) - (a^3*Hypergeome 
tric2F1[1, (-5 + m)/2, (-3 + m)/2, -Tan[c + d*x]^2]*(e*Tan[c + d*x])^(-5 + 
 m))/(d*e*(5 - m)) + (3*a^3*(Cos[c + d*x]^2)^((-4 + m)/2)*Hypergeometric2F 
1[(-5 + m)/2, (-4 + m)/2, (-3 + m)/2, Sin[c + d*x]^2]*Sec[c + d*x]*(e*Tan[ 
c + d*x])^(-5 + m))/(d*e*(5 - m)) + (a^3*(Cos[c + d*x]^2)^((-2 + m)/2)*Hyp 
ergeometric2F1[(-5 + m)/2, (-2 + m)/2, (-3 + m)/2, Sin[c + d*x]^2]*Sec[c + 
 d*x]^3*(e*Tan[c + d*x])^(-5 + m))/(d*e*(5 - m))))/a^6)
 

Defintions of rubi rules used

rule 25

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

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]

\[\int \frac {\left (e \tan \left (d x +c \right )\right )^{m}}{\left (a +a \sec \left (d x +c \right )\right )^{3}}d x\]

Input:

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^3,x)
 

Output:

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^3,x)
 

Fricas [F]

\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^3,x, algorithm="fricas")
 

Output:

integral((e*tan(d*x + c))^m/(a^3*sec(d*x + c)^3 + 3*a^3*sec(d*x + c)^2 + 3 
*a^3*sec(d*x + c) + a^3), x)
 

Sympy [F]

\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{m}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:

integrate((e*tan(d*x+c))**m/(a+a*sec(d*x+c))**3,x)
 

Output:

Integral((e*tan(c + d*x))**m/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec( 
c + d*x) + 1), x)/a**3
 

Maxima [F]

\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^3,x, algorithm="maxima")
 

Output:

integrate((e*tan(d*x + c))^m/(a*sec(d*x + c) + a)^3, x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^3,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,[0,1,4,0]%%%}+%%%{4,[0,1,2,0]%%%}+%%%{-7,[0,1,0,0]%%%} 
/ %%%{8,[
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m}{a^3\,{\left (\cos \left (c+d\,x\right )+1\right )}^3} \,d x \] Input:

int((e*tan(c + d*x))^m/(a + a/cos(c + d*x))^3,x)
 

Output:

int((cos(c + d*x)^3*(e*tan(c + d*x))^m)/(a^3*(cos(c + d*x) + 1)^3), x)
 

Reduce [F]

\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\frac {e^{m} \left (\int \frac {\tan \left (d x +c \right )^{m}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^3,x)
 

Output:

(e**m*int(tan(c + d*x)**m/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + 
 d*x) + 1),x))/a**3