Integrand size = 23, antiderivative size = 252 \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\frac {3 e^5 (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}+\frac {e^5 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {3 e^5 \cos ^2(c+d x)^{\frac {1}{2} (-4+m)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-5+m),\frac {1}{2} (-4+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {e^5 \cos ^2(c+d x)^{\frac {1}{2} (-2+m)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-5+m),\frac {1}{2} (-2+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)} \]
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)
\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx \]
Time = 0.68 (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 |
\(\displaystyle -\frac {e^6 \left (-\frac {a^3 (e \tan (c+d x))^{m-5} \operatorname {Hypergeometric2F1}\left (1,\frac {m-5}{2},\frac {m-3}{2},-\tan ^2(c+d x)\right )}{d e (5-m)}+\frac {a^3 \sec ^3(c+d x) \cos ^2(c+d x)^{\frac {m-2}{2}} (e \tan (c+d x))^{m-5} \operatorname {Hypergeometric2F1}\left (\frac {m-5}{2},\frac {m-2}{2},\frac {m-3}{2},\sin ^2(c+d x)\right )}{d e (5-m)}+\frac {3 a^3 \sec (c+d x) \cos ^2(c+d x)^{\frac {m-4}{2}} (e \tan (c+d x))^{m-5} \operatorname {Hypergeometric2F1}\left (\frac {m-5}{2},\frac {m-4}{2},\frac {m-3}{2},\sin ^2(c+d x)\right )}{d e (5-m)}-\frac {3 a^3 (e \tan (c+d x))^{m-5}}{d e (5-m)}\right )}{a^6}\) |
-((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)
3.3.14.3.1 Defintions of rubi rules used
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]
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]
\[\int \frac {\left (e \tan \left (d x +c \right )\right )^{m}}{\left (a +a \sec \left (d x +c \right )\right )^{3}}d x\]
\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
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)
\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{m}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
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
\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
\[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m}{a^3\,{\left (\cos \left (c+d\,x\right )+1\right )}^3} \,d x \]