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

Optimal result

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

[Out]

-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)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3973, 3971, 3557, 371, 2697, 2687, 32} \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx=-\frac {e^3 (e \tan (c+d x))^{m-3} \operatorname {Hypergeometric2F1}\left (1,\frac {m-3}{2},\frac {m-1}{2},-\tan ^2(c+d x)\right )}{a^2 d (3-m)}+\frac {2 e^3 \sec (c+d x) \cos ^2(c+d x)^{\frac {m-2}{2}} (e \tan (c+d x))^{m-3} \operatorname {Hypergeometric2F1}\left (\frac {m-3}{2},\frac {m-2}{2},\frac {m-1}{2},\sin ^2(c+d x)\right )}{a^2 d (3-m)}-\frac {e^3 (e \tan (c+d x))^{m-3}}{a^2 d (3-m)} \]

[In]

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

[Out]

-((e^3*(e*Tan[c + d*x])^(-3 + m))/(a^2*d*(3 - m))) - (e^3*Hypergeometric2F1[1, (-3 + m)/2, (-1 + m)/2, -Tan[c
+ d*x]^2]*(e*Tan[c + d*x])^(-3 + m))/(a^2*d*(3 - m)) + (2*e^3*(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))/(a^2*d*(3 - m))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2697

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] &&  !In
tegerQ[m/2]

Rule 3557

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[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 3971

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(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 3973

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[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] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {e^4 \int (-a+a \sec (c+d x))^2 (e \tan (c+d x))^{-4+m} \, dx}{a^4} \\ & = \frac {e^4 \int \left (a^2 (e \tan (c+d x))^{-4+m}-2 a^2 \sec (c+d x) (e \tan (c+d x))^{-4+m}+a^2 \sec ^2(c+d x) (e \tan (c+d x))^{-4+m}\right ) \, dx}{a^4} \\ & = \frac {e^4 \int (e \tan (c+d x))^{-4+m} \, dx}{a^2}+\frac {e^4 \int \sec ^2(c+d x) (e \tan (c+d x))^{-4+m} \, dx}{a^2}-\frac {\left (2 e^4\right ) \int \sec (c+d x) (e \tan (c+d x))^{-4+m} \, dx}{a^2} \\ & = \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)}+\frac {e^4 \text {Subst}\left (\int (e x)^{-4+m} \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac {e^5 \text {Subst}\left (\int \frac {x^{-4+m}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a^2 d} \\ & = -\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)} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.84 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.95 \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx=\frac {\left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^m \tan \left (\frac {1}{2} (c+d x)\right ) \left (-3 (3+m) \operatorname {Hypergeometric2F1}\left (m,\frac {1+m}{2},\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+(1+m) \operatorname {Hypergeometric2F1}\left (m,\frac {3+m}{2},\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (e \tan (c+d x))^m}{2 a^2 d (1+m) (3+m)}+\frac {i 2^{1-m} \left (-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (2^m \operatorname {Hypergeometric2F1}\left (1,m,1+m,-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )-\left (1+e^{2 i (c+d x)}\right )^m \operatorname {Hypergeometric2F1}\left (m,m,1+m,\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )\right ) \sec ^2(c+d x) \tan ^{-m}(c+d x) (e \tan (c+d x))^m}{d m (a+a \sec (c+d x))^2} \]

[In]

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

[Out]

((Cos[c + d*x]*Sec[(c + d*x)/2]^2)^m*Tan[(c + d*x)/2]*(-3*(3 + m)*Hypergeometric2F1[m, (1 + m)/2, (3 + m)/2, T
an[(c + d*x)/2]^2] + (1 + m)*Hypergeometric2F1[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)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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