3.2.37 \(\int \frac {(e \sin (c+d x))^m}{a+a \sec (c+d x)} \, dx\) [137]

3.2.37.1 Optimal result

Integrand size = 23, antiderivative size = 100 \[ \int \frac {(e \sin (c+d x))^m}{a+a \sec (c+d x)} \, dx=-\frac {e (e \sin (c+d x))^{-1+m}}{a d (1-m)}+\frac {e \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (-1+m),\frac {1+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-1+m}}{a d (1-m) \sqrt {\cos ^2(c+d x)}} \]

output

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

3.2.37.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 16.38 (sec) , antiderivative size = 383, normalized size of antiderivative = 3.83 \[ \int \frac {(e \sin (c+d x))^m}{a+a \sec (c+d x)} \, dx=\frac {i 2^{1-m} \left (1-e^{2 i (c+d x)}\right )^{-m} \left (-i e^{-i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )\right )^m \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (2 e^{i (c+d x)} (-2+m) m \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},1-m,\frac {3-m}{2},e^{2 i (c+d x)}\right )-4 e^{i (c+d x)} (-2+m) m \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},2-m,\frac {3-m}{2},e^{2 i (c+d x)}\right )-(-1+m) \left (2 (-2+m) \operatorname {Hypergeometric2F1}\left (1-m,-\frac {m}{2},1-\frac {m}{2},e^{2 i (c+d x)}\right )-2 e^{2 i (c+d x)} m \operatorname {Hypergeometric2F1}\left (2-m,1-\frac {m}{2},2-\frac {m}{2},e^{2 i (c+d x)}\right )-(-2+m) \left (2 \operatorname {Hypergeometric2F1}\left (2-m,-\frac {m}{2},1-\frac {m}{2},e^{2 i (c+d x)}\right )+\operatorname {Hypergeometric2F1}\left (-m,-\frac {m}{2},1-\frac {m}{2},e^{2 i (c+d x)}\right )\right )\right )\right ) \sec (c+d x) \sin ^{-m}(c+d x) (e \sin (c+d x))^m}{d (-2+m) (-1+m) m (a+a \sec (c+d x))} \]

input

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

output

(I*2^(1 - m)*(((-I)*(-1 + E^((2*I)*(c + d*x))))/E^(I*(c + d*x)))^m*Cos[c/2 
 + (d*x)/2]^2*(2*E^(I*(c + d*x))*(-2 + m)*m*Hypergeometric2F1[(1 - m)/2, 1 
 - m, (3 - m)/2, E^((2*I)*(c + d*x))] - 4*E^(I*(c + d*x))*(-2 + m)*m*Hyper 
geometric2F1[(1 - m)/2, 2 - m, (3 - m)/2, E^((2*I)*(c + d*x))] - (-1 + m)* 
(2*(-2 + m)*Hypergeometric2F1[1 - m, -1/2*m, 1 - m/2, E^((2*I)*(c + d*x))] 
 - 2*E^((2*I)*(c + d*x))*m*Hypergeometric2F1[2 - m, 1 - m/2, 2 - m/2, E^(( 
2*I)*(c + d*x))] - (-2 + m)*(2*Hypergeometric2F1[2 - m, -1/2*m, 1 - m/2, E 
^((2*I)*(c + d*x))] + Hypergeometric2F1[-m, -1/2*m, 1 - m/2, E^((2*I)*(c + 
 d*x))])))*Sec[c + d*x]*(e*Sin[c + d*x])^m)/(d*(1 - E^((2*I)*(c + d*x)))^m 
*(-2 + m)*(-1 + m)*m*(a + a*Sec[c + d*x])*Sin[c + d*x]^m)
 

3.2.37.3 Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4360, 25, 25, 3042, 3318, 3042, 3044, 15, 3057}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4360

\(\displaystyle \int -\frac {\cos (c+d x) (e \sin (c+d x))^m}{a (-\cos (c+d x))-a}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int -\frac {\cos (c+d x) (e \sin (c+d x))^m}{\cos (c+d x) a+a}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \int \frac {\cos (c+d x) (e \sin (c+d x))^m}{a \cos (c+d x)+a}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3044

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

\(\Big \downarrow \) 15

\(\displaystyle -\frac {e^2 \int \cos (c+d x)^2 (e \sin (c+d x))^{m-2}dx}{a}-\frac {e (e \sin (c+d x))^{m-1}}{a d (1-m)}\)

\(\Big \downarrow \) 3057

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

input

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

output

-((e*(e*Sin[c + d*x])^(-1 + m))/(a*d*(1 - m))) + (e*Cos[c + d*x]*Hypergeom 
etric2F1[-1/2, (-1 + m)/2, (1 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-1 
 + m))/(a*d*(1 - m)*Sqrt[Cos[c + d*x]^2])
 

3.2.37.3.1 Defintions of rubi rules used

rule 15

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

rule 25

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

rule 3042

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

rule 3044

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_ 
Symbol] :> Simp[1/(a*f)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a 
*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&  !(I 
ntegerQ[(m - 1)/2] && LtQ[0, m, n])
 

rule 3057

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[b^(2*IntPart[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*Frac 
Part[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^2)^Fr 
acPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[ 
e + f*x]^2], x] /; FreeQ[{a, b, e, f, m, n}, x]
 

rule 3318

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( 
n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a   Int 
[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Simp[g^2/(b*d)   Int 
[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, 
d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0]
 

rule 4360

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si 
n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
 

3.2.37.4 Maple [F]

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

input

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

output

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

3.2.37.5 Fricas [F]

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

input

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

output

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

3.2.37.6 Sympy [F]

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

input

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

output

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

3.2.37.7 Maxima [F]

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

input

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

output

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

3.2.37.8 Giac [F]

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

input

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

output

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

3.2.37.9 Mupad [F(-1)]

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

input

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

output

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