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

Optimal result

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

[Out]

-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)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3957, 2918, 2644, 30, 2657} \[ \int \frac {(e \sin (c+d x))^m}{a+a \sec (c+d x)} \, dx=\frac {e \cos (c+d x) (e \sin (c+d x))^{m-1} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {m-1}{2},\frac {m+1}{2},\sin ^2(c+d x)\right )}{a d (1-m) \sqrt {\cos ^2(c+d x)}}-\frac {e (e \sin (c+d x))^{m-1}}{a d (1-m)} \]

[In]

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

[Out]

-((e*(e*Sin[c + d*x])^(-1 + m))/(a*d*(1 - m))) + (e*Cos[c + d*x]*Hypergeometric2F1[-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])

Rule 30

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

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[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] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2657

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*FracPart[(n - 1)/2])*((a*Sin[e + f*x])^(m + 1)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(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 2918

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[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 3957

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

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

[In]

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

[Out]

(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*Hyp
ergeometric2F1[(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))] + Hypergeome
tric2F1[-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)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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