\(\int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx\) [138]

Optimal result

Integrand size = 23, antiderivative size = 207 \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx=\frac {2 e^3 (e \sin (c+d x))^{-3+m}}{a^2 d (3-m)}-\frac {e^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-3+m}}{a^2 d (3-m) \sqrt {\cos ^2(c+d x)}}-\frac {e^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-3+m}}{a^2 d (3-m) \sqrt {\cos ^2(c+d x)}}-\frac {2 e (e \sin (c+d x))^{-1+m}}{a^2 d (1-m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3957, 2954, 2952, 2657, 2644, 14} \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx=-\frac {e^3 \cos (c+d x) (e \sin (c+d x))^{m-3} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m-3}{2},\frac {m-1}{2},\sin ^2(c+d x)\right )}{a^2 d (3-m) \sqrt {\cos ^2(c+d x)}}-\frac {e^3 \cos (c+d x) (e \sin (c+d x))^{m-3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {m-3}{2},\frac {m-1}{2},\sin ^2(c+d x)\right )}{a^2 d (3-m) \sqrt {\cos ^2(c+d x)}}+\frac {2 e^3 (e \sin (c+d x))^{m-3}}{a^2 d (3-m)}-\frac {2 e (e \sin (c+d x))^{m-1}}{a^2 d (1-m)} \]

[In]

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

[Out]

(2*e^3*(e*Sin[c + d*x])^(-3 + m))/(a^2*d*(3 - m)) - (e^3*Cos[c + d*x]*Hypergeometric2F1[-3/2, (-3 + m)/2, (-1
+ m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-3 + m))/(a^2*d*(3 - m)*Sqrt[Cos[c + d*x]^2]) - (e^3*Cos[c + d*x]*Hy
pergeometric2F1[-1/2, (-3 + m)/2, (-1 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-3 + m))/(a^2*d*(3 - m)*Sqrt[C
os[c + d*x]^2]) - (2*e*(e*Sin[c + d*x])^(-1 + m))/(a^2*d*(1 - m))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2952

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

Rule 2954

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(a/g)^(2*m), Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e +
f*x])^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 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 ^2(c+d x) (e \sin (c+d x))^m}{(-a-a \cos (c+d x))^2} \, dx \\ & = \frac {e^4 \int \cos ^2(c+d x) (-a+a \cos (c+d x))^2 (e \sin (c+d x))^{-4+m} \, dx}{a^4} \\ & = \frac {e^4 \int \left (a^2 \cos ^2(c+d x) (e \sin (c+d x))^{-4+m}-2 a^2 \cos ^3(c+d x) (e \sin (c+d x))^{-4+m}+a^2 \cos ^4(c+d x) (e \sin (c+d x))^{-4+m}\right ) \, dx}{a^4} \\ & = \frac {e^4 \int \cos ^2(c+d x) (e \sin (c+d x))^{-4+m} \, dx}{a^2}+\frac {e^4 \int \cos ^4(c+d x) (e \sin (c+d x))^{-4+m} \, dx}{a^2}-\frac {\left (2 e^4\right ) \int \cos ^3(c+d x) (e \sin (c+d x))^{-4+m} \, dx}{a^2} \\ & = -\frac {e^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-3+m}}{a^2 d (3-m) \sqrt {\cos ^2(c+d x)}}-\frac {e^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-3+m}}{a^2 d (3-m) \sqrt {\cos ^2(c+d x)}}-\frac {\left (2 e^3\right ) \text {Subst}\left (\int x^{-4+m} \left (1-\frac {x^2}{e^2}\right ) \, dx,x,e \sin (c+d x)\right )}{a^2 d} \\ & = -\frac {e^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-3+m}}{a^2 d (3-m) \sqrt {\cos ^2(c+d x)}}-\frac {e^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-3+m}}{a^2 d (3-m) \sqrt {\cos ^2(c+d x)}}-\frac {\left (2 e^3\right ) \text {Subst}\left (\int \left (x^{-4+m}-\frac {x^{-2+m}}{e^2}\right ) \, dx,x,e \sin (c+d x)\right )}{a^2 d} \\ & = \frac {2 e^3 (e \sin (c+d x))^{-3+m}}{a^2 d (3-m)}-\frac {e^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-3+m}}{a^2 d (3-m) \sqrt {\cos ^2(c+d x)}}-\frac {e^3 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-3+m}}{a^2 d (3-m) \sqrt {\cos ^2(c+d x)}}-\frac {2 e (e \sin (c+d x))^{-1+m}}{a^2 d (1-m)} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.80 (sec) , antiderivative size = 612, normalized size of antiderivative = 2.96 \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx=-\frac {i 2^{4-m} \left (-i e^{-i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )\right )^m \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\left (-1+e^{2 i (c+d x)}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},1-\frac {m}{2},e^{2 i (c+d x)}\right )}{4 m}+\frac {e^{i (c+d x)} \left (1-e^{2 i (c+d x)}\right )^{-m} \left (\left (6-5 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},2-m,\frac {3-m}{2},e^{2 i (c+d x)}\right )+e^{i (c+d x)} (-1+m) \left (e^{i (c+d x)} (-2+m) \operatorname {Hypergeometric2F1}\left (2-m,\frac {3-m}{2},\frac {5-m}{2},e^{2 i (c+d x)}\right )-2 (-3+m) \operatorname {Hypergeometric2F1}\left (2-m,1-\frac {m}{2},2-\frac {m}{2},e^{2 i (c+d x)}\right )\right )\right )}{(-3+m) (-2+m) (-1+m)}+e^{2 i (c+d x)} \left (1-e^{2 i (c+d x)}\right )^{-m} \left (\frac {4 e^{i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {3-m}{2},4-m,\frac {5-m}{2},e^{2 i (c+d x)}\right )}{-3+m}+\frac {4 e^{3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (4-m,\frac {5-m}{2},\frac {7-m}{2},e^{2 i (c+d x)}\right )}{-5+m}-\frac {\operatorname {Hypergeometric2F1}\left (4-m,1-\frac {m}{2},2-\frac {m}{2},e^{2 i (c+d x)}\right )}{-2+m}-\frac {6 e^{2 i (c+d x)} \operatorname {Hypergeometric2F1}\left (4-m,2-\frac {m}{2},3-\frac {m}{2},e^{2 i (c+d x)}\right )}{-4+m}-\frac {e^{4 i (c+d x)} \operatorname {Hypergeometric2F1}\left (4-m,3-\frac {m}{2},4-\frac {m}{2},e^{2 i (c+d x)}\right )}{-6+m}\right )\right ) \sec ^2(c+d x) \sin ^{-m}(c+d x) (e \sin (c+d x))^m}{a^2 d (1+\sec (c+d x))^2} \]

[In]

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

[Out]

((-I)*2^(4 - m)*(((-I)*(-1 + E^((2*I)*(c + d*x))))/E^(I*(c + d*x)))^m*Cos[(c + d*x)/2]^4*(((-1 + E^((2*I)*(c +
d*x)))*Hypergeometric2F1[1, (2 + m)/2, 1 - m/2, E^((2*I)*(c + d*x))])/(4*m) + (E^(I*(c + d*x))*((6 - 5*m + m^
2)*Hypergeometric2F1[(1 - m)/2, 2 - m, (3 - m)/2, E^((2*I)*(c + d*x))] + E^(I*(c + d*x))*(-1 + m)*(E^(I*(c + d
*x))*(-2 + m)*Hypergeometric2F1[2 - m, (3 - m)/2, (5 - m)/2, E^((2*I)*(c + d*x))] - 2*(-3 + m)*Hypergeometric2
F1[2 - m, 1 - m/2, 2 - m/2, E^((2*I)*(c + d*x))])))/((1 - E^((2*I)*(c + d*x)))^m*(-3 + m)*(-2 + m)*(-1 + m)) +
(E^((2*I)*(c + d*x))*((4*E^(I*(c + d*x))*Hypergeometric2F1[(3 - m)/2, 4 - m, (5 - m)/2, E^((2*I)*(c + d*x))])
/(-3 + m) + (4*E^((3*I)*(c + d*x))*Hypergeometric2F1[4 - m, (5 - m)/2, (7 - m)/2, E^((2*I)*(c + d*x))])/(-5 +
m) - Hypergeometric2F1[4 - m, 1 - m/2, 2 - m/2, E^((2*I)*(c + d*x))]/(-2 + m) - (6*E^((2*I)*(c + d*x))*Hyperge
ometric2F1[4 - m, 2 - m/2, 3 - m/2, E^((2*I)*(c + d*x))])/(-4 + m) - (E^((4*I)*(c + d*x))*Hypergeometric2F1[4
- m, 3 - m/2, 4 - m/2, E^((2*I)*(c + d*x))])/(-6 + m)))/(1 - E^((2*I)*(c + d*x)))^m)*Sec[c + d*x]^2*(e*Sin[c +
d*x])^m)/(a^2*d*(1 + Sec[c + d*x])^2*Sin[c + d*x]^m)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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