\(\int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx\) [139]

Optimal result

Integrand size = 23, antiderivative size = 236 \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=-\frac {4 e^5 (e \sin (c+d x))^{-5+m}}{a^3 d (5-m)}+\frac {e^5 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt {\cos ^2(c+d x)}}+\frac {3 e^5 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt {\cos ^2(c+d x)}}+\frac {7 e^3 (e \sin (c+d x))^{-3+m}}{a^3 d (3-m)}-\frac {3 e (e \sin (c+d x))^{-1+m}}{a^3 d (1-m)} \]

[Out]

-4*e^5*(e*sin(d*x+c))^(-5+m)/a^3/d/(5-m)+7*e^3*(e*sin(d*x+c))^(-3+m)/a^3/d/(3-m)-3*e*(e*sin(d*x+c))^(-1+m)/a^3
/d/(1-m)+e^5*cos(d*x+c)*hypergeom([-5/2, -5/2+1/2*m],[-3/2+1/2*m],sin(d*x+c)^2)*(e*sin(d*x+c))^(-5+m)/a^3/d/(5
-m)/(cos(d*x+c)^2)^(1/2)+3*e^5*cos(d*x+c)*hypergeom([-3/2, -5/2+1/2*m],[-3/2+1/2*m],sin(d*x+c)^2)*(e*sin(d*x+c
))^(-5+m)/a^3/d/(5-m)/(cos(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3957, 2954, 2952, 2644, 14, 2657, 276} \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\frac {e^5 \cos (c+d x) (e \sin (c+d x))^{m-5} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {m-5}{2},\frac {m-3}{2},\sin ^2(c+d x)\right )}{a^3 d (5-m) \sqrt {\cos ^2(c+d x)}}+\frac {3 e^5 \cos (c+d x) (e \sin (c+d x))^{m-5} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {m-5}{2},\frac {m-3}{2},\sin ^2(c+d x)\right )}{a^3 d (5-m) \sqrt {\cos ^2(c+d x)}}-\frac {4 e^5 (e \sin (c+d x))^{m-5}}{a^3 d (5-m)}+\frac {7 e^3 (e \sin (c+d x))^{m-3}}{a^3 d (3-m)}-\frac {3 e (e \sin (c+d x))^{m-1}}{a^3 d (1-m)} \]

[In]

Int[(e*Sin[c + d*x])^m/(a + a*Sec[c + d*x])^3,x]

[Out]

(-4*e^5*(e*Sin[c + d*x])^(-5 + m))/(a^3*d*(5 - m)) + (e^5*Cos[c + d*x]*Hypergeometric2F1[-5/2, (-5 + m)/2, (-3
+ m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-5 + m))/(a^3*d*(5 - m)*Sqrt[Cos[c + d*x]^2]) + (3*e^5*Cos[c + d*x]
*Hypergeometric2F1[-3/2, (-5 + m)/2, (-3 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-5 + m))/(a^3*d*(5 - m)*Sqr
t[Cos[c + d*x]^2]) + (7*e^3*(e*Sin[c + d*x])^(-3 + m))/(a^3*d*(3 - m)) - (3*e*(e*Sin[c + d*x])^(-1 + m))/(a^3*
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 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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 ^3(c+d x) (e \sin (c+d x))^m}{(-a-a \cos (c+d x))^3} \, dx \\ & = -\frac {e^6 \int \cos ^3(c+d x) (-a+a \cos (c+d x))^3 (e \sin (c+d x))^{-6+m} \, dx}{a^6} \\ & = -\frac {e^6 \int \left (-a^3 \cos ^3(c+d x) (e \sin (c+d x))^{-6+m}+3 a^3 \cos ^4(c+d x) (e \sin (c+d x))^{-6+m}-3 a^3 \cos ^5(c+d x) (e \sin (c+d x))^{-6+m}+a^3 \cos ^6(c+d x) (e \sin (c+d x))^{-6+m}\right ) \, dx}{a^6} \\ & = \frac {e^6 \int \cos ^3(c+d x) (e \sin (c+d x))^{-6+m} \, dx}{a^3}-\frac {e^6 \int \cos ^6(c+d x) (e \sin (c+d x))^{-6+m} \, dx}{a^3}-\frac {\left (3 e^6\right ) \int \cos ^4(c+d x) (e \sin (c+d x))^{-6+m} \, dx}{a^3}+\frac {\left (3 e^6\right ) \int \cos ^5(c+d x) (e \sin (c+d x))^{-6+m} \, dx}{a^3} \\ & = \frac {e^5 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt {\cos ^2(c+d x)}}+\frac {3 e^5 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt {\cos ^2(c+d x)}}+\frac {e^5 \text {Subst}\left (\int x^{-6+m} \left (1-\frac {x^2}{e^2}\right ) \, dx,x,e \sin (c+d x)\right )}{a^3 d}+\frac {\left (3 e^5\right ) \text {Subst}\left (\int x^{-6+m} \left (1-\frac {x^2}{e^2}\right )^2 \, dx,x,e \sin (c+d x)\right )}{a^3 d} \\ & = \frac {e^5 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt {\cos ^2(c+d x)}}+\frac {3 e^5 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt {\cos ^2(c+d x)}}+\frac {e^5 \text {Subst}\left (\int \left (x^{-6+m}-\frac {x^{-4+m}}{e^2}\right ) \, dx,x,e \sin (c+d x)\right )}{a^3 d}+\frac {\left (3 e^5\right ) \text {Subst}\left (\int \left (x^{-6+m}-\frac {2 x^{-4+m}}{e^2}+\frac {x^{-2+m}}{e^4}\right ) \, dx,x,e \sin (c+d x)\right )}{a^3 d} \\ & = -\frac {4 e^5 (e \sin (c+d x))^{-5+m}}{a^3 d (5-m)}+\frac {e^5 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt {\cos ^2(c+d x)}}+\frac {3 e^5 \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{2} (-5+m),\frac {1}{2} (-3+m),\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt {\cos ^2(c+d x)}}+\frac {7 e^3 (e \sin (c+d x))^{-3+m}}{a^3 d (3-m)}-\frac {3 e (e \sin (c+d x))^{-1+m}}{a^3 d (1-m)} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 6.09 (sec) , antiderivative size = 995, normalized size of antiderivative = 4.22 \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, 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 ^6\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 )}{2 m}+\frac {3 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)}+6 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 )-4 e^{3 i (c+d x)} \left (1-e^{2 i (c+d x)}\right )^{-m} \left (-\frac {\operatorname {Hypergeometric2F1}\left (\frac {3-m}{2},6-m,\frac {5-m}{2},e^{2 i (c+d x)}\right )}{-3+m}-\frac {15 e^{2 i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {5-m}{2},6-m,\frac {7-m}{2},e^{2 i (c+d x)}\right )}{-5+m}-\frac {15 e^{4 i (c+d x)} \operatorname {Hypergeometric2F1}\left (6-m,\frac {7-m}{2},\frac {9-m}{2},e^{2 i (c+d x)}\right )}{-7+m}-\frac {e^{6 i (c+d x)} \operatorname {Hypergeometric2F1}\left (6-m,\frac {9-m}{2},\frac {11-m}{2},e^{2 i (c+d x)}\right )}{-9+m}+\frac {6 e^{i (c+d x)} \operatorname {Hypergeometric2F1}\left (6-m,2-\frac {m}{2},3-\frac {m}{2},e^{2 i (c+d x)}\right )}{-4+m}+\frac {20 e^{3 i (c+d x)} \operatorname {Hypergeometric2F1}\left (6-m,3-\frac {m}{2},4-\frac {m}{2},e^{2 i (c+d x)}\right )}{-6+m}+\frac {6 e^{5 i (c+d x)} \operatorname {Hypergeometric2F1}\left (6-m,4-\frac {m}{2},5-\frac {m}{2},e^{2 i (c+d x)}\right )}{-8+m}\right )\right ) \sec ^3(c+d x) \sin ^{-m}(c+d x) (e \sin (c+d x))^m}{a^3 d (1+\sec (c+d x))^3} \]

[In]

Integrate[(e*Sin[c + d*x])^m/(a + a*Sec[c + d*x])^3,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]^6*(((-1 + E^((2*I)*(c +
d*x)))*Hypergeometric2F1[1, (2 + m)/2, 1 - m/2, E^((2*I)*(c + d*x))])/(2*m) + (3*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)*Hypergeometri
c2F1[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))
+ (6*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))*Hyp
ergeometric2F1[4 - m, 2 - m/2, 3 - m/2, E^((2*I)*(c + d*x))])/(-4 + m) - (E^((4*I)*(c + d*x))*Hypergeometric2F
1[4 - m, 3 - m/2, 4 - m/2, E^((2*I)*(c + d*x))])/(-6 + m)))/(1 - E^((2*I)*(c + d*x)))^m - (4*E^((3*I)*(c + d*x
))*(-(Hypergeometric2F1[(3 - m)/2, 6 - m, (5 - m)/2, E^((2*I)*(c + d*x))]/(-3 + m)) - (15*E^((2*I)*(c + d*x))*
Hypergeometric2F1[(5 - m)/2, 6 - m, (7 - m)/2, E^((2*I)*(c + d*x))])/(-5 + m) - (15*E^((4*I)*(c + d*x))*Hyperg
eometric2F1[6 - m, (7 - m)/2, (9 - m)/2, E^((2*I)*(c + d*x))])/(-7 + m) - (E^((6*I)*(c + d*x))*Hypergeometric2
F1[6 - m, (9 - m)/2, (11 - m)/2, E^((2*I)*(c + d*x))])/(-9 + m) + (6*E^(I*(c + d*x))*Hypergeometric2F1[6 - m,
2 - m/2, 3 - m/2, E^((2*I)*(c + d*x))])/(-4 + m) + (20*E^((3*I)*(c + d*x))*Hypergeometric2F1[6 - m, 3 - m/2, 4
- m/2, E^((2*I)*(c + d*x))])/(-6 + m) + (6*E^((5*I)*(c + d*x))*Hypergeometric2F1[6 - m, 4 - m/2, 5 - m/2, E^(
(2*I)*(c + d*x))])/(-8 + m)))/(1 - E^((2*I)*(c + d*x)))^m)*Sec[c + d*x]^3*(e*Sin[c + d*x])^m)/(a^3*d*(1 + Sec[
c + d*x])^3*Sin[c + d*x]^m)

Maple [F]

\[\int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{\left (a +a \sec \left (d x +c \right )\right )^{3}}d x\]

[In]

int((e*sin(d*x+c))^m/(a+a*sec(d*x+c))^3,x)

[Out]

int((e*sin(d*x+c))^m/(a+a*sec(d*x+c))^3,x)

Fricas [F]

\[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate((e*sin(d*x+c))^m/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

integral((e*sin(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)

Sympy [F]

\[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\left (e \sin {\left (c + d x \right )}\right )^{m}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

[In]

integrate((e*sin(d*x+c))**m/(a+a*sec(d*x+c))**3,x)

[Out]

Integral((e*sin(c + d*x))**m/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

Maxima [F]

\[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate((e*sin(d*x+c))^m/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((e*sin(d*x + c))^m/(a*sec(d*x + c) + a)^3, x)

Giac [F]

\[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

integrate((e*sin(d*x+c))^m/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((e*sin(d*x + c))^m/(a*sec(d*x + c) + a)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3\,{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{a^3\,{\left (\cos \left (c+d\,x\right )+1\right )}^3} \,d x \]

[In]

int((e*sin(c + d*x))^m/(a + a/cos(c + d*x))^3,x)

[Out]

int((cos(c + d*x)^3*(e*sin(c + d*x))^m)/(a^3*(cos(c + d*x) + 1)^3), x)