3.138 \(\int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx\)
Optimal. Leaf size=207 \[ -\frac {e^3 \cos (c+d x) (e \sin (c+d x))^{m-3} \, _2F_1\left (-\frac {3}{2},\frac {m-3}{2};\frac {m-1}{2};\sin ^2(c+d x)\right )}{a^2 d (3-m) \sqrt {\cos ^2(c+d x)}}-\frac {e^3 \cos (c+d x) (e \sin (c+d x))^{m-3} \, _2F_1\left (-\frac {1}{2},\frac {m-3}{2};\frac {m-1}{2};\sin ^2(c+d x)\right )}{a^2 d (3-m) \sqrt {\cos ^2(c+d x)}}+\frac {2 e^3 (e \sin (c+d x))^{m-3}}{a^2 d (3-m)}-\frac {2 e (e \sin (c+d x))^{m-1}}{a^2 d (1-m)} \]
[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] time = 0.53, antiderivative size = 207, normalized size of antiderivative = 1.00,
number of steps used = 9, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used =
{3872, 2875, 2873, 2577, 2564, 14} \[ -\frac {e^3 \cos (c+d x) (e \sin (c+d x))^{m-3} \, _2F_1\left (-\frac {3}{2},\frac {m-3}{2};\frac {m-1}{2};\sin ^2(c+d x)\right )}{a^2 d (3-m) \sqrt {\cos ^2(c+d x)}}-\frac {e^3 \cos (c+d x) (e \sin (c+d x))^{m-3} \, _2F_1\left (-\frac {1}{2},\frac {m-3}{2};\frac {m-1}{2};\sin ^2(c+d x)\right )}{a^2 d (3-m) \sqrt {\cos ^2(c+d x)}}+\frac {2 e^3 (e \sin (c+d x))^{m-3}}{a^2 d (3-m)}-\frac {2 e (e \sin (c+d x))^{m-1}}{a^2 d (1-m)} \]
Antiderivative was successfully verified.
[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 2564
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 2577
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)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]
Rule 2873
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 2875
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 3872
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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*} \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx &=\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) \, _2F_1\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) \, _2F_1\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 ) \operatorname {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) \, _2F_1\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) \, _2F_1\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 ) \operatorname {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) \, _2F_1\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) \, _2F_1\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] time = 22.70, size = 2833, normalized size = 13.69 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(e*Sin[c + d*x])^m/(a + a*Sec[c + d*x])^2,x]
[Out]
(2^(2 - m)*E^(I*(c + d*x))*(((-I)*(-1 + E^((2*I)*(c + d*x))))/E^(I*(c + d*x)))^(1 + m)*Cos[c/2 + (d*x)/2]^4*Hy
pergeometric2F1[1, (2 + m)/2, 1 - m/2, E^((2*I)*(c + d*x))]*Sec[c + d*x]^2*(e*Sin[c + d*x])^m)/(d*m*(a + a*Sec
[c + d*x])^2*Sin[c + d*x]^m) + ((AppellF1[(1 + m)/2, 1 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)
/4]^2] - 6*AppellF1[(1 + m)/2, 2 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] + 12*AppellF1[(
1 + m)/2, 3 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] - 8*AppellF1[(1 + m)/2, 4 - m, 2*m,
(3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2])*(Sec[(c + d*x)/4]^2)^(2*m)*Sec[(c + d*x)/2]^4*Sin[c + d*x
]^m*(e*Sin[c + d*x])^m*Tan[(c + d*x)/4])/(4*a^2*d*(1 + m)*(1 - Tan[(c + d*x)/4]^2)^m*((m*(AppellF1[(1 + m)/2,
1 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] - 6*AppellF1[(1 + m)/2, 2 - m, 2*m, (3 + m)/2,
Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] + 12*AppellF1[(1 + m)/2, 3 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2,
-Tan[(c + d*x)/4]^2] - 8*AppellF1[(1 + m)/2, 4 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2])*
(Sec[(c + d*x)/4]^2)^(1 + 2*m)*Sin[c + d*x]^m*Tan[(c + d*x)/4]^2*(1 - Tan[(c + d*x)/4]^2)^(-1 - m))/(2*(1 + m)
) + ((AppellF1[(1 + m)/2, 1 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] - 6*AppellF1[(1 + m)
/2, 2 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] + 12*AppellF1[(1 + m)/2, 3 - m, 2*m, (3 +
m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] - 8*AppellF1[(1 + m)/2, 4 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]
^2, -Tan[(c + d*x)/4]^2])*(Sec[(c + d*x)/4]^2)^(1 + 2*m)*Sin[c + d*x]^m)/(4*(1 + m)*(1 - Tan[(c + d*x)/4]^2)^m
) + (m*(AppellF1[(1 + m)/2, 1 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] - 6*AppellF1[(1 +
m)/2, 2 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] + 12*AppellF1[(1 + m)/2, 3 - m, 2*m, (3
+ m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] - 8*AppellF1[(1 + m)/2, 4 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/
4]^2, -Tan[(c + d*x)/4]^2])*Cos[c + d*x]*(Sec[(c + d*x)/4]^2)^(2*m)*Sin[c + d*x]^(-1 + m)*Tan[(c + d*x)/4])/((
1 + m)*(1 - Tan[(c + d*x)/4]^2)^m) + (m*(AppellF1[(1 + m)/2, 1 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(
c + d*x)/4]^2] - 6*AppellF1[(1 + m)/2, 2 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] + 12*Ap
pellF1[(1 + m)/2, 3 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] - 8*AppellF1[(1 + m)/2, 4 -
m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2])*(Sec[(c + d*x)/4]^2)^(2*m)*Sin[c + d*x]^m*Tan[(c
+ d*x)/4]^2)/((1 + m)*(1 - Tan[(c + d*x)/4]^2)^m) + ((Sec[(c + d*x)/4]^2)^(2*m)*Sin[c + d*x]^m*Tan[(c + d*x)/4
]*(-((m*(1 + m)*AppellF1[1 + (1 + m)/2, 1 - m, 1 + 2*m, 1 + (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2
]*Sec[(c + d*x)/4]^2*Tan[(c + d*x)/4])/(3 + m)) + ((1 - m)*(1 + m)*AppellF1[1 + (1 + m)/2, 2 - m, 2*m, 1 + (3
+ m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2]*Sec[(c + d*x)/4]^2*Tan[(c + d*x)/4])/(2*(3 + m)) - 6*(-((m*(1
+ m)*AppellF1[1 + (1 + m)/2, 2 - m, 1 + 2*m, 1 + (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2]*Sec[(c +
d*x)/4]^2*Tan[(c + d*x)/4])/(3 + m)) + ((2 - m)*(1 + m)*AppellF1[1 + (1 + m)/2, 3 - m, 2*m, 1 + (3 + m)/2, Ta
n[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2]*Sec[(c + d*x)/4]^2*Tan[(c + d*x)/4])/(2*(3 + m))) + 12*(-((m*(1 + m)*Ap
pellF1[1 + (1 + m)/2, 3 - m, 1 + 2*m, 1 + (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2]*Sec[(c + d*x)/4]
^2*Tan[(c + d*x)/4])/(3 + m)) + ((3 - m)*(1 + m)*AppellF1[1 + (1 + m)/2, 4 - m, 2*m, 1 + (3 + m)/2, Tan[(c + d
*x)/4]^2, -Tan[(c + d*x)/4]^2]*Sec[(c + d*x)/4]^2*Tan[(c + d*x)/4])/(2*(3 + m))) - 8*(-((m*(1 + m)*AppellF1[1
+ (1 + m)/2, 4 - m, 1 + 2*m, 1 + (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2]*Sec[(c + d*x)/4]^2*Tan[(c
+ d*x)/4])/(3 + m)) + ((4 - m)*(1 + m)*AppellF1[1 + (1 + m)/2, 5 - m, 2*m, 1 + (3 + m)/2, Tan[(c + d*x)/4]^2,
-Tan[(c + d*x)/4]^2]*Sec[(c + d*x)/4]^2*Tan[(c + d*x)/4])/(2*(3 + m)))))/((1 + m)*(1 - Tan[(c + d*x)/4]^2)^m)
)) - (2*(3 + m)*(AppellF1[(1 + m)/2, 1 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] - 2*Appel
lF1[(1 + m)/2, 2 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2])*Sec[(c + d*x)/2]*(e*Sin[c + d*
x])^m*Tan[(c + d*x)/2])/(a^2*d*(1 + m)*((3 + m)*AppellF1[(1 + m)/2, 1 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2,
-Tan[(c + d*x)/4]^2] - 2*((3 + m)*AppellF1[(1 + m)/2, 2 - m, 2*m, (3 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*
x)/4]^2] + (2*m*AppellF1[(3 + m)/2, 1 - m, 1 + 2*m, (5 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] + (-1
+ m)*AppellF1[(3 + m)/2, 2 - m, 2*m, (5 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] - 4*m*AppellF1[(3 + m
)/2, 2 - m, 1 + 2*m, (5 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] + 4*AppellF1[(3 + m)/2, 3 - m, 2*m, (
5 + m)/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] - 2*m*AppellF1[(3 + m)/2, 3 - m, 2*m, (5 + m)/2, Tan[(c + d
*x)/4]^2, -Tan[(c + d*x)/4]^2])*Tan[(c + d*x)/4]^2)))
________________________________________________________________________________________
fricas [F] time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (e \sin \left (d x + c\right )\right )^{m}}{a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
________________________________________________________________________________________
maple [F] time = 1.46, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{\left (a +a \sec \left (d x +c \right )\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[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)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\left (e \sin {\left (c + d x \right )}\right )^{m}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[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
________________________________________________________________________________________