3.262 \(\int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^3} \, dx\)
Optimal. Leaf size=580 \[ -\frac {b^3 e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} F_1\left (3-m;\frac {1-m}{2},\frac {1-m}{2};4-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^4 d (3-m) (a \cos (c+d x)+b)^2}+\frac {3 b^2 e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} F_1\left (2-m;\frac {1-m}{2},\frac {1-m}{2};3-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^4 d (2-m) (a \cos (c+d x)+b)}-\frac {3 b e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} F_1\left (1-m;\frac {1-m}{2},\frac {1-m}{2};2-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^4 d (1-m)}+\frac {\cos (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{a^3 d e (m+1) \sqrt {\cos ^2(c+d x)}} \]
[Out]
-3*b*e*AppellF1(1-m,1/2-1/2*m,1/2-1/2*m,2-m,(-a+b)/(b+a*cos(d*x+c)),(a+b)/(b+a*cos(d*x+c)))*(-a*(1-cos(d*x+c))
/(b+a*cos(d*x+c)))^(1/2-1/2*m)*(a*(1+cos(d*x+c))/(b+a*cos(d*x+c)))^(1/2-1/2*m)*(e*sin(d*x+c))^(-1+m)/a^4/d/(1-
m)-b^3*e*AppellF1(3-m,1/2-1/2*m,1/2-1/2*m,4-m,(-a+b)/(b+a*cos(d*x+c)),(a+b)/(b+a*cos(d*x+c)))*(-a*(1-cos(d*x+c
))/(b+a*cos(d*x+c)))^(1/2-1/2*m)*(a*(1+cos(d*x+c))/(b+a*cos(d*x+c)))^(1/2-1/2*m)*(e*sin(d*x+c))^(-1+m)/a^4/d/(
3-m)/(b+a*cos(d*x+c))^2+3*b^2*e*AppellF1(2-m,1/2-1/2*m,1/2-1/2*m,3-m,(-a+b)/(b+a*cos(d*x+c)),(a+b)/(b+a*cos(d*
x+c)))*(-a*(1-cos(d*x+c))/(b+a*cos(d*x+c)))^(1/2-1/2*m)*(a*(1+cos(d*x+c))/(b+a*cos(d*x+c)))^(1/2-1/2*m)*(e*sin
(d*x+c))^(-1+m)/a^4/d/(2-m)/(b+a*cos(d*x+c))+cos(d*x+c)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],sin(d*x+c)^2)*(
e*sin(d*x+c))^(1+m)/a^3/d/e/(1+m)/(cos(d*x+c)^2)^(1/2)
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Rubi [A] time = 0.59, antiderivative size = 580, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used =
{3872, 2912, 2643, 2703} \[ \frac {3 b^2 e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} F_1\left (2-m;\frac {1-m}{2},\frac {1-m}{2};3-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^4 d (2-m) (a \cos (c+d x)+b)}-\frac {b^3 e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} F_1\left (3-m;\frac {1-m}{2},\frac {1-m}{2};4-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^4 d (3-m) (a \cos (c+d x)+b)^2}-\frac {3 b e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} F_1\left (1-m;\frac {1-m}{2},\frac {1-m}{2};2-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^4 d (1-m)}+\frac {\cos (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{a^3 d e (m+1) \sqrt {\cos ^2(c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[(e*Sin[c + d*x])^m/(a + b*Sec[c + d*x])^3,x]
[Out]
(-3*b*e*AppellF1[1 - m, (1 - m)/2, (1 - m)/2, 2 - m, -((a - b)/(b + a*Cos[c + d*x])), (a + b)/(b + a*Cos[c + d
*x])]*(-((a*(1 - Cos[c + d*x]))/(b + a*Cos[c + d*x])))^((1 - m)/2)*((a*(1 + Cos[c + d*x]))/(b + a*Cos[c + d*x]
))^((1 - m)/2)*(e*Sin[c + d*x])^(-1 + m))/(a^4*d*(1 - m)) - (b^3*e*AppellF1[3 - m, (1 - m)/2, (1 - m)/2, 4 - m
, -((a - b)/(b + a*Cos[c + d*x])), (a + b)/(b + a*Cos[c + d*x])]*(-((a*(1 - Cos[c + d*x]))/(b + a*Cos[c + d*x]
)))^((1 - m)/2)*((a*(1 + Cos[c + d*x]))/(b + a*Cos[c + d*x]))^((1 - m)/2)*(e*Sin[c + d*x])^(-1 + m))/(a^4*d*(3
- m)*(b + a*Cos[c + d*x])^2) + (3*b^2*e*AppellF1[2 - m, (1 - m)/2, (1 - m)/2, 3 - m, -((a - b)/(b + a*Cos[c +
d*x])), (a + b)/(b + a*Cos[c + d*x])]*(-((a*(1 - Cos[c + d*x]))/(b + a*Cos[c + d*x])))^((1 - m)/2)*((a*(1 + C
os[c + d*x]))/(b + a*Cos[c + d*x]))^((1 - m)/2)*(e*Sin[c + d*x])^(-1 + m))/(a^4*d*(2 - m)*(b + a*Cos[c + d*x])
) + (Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(1 + m))/(a^3*
d*e*(1 + m)*Sqrt[Cos[c + d*x]^2])
Rule 2643
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] && !IntegerQ[2*n]
Rule 2703
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*AppellF1[-p - m, (1 - p)/2, (1 - p)/2, 1 - p - m, (a + b)/(
a + b*Sin[e + f*x]), (a - b)/(a + b*Sin[e + f*x])])/(b*f*(m + p)*(-((b*(1 - Sin[e + f*x]))/(a + b*Sin[e + f*x]
)))^((p - 1)/2)*((b*(1 + Sin[e + f*x]))/(a + b*Sin[e + f*x]))^((p - 1)/2)), x] /; FreeQ[{a, b, e, f, g, p}, x]
&& NeQ[a^2 - b^2, 0] && ILtQ[m, 0] && !IGtQ[m + p + 1, 0]
Rule 2912
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] && NeQ[a^2 - b^2, 0] && IntegerQ[m] && (GtQ[m, 0] || IntegerQ[n])
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+b \sec (c+d x))^3} \, dx &=-\int \frac {\cos ^3(c+d x) (e \sin (c+d x))^m}{(-b-a \cos (c+d x))^3} \, dx\\ &=-\int \left (-\frac {(e \sin (c+d x))^m}{a^3}+\frac {b^3 (e \sin (c+d x))^m}{a^3 (b+a \cos (c+d x))^3}-\frac {3 b^2 (e \sin (c+d x))^m}{a^3 (b+a \cos (c+d x))^2}+\frac {3 b (e \sin (c+d x))^m}{a^3 (b+a \cos (c+d x))}\right ) \, dx\\ &=\frac {\int (e \sin (c+d x))^m \, dx}{a^3}-\frac {(3 b) \int \frac {(e \sin (c+d x))^m}{b+a \cos (c+d x)} \, dx}{a^3}+\frac {\left (3 b^2\right ) \int \frac {(e \sin (c+d x))^m}{(b+a \cos (c+d x))^2} \, dx}{a^3}-\frac {b^3 \int \frac {(e \sin (c+d x))^m}{(b+a \cos (c+d x))^3} \, dx}{a^3}\\ &=-\frac {3 b e F_1\left (1-m;\frac {1-m}{2},\frac {1-m}{2};2-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^4 d (1-m)}-\frac {b^3 e F_1\left (3-m;\frac {1-m}{2},\frac {1-m}{2};4-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^4 d (3-m) (b+a \cos (c+d x))^2}+\frac {3 b^2 e F_1\left (2-m;\frac {1-m}{2},\frac {1-m}{2};3-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^4 d (2-m) (b+a \cos (c+d x))}+\frac {\cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{a^3 d e (1+m) \sqrt {\cos ^2(c+d x)}}\\ \end {align*}
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Mathematica [B] time = 17.97, size = 2700, normalized size = 4.66 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(e*Sin[c + d*x])^m/(a + b*Sec[c + d*x])^3,x]
[Out]
(-6*b*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]*(b + a*C
os[c + d*x])^2*Sec[c + d*x]^3*(e*Sin[c + d*x])^m*Tan[(c + d*x)/2])/(a^3*d*(a + b*Sec[c + d*x])^3*(AppellF1[(1
+ m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]*Sec[(c + d*x)/2]^2 + 2*m*A
ppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]*Cot[c + d*x]*Ta
n[(c + d*x)/2] + 2*m*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a
+ b)]*Tan[(c + d*x)/2]^2 - (2*(1 + m)*((-a + b)*AppellF1[(3 + m)/2, m, 2, (5 + m)/2, -Tan[(c + d*x)/2]^2, ((a
- b)*Tan[(c + d*x)/2]^2)/(a + b)] + (a + b)*m*AppellF1[(3 + m)/2, 1 + m, 1, (5 + m)/2, -Tan[(c + d*x)/2]^2, (
(a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]^2)/((a + b)*(3 + m)))) + (6*b^2*((a
+ b)*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a*App
ellF1[(1 + m)/2, m, 2, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*(b + a*Cos[c + d
*x])*Sec[c + d*x]^3*(e*Sin[c + d*x])^m*Tan[(c + d*x)/2])/(a^3*d*(a + b*Sec[c + d*x])^3*(((a + b)*AppellF1[(1 +
m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a*AppellF1[(1 + m)/2, m
, 2, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[(c + d*x)/2]^2 + 2*m*((a + b)*
AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a*AppellF1
[(1 + m)/2, m, 2, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*Cot[c + d*x]*Tan[(c +
d*x)/2] + 2*m*((a + b)*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)
/(a + b)] - 2*a*AppellF1[(1 + m)/2, m, 2, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)
])*Tan[(c + d*x)/2]^2 - (2*(1 + m)*((-a^2 + b^2)*AppellF1[(3 + m)/2, m, 2, (5 + m)/2, -Tan[(c + d*x)/2]^2, ((a
- b)*Tan[(c + d*x)/2]^2)/(a + b)] + 4*a*(a - b)*AppellF1[(3 + m)/2, m, 3, (5 + m)/2, -Tan[(c + d*x)/2]^2, ((a
- b)*Tan[(c + d*x)/2]^2)/(a + b)] + (a + b)*m*((a + b)*AppellF1[(3 + m)/2, 1 + m, 1, (5 + m)/2, -Tan[(c + d*x
)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a*AppellF1[(3 + m)/2, 1 + m, 2, (5 + m)/2, -Tan[(c + d*x)/2]
^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]))*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]^2)/((a + b)*(3 + m)))) - (2*b^
3*((a + b)^2*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] -
4*a*(a + b)*AppellF1[(1 + m)/2, m, 2, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] +
4*a^2*AppellF1[(1 + m)/2, m, 3, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[c
+ d*x]^3*(e*Sin[c + d*x])^m*Tan[(c + d*x)/2])/(a^3*d*(a + b*Sec[c + d*x])^3*(((a + b)^2*AppellF1[(1 + m)/2, m,
1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] - 4*a*(a + b)*AppellF1[(1 + m)/2, m,
2, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] + 4*a^2*AppellF1[(1 + m)/2, m, 3, (3
+ m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[(c + d*x)/2]^2 + 2*m*((a + b)^2*Appel
lF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] - 4*a*(a + b)*Appel
lF1[(1 + m)/2, m, 2, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] + 4*a^2*AppellF1[(1
+ m)/2, m, 3, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*Cot[c + d*x]*Tan[(c + d*
x)/2] + 2*m*((a + b)^2*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/
(a + b)] - 4*a*(a + b)*AppellF1[(1 + m)/2, m, 2, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/
(a + b)] + 4*a^2*AppellF1[(1 + m)/2, m, 3, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b
)])*Tan[(c + d*x)/2]^2 + (2*(1 + m)*((a + b)^2*((a - b)*AppellF1[(3 + m)/2, m, 2, (5 + m)/2, -Tan[(c + d*x)/2]
^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] - (a + b)*m*AppellF1[(3 + m)/2, 1 + m, 1, (5 + m)/2, -Tan[(c + d*x)/
2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]) - 4*a*(a + b)*(2*(a - b)*AppellF1[(3 + m)/2, m, 3, (5 + m)/2, -Ta
n[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] - (a + b)*m*AppellF1[(3 + m)/2, 1 + m, 2, (5 + m)/2, -
Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]) + 4*a^2*(3*(a - b)*AppellF1[(3 + m)/2, m, 4, (5 + m
)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] - (a + b)*m*AppellF1[(3 + m)/2, 1 + m, 3, (5 +
m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]))*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]^2)/((a
+ b)*(3 + m)))) - ((b + a*Cos[c + d*x])^3*Hypergeometric2F1[1/2, (1 - m)/2, 3/2, Cos[c + d*x]^2]*Sec[c + d*x]
*(e*Sin[c + d*x])^m*(Sin[c + d*x]^2)^((-1 - m)/2)*Tan[c + d*x])/(a^3*d*(a + b*Sec[c + d*x])^3)
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fricas [F] time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (e \sin \left (d x + c\right )\right )^{m}}{b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*sin(d*x+c))^m/(a+b*sec(d*x+c))^3,x, algorithm="fricas")
[Out]
integral((e*sin(d*x + c))^m/(b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*x + c) + a^3), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*sin(d*x+c))^m/(a+b*sec(d*x+c))^3,x, algorithm="giac")
[Out]
integrate((e*sin(d*x + c))^m/(b*sec(d*x + c) + a)^3, x)
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maple [F] time = 1.22, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{\left (a +b \sec \left (d x +c \right )\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*sin(d*x+c))^m/(a+b*sec(d*x+c))^3,x)
[Out]
int((e*sin(d*x+c))^m/(a+b*sec(d*x+c))^3,x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*sin(d*x+c))^m/(a+b*sec(d*x+c))^3,x, algorithm="maxima")
[Out]
integrate((e*sin(d*x + c))^m/(b*sec(d*x + c) + a)^3, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^3\,{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{{\left (b+a\,\cos \left (c+d\,x\right )\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*sin(c + d*x))^m/(a + b/cos(c + d*x))^3,x)
[Out]
int((cos(c + d*x)^3*(e*sin(c + d*x))^m)/(b + a*cos(c + d*x))^3, x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sin {\left (c + d x \right )}\right )^{m}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*sin(d*x+c))**m/(a+b*sec(d*x+c))**3,x)
[Out]
Integral((e*sin(c + d*x))**m/(a + b*sec(c + d*x))**3, x)
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