3.5 \(\int \frac {(a+a \sin (e+f x))^m (A+C \sin ^2(e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=202 \[ \frac {(A (1-2 m)-C (2 m+7)) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1)\right )}{4 c f (2 m+1) \sqrt {c-c \sin (e+f x)}}+\frac {(2 A m+A+C (2 m+9)) \cos (e+f x) (a \sin (e+f x)+a)^m}{4 c f (2 m+1) \sqrt {c-c \sin (e+f x)}}+\frac {(A+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{4 a f (c-c \sin (e+f x))^{3/2}} \]
[Out]
1/4*(A+C)*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)/a/f/(c-c*sin(f*x+e))^(3/2)+1/4*(A+2*A*m+C*(9+2*m))*cos(f*x+e)*(a+a
*sin(f*x+e))^m/c/f/(1+2*m)/(c-c*sin(f*x+e))^(1/2)+1/4*(A*(1-2*m)-C*(7+2*m))*cos(f*x+e)*hypergeom([1, 1/2+m],[3
/2+m],1/2+1/2*sin(f*x+e))*(a+a*sin(f*x+e))^m/c/f/(1+2*m)/(c-c*sin(f*x+e))^(1/2)
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Rubi [A] time = 0.59, antiderivative size = 202, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used =
{3036, 2973, 2745, 2667, 68} \[ \frac {(A (1-2 m)-C (2 m+7)) \cos (e+f x) (a \sin (e+f x)+a)^m \, _2F_1\left (1,m+\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1)\right )}{4 c f (2 m+1) \sqrt {c-c \sin (e+f x)}}+\frac {(2 A m+A+C (2 m+9)) \cos (e+f x) (a \sin (e+f x)+a)^m}{4 c f (2 m+1) \sqrt {c-c \sin (e+f x)}}+\frac {(A+C) \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{4 a f (c-c \sin (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[((a + a*Sin[e + f*x])^m*(A + C*Sin[e + f*x]^2))/(c - c*Sin[e + f*x])^(3/2),x]
[Out]
((A + C)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(4*a*f*(c - c*Sin[e + f*x])^(3/2)) + ((A + 2*A*m + C*(9 +
2*m))*Cos[e + f*x]*(a + a*Sin[e + f*x])^m)/(4*c*f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]]) + ((A*(1 - 2*m) - C*(7 +
2*m))*Cos[e + f*x]*Hypergeometric2F1[1, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2]*(a + a*Sin[e + f*x])^m)/(4*c*
f*(1 + 2*m)*Sqrt[c - c*Sin[e + f*x]])
Rule 68
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[m] && IntegerQ[n]
Rule 2667
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/2])
Rule 2745
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[(a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*(c + d*Sin[e + f*x])^FracPart[m])/Cos[e + f*x]^(2
*FracPart[m]), Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, m, n},
x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] || !FractionQ[n])
Rule 2973
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(B*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(f*(
m + n + 1)), x] - Dist[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&
!LtQ[m, -2^(-1)] && NeQ[m + n + 1, 0]
Rule 3036
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((a*A + a*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f
*x])^(n + 1))/(2*b*c*f*(2*m + 1)), x] - Dist[1/(2*b*c*d*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Si
n[e + f*x])^n*Simp[A*(c^2*(m + 1) + d^2*(2*m + n + 2)) - C*(c^2*m - d^2*(n + 1)) + d*(A*c*(m + n + 2) - c*C*(3
*m - n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2
- b^2, 0] && (LtQ[m, -2^(-1)] || (EqQ[m + n + 2, 0] && NeQ[2*m + 1, 0]))
Rubi steps
\begin {align*} \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c-c \sin (e+f x))^{3/2}} \, dx &=\frac {(A+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{4 a f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {(a+a \sin (e+f x))^m \left (-\frac {1}{2} a^2 (A (3-2 m)-C (5+2 m))+\frac {1}{2} a^2 (A+2 A m+C (9+2 m)) \sin (e+f x)\right )}{\sqrt {c-c \sin (e+f x)}} \, dx}{4 a^2 c}\\ &=\frac {(A+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{4 a f (c-c \sin (e+f x))^{3/2}}+\frac {(A+2 A m+C (9+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{4 c f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {(A (1-2 m)-C (7+2 m)) \int \frac {(a+a \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx}{4 c}\\ &=\frac {(A+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{4 a f (c-c \sin (e+f x))^{3/2}}+\frac {(A+2 A m+C (9+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{4 c f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {((A (1-2 m)-C (7+2 m)) \cos (e+f x)) \int \sec (e+f x) (a+a \sin (e+f x))^{\frac {1}{2}+m} \, dx}{4 c \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {(A+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{4 a f (c-c \sin (e+f x))^{3/2}}+\frac {(A+2 A m+C (9+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{4 c f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {(a (A (1-2 m)-C (7+2 m)) \cos (e+f x)) \operatorname {Subst}\left (\int \frac {(a+x)^{-\frac {1}{2}+m}}{a-x} \, dx,x,a \sin (e+f x)\right )}{4 c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {(A+C) \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{4 a f (c-c \sin (e+f x))^{3/2}}+\frac {(A+2 A m+C (9+2 m)) \cos (e+f x) (a+a \sin (e+f x))^m}{4 c f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {(A (1-2 m)-C (7+2 m)) \cos (e+f x) \, _2F_1\left (1,\frac {1}{2}+m;\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{4 c f (1+2 m) \sqrt {c-c \sin (e+f x)}}\\ \end {align*}
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Mathematica [C] time = 7.13, size = 4061, normalized size = 20.10 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((a + a*Sin[e + f*x])^m*(A + C*Sin[e + f*x]^2))/(c - c*Sin[e + f*x])^(3/2),x]
[Out]
(2^(-1/2 - 2*m)*C*(-(4^m*Hypergeometric2F1[1, 2*m, 1 + 2*m, Cos[(-e + Pi/2 - f*x)/2]]) + Hypergeometric2F1[2*m
, 2*m, 1 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2]*(Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m))*(Cos[(e + f*x)/2] - Si
n[(e + f*x)/2])^3*(a + a*Sin[e + f*x])^m)/(f*m*(c - c*Sin[e + f*x])^(3/2)) - (C*(((-1/2*I)*(((-I)*2^(1 - 2*m)*
((1 + E^(I*(-e + Pi/2 - f*x)))/E^((I/2)*(-e + Pi/2 - f*x)))^(1 + 2*m)*Hypergeometric2F1[1, 1/2 + m, 1/2 - m, -
E^(I*(-e + Pi/2 - f*x))])/(1 + 2*m) + (I*2^(1 - 2*m)*(1 + E^(I*(-e + Pi/2 - f*x)))^2*((1 + E^(I*(-e + Pi/2 - f
*x)))/E^((I/2)*(-e + Pi/2 - f*x)))^(-1 + 2*m)*Hypergeometric2F1[1, 3/2 + m, 3/2 - m, -E^(I*(-e + Pi/2 - f*x))]
)/(-1 + 2*m)))/Sqrt[2] - (Sqrt[2]*Cos[(-e + Pi/2 - f*x)/2]^(2 + 2*m)*Hypergeometric2F1[1/2, (2 + 2*m)/2, (4 +
2*m)/2, Cos[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2])/((2 + 2*m)*Sqrt[Sin[(-e + Pi/2 - f*x)/2]^2]))*(C
os[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(a + a*Sin[e + f*x])^m)/(f*Cos[(-e + Pi/2 - f*x)/2]^(2*m)*(c - c*Sin[e +
f*x])^(3/2)) + (C*(((I/2)*(((-I)*2^(1 - 2*m)*((1 + E^(I*(-e + Pi/2 - f*x)))/E^((I/2)*(-e + Pi/2 - f*x)))^(1 +
2*m)*Hypergeometric2F1[1, 1/2 + m, 1/2 - m, -E^(I*(-e + Pi/2 - f*x))])/(1 + 2*m) + (I*2^(1 - 2*m)*(1 + E^(I*(
-e + Pi/2 - f*x)))^2*((1 + E^(I*(-e + Pi/2 - f*x)))/E^((I/2)*(-e + Pi/2 - f*x)))^(-1 + 2*m)*Hypergeometric2F1[
1, 3/2 + m, 3/2 - m, -E^(I*(-e + Pi/2 - f*x))])/(-1 + 2*m)))/Sqrt[2] - (Sqrt[2]*Cos[(-e + Pi/2 - f*x)/2]^(2 +
2*m)*Hypergeometric2F1[1/2, (2 + 2*m)/2, (4 + 2*m)/2, Cos[(-e + Pi/2 - f*x)/2]^2]*Sin[(-e + Pi/2 - f*x)/2])/((
2 + 2*m)*Sqrt[Sin[(-e + Pi/2 - f*x)/2]^2]))*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(a + a*Sin[e + f*x])^m)/(f
*Cos[(-e + Pi/2 - f*x)/2]^(2*m)*(c - c*Sin[e + f*x])^(3/2)) - ((A + C)*(Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(Cos
[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(a + a*Sin[e + f*x])^m*(AppellF1[1, -2*m, 2*m, 2, Tan[(-e + Pi/2 - f*x)/4]
^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Tan[(-e + Pi/2 - f*x)/4]^2 - (AppellF1[1,
-2*m, 2*m, 2, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^2*(Csc[(-e + P
i/2 - f*x)/4]^2)^(2*m)*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^(2*m) + (2^(1
- 2*m)*AppellF1[1 + 2*m, 2*m, 1, 2 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2, 1 - Tan[(-e + Pi/2 - f*x)/4]^2]*
(-1 + Tan[(-e + Pi/2 - f*x)/4]^2)*(1 - Tan[(-e + Pi/2 - f*x)/4]^4)^(2*m))/(1 + 2*m)))/(8*Sqrt[2]*f*(c - c*Sin[
e + f*x])^(3/2)*(Cos[Pi/4 + (e - Pi/2 + f*x)/2] - Sin[Pi/4 + (e - Pi/2 + f*x)/2])^3*(-1/8*(m*Cos[(-e + Pi/2 -
f*x)/4]*(Cos[(-e + Pi/2 - f*x)/4]^2)^(-1 + 2*m)*Sin[(-e + Pi/2 - f*x)/4]*(AppellF1[1, -2*m, 2*m, 2, Tan[(-e +
Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Tan[(-e + Pi/2 - f*x)/4]^2 -
(AppellF1[1, -2*m, 2*m, 2, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^
2*(Csc[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^
(2*m) + (2^(1 - 2*m)*AppellF1[1 + 2*m, 2*m, 1, 2 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2, 1 - Tan[(-e + Pi/2
- f*x)/4]^2]*(-1 + Tan[(-e + Pi/2 - f*x)/4]^2)*(1 - Tan[(-e + Pi/2 - f*x)/4]^4)^(2*m))/(1 + 2*m)))/Sqrt[2] +
((Cos[(-e + Pi/2 - f*x)/4]^2)^(2*m)*((AppellF1[1, -2*m, 2*m, 2, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 -
f*x)/4]^2]*(Sec[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)*Tan[(-e + Pi/2 - f*x)/4])/2 + m*AppellF1[1, -2*m, 2*m, 2, Ta
n[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*(Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Tan[(-e + Pi/2 - f*x
)/4]^3 + (Sec[(-e + Pi/2 - f*x)/4]^2)^(2*m)*Tan[(-e + Pi/2 - f*x)/4]^2*(-1/2*(m*AppellF1[2, 1 - 2*m, 2*m, 3, T
an[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]) -
(m*AppellF1[2, -2*m, 1 + 2*m, 3, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*
x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/2) + (m*AppellF1[1, -2*m, 2*m, 2, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/
2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^3*(Csc[(-e + Pi/2 - f*x)/4]^2)^(2*m)*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^
(2*m))/(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^(2*m) + m*AppellF1[1, -2*m, 2*m, 2, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(
-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]^3*(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^(-1 - 2*m)*(Csc[(-e + Pi/2
- f*x)/4]^2)^(1 + 2*m)*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(2*m) + (AppellF1[1, -2*m, 2*m, 2, Cot[(-e + Pi/2 - f*
x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]*(Csc[(-e + Pi/2 - f*x)/4]^2)^(1 + 2*m)*(1 - Tan
[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(2*(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^(2*m)) - (Cot[(-e + Pi/2 - f*x)/4]^2*(Csc[
(-e + Pi/2 - f*x)/4]^2)^(2*m)*((m*AppellF1[2, 1 - 2*m, 2*m, 3, Cot[(-e + Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f
*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]*Csc[(-e + Pi/2 - f*x)/4]^2)/2 + (m*AppellF1[2, -2*m, 1 + 2*m, 3, Cot[(-e +
Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Cot[(-e + Pi/2 - f*x)/4]*Csc[(-e + Pi/2 - f*x)/4]^2)/2)*(1 - Ta
n[(-e + Pi/2 - f*x)/4]^2)^(2*m))/(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^(2*m) + (m*AppellF1[1, -2*m, 2*m, 2, Cot[(-e
+ Pi/2 - f*x)/4]^2, -Cot[(-e + Pi/2 - f*x)/4]^2]*Csc[(-e + Pi/2 - f*x)/4]*(Csc[(-e + Pi/2 - f*x)/4]^2)^(2*m)*
Sec[(-e + Pi/2 - f*x)/4]*(1 - Tan[(-e + Pi/2 - f*x)/4]^2)^(-1 + 2*m))/(1 - Cot[(-e + Pi/2 - f*x)/4]^2)^(2*m) +
(AppellF1[1 + 2*m, 2*m, 1, 2 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2, 1 - Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(
-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]*(1 - Tan[(-e + Pi/2 - f*x)/4]^4)^(2*m))/(2^(2*m)*(1 + 2*m)) + (
2^(1 - 2*m)*(-1/2*((1 + 2*m)*AppellF1[2 + 2*m, 2*m, 2, 3 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2, 1 - Tan[(-
e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/(2 + 2*m) - (m*(1 + 2*m)*AppellF1[2
+ 2*m, 1 + 2*m, 1, 3 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]^2)/2, 1 - Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/
2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4])/(2*(2 + 2*m)))*(-1 + Tan[(-e + Pi/2 - f*x)/4]^2)*(1 - Tan[(-e + Pi/2 -
f*x)/4]^4)^(2*m))/(1 + 2*m) - (2^(2 - 2*m)*m*AppellF1[1 + 2*m, 2*m, 1, 2 + 2*m, (1 - Tan[(-e + Pi/2 - f*x)/4]
^2)/2, 1 - Tan[(-e + Pi/2 - f*x)/4]^2]*Sec[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 - f*x)/4]^3*(-1 + Tan[(-e + P
i/2 - f*x)/4]^2)*(1 - Tan[(-e + Pi/2 - f*x)/4]^4)^(-1 + 2*m))/(1 + 2*m)))/(8*Sqrt[2])))
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (f x + e\right )^{2} - A - C\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c^{2} \sin \left (f x + e\right ) - 2 \, c^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")
[Out]
integral((C*cos(f*x + e)^2 - A - C)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(c^2*cos(f*x + e)^2 + 2*c
^2*sin(f*x + e) - 2*c^2), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")
[Out]
integrate((C*sin(f*x + e)^2 + A)*(a*sin(f*x + e) + a)^m/(-c*sin(f*x + e) + c)^(3/2), x)
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maple [F] time = 3.30, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +C \left (\sin ^{2}\left (f x +e \right )\right )\right )}{\left (c -c \sin \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(3/2),x)
[Out]
int((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(3/2),x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")
[Out]
integrate((C*sin(f*x + e)^2 + A)*(a*sin(f*x + e) + a)^m/(-c*sin(f*x + e) + c)^(3/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (C\,{\sin \left (e+f\,x\right )}^2+A\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + C*sin(e + f*x)^2)*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(3/2),x)
[Out]
int(((A + C*sin(e + f*x)^2)*(a + a*sin(e + f*x))^m)/(c - c*sin(e + f*x))^(3/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + C \sin ^{2}{\left (e + f x \right )}\right )}{\left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))**m*(A+C*sin(f*x+e)**2)/(c-c*sin(f*x+e))**(3/2),x)
[Out]
Integral((a*(sin(e + f*x) + 1))**m*(A + C*sin(e + f*x)**2)/(-c*(sin(e + f*x) - 1))**(3/2), x)
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