3.61 \(\int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^2} \, dx\)
Optimal. Leaf size=180 \[ \frac{35 c^4 (A-2 B) \cos ^3(e+f x)}{3 a^2 f}-\frac{a^4 c^4 (A-B) \cos ^9(e+f x)}{3 f (a \sin (e+f x)+a)^6}+\frac{2 a^2 c^4 (A-2 B) \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^4}+\frac{35 c^4 (A-2 B) \sin (e+f x) \cos (e+f x)}{2 a^2 f}+\frac{35 c^4 x (A-2 B)}{2 a^2}+\frac{14 c^4 (A-2 B) \cos ^5(e+f x)}{f (a \sin (e+f x)+a)^2} \]
[Out]
(35*(A - 2*B)*c^4*x)/(2*a^2) + (35*(A - 2*B)*c^4*Cos[e + f*x]^3)/(3*a^2*f) + (35*(A - 2*B)*c^4*Cos[e + f*x]*Si
n[e + f*x])/(2*a^2*f) - (a^4*(A - B)*c^4*Cos[e + f*x]^9)/(3*f*(a + a*Sin[e + f*x])^6) + (2*a^2*(A - 2*B)*c^4*C
os[e + f*x]^7)/(f*(a + a*Sin[e + f*x])^4) + (14*(A - 2*B)*c^4*Cos[e + f*x]^5)/(f*(a + a*Sin[e + f*x])^2)
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Rubi [A] time = 0.36157, antiderivative size = 180, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used =
{2967, 2859, 2680, 2682, 2635, 8} \[ \frac{35 c^4 (A-2 B) \cos ^3(e+f x)}{3 a^2 f}-\frac{a^4 c^4 (A-B) \cos ^9(e+f x)}{3 f (a \sin (e+f x)+a)^6}+\frac{2 a^2 c^4 (A-2 B) \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^4}+\frac{35 c^4 (A-2 B) \sin (e+f x) \cos (e+f x)}{2 a^2 f}+\frac{35 c^4 x (A-2 B)}{2 a^2}+\frac{14 c^4 (A-2 B) \cos ^5(e+f x)}{f (a \sin (e+f x)+a)^2} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^4)/(a + a*Sin[e + f*x])^2,x]
[Out]
(35*(A - 2*B)*c^4*x)/(2*a^2) + (35*(A - 2*B)*c^4*Cos[e + f*x]^3)/(3*a^2*f) + (35*(A - 2*B)*c^4*Cos[e + f*x]*Si
n[e + f*x])/(2*a^2*f) - (a^4*(A - B)*c^4*Cos[e + f*x]^9)/(3*f*(a + a*Sin[e + f*x])^6) + (2*a^2*(A - 2*B)*c^4*C
os[e + f*x]^7)/(f*(a + a*Sin[e + f*x])^4) + (14*(A - 2*B)*c^4*Cos[e + f*x]^5)/(f*(a + a*Sin[e + f*x])^2)
Rule 2967
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] && !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Rule 2859
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]
Rule 2680
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] && !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]
Rule 2682
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e
+ f*x])^(p - 1))/(b*f*(p - 1)), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]
Rule 2635
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rubi steps
\begin{align*} \int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^2} \, dx &=\left (a^4 c^4\right ) \int \frac{\cos ^8(e+f x) (A+B \sin (e+f x))}{(a+a \sin (e+f x))^6} \, dx\\ &=-\frac{a^4 (A-B) c^4 \cos ^9(e+f x)}{3 f (a+a \sin (e+f x))^6}-\left (a^3 (A-2 B) c^4\right ) \int \frac{\cos ^8(e+f x)}{(a+a \sin (e+f x))^5} \, dx\\ &=-\frac{a^4 (A-B) c^4 \cos ^9(e+f x)}{3 f (a+a \sin (e+f x))^6}+\frac{2 a^2 (A-2 B) c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}+\left (7 a (A-2 B) c^4\right ) \int \frac{\cos ^6(e+f x)}{(a+a \sin (e+f x))^3} \, dx\\ &=-\frac{a^4 (A-B) c^4 \cos ^9(e+f x)}{3 f (a+a \sin (e+f x))^6}+\frac{2 a^2 (A-2 B) c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}+\frac{14 (A-2 B) c^4 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^2}+\frac{\left (35 (A-2 B) c^4\right ) \int \frac{\cos ^4(e+f x)}{a+a \sin (e+f x)} \, dx}{a}\\ &=\frac{35 (A-2 B) c^4 \cos ^3(e+f x)}{3 a^2 f}-\frac{a^4 (A-B) c^4 \cos ^9(e+f x)}{3 f (a+a \sin (e+f x))^6}+\frac{2 a^2 (A-2 B) c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}+\frac{14 (A-2 B) c^4 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^2}+\frac{\left (35 (A-2 B) c^4\right ) \int \cos ^2(e+f x) \, dx}{a^2}\\ &=\frac{35 (A-2 B) c^4 \cos ^3(e+f x)}{3 a^2 f}+\frac{35 (A-2 B) c^4 \cos (e+f x) \sin (e+f x)}{2 a^2 f}-\frac{a^4 (A-B) c^4 \cos ^9(e+f x)}{3 f (a+a \sin (e+f x))^6}+\frac{2 a^2 (A-2 B) c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}+\frac{14 (A-2 B) c^4 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^2}+\frac{\left (35 (A-2 B) c^4\right ) \int 1 \, dx}{2 a^2}\\ &=\frac{35 (A-2 B) c^4 x}{2 a^2}+\frac{35 (A-2 B) c^4 \cos ^3(e+f x)}{3 a^2 f}+\frac{35 (A-2 B) c^4 \cos (e+f x) \sin (e+f x)}{2 a^2 f}-\frac{a^4 (A-B) c^4 \cos ^9(e+f x)}{3 f (a+a \sin (e+f x))^6}+\frac{2 a^2 (A-2 B) c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^4}+\frac{14 (A-2 B) c^4 \cos ^5(e+f x)}{f (a+a \sin (e+f x))^2}\\ \end{align*}
Mathematica [A] time = 1.25873, size = 311, normalized size = 1.73 \[ \frac{(c-c \sin (e+f x))^4 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (128 (A-B) \sin \left (\frac{1}{2} (e+f x)\right )+210 (A-2 B) (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+3 (24 A-71 B) \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3-3 (A-6 B) \sin (2 (e+f x)) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3-128 (5 A-8 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-64 (A-B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+B \cos (3 (e+f x)) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3\right )}{12 a^2 f (\sin (e+f x)+1)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^8} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^4)/(a + a*Sin[e + f*x])^2,x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c - c*Sin[e + f*x])^4*(128*(A - B)*Sin[(e + f*x)/2] - 64*(A - B)*(Cos[
(e + f*x)/2] + Sin[(e + f*x)/2]) - 128*(5*A - 8*B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 +
210*(A - 2*B)*(e + f*x)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 3*(24*A - 71*B)*Cos[e + f*x]*(Cos[(e + f*x)/
2] + Sin[(e + f*x)/2])^3 + B*Cos[3*(e + f*x)]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 - 3*(A - 6*B)*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2])^3*Sin[2*(e + f*x)]))/(12*a^2*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^8*(1 + Sin[e
+ f*x])^2)
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Maple [B] time = 0.148, size = 549, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^2,x)
[Out]
1/f*c^4/a^2/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^5*A-6/f*c^4/a^2/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f
*x+1/2*e)^5*B+12/f*c^4/a^2/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^4*A-34/f*c^4/a^2/(1+tan(1/2*f*x+1/2*e
)^2)^3*tan(1/2*f*x+1/2*e)^4*B+24/f*c^4/a^2/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^2*A-72/f*c^4/a^2/(1+t
an(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)^2*B-1/f*c^4/a^2/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)*A+6/f*
c^4/a^2/(1+tan(1/2*f*x+1/2*e)^2)^3*tan(1/2*f*x+1/2*e)*B+12/f*c^4/a^2/(1+tan(1/2*f*x+1/2*e)^2)^3*A-106/3/f*c^4/
a^2/(1+tan(1/2*f*x+1/2*e)^2)^3*B+35/f*c^4/a^2*arctan(tan(1/2*f*x+1/2*e))*A-70/f*c^4/a^2*arctan(tan(1/2*f*x+1/2
*e))*B+32/f*c^4/a^2/(tan(1/2*f*x+1/2*e)+1)^2*A-32/f*c^4/a^2/(tan(1/2*f*x+1/2*e)+1)^2*B+32/f*c^4/a^2/(tan(1/2*f
*x+1/2*e)+1)*A-64/f*c^4/a^2/(tan(1/2*f*x+1/2*e)+1)*B-64/3/f*c^4/a^2/(tan(1/2*f*x+1/2*e)+1)^3*A+64/3/f*c^4/a^2/
(tan(1/2*f*x+1/2*e)+1)^3*B
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Maxima [B] time = 1.65923, size = 2827, normalized size = 15.71 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^2,x, algorithm="maxima")
[Out]
1/3*(A*c^4*((75*sin(f*x + e)/(cos(f*x + e) + 1) + 97*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 126*sin(f*x + e)^3/
(cos(f*x + e) + 1)^3 + 98*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 63*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 21*si
n(f*x + e)^6/(cos(f*x + e) + 1)^6 + 32)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 5*a^2*sin(f*x + e)^2/(c
os(f*x + e) + 1)^2 + 7*a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 7*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 5
*a^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 3*a^2*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a^2*sin(f*x + e)^7/(cos
(f*x + e) + 1)^7) + 21*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) - 4*B*c^4*((75*sin(f*x + e)/(cos(f*x + e)
+ 1) + 97*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 126*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 98*sin(f*x + e)^4/(c
os(f*x + e) + 1)^4 + 63*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 21*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 32)/(a^
2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 5*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 7*a^2*sin(f*x + e)^3/(
cos(f*x + e) + 1)^3 + 7*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 5*a^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 +
3*a^2*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a^2*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + 21*arctan(sin(f*x + e)/
(cos(f*x + e) + 1))/a^2) - 2*B*c^4*((57*sin(f*x + e)/(cos(f*x + e) + 1) + 99*sin(f*x + e)^2/(cos(f*x + e) + 1)
^2 + 155*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 153*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 135*sin(f*x + e)^5/(c
os(f*x + e) + 1)^5 + 85*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 45*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 15*sin(
f*x + e)^8/(cos(f*x + e) + 1)^8 + 24)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 6*a^2*sin(f*x + e)^2/(cos
(f*x + e) + 1)^2 + 10*a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 12*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 1
2*a^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 10*a^2*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 6*a^2*sin(f*x + e)^7/
(cos(f*x + e) + 1)^7 + 3*a^2*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + a^2*sin(f*x + e)^9/(cos(f*x + e) + 1)^9) +
15*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) + 16*A*c^4*((12*sin(f*x + e)/(cos(f*x + e) + 1) + 11*sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 + 9*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 5
)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 4*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4*a^2*sin(f*x + e
)^3/(cos(f*x + e) + 1)^3 + 3*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5
) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) - 24*B*c^4*((12*sin(f*x + e)/(cos(f*x + e) + 1) + 11*sin(f*
x + e)^2/(cos(f*x + e) + 1)^2 + 9*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4
+ 5)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 4*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4*a^2*sin(f*x
+ e)^3/(cos(f*x + e) + 1)^3 + 3*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^2*sin(f*x + e)^5/(cos(f*x + e) + 1
)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) + 12*A*c^4*((9*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f
*x + e)^2/(cos(f*x + e) + 1)^2 + 4)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f
*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) - 8
*B*c^4*((9*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4)/(a^2 + 3*a^2*sin(f*x +
e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)
+ 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) - 2*A*c^4*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)
^2/(cos(f*x + e) + 1)^2 + 2)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 8*A*c^4*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 1)/(a^2 + 3*
a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x
+ e) + 1)^3) - 2*B*c^4*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 1)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) +
3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3))/f
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Fricas [A] time = 1.82285, size = 786, normalized size = 4.37 \begin{align*} \frac{2 \, B c^{4} \cos \left (f x + e\right )^{5} -{\left (3 \, A - 16 \, B\right )} c^{4} \cos \left (f x + e\right )^{4} + 2 \,{\left (15 \, A - 38 \, B\right )} c^{4} \cos \left (f x + e\right )^{3} - 210 \,{\left (A - 2 \, B\right )} c^{4} f x + 32 \,{\left (A - B\right )} c^{4} +{\left (105 \,{\left (A - 2 \, B\right )} c^{4} f x -{\left (193 \, A - 346 \, B\right )} c^{4}\right )} \cos \left (f x + e\right )^{2} -{\left (105 \,{\left (A - 2 \, B\right )} c^{4} f x + 2 \,{\left (97 \, A - 202 \, B\right )} c^{4}\right )} \cos \left (f x + e\right ) -{\left (2 \, B c^{4} \cos \left (f x + e\right )^{4} +{\left (3 \, A - 14 \, B\right )} c^{4} \cos \left (f x + e\right )^{3} + 210 \,{\left (A - 2 \, B\right )} c^{4} f x + 3 \,{\left (11 \, A - 30 \, B\right )} c^{4} \cos \left (f x + e\right )^{2} + 32 \,{\left (A - B\right )} c^{4} +{\left (105 \,{\left (A - 2 \, B\right )} c^{4} f x + 2 \,{\left (113 \, A - 218 \, B\right )} c^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f -{\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^2,x, algorithm="fricas")
[Out]
1/6*(2*B*c^4*cos(f*x + e)^5 - (3*A - 16*B)*c^4*cos(f*x + e)^4 + 2*(15*A - 38*B)*c^4*cos(f*x + e)^3 - 210*(A -
2*B)*c^4*f*x + 32*(A - B)*c^4 + (105*(A - 2*B)*c^4*f*x - (193*A - 346*B)*c^4)*cos(f*x + e)^2 - (105*(A - 2*B)*
c^4*f*x + 2*(97*A - 202*B)*c^4)*cos(f*x + e) - (2*B*c^4*cos(f*x + e)^4 + (3*A - 14*B)*c^4*cos(f*x + e)^3 + 210
*(A - 2*B)*c^4*f*x + 3*(11*A - 30*B)*c^4*cos(f*x + e)^2 + 32*(A - B)*c^4 + (105*(A - 2*B)*c^4*f*x + 2*(113*A -
218*B)*c^4)*cos(f*x + e))*sin(f*x + e))/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*f - (a^2*f*cos(f*x
+ e) + 2*a^2*f)*sin(f*x + e))
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**4/(a+a*sin(f*x+e))**2,x)
[Out]
Timed out
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Giac [B] time = 1.22226, size = 498, normalized size = 2.77 \begin{align*} \frac{\frac{105 \,{\left (A c^{4} - 2 \, B c^{4}\right )}{\left (f x + e\right )}}{a^{2}} + \frac{2 \,{\left (99 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} - 210 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} + 333 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} - 636 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 533 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 1160 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 1047 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 1980 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 921 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1980 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 1107 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2140 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 651 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1344 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 393 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 780 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 164 \, A c^{4} - 330 \, B c^{4}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3} a^{2}}}{6 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^2,x, algorithm="giac")
[Out]
1/6*(105*(A*c^4 - 2*B*c^4)*(f*x + e)/a^2 + 2*(99*A*c^4*tan(1/2*f*x + 1/2*e)^8 - 210*B*c^4*tan(1/2*f*x + 1/2*e)
^8 + 333*A*c^4*tan(1/2*f*x + 1/2*e)^7 - 636*B*c^4*tan(1/2*f*x + 1/2*e)^7 + 533*A*c^4*tan(1/2*f*x + 1/2*e)^6 -
1160*B*c^4*tan(1/2*f*x + 1/2*e)^6 + 1047*A*c^4*tan(1/2*f*x + 1/2*e)^5 - 1980*B*c^4*tan(1/2*f*x + 1/2*e)^5 + 92
1*A*c^4*tan(1/2*f*x + 1/2*e)^4 - 1980*B*c^4*tan(1/2*f*x + 1/2*e)^4 + 1107*A*c^4*tan(1/2*f*x + 1/2*e)^3 - 2140*
B*c^4*tan(1/2*f*x + 1/2*e)^3 + 651*A*c^4*tan(1/2*f*x + 1/2*e)^2 - 1344*B*c^4*tan(1/2*f*x + 1/2*e)^2 + 393*A*c^
4*tan(1/2*f*x + 1/2*e) - 780*B*c^4*tan(1/2*f*x + 1/2*e) + 164*A*c^4 - 330*B*c^4)/((tan(1/2*f*x + 1/2*e)^3 + ta
n(1/2*f*x + 1/2*e)^2 + tan(1/2*f*x + 1/2*e) + 1)^3*a^2))/f