3.52 \(\int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))^4}{a+a \sin (e+f x)} \, dx\)
Optimal. Leaf size=190 \[ -\frac{a^4 c^4 (A-B) \cos ^9(e+f x)}{f (a \sin (e+f x)+a)^5}-\frac{2 a^2 c^4 (4 A-5 B) \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^3}-\frac{35 c^4 (4 A-5 B) \cos ^3(e+f x)}{12 a f}-\frac{7 c^4 (4 A-5 B) \cos ^5(e+f x)}{4 f (a \sin (e+f x)+a)}-\frac{35 c^4 (4 A-5 B) \sin (e+f x) \cos (e+f x)}{8 a f}-\frac{35 c^4 x (4 A-5 B)}{8 a} \]
[Out]
(-35*(4*A - 5*B)*c^4*x)/(8*a) - (35*(4*A - 5*B)*c^4*Cos[e + f*x]^3)/(12*a*f) - (35*(4*A - 5*B)*c^4*Cos[e + f*x
]*Sin[e + f*x])/(8*a*f) - (a^4*(A - B)*c^4*Cos[e + f*x]^9)/(f*(a + a*Sin[e + f*x])^5) - (2*a^2*(4*A - 5*B)*c^4
*Cos[e + f*x]^7)/(f*(a + a*Sin[e + f*x])^3) - (7*(4*A - 5*B)*c^4*Cos[e + f*x]^5)/(4*f*(a + a*Sin[e + f*x]))
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Rubi [A] time = 0.362638, antiderivative size = 190, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used =
{2967, 2859, 2680, 2679, 2682, 2635, 8} \[ -\frac{a^4 c^4 (A-B) \cos ^9(e+f x)}{f (a \sin (e+f x)+a)^5}-\frac{2 a^2 c^4 (4 A-5 B) \cos ^7(e+f x)}{f (a \sin (e+f x)+a)^3}-\frac{35 c^4 (4 A-5 B) \cos ^3(e+f x)}{12 a f}-\frac{7 c^4 (4 A-5 B) \cos ^5(e+f x)}{4 f (a \sin (e+f x)+a)}-\frac{35 c^4 (4 A-5 B) \sin (e+f x) \cos (e+f x)}{8 a f}-\frac{35 c^4 x (4 A-5 B)}{8 a} \]
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]),x]
[Out]
(-35*(4*A - 5*B)*c^4*x)/(8*a) - (35*(4*A - 5*B)*c^4*Cos[e + f*x]^3)/(12*a*f) - (35*(4*A - 5*B)*c^4*Cos[e + f*x
]*Sin[e + f*x])/(8*a*f) - (a^4*(A - B)*c^4*Cos[e + f*x]^9)/(f*(a + a*Sin[e + f*x])^5) - (2*a^2*(4*A - 5*B)*c^4
*Cos[e + f*x]^7)/(f*(a + a*Sin[e + f*x])^3) - (7*(4*A - 5*B)*c^4*Cos[e + f*x]^5)/(4*f*(a + a*Sin[e + f*x]))
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 2679
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))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(a*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
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)} \, 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))^5} \, dx\\ &=-\frac{a^4 (A-B) c^4 \cos ^9(e+f x)}{f (a+a \sin (e+f x))^5}-\left (a^3 (4 A-5 B) c^4\right ) \int \frac{\cos ^8(e+f x)}{(a+a \sin (e+f x))^4} \, dx\\ &=-\frac{a^4 (A-B) c^4 \cos ^9(e+f x)}{f (a+a \sin (e+f x))^5}-\frac{2 a^2 (4 A-5 B) c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^3}-\left (7 a (4 A-5 B) c^4\right ) \int \frac{\cos ^6(e+f x)}{(a+a \sin (e+f x))^2} \, dx\\ &=-\frac{a^4 (A-B) c^4 \cos ^9(e+f x)}{f (a+a \sin (e+f x))^5}-\frac{2 a^2 (4 A-5 B) c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^3}-\frac{7 (4 A-5 B) c^4 \cos ^5(e+f x)}{4 f (a+a \sin (e+f x))}-\frac{1}{4} \left (35 (4 A-5 B) c^4\right ) \int \frac{\cos ^4(e+f x)}{a+a \sin (e+f x)} \, dx\\ &=-\frac{35 (4 A-5 B) c^4 \cos ^3(e+f x)}{12 a f}-\frac{a^4 (A-B) c^4 \cos ^9(e+f x)}{f (a+a \sin (e+f x))^5}-\frac{2 a^2 (4 A-5 B) c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^3}-\frac{7 (4 A-5 B) c^4 \cos ^5(e+f x)}{4 f (a+a \sin (e+f x))}-\frac{\left (35 (4 A-5 B) c^4\right ) \int \cos ^2(e+f x) \, dx}{4 a}\\ &=-\frac{35 (4 A-5 B) c^4 \cos ^3(e+f x)}{12 a f}-\frac{35 (4 A-5 B) c^4 \cos (e+f x) \sin (e+f x)}{8 a f}-\frac{a^4 (A-B) c^4 \cos ^9(e+f x)}{f (a+a \sin (e+f x))^5}-\frac{2 a^2 (4 A-5 B) c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^3}-\frac{7 (4 A-5 B) c^4 \cos ^5(e+f x)}{4 f (a+a \sin (e+f x))}-\frac{\left (35 (4 A-5 B) c^4\right ) \int 1 \, dx}{8 a}\\ &=-\frac{35 (4 A-5 B) c^4 x}{8 a}-\frac{35 (4 A-5 B) c^4 \cos ^3(e+f x)}{12 a f}-\frac{35 (4 A-5 B) c^4 \cos (e+f x) \sin (e+f x)}{8 a f}-\frac{a^4 (A-B) c^4 \cos ^9(e+f x)}{f (a+a \sin (e+f x))^5}-\frac{2 a^2 (4 A-5 B) c^4 \cos ^7(e+f x)}{f (a+a \sin (e+f x))^3}-\frac{7 (4 A-5 B) c^4 \cos ^5(e+f x)}{4 f (a+a \sin (e+f x))}\\ \end{align*}
Mathematica [A] time = 2.29211, size = 274, normalized size = 1.44 \[ \frac{(c-c \sin (e+f x))^4 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (3072 (A-B) \sin \left (\frac{1}{2} (e+f x)\right )-420 (4 A-5 B) (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )-24 (47 A-75 B) \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+8 (A-5 B) \cos (3 (e+f x)) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+24 (5 A-12 B) \sin (2 (e+f x)) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )+3 B \sin (4 (e+f x)) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{96 a f (\sin (e+f x)+1) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^8} \]
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]),x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c - c*Sin[e + f*x])^4*(3072*(A - B)*Sin[(e + f*x)/2] - 420*(4*A - 5*B)
*(e + f*x)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - 24*(47*A - 75*B)*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e +
f*x)/2]) + 8*(A - 5*B)*Cos[3*(e + f*x)]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 24*(5*A - 12*B)*(Cos[(e + f*x)
/2] + Sin[(e + f*x)/2])*Sin[2*(e + f*x)] + 3*B*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sin[4*(e + f*x)]))/(96*a*
f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^8*(1 + Sin[e + f*x]))
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Maple [B] time = 0.134, size = 678, normalized size = 3.6 \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)),x)
[Out]
-5/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^4*tan(1/2*f*x+1/2*e)^7*A+47/4/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^4*tan(1/2*f
*x+1/2*e)^7*B-22/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^4*tan(1/2*f*x+1/2*e)^6*A+30/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)
^4*tan(1/2*f*x+1/2*e)^6*B-5/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^4*tan(1/2*f*x+1/2*e)^5*A+55/4/f*c^4/a/(1+tan(1/2*
f*x+1/2*e)^2)^4*tan(1/2*f*x+1/2*e)^5*B-70/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^4*tan(1/2*f*x+1/2*e)^4*A+110/f*c^4/
a/(1+tan(1/2*f*x+1/2*e)^2)^4*tan(1/2*f*x+1/2*e)^4*B+5/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^4*tan(1/2*f*x+1/2*e)^3*
A-55/4/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^4*tan(1/2*f*x+1/2*e)^3*B-214/3/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^4*tan(
1/2*f*x+1/2*e)^2*A+350/3/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^4*tan(1/2*f*x+1/2*e)^2*B+5/f*c^4/a/(1+tan(1/2*f*x+1/
2*e)^2)^4*tan(1/2*f*x+1/2*e)*A-47/4/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^4*tan(1/2*f*x+1/2*e)*B-70/3/f*c^4/a/(1+ta
n(1/2*f*x+1/2*e)^2)^4*A+110/3/f*c^4/a/(1+tan(1/2*f*x+1/2*e)^2)^4*B-35/f*c^4/a*arctan(tan(1/2*f*x+1/2*e))*A+175
/4/f*c^4/a*arctan(tan(1/2*f*x+1/2*e))*B-32/f*c^4/a/(tan(1/2*f*x+1/2*e)+1)*A+32/f*c^4/a/(tan(1/2*f*x+1/2*e)+1)*
B
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Maxima [B] time = 1.57989, size = 2425, normalized size = 12.76 \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)),x, algorithm="maxima")
[Out]
1/12*(B*c^4*((19*sin(f*x + e)/(cos(f*x + e) + 1) + 211*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 91*sin(f*x + e)^3
/(cos(f*x + e) + 1)^3 + 219*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 165*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 16
5*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 45*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 45*sin(f*x + e)^8/(cos(f*x +
e) + 1)^8 + 64)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + 4*a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4*a*sin(f*x
+ e)^3/(cos(f*x + e) + 1)^3 + 6*a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 6*a*sin(f*x + e)^5/(cos(f*x + e) + 1)
^5 + 4*a*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 4*a*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + a*sin(f*x + e)^8/(cos
(f*x + e) + 1)^8 + a*sin(f*x + e)^9/(cos(f*x + e) + 1)^9) + 45*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) - 4*
A*c^4*((7*sin(f*x + e)/(cos(f*x + e) + 1) + 39*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 24*sin(f*x + e)^3/(cos(f*
x + e) + 1)^3 + 24*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 9*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 9*sin(f*x + e
)^6/(cos(f*x + e) + 1)^6 + 16)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a*sin(f*x + e)^2/(cos(f*x + e) + 1)^
2 + 3*a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 3*a*sin(f*x + e)^5/(co
s(f*x + e) + 1)^5 + a*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + 9*arctan(
sin(f*x + e)/(cos(f*x + e) + 1))/a) + 16*B*c^4*((7*sin(f*x + e)/(cos(f*x + e) + 1) + 39*sin(f*x + e)^2/(cos(f*
x + e) + 1)^2 + 24*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 24*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 9*sin(f*x +
e)^5/(cos(f*x + e) + 1)^5 + 9*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 16)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1)
+ 3*a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a*sin(f*x + e)^4/(cos
(f*x + e) + 1)^4 + 3*a*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + a*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a*sin(f*x
+ e)^7/(cos(f*x + e) + 1)^7) + 9*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) - 48*A*c^4*((sin(f*x + e)/(cos(f*
x + e) + 1) + 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^4
/(cos(f*x + e) + 1)^4 + 4)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + 2*a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 +
2*a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a*sin(f*x + e)^5/(cos(f*x +
e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) + 72*B*c^4*((sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin
(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^4/(cos(f*x + e) + 1)
^4 + 4)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + 2*a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2*a*sin(f*x + e)^3/
(cos(f*x + e) + 1)^3 + a*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arct
an(sin(f*x + e)/(cos(f*x + e) + 1))/a) - 144*A*c^4*((sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 + 2)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + a*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a*sin(f*x +
e)^3/(cos(f*x + e) + 1)^3) + arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) + 96*B*c^4*((sin(f*x + e)/(cos(f*x +
e) + 1) + sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2)/(a + a*sin(f*x + e)/(cos(f*x + e) + 1) + a*sin(f*x + e)^2/(
cos(f*x + e) + 1)^2 + a*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a) - 96
*A*c^4*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a + 1/(a + a*sin(f*x + e)/(cos(f*x + e) + 1))) + 24*B*c^4*(arc
tan(sin(f*x + e)/(cos(f*x + e) + 1))/a + 1/(a + a*sin(f*x + e)/(cos(f*x + e) + 1))) - 24*A*c^4/(a + a*sin(f*x
+ e)/(cos(f*x + e) + 1)))/f
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Fricas [A] time = 1.50523, size = 651, normalized size = 3.43 \begin{align*} -\frac{6 \, B c^{4} \cos \left (f x + e\right )^{5} - 8 \,{\left (A - 5 \, B\right )} c^{4} \cos \left (f x + e\right )^{4} +{\left (52 \, A - 113 \, B\right )} c^{4} \cos \left (f x + e\right )^{3} + 105 \,{\left (4 \, A - 5 \, B\right )} c^{4} f x + 96 \,{\left (3 \, A - 5 \, B\right )} c^{4} \cos \left (f x + e\right )^{2} + 384 \,{\left (A - B\right )} c^{4} + 3 \,{\left (35 \,{\left (4 \, A - 5 \, B\right )} c^{4} f x +{\left (204 \, A - 239 \, B\right )} c^{4}\right )} \cos \left (f x + e\right ) -{\left (6 \, B c^{4} \cos \left (f x + e\right )^{4} + 2 \,{\left (4 \, A - 17 \, B\right )} c^{4} \cos \left (f x + e\right )^{3} - 105 \,{\left (4 \, A - 5 \, B\right )} c^{4} f x + 3 \,{\left (20 \, A - 49 \, B\right )} c^{4} \cos \left (f x + e\right )^{2} - 3 \,{\left (76 \, A - 111 \, B\right )} c^{4} \cos \left (f x + e\right ) + 384 \,{\left (A - B\right )} c^{4}\right )} \sin \left (f x + e\right )}{24 \,{\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\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)),x, algorithm="fricas")
[Out]
-1/24*(6*B*c^4*cos(f*x + e)^5 - 8*(A - 5*B)*c^4*cos(f*x + e)^4 + (52*A - 113*B)*c^4*cos(f*x + e)^3 + 105*(4*A
- 5*B)*c^4*f*x + 96*(3*A - 5*B)*c^4*cos(f*x + e)^2 + 384*(A - B)*c^4 + 3*(35*(4*A - 5*B)*c^4*f*x + (204*A - 23
9*B)*c^4)*cos(f*x + e) - (6*B*c^4*cos(f*x + e)^4 + 2*(4*A - 17*B)*c^4*cos(f*x + e)^3 - 105*(4*A - 5*B)*c^4*f*x
+ 3*(20*A - 49*B)*c^4*cos(f*x + e)^2 - 3*(76*A - 111*B)*c^4*cos(f*x + e) + 384*(A - B)*c^4)*sin(f*x + e))/(a*
f*cos(f*x + e) + a*f*sin(f*x + e) + a*f)
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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)),x)
[Out]
Timed out
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Giac [A] time = 1.23664, size = 463, normalized size = 2.44 \begin{align*} -\frac{\frac{105 \,{\left (4 \, A c^{4} - 5 \, B c^{4}\right )}{\left (f x + e\right )}}{a} + \frac{768 \,{\left (A c^{4} - B c^{4}\right )}}{a{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}} + \frac{2 \,{\left (60 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} - 141 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 264 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 360 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} + 60 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 165 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 840 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1320 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 60 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 165 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 856 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1400 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 60 \, A c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 141 \, B c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 280 \, A c^{4} - 440 \, B c^{4}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{4} a}}{24 \, 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)),x, algorithm="giac")
[Out]
-1/24*(105*(4*A*c^4 - 5*B*c^4)*(f*x + e)/a + 768*(A*c^4 - B*c^4)/(a*(tan(1/2*f*x + 1/2*e) + 1)) + 2*(60*A*c^4*
tan(1/2*f*x + 1/2*e)^7 - 141*B*c^4*tan(1/2*f*x + 1/2*e)^7 + 264*A*c^4*tan(1/2*f*x + 1/2*e)^6 - 360*B*c^4*tan(1
/2*f*x + 1/2*e)^6 + 60*A*c^4*tan(1/2*f*x + 1/2*e)^5 - 165*B*c^4*tan(1/2*f*x + 1/2*e)^5 + 840*A*c^4*tan(1/2*f*x
+ 1/2*e)^4 - 1320*B*c^4*tan(1/2*f*x + 1/2*e)^4 - 60*A*c^4*tan(1/2*f*x + 1/2*e)^3 + 165*B*c^4*tan(1/2*f*x + 1/
2*e)^3 + 856*A*c^4*tan(1/2*f*x + 1/2*e)^2 - 1400*B*c^4*tan(1/2*f*x + 1/2*e)^2 - 60*A*c^4*tan(1/2*f*x + 1/2*e)
+ 141*B*c^4*tan(1/2*f*x + 1/2*e) + 280*A*c^4 - 440*B*c^4)/((tan(1/2*f*x + 1/2*e)^2 + 1)^4*a))/f