3.73 \(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx\)
Optimal. Leaf size=110 \[ \frac {2 B c^2 \cos (e+f x)}{f \left (a^3 \sin (e+f x)+a^3\right )}+\frac {B c^2 x}{a^3}-\frac {a^2 c^2 (A-B) \cos ^5(e+f x)}{5 f (a \sin (e+f x)+a)^5}-\frac {2 B c^2 \cos ^3(e+f x)}{3 f (a \sin (e+f x)+a)^3} \]
[Out]
B*c^2*x/a^3-1/5*a^2*(A-B)*c^2*cos(f*x+e)^5/f/(a+a*sin(f*x+e))^5-2/3*B*c^2*cos(f*x+e)^3/f/(a+a*sin(f*x+e))^3+2*
B*c^2*cos(f*x+e)/f/(a^3+a^3*sin(f*x+e))
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Rubi [A] time = 0.26, antiderivative size = 110, normalized size of antiderivative = 1.00,
number of steps used = 5, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used =
{2967, 2859, 2680, 8} \[ -\frac {a^2 c^2 (A-B) \cos ^5(e+f x)}{5 f (a \sin (e+f x)+a)^5}+\frac {2 B c^2 \cos (e+f x)}{f \left (a^3 \sin (e+f x)+a^3\right )}+\frac {B c^2 x}{a^3}-\frac {2 B c^2 \cos ^3(e+f x)}{3 f (a \sin (e+f x)+a)^3} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^2)/(a + a*Sin[e + f*x])^3,x]
[Out]
(B*c^2*x)/a^3 - (a^2*(A - B)*c^2*Cos[e + f*x]^5)/(5*f*(a + a*Sin[e + f*x])^5) - (2*B*c^2*Cos[e + f*x]^3)/(3*f*
(a + a*Sin[e + f*x])^3) + (2*B*c^2*Cos[e + f*x])/(f*(a^3 + a^3*Sin[e + f*x]))
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
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 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 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]))
Rubi steps
\begin {align*} \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx &=\left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x) (A+B \sin (e+f x))}{(a+a \sin (e+f x))^5} \, dx\\ &=-\frac {a^2 (A-B) c^2 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^5}+\left (a B c^2\right ) \int \frac {\cos ^4(e+f x)}{(a+a \sin (e+f x))^4} \, dx\\ &=-\frac {a^2 (A-B) c^2 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^5}-\frac {2 B c^2 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^3}-\frac {\left (B c^2\right ) \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^2} \, dx}{a}\\ &=-\frac {a^2 (A-B) c^2 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^5}-\frac {2 B c^2 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^3}+\frac {2 B c^2 \cos (e+f x)}{f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {\left (B c^2\right ) \int 1 \, dx}{a^3}\\ &=\frac {B c^2 x}{a^3}-\frac {a^2 (A-B) c^2 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^5}-\frac {2 B c^2 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^3}+\frac {2 B c^2 \cos (e+f x)}{f \left (a^3+a^3 \sin (e+f x)\right )}\\ \end {align*}
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Mathematica [B] time = 0.70, size = 272, normalized size = 2.47 \[ \frac {(c-c \sin (e+f x))^2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (24 (A-B) \sin \left (\frac {1}{2} (e+f x)\right )+2 (3 A-43 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4+4 (3 A-8 B) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3-8 (3 A-8 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2-12 (A-B) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )+15 B (e+f x) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5\right )}{15 a^3 f (\sin (e+f x)+1)^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^2)/(a + a*Sin[e + f*x])^3,x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(24*(A - B)*Sin[(e + f*x)/2] - 12*(A - B)*(Cos[(e + f*x)/2] + Sin[(e +
f*x)/2]) - 8*(3*A - 8*B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 4*(3*A - 8*B)*(Cos[(e + f*
x)/2] + Sin[(e + f*x)/2])^3 + 2*(3*A - 43*B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 + 15*B*(
e + f*x)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)*(c - c*Sin[e + f*x])^2)/(15*a^3*f*(Cos[(e + f*x)/2] - Sin[(e
+ f*x)/2])^4*(1 + Sin[e + f*x])^3)
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fricas [B] time = 0.42, size = 279, normalized size = 2.54 \[ -\frac {60 \, B c^{2} f x - {\left (15 \, B c^{2} f x - {\left (3 \, A - 43 \, B\right )} c^{2}\right )} \cos \left (f x + e\right )^{3} - 12 \, {\left (A - B\right )} c^{2} - {\left (45 \, B c^{2} f x - {\left (9 \, A + 11 \, B\right )} c^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, B c^{2} f x - {\left (A - 11 \, B\right )} c^{2}\right )} \cos \left (f x + e\right ) + {\left (60 \, B c^{2} f x + 12 \, {\left (A - B\right )} c^{2} - {\left (15 \, B c^{2} f x + {\left (3 \, A - 43 \, B\right )} c^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, B c^{2} f x + {\left (A + 9 \, B\right )} c^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
-1/15*(60*B*c^2*f*x - (15*B*c^2*f*x - (3*A - 43*B)*c^2)*cos(f*x + e)^3 - 12*(A - B)*c^2 - (45*B*c^2*f*x - (9*A
+ 11*B)*c^2)*cos(f*x + e)^2 + 6*(5*B*c^2*f*x - (A - 11*B)*c^2)*cos(f*x + e) + (60*B*c^2*f*x + 12*(A - B)*c^2
- (15*B*c^2*f*x + (3*A - 43*B)*c^2)*cos(f*x + e)^2 + 6*(5*B*c^2*f*x + (A + 9*B)*c^2)*cos(f*x + e))*sin(f*x + e
))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2
*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(f*x + e))
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giac [A] time = 0.21, size = 159, normalized size = 1.45 \[ \frac {\frac {15 \, {\left (f x + e\right )} B c^{2}}{a^{3}} - \frac {2 \, {\left (15 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 60 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 170 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 100 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, A c^{2} - 23 \, B c^{2}\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{15 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="giac")
[Out]
1/15*(15*(f*x + e)*B*c^2/a^3 - 2*(15*A*c^2*tan(1/2*f*x + 1/2*e)^4 - 15*B*c^2*tan(1/2*f*x + 1/2*e)^4 - 60*B*c^2
*tan(1/2*f*x + 1/2*e)^3 + 30*A*c^2*tan(1/2*f*x + 1/2*e)^2 - 170*B*c^2*tan(1/2*f*x + 1/2*e)^2 - 100*B*c^2*tan(1
/2*f*x + 1/2*e) + 3*A*c^2 - 23*B*c^2)/(a^3*(tan(1/2*f*x + 1/2*e) + 1)^5))/f
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maple [B] time = 0.47, size = 249, normalized size = 2.26 \[ \frac {2 c^{2} B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a^{3} f}+\frac {16 c^{2} A}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {16 c^{2} B}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 c^{2} A}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {2 c^{2} B}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {32 c^{2} A}{5 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {32 c^{2} B}{5 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {16 c^{2} A}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {32 c^{2} B}{3 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {8 c^{2} A}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x)
[Out]
2*c^2/a^3/f*B*arctan(tan(1/2*f*x+1/2*e))+16*c^2/a^3/f/(tan(1/2*f*x+1/2*e)+1)^4*A-16*c^2/a^3/f/(tan(1/2*f*x+1/2
*e)+1)^4*B-2*c^2/a^3/f/(tan(1/2*f*x+1/2*e)+1)*A+2*c^2/a^3/f/(tan(1/2*f*x+1/2*e)+1)*B-32/5*c^2/a^3/f/(tan(1/2*f
*x+1/2*e)+1)^5*A+32/5*c^2/a^3/f/(tan(1/2*f*x+1/2*e)+1)^5*B-16*c^2/a^3/f/(tan(1/2*f*x+1/2*e)+1)^3*A+32/3*c^2/a^
3/f/(tan(1/2*f*x+1/2*e)+1)^3*B+8*c^2/a^3/f*A/(tan(1/2*f*x+1/2*e)+1)^2
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maxima [B] time = 0.47, size = 1134, normalized size = 10.31 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="maxima")
[Out]
2/15*(B*c^2*((95*sin(f*x + e)/(cos(f*x + e) + 1) + 145*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 75*sin(f*x + e)^3
/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 22)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) +
1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan(sin(f*x + e)/(cos(f*x + e)
+ 1))/a^3) - A*c^2*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x +
e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 7)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x +
e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(
f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 2*A*c^2*(5*sin(f*x + e)/(cos(f*x
+ e) + 1) + 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*
sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f
*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 4*B*c^2*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 10*si
n(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(c
os(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 +
a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 6*A*c^2*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x + e)^2/(cos(
f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10
*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(
cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 3*B*c^2*(5*sin(f*x + e)/(cos(f*x + e) + 1) +
5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(
cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 +
5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f
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mupad [B] time = 15.08, size = 230, normalized size = 2.09 \[ \frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {c^2\,\left (120\,B+150\,B\,\left (e+f\,x\right )\right )}{15}-10\,B\,c^2\,\left (e+f\,x\right )\right )+\frac {c^2\,\left (46\,B-6\,A+15\,B\,\left (e+f\,x\right )\right )}{15}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {c^2\,\left (30\,B-30\,A+75\,B\,\left (e+f\,x\right )\right )}{15}-5\,B\,c^2\,\left (e+f\,x\right )\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {c^2\,\left (340\,B-60\,A+150\,B\,\left (e+f\,x\right )\right )}{15}-10\,B\,c^2\,\left (e+f\,x\right )\right )+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {c^2\,\left (200\,B+75\,B\,\left (e+f\,x\right )\right )}{15}-5\,B\,c^2\,\left (e+f\,x\right )\right )-B\,c^2\,\left (e+f\,x\right )}{a^3\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}^5}+\frac {B\,c^2\,x}{a^3} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^2)/(a + a*sin(e + f*x))^3,x)
[Out]
(tan(e/2 + (f*x)/2)^3*((c^2*(120*B + 150*B*(e + f*x)))/15 - 10*B*c^2*(e + f*x)) + (c^2*(46*B - 6*A + 15*B*(e +
f*x)))/15 + tan(e/2 + (f*x)/2)^4*((c^2*(30*B - 30*A + 75*B*(e + f*x)))/15 - 5*B*c^2*(e + f*x)) + tan(e/2 + (f
*x)/2)^2*((c^2*(340*B - 60*A + 150*B*(e + f*x)))/15 - 10*B*c^2*(e + f*x)) + tan(e/2 + (f*x)/2)*((c^2*(200*B +
75*B*(e + f*x)))/15 - 5*B*c^2*(e + f*x)) - B*c^2*(e + f*x))/(a^3*f*(tan(e/2 + (f*x)/2) + 1)^5) + (B*c^2*x)/a^3
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sympy [A] time = 26.03, size = 1647, normalized size = 14.97 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**2/(a+a*sin(f*x+e))**3,x)
[Out]
Piecewise((-30*A*c**2*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150
*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*A*
c**2*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 +
f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 6*A*c**2/(15*a**3*f*ta
n(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/
2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 15*B*c**2*f*x*tan(e/2 + f*x/2)**5/(15*a**3*f*tan(e/2 + f*x/2
)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a*
*3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 75*B*c**2*f*x*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**
3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2
+ f*x/2) + 15*a**3*f) + 150*B*c**2*f*x*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2
+ f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 1
5*a**3*f) + 150*B*c**2*f*x*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4
+ 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) +
75*B*c**2*f*x*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan
(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 15*B*c**2*f*x/(1
5*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan
(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 30*B*c**2*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2
+ f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2
+ 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 120*B*c**2*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75
*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(
e/2 + f*x/2) + 15*a**3*f) + 340*B*c**2*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2
+ f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 1
5*a**3*f) + 200*B*c**2*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a
**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 46*B*c*
*2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*
f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f), Ne(f, 0)), (x*(A + B*sin(e))*(-c*sin(e) + c)*
*2/(a*sin(e) + a)**3, True))
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