3.32 \(\int \frac{(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx\)
Optimal. Leaf size=109 \[ -\frac{a^2 (A+4 B) \cos (e+f x)}{c^2 f}+\frac{a^2 c^2 (A+B) \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}+\frac{a^2 x (A+4 B)}{c^2}-\frac{2 a^2 (A+4 B) \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2} \]
[Out]
(a^2*(A + 4*B)*x)/c^2 - (a^2*(A + 4*B)*Cos[e + f*x])/(c^2*f) + (a^2*(A + B)*c^2*Cos[e + f*x]^5)/(3*f*(c - c*Si
n[e + f*x])^4) - (2*a^2*(A + 4*B)*Cos[e + f*x]^3)/(3*f*(c - c*Sin[e + f*x])^2)
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Rubi [A] time = 0.283956, antiderivative size = 109, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used =
{2967, 2859, 2680, 2682, 8} \[ -\frac{a^2 (A+4 B) \cos (e+f x)}{c^2 f}+\frac{a^2 c^2 (A+B) \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}+\frac{a^2 x (A+4 B)}{c^2}-\frac{2 a^2 (A+4 B) \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2} \]
Antiderivative was successfully verified.
[In]
Int[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^2,x]
[Out]
(a^2*(A + 4*B)*x)/c^2 - (a^2*(A + 4*B)*Cos[e + f*x])/(c^2*f) + (a^2*(A + B)*c^2*Cos[e + f*x]^5)/(3*f*(c - c*Si
n[e + f*x])^4) - (2*a^2*(A + 4*B)*Cos[e + f*x]^3)/(3*f*(c - c*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 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx &=\left (a^2 c^2\right ) \int \frac{\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx\\ &=\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac{1}{3} \left (a^2 (A+4 B) c\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^3} \, dx\\ &=\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac{2 a^2 (A+4 B) \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2}+\frac{\left (a^2 (A+4 B)\right ) \int \frac{\cos ^2(e+f x)}{c-c \sin (e+f x)} \, dx}{c}\\ &=-\frac{a^2 (A+4 B) \cos (e+f x)}{c^2 f}+\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac{2 a^2 (A+4 B) \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2}+\frac{\left (a^2 (A+4 B)\right ) \int 1 \, dx}{c^2}\\ &=\frac{a^2 (A+4 B) x}{c^2}-\frac{a^2 (A+4 B) \cos (e+f x)}{c^2 f}+\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac{2 a^2 (A+4 B) \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2}\\ \end{align*}
Mathematica [B] time = 0.606948, size = 238, normalized size = 2.18 \[ \frac{a^2 (\sin (e+f x)+1)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (8 (A+B) \sin \left (\frac{1}{2} (e+f x)\right )+3 (A+4 B) (e+f x) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3-8 (2 A+5 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+4 (A+B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-3 B \cos (e+f x) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3\right )}{3 f (c-c \sin (e+f x))^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^2,x]
[Out]
(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(4*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) + 3*(A + 4*B)*(e +
f*x)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - 3*B*Cos[e + f*x]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 + 8*(
A + B)*Sin[(e + f*x)/2] - 8*(2*A + 5*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2])*(1 + Sin[e +
f*x])^2)/(3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(c - c*Sin[e + f*x])^2)
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Maple [A] time = 0.116, size = 198, normalized size = 1.8 \begin{align*} -{\frac{16\,A{a}^{2}}{3\,f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-3}}-{\frac{16\,B{a}^{2}}{3\,f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-3}}-8\,{\frac{A{a}^{2}}{f{c}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-8\,{\frac{B{a}^{2}}{f{c}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}+8\,{\frac{B{a}^{2}}{f{c}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}-2\,{\frac{B{a}^{2}}{f{c}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}+2\,{\frac{{a}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) A}{f{c}^{2}}}+8\,{\frac{{a}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) B}{f{c}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x)
[Out]
-16/3/f*a^2/c^2/(tan(1/2*f*x+1/2*e)-1)^3*A-16/3/f*a^2/c^2/(tan(1/2*f*x+1/2*e)-1)^3*B-8/f*a^2/c^2/(tan(1/2*f*x+
1/2*e)-1)^2*A-8/f*a^2/c^2/(tan(1/2*f*x+1/2*e)-1)^2*B+8/f*a^2/c^2*B/(tan(1/2*f*x+1/2*e)-1)-2/f*a^2/c^2*B/(1+tan
(1/2*f*x+1/2*e)^2)+2/f*a^2/c^2*arctan(tan(1/2*f*x+1/2*e))*A+8/f*a^2/c^2*arctan(tan(1/2*f*x+1/2*e))*B
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Maxima [B] time = 1.53119, size = 1133, normalized size = 10.39 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algorithm="maxima")
[Out]
2/3*(2*B*a^2*((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)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1)
+ 4*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4*c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*c^2*sin(f*x + e)^
4/(cos(f*x + e) + 1)^4 - c^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/
c^2) + A*a^2*((9*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4)/(c^2 - 3*c^2*sin
(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) +
1)^3) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) + 2*B*a^2*((9*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f
*x + e)^2/(cos(f*x + e) + 1)^2 - 4)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f
*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) - A
*a^2*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2)/(c^2 - 3*c^2*sin(f*x + e)
/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 2
*A*a^2*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 1)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + B*a^2*(3*sin(f*x + e)/(cos(f*x + e) + 1
) - 1)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x
+ e)^3/(cos(f*x + e) + 1)^3))/f
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Fricas [B] time = 1.43296, size = 568, normalized size = 5.21 \begin{align*} -\frac{3 \, B a^{2} \cos \left (f x + e\right )^{3} + 6 \,{\left (A + 4 \, B\right )} a^{2} f x + 4 \,{\left (A + B\right )} a^{2} -{\left (3 \,{\left (A + 4 \, B\right )} a^{2} f x +{\left (8 \, A + 23 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )^{2} +{\left (3 \,{\left (A + 4 \, B\right )} a^{2} f x - 2 \,{\left (2 \, A + 11 \, B\right )} a^{2}\right )} \cos \left (f x + e\right ) -{\left (6 \,{\left (A + 4 \, B\right )} a^{2} f x - 3 \, B a^{2} \cos \left (f x + e\right )^{2} - 4 \,{\left (A + B\right )} a^{2} +{\left (3 \,{\left (A + 4 \, B\right )} a^{2} f x - 2 \,{\left (4 \, A + 13 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \,{\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f +{\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{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+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algorithm="fricas")
[Out]
-1/3*(3*B*a^2*cos(f*x + e)^3 + 6*(A + 4*B)*a^2*f*x + 4*(A + B)*a^2 - (3*(A + 4*B)*a^2*f*x + (8*A + 23*B)*a^2)*
cos(f*x + e)^2 + (3*(A + 4*B)*a^2*f*x - 2*(2*A + 11*B)*a^2)*cos(f*x + e) - (6*(A + 4*B)*a^2*f*x - 3*B*a^2*cos(
f*x + e)^2 - 4*(A + B)*a^2 + (3*(A + 4*B)*a^2*f*x - 2*(4*A + 13*B)*a^2)*cos(f*x + e))*sin(f*x + e))/(c^2*f*cos
(f*x + e)^2 - c^2*f*cos(f*x + e) - 2*c^2*f + (c^2*f*cos(f*x + e) + 2*c^2*f)*sin(f*x + e))
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Sympy [A] time = 26.4778, size = 2474, normalized size = 22.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**2,x)
[Out]
Piecewise((3*A*a**2*f*x*tan(e/2 + f*x/2)**5/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*
c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 9*A*a**2*
f*x*tan(e/2 + f*x/2)**4/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x
/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 12*A*a**2*f*x*tan(e/2 + f*x/2
)**3/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*
tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 12*A*a**2*f*x*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(
e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2
+ 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 9*A*a**2*f*x*tan(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**
2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f
*x/2) - 3*c**2*f) - 3*A*a**2*f*x/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(
e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 24*A*a**2*tan(e/2 +
f*x/2)**3/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c*
*2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 8*A*a**2*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(
e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2
+ 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 24*A*a**2*tan(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f
*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/
2) - 3*c**2*f) + 8*A*a**2/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f
*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 12*B*a**2*f*x*tan(e/2 + f*x
/2)**5/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*
f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 36*B*a**2*f*x*tan(e/2 + f*x/2)**4/(3*c**2*f*ta
n(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)*
*2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 48*B*a**2*f*x*tan(e/2 + f*x/2)**3/(3*c**2*f*tan(e/2 + f*x/2)**5 -
9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e
/2 + f*x/2) - 3*c**2*f) - 48*B*a**2*f*x*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 +
f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**
2*f) + 36*B*a**2*f*x*tan(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f
*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 12*B*a**2*f*x/(
3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/
2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 24*B*a**2*tan(e/2 + f*x/2)**4/(3*c**2*f*tan(e/2 + f*x/
2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*
f*tan(e/2 + f*x/2) - 3*c**2*f) - 78*B*a**2*tan(e/2 + f*x/2)**3/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/
2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*
c**2*f) + 74*B*a**2*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2
*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 90*B*a**2*tan
(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 1
2*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 38*B*a**2/(3*c**2*f*tan(e/2 + f*x/2)**5
- 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan
(e/2 + f*x/2) - 3*c**2*f), Ne(f, 0)), (x*(A + B*sin(e))*(a*sin(e) + a)**2/(-c*sin(e) + c)**2, True))
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Giac [A] time = 1.19942, size = 182, normalized size = 1.67 \begin{align*} \frac{\frac{3 \,{\left (A a^{2} + 4 \, B a^{2}\right )}{\left (f x + e\right )}}{c^{2}} - \frac{6 \, B a^{2}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )} c^{2}} + \frac{8 \,{\left (3 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 3 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 9 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + A a^{2} + 4 \, B a^{2}\right )}}{c^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{3}}}{3 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algorithm="giac")
[Out]
1/3*(3*(A*a^2 + 4*B*a^2)*(f*x + e)/c^2 - 6*B*a^2/((tan(1/2*f*x + 1/2*e)^2 + 1)*c^2) + 8*(3*B*a^2*tan(1/2*f*x +
1/2*e)^2 - 3*A*a^2*tan(1/2*f*x + 1/2*e) - 9*B*a^2*tan(1/2*f*x + 1/2*e) + A*a^2 + 4*B*a^2)/(c^2*(tan(1/2*f*x +
1/2*e) - 1)^3))/f