3.31 \(\int \frac{(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx\)
Optimal. Leaf size=117 \[ \frac{a^2 c^2 (A+B) \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac{3 a^2 (2 A+3 B) \cos (e+f x)}{2 c f}+\frac{a^2 (2 A+3 B) \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))}-\frac{3 a^2 x (2 A+3 B)}{2 c} \]
[Out]
(-3*a^2*(2*A + 3*B)*x)/(2*c) + (3*a^2*(2*A + 3*B)*Cos[e + f*x])/(2*c*f) + (a^2*(A + B)*c^2*Cos[e + f*x]^5)/(f*
(c - c*Sin[e + f*x])^3) + (a^2*(2*A + 3*B)*Cos[e + f*x]^3)/(2*f*(c - c*Sin[e + f*x]))
________________________________________________________________________________________
Rubi [A] time = 0.288545, antiderivative size = 117, 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, 2679, 2682, 8} \[ \frac{a^2 c^2 (A+B) \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac{3 a^2 (2 A+3 B) \cos (e+f x)}{2 c f}+\frac{a^2 (2 A+3 B) \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))}-\frac{3 a^2 x (2 A+3 B)}{2 c} \]
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]),x]
[Out]
(-3*a^2*(2*A + 3*B)*x)/(2*c) + (3*a^2*(2*A + 3*B)*Cos[e + f*x])/(2*c*f) + (a^2*(A + B)*c^2*Cos[e + f*x]^5)/(f*
(c - c*Sin[e + f*x])^3) + (a^2*(2*A + 3*B)*Cos[e + f*x]^3)/(2*f*(c - c*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 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 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)} \, 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))^3} \, dx\\ &=\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}-\left (a^2 (2 A+3 B) c\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^2} \, dx\\ &=\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac{a^2 (2 A+3 B) \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))}-\frac{1}{2} \left (3 a^2 (2 A+3 B)\right ) \int \frac{\cos ^2(e+f x)}{c-c \sin (e+f x)} \, dx\\ &=\frac{3 a^2 (2 A+3 B) \cos (e+f x)}{2 c f}+\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac{a^2 (2 A+3 B) \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))}-\frac{\left (3 a^2 (2 A+3 B)\right ) \int 1 \, dx}{2 c}\\ &=-\frac{3 a^2 (2 A+3 B) x}{2 c}+\frac{3 a^2 (2 A+3 B) \cos (e+f x)}{2 c f}+\frac{a^2 (A+B) c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac{a^2 (2 A+3 B) \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))}\\ \end{align*}
Mathematica [A] time = 1.23508, size = 191, normalized size = 1.63 \[ \frac{a^2 (\sin (e+f x)+1)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right ) (6 (2 A+3 B) (e+f x)-4 (A+3 B) \cos (e+f x)-B \sin (2 (e+f x)))-\sin \left (\frac{1}{2} (e+f x)\right ) (-4 (A+3 B) \cos (e+f x)+4 A (3 e+3 f x+8)+2 B (9 e+9 f x+16)-B \sin (2 (e+f x)))\right )}{4 c f (\sin (e+f x)-1) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4} \]
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]),x]
[Out]
(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2*(Cos[(e + f*x)/2]*(6*(2*A + 3*B)*(e + f*x) - 4
*(A + 3*B)*Cos[e + f*x] - B*Sin[2*(e + f*x)]) - Sin[(e + f*x)/2]*(4*A*(8 + 3*e + 3*f*x) + 2*B*(16 + 9*e + 9*f*
x) - 4*(A + 3*B)*Cos[e + f*x] - B*Sin[2*(e + f*x)])))/(4*c*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(-1 + Sin
[e + f*x]))
________________________________________________________________________________________
Maple [B] time = 0.113, size = 299, normalized size = 2.6 \begin{align*} -8\,{\frac{A{a}^{2}}{cf \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}-8\,{\frac{B{a}^{2}}{cf \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}-{\frac{B{a}^{2}}{cf} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+2\,{\frac{{a}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}A}{cf \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}+6\,{\frac{{a}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}B}{cf \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{B{a}^{2}}{cf}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+2\,{\frac{A{a}^{2}}{cf \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}+6\,{\frac{B{a}^{2}}{cf \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-9\,{\frac{{a}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) B}{cf}}-6\,{\frac{{a}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) A}{cf}} \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)),x)
[Out]
-8/f*a^2/c/(tan(1/2*f*x+1/2*e)-1)*A-8/f*a^2/c/(tan(1/2*f*x+1/2*e)-1)*B-1/f*a^2/c/(1+tan(1/2*f*x+1/2*e)^2)^2*ta
n(1/2*f*x+1/2*e)^3*B+2/f*a^2/c/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^2*A+6/f*a^2/c/(1+tan(1/2*f*x+1/2*
e)^2)^2*tan(1/2*f*x+1/2*e)^2*B+1/f*a^2/c/(1+tan(1/2*f*x+1/2*e)^2)^2*B*tan(1/2*f*x+1/2*e)+2/f*a^2/c/(1+tan(1/2*
f*x+1/2*e)^2)^2*A+6/f*a^2/c/(1+tan(1/2*f*x+1/2*e)^2)^2*B-9/f*a^2/c*arctan(tan(1/2*f*x+1/2*e))*B-6/f*a^2/c*arct
an(tan(1/2*f*x+1/2*e))*A
________________________________________________________________________________________
Maxima [B] time = 1.48282, size = 842, normalized size = 7.2 \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)),x, algorithm="maxima")
[Out]
-(2*A*a^2*((sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2)/(c - c*sin(f*x + e)/(co
s(f*x + e) + 1) + c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + arctan(sin(
f*x + e)/(cos(f*x + e) + 1))/c) + 4*B*a^2*((sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^2/(cos(f*x + e) + 1
)^2 - 2)/(c - c*sin(f*x + e)/(cos(f*x + e) + 1) + c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c*sin(f*x + e)^3/(co
s(f*x + e) + 1)^3) + arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c) + B*a^2*((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)/(c - c*sin(f*x + e)/(cos(f*x + e) + 1) + 2*c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2*c*sin(f*x + e)
^3/(cos(f*x + e) + 1)^3 + c*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*a
rctan(sin(f*x + e)/(cos(f*x + e) + 1))/c) + 4*A*a^2*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c - 1/(c - c*sin(
f*x + e)/(cos(f*x + e) + 1))) + 2*B*a^2*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c - 1/(c - c*sin(f*x + e)/(co
s(f*x + e) + 1))) - 2*A*a^2/(c - c*sin(f*x + e)/(cos(f*x + e) + 1)))/f
________________________________________________________________________________________
Fricas [A] time = 1.39811, size = 423, normalized size = 3.62 \begin{align*} \frac{B a^{2} \cos \left (f x + e\right )^{3} - 3 \,{\left (2 \, A + 3 \, B\right )} a^{2} f x + 2 \,{\left (A + 3 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} + 8 \,{\left (A + B\right )} a^{2} -{\left (3 \,{\left (2 \, A + 3 \, B\right )} a^{2} f x -{\left (10 \, A + 13 \, B\right )} a^{2}\right )} \cos \left (f x + e\right ) +{\left (3 \,{\left (2 \, A + 3 \, B\right )} a^{2} f x + B a^{2} \cos \left (f x + e\right )^{2} -{\left (2 \, A + 5 \, B\right )} a^{2} \cos \left (f x + e\right ) + 8 \,{\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )}{2 \,{\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\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)),x, algorithm="fricas")
[Out]
1/2*(B*a^2*cos(f*x + e)^3 - 3*(2*A + 3*B)*a^2*f*x + 2*(A + 3*B)*a^2*cos(f*x + e)^2 + 8*(A + B)*a^2 - (3*(2*A +
3*B)*a^2*f*x - (10*A + 13*B)*a^2)*cos(f*x + e) + (3*(2*A + 3*B)*a^2*f*x + B*a^2*cos(f*x + e)^2 - (2*A + 5*B)*
a^2*cos(f*x + e) + 8*(A + B)*a^2)*sin(f*x + e))/(c*f*cos(f*x + e) - c*f*sin(f*x + e) + c*f)
________________________________________________________________________________________
Sympy [A] time = 9.73573, size = 2365, normalized size = 20.21 \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)),x)
[Out]
Piecewise((-6*A*a**2*f*x*tan(e/2 + f*x/2)**5/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*ta
n(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) + 6*A*a**2*f*x*tan(e/2 + f*x/2
)**4/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/
2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 12*A*a**2*f*x*tan(e/2 + f*x/2)**3/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f
*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f)
+ 12*A*a**2*f*x*tan(e/2 + f*x/2)**2/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 +
f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 6*A*a**2*f*x*tan(e/2 + f*x/2)/(2*c*f
*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c
*f*tan(e/2 + f*x/2) - 2*c*f) + 6*A*a**2*f*x/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan
(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 16*A*a**2*tan(e/2 + f*x/2)**5
/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**
2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 28*A*a**2*tan(e/2 + f*x/2)**3/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2
+ f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 4*A*a
**2*tan(e/2 + f*x/2)**2/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4
*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 12*A*a**2*tan(e/2 + f*x/2)/(2*c*f*tan(e/2 + f*x/2
)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x
/2) - 2*c*f) - 4*A*a**2/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4
*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 9*B*a**2*f*x*tan(e/2 + f*x/2)**5/(2*c*f*tan(e/2 +
f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2
+ f*x/2) - 2*c*f) + 9*B*a**2*f*x*tan(e/2 + f*x/2)**4/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 +
4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 18*B*a**2*f*x*tan(e
/2 + f*x/2)**3/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(
e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) + 18*B*a**2*f*x*tan(e/2 + f*x/2)**2/(2*c*f*tan(e/2 + f*x/2)*
*5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2
) - 2*c*f) - 9*B*a**2*f*x*tan(e/2 + f*x/2)/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(
e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) + 9*B*a**2*f*x/(2*c*f*tan(e/2 +
f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2
+ f*x/2) - 2*c*f) - 18*B*a**2*tan(e/2 + f*x/2)**5/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c
*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 22*B*a**2*tan(e/2 + f*x
/2)**3/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*
x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 6*B*a**2*tan(e/2 + f*x/2)**2/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*ta
n(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) -
8*B*a**2*tan(e/2 + f*x/2)/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 -
4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 10*B*a**2/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*ta
n(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f), N
e(f, 0)), (x*(A + B*sin(e))*(a*sin(e) + a)**2/(-c*sin(e) + c), True))
________________________________________________________________________________________
Giac [A] time = 1.17209, size = 220, normalized size = 1.88 \begin{align*} -\frac{\frac{3 \,{\left (2 \, A a^{2} + 3 \, B a^{2}\right )}{\left (f x + e\right )}}{c} + \frac{16 \,{\left (A a^{2} + B a^{2}\right )}}{c{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}} + \frac{2 \,{\left (B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 2 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 6 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, A a^{2} - 6 \, B a^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{2} c}}{2 \, 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)),x, algorithm="giac")
[Out]
-1/2*(3*(2*A*a^2 + 3*B*a^2)*(f*x + e)/c + 16*(A*a^2 + B*a^2)/(c*(tan(1/2*f*x + 1/2*e) - 1)) + 2*(B*a^2*tan(1/2
*f*x + 1/2*e)^3 - 2*A*a^2*tan(1/2*f*x + 1/2*e)^2 - 6*B*a^2*tan(1/2*f*x + 1/2*e)^2 - B*a^2*tan(1/2*f*x + 1/2*e)
- 2*A*a^2 - 6*B*a^2)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*c))/f