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]

-3/2*a^2*(2*A+3*B)*x/c+3/2*a^2*(2*A+3*B)*cos(f*x+e)/c/f+a^2*(A+B)*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^3+1/2*a^
2*(2*A+3*B)*cos(f*x+e)^3/f/(c-c*sin(f*x+e))

________________________________________________________________________________________

Rubi [A]  time = 0.29, antiderivative size = 117, normalized size of antiderivative = 1.00, 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 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+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 = 1.30, 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]))

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 177, normalized size = 1.51 \[ \frac {B a^{2} \cos \left (f x + e\right )^{3} - 3 \, {\left (2 \, A + 3 \, B\right )} a^{2} f x + 2 \, {\left (A + 3 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} + 8 \, {\left (A + B\right )} a^{2} - {\left (3 \, {\left (2 \, A + 3 \, B\right )} a^{2} f x - {\left (10 \, A + 13 \, B\right )} a^{2}\right )} \cos \left (f x + e\right ) + {\left (3 \, {\left (2 \, A + 3 \, B\right )} a^{2} f x + B a^{2} \cos \left (f x + e\right )^{2} - {\left (2 \, A + 5 \, B\right )} a^{2} \cos \left (f x + e\right ) + 8 \, {\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )}{2 \, {\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \]

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)

________________________________________________________________________________________

giac [A]  time = 0.18, size = 163, normalized size = 1.39 \[ -\frac {\frac {3 \, {\left (2 \, A a^{2} + 3 \, B a^{2}\right )} {\left (f x + e\right )}}{c} + \frac {16 \, {\left (A a^{2} + B a^{2}\right )}}{c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}} + \frac {2 \, {\left (B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 6 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, A a^{2} - 6 \, B a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} c}}{2 \, f} \]

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

________________________________________________________________________________________

maple [B]  time = 0.43, size = 299, normalized size = 2.56 \[ -\frac {8 a^{2} A}{f c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {8 a^{2} B}{f c \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {a^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) B}{f c \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {2 a^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) A}{f c \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {6 a^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) B}{f c \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {a^{2} B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f c \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {2 a^{2} A}{f c \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {6 a^{2} B}{f c \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}-\frac {9 a^{2} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) B}{f c}-\frac {6 a^{2} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) A}{f c} \]

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 = 0.48, size = 624, normalized size = 5.33 \[ -\frac {2 \, A a^{2} {\left (\frac {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 2}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c}\right )} + 4 \, B a^{2} {\left (\frac {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 2}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c}\right )} + B a^{2} {\left (\frac {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {3 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 4}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {c \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {3 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c}\right )} + 4 \, A a^{2} {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} - \frac {1}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} + 2 \, B a^{2} {\left (\frac {\arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} - \frac {1}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - \frac {2 \, A a^{2}}{c - \frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}}{f} \]

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

________________________________________________________________________________________

mupad [B]  time = 14.70, size = 244, normalized size = 2.09 \[ \frac {10\,A\,a^2-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,A\,a^2+5\,B\,a^2\right )+14\,B\,a^2-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,A\,a^2+7\,B\,a^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (8\,A\,a^2+9\,B\,a^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (18\,A\,a^2+21\,B\,a^2\right )}{f\,\left (-c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-2\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+2\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+c\right )}-\frac {3\,a^2\,\mathrm {atan}\left (\frac {3\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,A+3\,B\right )}{6\,A\,a^2+9\,B\,a^2}\right )\,\left (2\,A+3\,B\right )}{c\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x)),x)

[Out]

(10*A*a^2 - tan(e/2 + (f*x)/2)*(2*A*a^2 + 5*B*a^2) + 14*B*a^2 - tan(e/2 + (f*x)/2)^3*(2*A*a^2 + 7*B*a^2) + tan
(e/2 + (f*x)/2)^4*(8*A*a^2 + 9*B*a^2) + tan(e/2 + (f*x)/2)^2*(18*A*a^2 + 21*B*a^2))/(f*(c - c*tan(e/2 + (f*x)/
2) + 2*c*tan(e/2 + (f*x)/2)^2 - 2*c*tan(e/2 + (f*x)/2)^3 + c*tan(e/2 + (f*x)/2)^4 - c*tan(e/2 + (f*x)/2)^5)) -
(3*a^2*atan((3*a^2*tan(e/2 + (f*x)/2)*(2*A + 3*B))/(6*A*a^2 + 9*B*a^2))*(2*A + 3*B))/(c*f)

________________________________________________________________________________________

sympy [A]  time = 8.09, size = 2365, normalized size = 20.21 \[ \text {result too large to display} \]

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)**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) + 4*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) - 36*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) + 4*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) - 20*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)**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) + 14*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) - 42*B*a**2*tan(e/2 + f*x/2)**2/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*t
an(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*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) - 28*B*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),
Ne(f, 0)), (x*(A + B*sin(e))*(a*sin(e) + a)**2/(-c*sin(e) + c), True))

________________________________________________________________________________________