3.45 \(\int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx\)
Optimal. Leaf size=163 \[ \frac {a^3 c^3 (A+B) \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {5 a^3 (2 A+5 B) \cos (e+f x)}{2 c^2 f}-\frac {5 a^3 (2 A+5 B) \cos ^3(e+f x)}{6 f \left (c^2-c^2 \sin (e+f x)\right )}+\frac {5 a^3 x (2 A+5 B)}{2 c^2}-\frac {2 a^3 c (2 A+5 B) \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^3} \]
[Out]
5/2*a^3*(2*A+5*B)*x/c^2-5/2*a^3*(2*A+5*B)*cos(f*x+e)/c^2/f+1/3*a^3*(A+B)*c^3*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^5
-2/3*a^3*(2*A+5*B)*c*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^3-5/6*a^3*(2*A+5*B)*cos(f*x+e)^3/f/(c^2-c^2*sin(f*x+e))
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Rubi [A] time = 0.35, antiderivative size = 163, normalized size of antiderivative = 1.00,
number of steps used = 6, number of rules used = 6, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used =
{2967, 2859, 2680, 2679, 2682, 8} \[ -\frac {5 a^3 (2 A+5 B) \cos (e+f x)}{2 c^2 f}+\frac {a^3 c^3 (A+B) \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {5 a^3 (2 A+5 B) \cos ^3(e+f x)}{6 f \left (c^2-c^2 \sin (e+f x)\right )}+\frac {5 a^3 x (2 A+5 B)}{2 c^2}-\frac {2 a^3 c (2 A+5 B) \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^3} \]
Antiderivative was successfully verified.
[In]
Int[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^2,x]
[Out]
(5*a^3*(2*A + 5*B)*x)/(2*c^2) - (5*a^3*(2*A + 5*B)*Cos[e + f*x])/(2*c^2*f) + (a^3*(A + B)*c^3*Cos[e + f*x]^7)/
(3*f*(c - c*Sin[e + f*x])^5) - (2*a^3*(2*A + 5*B)*c*Cos[e + f*x]^5)/(3*f*(c - c*Sin[e + f*x])^3) - (5*a^3*(2*A
+ 5*B)*Cos[e + f*x]^3)/(6*f*(c^2 - c^2*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 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 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))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx &=\left (a^3 c^3\right ) \int \frac {\cos ^6(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^5} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {1}{3} \left (a^3 (2 A+5 B) c^2\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^4} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {2 a^3 (2 A+5 B) c \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^3}+\frac {1}{3} \left (5 a^3 (2 A+5 B)\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^2} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {2 a^3 (2 A+5 B) c \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^3}-\frac {5 a^3 (2 A+5 B) \cos ^3(e+f x)}{6 f \left (c^2-c^2 \sin (e+f x)\right )}+\frac {\left (5 a^3 (2 A+5 B)\right ) \int \frac {\cos ^2(e+f x)}{c-c \sin (e+f x)} \, dx}{2 c}\\ &=-\frac {5 a^3 (2 A+5 B) \cos (e+f x)}{2 c^2 f}+\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {2 a^3 (2 A+5 B) c \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^3}-\frac {5 a^3 (2 A+5 B) \cos ^3(e+f x)}{6 f \left (c^2-c^2 \sin (e+f x)\right )}+\frac {\left (5 a^3 (2 A+5 B)\right ) \int 1 \, dx}{2 c^2}\\ &=\frac {5 a^3 (2 A+5 B) x}{2 c^2}-\frac {5 a^3 (2 A+5 B) \cos (e+f x)}{2 c^2 f}+\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{3 f (c-c \sin (e+f x))^5}-\frac {2 a^3 (2 A+5 B) c \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^3}-\frac {5 a^3 (2 A+5 B) \cos ^3(e+f x)}{6 f \left (c^2-c^2 \sin (e+f x)\right )}\\ \end {align*}
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Mathematica [A] time = 0.88, size = 280, normalized size = 1.72 \[ \frac {a^3 (\sin (e+f x)+1)^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (64 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )+30 (2 A+5 B) (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-12 (A+5 B) \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-32 (7 A+13 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+32 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-3 B \sin (2 (e+f x)) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3\right )}{12 f (c-c \sin (e+f x))^2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^6} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + a*Sin[e + f*x])^3*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^2,x]
[Out]
(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3*(32*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/
2]) + 30*(2*A + 5*B)*(e + f*x)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - 12*(A + 5*B)*Cos[e + f*x]*(Cos[(e + f
*x)/2] - Sin[(e + f*x)/2])^3 + 64*(A + B)*Sin[(e + f*x)/2] - 32*(7*A + 13*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)
/2])^2*Sin[(e + f*x)/2] - 3*B*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sin[2*(e + f*x)]))/(12*f*(Cos[(e + f*x)/
2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^2)
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fricas [A] time = 0.45, size = 286, normalized size = 1.75 \[ \frac {3 \, B a^{3} \cos \left (f x + e\right )^{4} - 6 \, {\left (A + 4 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} - 30 \, {\left (2 \, A + 5 \, B\right )} a^{3} f x - 16 \, {\left (A + B\right )} a^{3} + {\left (15 \, {\left (2 \, A + 5 \, B\right )} a^{3} f x + {\left (62 \, A + 131 \, B\right )} a^{3}\right )} \cos \left (f x + e\right )^{2} - {\left (15 \, {\left (2 \, A + 5 \, B\right )} a^{3} f x - 2 \, {\left (26 \, A + 71 \, B\right )} a^{3}\right )} \cos \left (f x + e\right ) - {\left (3 \, B a^{3} \cos \left (f x + e\right )^{3} - 30 \, {\left (2 \, A + 5 \, B\right )} a^{3} f x + 3 \, {\left (2 \, A + 9 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 16 \, {\left (A + B\right )} a^{3} - {\left (15 \, {\left (2 \, A + 5 \, B\right )} a^{3} f x - 2 \, {\left (34 \, A + 79 \, B\right )} a^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algorithm="fricas")
[Out]
1/6*(3*B*a^3*cos(f*x + e)^4 - 6*(A + 4*B)*a^3*cos(f*x + e)^3 - 30*(2*A + 5*B)*a^3*f*x - 16*(A + B)*a^3 + (15*(
2*A + 5*B)*a^3*f*x + (62*A + 131*B)*a^3)*cos(f*x + e)^2 - (15*(2*A + 5*B)*a^3*f*x - 2*(26*A + 71*B)*a^3)*cos(f
*x + e) - (3*B*a^3*cos(f*x + e)^3 - 30*(2*A + 5*B)*a^3*f*x + 3*(2*A + 9*B)*a^3*cos(f*x + e)^2 + 16*(A + B)*a^3
- (15*(2*A + 5*B)*a^3*f*x - 2*(34*A + 79*B)*a^3)*cos(f*x + e))*sin(f*x + e))/(c^2*f*cos(f*x + e)^2 - c^2*f*co
s(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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giac [A] time = 0.18, size = 233, normalized size = 1.43 \[ \frac {\frac {15 \, {\left (2 \, A a^{3} + 5 \, B a^{3}\right )} {\left (f x + e\right )}}{c^{2}} + \frac {6 \, {\left (B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 10 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, A a^{3} - 10 \, B a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} c^{2}} + \frac {16 \, {\left (3 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 9 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 12 \, A a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 24 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, A a^{3} + 11 \, B a^{3}\right )}}{c^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}}{6 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algorithm="giac")
[Out]
1/6*(15*(2*A*a^3 + 5*B*a^3)*(f*x + e)/c^2 + 6*(B*a^3*tan(1/2*f*x + 1/2*e)^3 - 2*A*a^3*tan(1/2*f*x + 1/2*e)^2 -
10*B*a^3*tan(1/2*f*x + 1/2*e)^2 - B*a^3*tan(1/2*f*x + 1/2*e) - 2*A*a^3 - 10*B*a^3)/((tan(1/2*f*x + 1/2*e)^2 +
1)^2*c^2) + 16*(3*A*a^3*tan(1/2*f*x + 1/2*e)^2 + 9*B*a^3*tan(1/2*f*x + 1/2*e)^2 - 12*A*a^3*tan(1/2*f*x + 1/2*
e) - 24*B*a^3*tan(1/2*f*x + 1/2*e) + 5*A*a^3 + 11*B*a^3)/(c^2*(tan(1/2*f*x + 1/2*e) - 1)^3))/f
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maple [B] time = 0.47, size = 399, normalized size = 2.45 \[ \frac {8 a^{3} A}{c^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {24 a^{3} B}{c^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {32 a^{3} A}{3 c^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {32 a^{3} B}{3 c^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {16 a^{3} A}{c^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {16 a^{3} B}{c^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\frac {a^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) B}{c^{2} f \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}-\frac {2 a^{3} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) A}{c^{2} f \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}-\frac {10 a^{3} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) B}{c^{2} f \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}-\frac {a^{3} B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c^{2} f \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}-\frac {2 a^{3} A}{c^{2} f \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}-\frac {10 a^{3} B}{c^{2} f \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {25 a^{3} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) B}{c^{2} f}+\frac {10 a^{3} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) A}{c^{2} f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x)
[Out]
8*a^3/c^2/f/(tan(1/2*f*x+1/2*e)-1)*A+24*a^3/c^2/f/(tan(1/2*f*x+1/2*e)-1)*B-32/3*a^3/c^2/f/(tan(1/2*f*x+1/2*e)-
1)^3*A-32/3*a^3/c^2/f/(tan(1/2*f*x+1/2*e)-1)^3*B-16*a^3/c^2/f/(tan(1/2*f*x+1/2*e)-1)^2*A-16*a^3/c^2/f/(tan(1/2
*f*x+1/2*e)-1)^2*B+a^3/c^2/f/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^3*B-2*a^3/c^2/f/(1+tan(1/2*f*x+1/2*
e)^2)^2*tan(1/2*f*x+1/2*e)^2*A-10*a^3/c^2/f/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^2*B-a^3/c^2/f/(1+tan
(1/2*f*x+1/2*e)^2)^2*B*tan(1/2*f*x+1/2*e)-2*a^3/c^2/f/(1+tan(1/2*f*x+1/2*e)^2)^2*A-10*a^3/c^2/f/(1+tan(1/2*f*x
+1/2*e)^2)^2*B+25*a^3/c^2/f*arctan(tan(1/2*f*x+1/2*e))*B+10*a^3/c^2/f*arctan(tan(1/2*f*x+1/2*e))*A
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maxima [B] time = 0.61, size = 1386, normalized size = 8.50 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algorithm="maxima")
[Out]
1/3*(B*a^3*((75*sin(f*x + e)/(cos(f*x + e) + 1) - 97*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 126*sin(f*x + e)^3/
(cos(f*x + e) + 1)^3 - 98*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 63*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 21*si
n(f*x + e)^6/(cos(f*x + e) + 1)^6 - 32)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 5*c^2*sin(f*x + e)^2/(c
os(f*x + e) + 1)^2 - 7*c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 7*c^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 5
*c^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 3*c^2*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^2*sin(f*x + e)^7/(cos
(f*x + e) + 1)^7) + 21*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) + 4*A*a^3*((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) + 12*B*a^3*((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/(c
os(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) + 6*A*a^3*((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) + 6*B*a^3*((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*A*a^3*(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) + 6*A*a^3*(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) + 2*B*a^3*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 1)/(c^2 - 3*c^2*sin(f*x + e)/(co
s(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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mupad [B] time = 14.00, size = 341, normalized size = 2.09 \[ \frac {5\,a^3\,\mathrm {atan}\left (\frac {5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,A+5\,B\right )}{10\,A\,a^3+25\,B\,a^3}\right )\,\left (2\,A+5\,B\right )}{c^2\,f}-\frac {\frac {46\,A\,a^3}{3}-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (38\,A\,a^3+93\,B\,a^3\right )+\frac {118\,B\,a^3}{3}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (8\,A\,a^3+25\,B\,a^3\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (34\,A\,a^3+77\,B\,a^3\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (72\,A\,a^3+166\,B\,a^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {106\,A\,a^3}{3}+\frac {328\,B\,a^3}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {128\,A\,a^3}{3}+\frac {359\,B\,a^3}{3}\right )}{f\,\left (-c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+3\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6-5\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+7\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-7\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+5\,c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-3\,c^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+c^2\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^3)/(c - c*sin(e + f*x))^2,x)
[Out]
(5*a^3*atan((5*a^3*tan(e/2 + (f*x)/2)*(2*A + 5*B))/(10*A*a^3 + 25*B*a^3))*(2*A + 5*B))/(c^2*f) - ((46*A*a^3)/3
- tan(e/2 + (f*x)/2)*(38*A*a^3 + 93*B*a^3) + (118*B*a^3)/3 + tan(e/2 + (f*x)/2)^6*(8*A*a^3 + 25*B*a^3) - tan(
e/2 + (f*x)/2)^5*(34*A*a^3 + 77*B*a^3) - tan(e/2 + (f*x)/2)^3*(72*A*a^3 + 166*B*a^3) + tan(e/2 + (f*x)/2)^4*((
106*A*a^3)/3 + (328*B*a^3)/3) + tan(e/2 + (f*x)/2)^2*((128*A*a^3)/3 + (359*B*a^3)/3))/(f*(5*c^2*tan(e/2 + (f*x
)/2)^2 - 7*c^2*tan(e/2 + (f*x)/2)^3 + 7*c^2*tan(e/2 + (f*x)/2)^4 - 5*c^2*tan(e/2 + (f*x)/2)^5 + 3*c^2*tan(e/2
+ (f*x)/2)^6 - c^2*tan(e/2 + (f*x)/2)^7 + c^2 - 3*c^2*tan(e/2 + (f*x)/2)))
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sympy [A] time = 29.89, size = 4665, normalized size = 28.62 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))**3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**2,x)
[Out]
Piecewise((30*A*a**3*f*x*tan(e/2 + f*x/2)**7/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 3
0*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e
/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) - 90*A*a**3*f*x*tan(e/2 + f*x/2)**6/(6*c**2*f*tan(e/2
+ f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 +
42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) + 150*A
*a**3*f*x*tan(e/2 + f*x/2)**5/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/
2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2
+ 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) - 210*A*a**3*f*x*tan(e/2 + f*x/2)**4/(6*c**2*f*tan(e/2 + f*x/2)**7 -
18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(
e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) + 210*A*a**3*f*x*tan(
e/2 + f*x/2)**3/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5
- 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*ta
n(e/2 + f*x/2) - 6*c**2*f) - 150*A*a**3*f*x*tan(e/2 + f*x/2)**2/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(
e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**
3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) + 90*A*a**3*f*x*tan(e/2 + f*x/2)/(6
*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/
2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6
*c**2*f) - 30*A*a**3*f*x/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f
*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*
c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) + 48*A*a**3*tan(e/2 + f*x/2)**6/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*
tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/
2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) - 204*A*a**3*tan(e/2 + f*x/2)**
5/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*ta
n(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2)
- 6*c**2*f) + 212*A*a**3*tan(e/2 + f*x/2)**4/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 +
30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(
e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) - 432*A*a**3*tan(e/2 + f*x/2)**3/(6*c**2*f*tan(e/2 +
f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42
*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) + 256*A*a
**3*tan(e/2 + f*x/2)**2/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*
x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c
**2*f*tan(e/2 + f*x/2) - 6*c**2*f) - 228*A*a**3*tan(e/2 + f*x/2)/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan
(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)*
*3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) + 92*A*a**3/(6*c**2*f*tan(e/2 + f*
x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c
**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) + 75*B*a**3
*f*x*tan(e/2 + f*x/2)**7/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f
*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*
c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) - 225*B*a**3*f*x*tan(e/2 + f*x/2)**6/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c*
*2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 +
f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) + 375*B*a**3*f*x*tan(e/2 +
f*x/2)**5/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*
c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2
+ f*x/2) - 6*c**2*f) - 525*B*a**3*f*x*tan(e/2 + f*x/2)**4/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 +
f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 3
0*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) + 525*B*a**3*f*x*tan(e/2 + f*x/2)**3/(6*
c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2
+ f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*
c**2*f) - 375*B*a**3*f*x*tan(e/2 + f*x/2)**2/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 3
0*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e
/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) + 225*B*a**3*f*x*tan(e/2 + f*x/2)/(6*c**2*f*tan(e/2 +
f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42
*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) - 75*B*a*
*3*f*x/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2
*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f
*x/2) - 6*c**2*f) + 150*B*a**3*tan(e/2 + f*x/2)**6/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)*
*6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f
*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) - 462*B*a**3*tan(e/2 + f*x/2)**5/(6*c**2*f*tan(e
/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4
+ 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) + 65
6*B*a**3*tan(e/2 + f*x/2)**4/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2
+ f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 +
18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) - 996*B*a**3*tan(e/2 + f*x/2)**3/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c*
*2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 +
f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) + 718*B*a**3*tan(e/2 + f*x
/2)**2/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2
*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f
*x/2) - 6*c**2*f) - 558*B*a**3*tan(e/2 + f*x/2)/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**2*f*tan(e/2 + f*x/2)**6
+ 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 + f*x/2)**3 - 30*c**2*f*ta
n(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f) + 236*B*a**3/(6*c**2*f*tan(e/2 + f*x/2)**7 - 18*c**
2*f*tan(e/2 + f*x/2)**6 + 30*c**2*f*tan(e/2 + f*x/2)**5 - 42*c**2*f*tan(e/2 + f*x/2)**4 + 42*c**2*f*tan(e/2 +
f*x/2)**3 - 30*c**2*f*tan(e/2 + f*x/2)**2 + 18*c**2*f*tan(e/2 + f*x/2) - 6*c**2*f), Ne(f, 0)), (x*(A + B*sin(e
))*(a*sin(e) + a)**3/(-c*sin(e) + c)**2, True))
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