3.68 \(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4} \, dx\)

Optimal. Leaf size=135 \[ \frac {4 (5 A-2 B) \tan ^3(e+f x)}{105 a^2 c^4 f}+\frac {4 (5 A-2 B) \tan (e+f x)}{35 a^2 c^4 f}+\frac {(5 A-2 B) \sec ^3(e+f x)}{35 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {(A+B) \sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2} \]

[Out]

1/7*(A+B)*sec(f*x+e)^3/a^2/f/(c^2-c^2*sin(f*x+e))^2+1/35*(5*A-2*B)*sec(f*x+e)^3/a^2/f/(c^4-c^4*sin(f*x+e))+4/3
5*(5*A-2*B)*tan(f*x+e)/a^2/c^4/f+4/105*(5*A-2*B)*tan(f*x+e)^3/a^2/c^4/f

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Rubi [A]  time = 0.27, antiderivative size = 135, 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, 2672, 3767} \[ \frac {4 (5 A-2 B) \tan ^3(e+f x)}{105 a^2 c^4 f}+\frac {4 (5 A-2 B) \tan (e+f x)}{35 a^2 c^4 f}+\frac {(5 A-2 B) \sec ^3(e+f x)}{35 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {(A+B) \sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^4),x]

[Out]

((A + B)*Sec[e + f*x]^3)/(7*a^2*f*(c^2 - c^2*Sin[e + f*x])^2) + ((5*A - 2*B)*Sec[e + f*x]^3)/(35*a^2*f*(c^4 -
c^4*Sin[e + f*x])) + (4*(5*A - 2*B)*Tan[e + f*x])/(35*a^2*c^4*f) + (4*(5*A - 2*B)*Tan[e + f*x]^3)/(105*a^2*c^4
*f)

Rule 2672

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 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 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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps

\begin {align*} \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4} \, dx &=\frac {\int \frac {\sec ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^2} \, dx}{a^2 c^2}\\ &=\frac {(A+B) \sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {(5 A-2 B) \int \frac {\sec ^4(e+f x)}{c-c \sin (e+f x)} \, dx}{7 a^2 c^3}\\ &=\frac {(A+B) \sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {(5 A-2 B) \sec ^3(e+f x)}{35 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {(4 (5 A-2 B)) \int \sec ^4(e+f x) \, dx}{35 a^2 c^4}\\ &=\frac {(A+B) \sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {(5 A-2 B) \sec ^3(e+f x)}{35 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}-\frac {(4 (5 A-2 B)) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{35 a^2 c^4 f}\\ &=\frac {(A+B) \sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac {(5 A-2 B) \sec ^3(e+f x)}{35 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac {4 (5 A-2 B) \tan (e+f x)}{35 a^2 c^4 f}+\frac {4 (5 A-2 B) \tan ^3(e+f x)}{105 a^2 c^4 f}\\ \end {align*}

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Mathematica [B]  time = 0.96, size = 285, normalized size = 2.11 \[ -\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) (42 (25 A+4 B) \cos (e+f x)-512 (5 A-2 B) \cos (2 (e+f x))-4480 A \sin (e+f x)-600 A \sin (2 (e+f x))-960 A \sin (3 (e+f x))-300 A \sin (4 (e+f x))+320 A \sin (5 (e+f x))+225 A \cos (3 (e+f x))-1280 A \cos (4 (e+f x))-75 A \cos (5 (e+f x))+1792 B \sin (e+f x)-96 B \sin (2 (e+f x))+384 B \sin (3 (e+f x))-48 B \sin (4 (e+f x))-128 B \sin (5 (e+f x))+36 B \cos (3 (e+f x))+512 B \cos (4 (e+f x))-12 B \cos (5 (e+f x))-2688 B)}{13440 a^2 c^4 f (\sin (e+f x)-1)^4 (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^4),x]

[Out]

-1/13440*((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-2688*B + 42*(25*A + 4*
B)*Cos[e + f*x] - 512*(5*A - 2*B)*Cos[2*(e + f*x)] + 225*A*Cos[3*(e + f*x)] + 36*B*Cos[3*(e + f*x)] - 1280*A*C
os[4*(e + f*x)] + 512*B*Cos[4*(e + f*x)] - 75*A*Cos[5*(e + f*x)] - 12*B*Cos[5*(e + f*x)] - 4480*A*Sin[e + f*x]
+ 1792*B*Sin[e + f*x] - 600*A*Sin[2*(e + f*x)] - 96*B*Sin[2*(e + f*x)] - 960*A*Sin[3*(e + f*x)] + 384*B*Sin[3
*(e + f*x)] - 300*A*Sin[4*(e + f*x)] - 48*B*Sin[4*(e + f*x)] + 320*A*Sin[5*(e + f*x)] - 128*B*Sin[5*(e + f*x)]
))/(a^2*c^4*f*(-1 + Sin[e + f*x])^4*(1 + Sin[e + f*x])^2)

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fricas [A]  time = 0.43, size = 151, normalized size = 1.12 \[ -\frac {16 \, {\left (5 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (5 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{2} - {\left (8 \, {\left (5 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{4} - 12 \, {\left (5 \, A - 2 \, B\right )} \cos \left (f x + e\right )^{2} - 25 \, A + 10 \, B\right )} \sin \left (f x + e\right ) - 10 \, A + 25 \, B}{105 \, {\left (a^{2} c^{4} f \cos \left (f x + e\right )^{5} + 2 \, a^{2} c^{4} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - 2 \, a^{2} c^{4} f \cos \left (f x + e\right )^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x, algorithm="fricas")

[Out]

-1/105*(16*(5*A - 2*B)*cos(f*x + e)^4 - 8*(5*A - 2*B)*cos(f*x + e)^2 - (8*(5*A - 2*B)*cos(f*x + e)^4 - 12*(5*A
- 2*B)*cos(f*x + e)^2 - 25*A + 10*B)*sin(f*x + e) - 10*A + 25*B)/(a^2*c^4*f*cos(f*x + e)^5 + 2*a^2*c^4*f*cos(
f*x + e)^3*sin(f*x + e) - 2*a^2*c^4*f*cos(f*x + e)^3)

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giac [B]  time = 0.23, size = 295, normalized size = 2.19 \[ -\frac {\frac {35 \, {\left (9 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 6 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 9 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 8 \, A - 5 \, B\right )}}{a^{2} c^{4} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} + \frac {1365 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 210 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 5775 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 105 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 12250 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 175 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 14350 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 910 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 10185 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 756 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3955 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 427 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 760 \, A - 31 \, B}{a^{2} c^{4} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{7}}}{840 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x, algorithm="giac")

[Out]

-1/840*(35*(9*A*tan(1/2*f*x + 1/2*e)^2 - 6*B*tan(1/2*f*x + 1/2*e)^2 + 15*A*tan(1/2*f*x + 1/2*e) - 9*B*tan(1/2*
f*x + 1/2*e) + 8*A - 5*B)/(a^2*c^4*(tan(1/2*f*x + 1/2*e) + 1)^3) + (1365*A*tan(1/2*f*x + 1/2*e)^6 + 210*B*tan(
1/2*f*x + 1/2*e)^6 - 5775*A*tan(1/2*f*x + 1/2*e)^5 - 105*B*tan(1/2*f*x + 1/2*e)^5 + 12250*A*tan(1/2*f*x + 1/2*
e)^4 - 175*B*tan(1/2*f*x + 1/2*e)^4 - 14350*A*tan(1/2*f*x + 1/2*e)^3 + 910*B*tan(1/2*f*x + 1/2*e)^3 + 10185*A*
tan(1/2*f*x + 1/2*e)^2 - 756*B*tan(1/2*f*x + 1/2*e)^2 - 3955*A*tan(1/2*f*x + 1/2*e) + 427*B*tan(1/2*f*x + 1/2*
e) + 760*A - 31*B)/(a^2*c^4*(tan(1/2*f*x + 1/2*e) - 1)^7))/f

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maple [A]  time = 0.50, size = 233, normalized size = 1.73 \[ \frac {-\frac {2 \left (2 A +2 B \right )}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {6 A +6 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {10 A +8 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {2 \left (10 A +9 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {2 \left (\frac {13 A}{16}+\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {\frac {23 A}{8}+\frac {11 B}{8}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {2 \left (\frac {55 A}{8}+\frac {35 B}{8}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {-\frac {A}{8}+\frac {B}{8}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {A}{8}-\frac {B}{8}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {3 A}{16}-\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{f \,a^{2} c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x)

[Out]

2/f/a^2/c^4*(-1/7*(2*A+2*B)/(tan(1/2*f*x+1/2*e)-1)^7-1/6*(6*A+6*B)/(tan(1/2*f*x+1/2*e)-1)^6-1/4*(10*A+8*B)/(ta
n(1/2*f*x+1/2*e)-1)^4-1/5*(10*A+9*B)/(tan(1/2*f*x+1/2*e)-1)^5-(13/16*A+1/8*B)/(tan(1/2*f*x+1/2*e)-1)-1/2*(23/8
*A+11/8*B)/(tan(1/2*f*x+1/2*e)-1)^2-1/3*(55/8*A+35/8*B)/(tan(1/2*f*x+1/2*e)-1)^3-1/2*(-1/8*A+1/8*B)/(tan(1/2*f
*x+1/2*e)+1)^2-1/3*(1/8*A-1/8*B)/(tan(1/2*f*x+1/2*e)+1)^3-(3/16*A-1/8*B)/(tan(1/2*f*x+1/2*e)+1))

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maxima [B]  time = 0.37, size = 835, normalized size = 6.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^4,x, algorithm="maxima")

[Out]

-2/105*(B*(36*sin(f*x + e)/(cos(f*x + e) + 1) - 132*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 68*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 - 14*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 84*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 140*sin
(f*x + e)^6/(cos(f*x + e) + 1)^6 + 140*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 105*sin(f*x + e)^8/(cos(f*x + e)
+ 1)^8 - 9)/(a^2*c^4 - 4*a^2*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)
^2 + 8*a^2*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 14*a^2*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 14*a^2*c
^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 8*a^2*c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 3*a^2*c^4*sin(f*x + e
)^8/(cos(f*x + e) + 1)^8 + 4*a^2*c^4*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - a^2*c^4*sin(f*x + e)^10/(cos(f*x +
e) + 1)^10) + 5*A*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 24*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 76*sin(f*x + e
)^3/(cos(f*x + e) + 1)^3 + 28*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 42*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 5
6*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 28*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 42*sin(f*x + e)^8/(cos(f*x +
e) + 1)^8 - 21*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - 6)/(a^2*c^4 - 4*a^2*c^4*sin(f*x + e)/(cos(f*x + e) + 1) +
3*a^2*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 8*a^2*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 14*a^2*c^4*si
n(f*x + e)^4/(cos(f*x + e) + 1)^4 + 14*a^2*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 8*a^2*c^4*sin(f*x + e)^7/
(cos(f*x + e) + 1)^7 - 3*a^2*c^4*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 4*a^2*c^4*sin(f*x + e)^9/(cos(f*x + e)
+ 1)^9 - a^2*c^4*sin(f*x + e)^10/(cos(f*x + e) + 1)^10))/f

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mupad [B]  time = 12.75, size = 197, normalized size = 1.46 \[ -\frac {\left (\frac {32\,A}{21}-\frac {64\,B}{105}-\frac {16\,A\,\sin \left (e+f\,x\right )}{21}+\frac {32\,B\,\sin \left (e+f\,x\right )}{105}\right )\,{\cos \left (e+f\,x\right )}^4+\left (\frac {8\,A}{7}+\frac {12\,B}{35}-\frac {8\,A\,\sin \left (e+f\,x\right )}{7}-\frac {12\,B\,\sin \left (e+f\,x\right )}{35}+\frac {\left (4\,\sin \left (e+f\,x\right )-4\right )\,\left (\frac {4\,A}{7}+\frac {6\,B}{35}\right )}{2}\right )\,{\cos \left (e+f\,x\right )}^3+\left (\frac {32\,B}{105}-\frac {16\,A}{21}+\frac {8\,A\,\sin \left (e+f\,x\right )}{7}-\frac {16\,B\,\sin \left (e+f\,x\right )}{35}\right )\,{\cos \left (e+f\,x\right )}^2-\frac {4\,A}{21}+\frac {10\,B}{21}+\frac {10\,A\,\sin \left (e+f\,x\right )}{21}-\frac {4\,B\,\sin \left (e+f\,x\right )}{21}}{a^2\,c^4\,f\,\left (4\,{\cos \left (e+f\,x\right )}^3\,\sin \left (e+f\,x\right )-4\,{\cos \left (e+f\,x\right )}^3+2\,{\cos \left (e+f\,x\right )}^5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^2*(c - c*sin(e + f*x))^4),x)

[Out]

-((10*B)/21 - (4*A)/21 + (10*A*sin(e + f*x))/21 - (4*B*sin(e + f*x))/21 + cos(e + f*x)^3*((8*A)/7 + (12*B)/35
- (8*A*sin(e + f*x))/7 - (12*B*sin(e + f*x))/35 + ((4*sin(e + f*x) - 4)*((4*A)/7 + (6*B)/35))/2) - cos(e + f*x
)^2*((16*A)/21 - (32*B)/105 - (8*A*sin(e + f*x))/7 + (16*B*sin(e + f*x))/35) + cos(e + f*x)^4*((32*A)/21 - (64
*B)/105 - (16*A*sin(e + f*x))/21 + (32*B*sin(e + f*x))/105))/(a^2*c^4*f*(4*cos(e + f*x)^3*sin(e + f*x) - 4*cos
(e + f*x)^3 + 2*cos(e + f*x)^5))

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sympy [A]  time = 49.83, size = 4228, normalized size = 31.32 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**4,x)

[Out]

Piecewise((-210*A*tan(e/2 + f*x/2)**9/(105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 + f*x/2)
**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e/2 + f
*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*tan(e/
2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2) - 105*a**2*c**4*f) + 420*A*tan(e/2 + f*x/2)**8/(105*a**2*c**4
*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840*a**2
*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 84
0*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2) - 1
05*a**2*c**4*f) - 280*A*tan(e/2 + f*x/2)**7/(105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 +
f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e
/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*
tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2) - 105*a**2*c**4*f) - 560*A*tan(e/2 + f*x/2)**6/(105*a**
2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 84
0*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**
4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/
2) - 105*a**2*c**4*f) + 420*A*tan(e/2 + f*x/2)**5/(105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(
e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f
*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c
**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2) - 105*a**2*c**4*f) + 280*A*tan(e/2 + f*x/2)**4/(1
05*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**
8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*
x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2
+ f*x/2) - 105*a**2*c**4*f) - 760*A*tan(e/2 + f*x/2)**3/(105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*
f*tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*
c**4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*
a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2) - 105*a**2*c**4*f) + 240*A*tan(e/2 + f*x/2)
**2/(105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*
x/2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/
2 + f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*ta
n(e/2 + f*x/2) - 105*a**2*c**4*f) + 30*A*tan(e/2 + f*x/2)/(105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**
4*f*tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**
2*c**4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 31
5*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2) - 105*a**2*c**4*f) - 60*A/(105*a**2*c**4*
f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840*a**2*
c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 840
*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2) - 10
5*a**2*c**4*f) - 210*B*tan(e/2 + f*x/2)**8/(105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 + f
*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e/
2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*t
an(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2) - 105*a**2*c**4*f) + 280*B*tan(e/2 + f*x/2)**7/(105*a**2
*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840
*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4
- 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2
) - 105*a**2*c**4*f) - 280*B*tan(e/2 + f*x/2)**6/(105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e
/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*
tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c*
*4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2) - 105*a**2*c**4*f) - 168*B*tan(e/2 + f*x/2)**5/(10
5*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8
+ 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x
/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 +
f*x/2) - 105*a**2*c**4*f) - 28*B*tan(e/2 + f*x/2)**4/(105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*
tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c*
*4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a*
*2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2) - 105*a**2*c**4*f) + 136*B*tan(e/2 + f*x/2)**
3/(105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/
2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2
+ f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(
e/2 + f*x/2) - 105*a**2*c**4*f) - 264*B*tan(e/2 + f*x/2)**2/(105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c
**4*f*tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a
**2*c**4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 -
315*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2) - 105*a**2*c**4*f) + 72*B*tan(e/2 + f*x
/2)/(105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 + f*x/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*
x/2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e/2 + f*x/2)**6 + 1470*a**2*c**4*f*tan(e/
2 + f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*tan(e/2 + f*x/2)**2 + 420*a**2*c**4*f*ta
n(e/2 + f*x/2) - 105*a**2*c**4*f) - 18*B/(105*a**2*c**4*f*tan(e/2 + f*x/2)**10 - 420*a**2*c**4*f*tan(e/2 + f*x
/2)**9 + 315*a**2*c**4*f*tan(e/2 + f*x/2)**8 + 840*a**2*c**4*f*tan(e/2 + f*x/2)**7 - 1470*a**2*c**4*f*tan(e/2
+ f*x/2)**6 + 1470*a**2*c**4*f*tan(e/2 + f*x/2)**4 - 840*a**2*c**4*f*tan(e/2 + f*x/2)**3 - 315*a**2*c**4*f*tan
(e/2 + f*x/2)**2 + 420*a**2*c**4*f*tan(e/2 + f*x/2) - 105*a**2*c**4*f), Ne(f, 0)), (x*(A + B*sin(e))/((a*sin(e
) + a)**2*(-c*sin(e) + c)**4), True))

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