∫ (A+B sin(e+fx))(c+d sin(e+fx))2 ___________________________________________________

3.280 \(\int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=164 \[ -\frac {\left (2 A \left (c^2+3 c d+2 d^2\right )+B \left (3 c^2+14 c d-29 d^2\right )\right ) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}+\frac {B d^2 x}{a^3}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}-\frac {(c-d) (2 A (c+d)+B (3 c-7 d)) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2} \]

[Out]

B*d^2*x/a^3-1/15*(c-d)*(B*(3*c-7*d)+2*A*(c+d))*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^2-1/15*(B*(3*c^2+14*c*d-29*d^2)

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

*x+e))^3

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Rubi [A] time = 0.46, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2977, 2968, 3019, 2735, 2648} \[ -\frac {\left (2 A \left (c^2+3 c d+2 d^2\right )+B \left (3 c^2+14 c d-29 d^2\right )\right ) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}+\frac {B d^2 x}{a^3}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a \sin (e+f x)+a)^3}-\frac {(c-d) (2 A (c+d)+B (3 c-7 d)) \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully veriﬁed.

[In]

Int[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/(a + a*Sin[e + f*x])^3,x]

[Out]

(B*d^2*x)/a^3 - ((c - d)*(B*(3*c - 7*d) + 2*A*(c + d))*Cos[e + f*x])/(15*a*f*(a + a*Sin[e + f*x])^2) - ((B*(3*

c^2 + 14*c*d - 29*d^2) + 2*A*(c^2 + 3*c*d + 2*d^2))*Cos[e + f*x])/(15*f*(a^3 + a^3*Sin[e + f*x])) - ((A - B)*C

os[e + f*x]*(c + d*Sin[e + f*x])^2)/(5*f*(a + a*Sin[e + f*x])^3)

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]

/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d

, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d

, 0]

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(

e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x

]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_

.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]

)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -

1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],

x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ

[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 3019

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_

.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D

ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e

+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {(A+B \sin (e+f x)) (c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx &=-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}+\frac {\int \frac {(c+d \sin (e+f x)) (a (B (3 c-2 d)+2 A (c+d))+5 a B d \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}+\frac {\int \frac {a c (B (3 c-2 d)+2 A (c+d))+(5 a B c d+a d (B (3 c-2 d)+2 A (c+d))) \sin (e+f x)+5 a B d^2 \sin ^2(e+f x)}{(a+a \sin (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac {(c-d) (B (3 c-7 d)+2 A (c+d)) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}-\frac {\int \frac {-a^2 \left (B \left (3 c^2+14 c d-14 d^2\right )+2 A \left (c^2+3 c d+2 d^2\right )\right )-15 a^2 B d^2 \sin (e+f x)}{a+a \sin (e+f x)} \, dx}{15 a^4}\\ &=\frac {B d^2 x}{a^3}-\frac {(c-d) (B (3 c-7 d)+2 A (c+d)) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}+\frac {\left (B \left (3 c^2+14 c d-29 d^2\right )+2 A \left (c^2+3 c d+2 d^2\right )\right ) \int \frac {1}{a+a \sin (e+f x)} \, dx}{15 a^2}\\ &=\frac {B d^2 x}{a^3}-\frac {(c-d) (B (3 c-7 d)+2 A (c+d)) \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {\left (B \left (3 c^2+14 c d-29 d^2\right )+2 A \left (c^2+3 c d+2 d^2\right )\right ) \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {(A-B) \cos (e+f x) (c+d \sin (e+f x))^2}{5 f (a+a \sin (e+f x))^3}\\ \end {align*}

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Mathematica [B] time = 0.90, size = 514, normalized size = 3.13 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (30 \cos \left (\frac {1}{2} (e+f x)\right ) \left (2 A d (c+d)+B \left (c^2+4 c d+d^2 (5 e+5 f x-9)\right )\right )-5 \cos \left (\frac {3}{2} (e+f x)\right ) \left (4 A \left (c^2+3 c d+2 d^2\right )+B \left (6 c^2+16 c d+d^2 (15 e+15 f x-46)\right )\right )+40 A c^2 \sin \left (\frac {1}{2} (e+f x)\right )-4 A c^2 \sin \left (\frac {5}{2} (e+f x)\right )+60 A c d \sin \left (\frac {1}{2} (e+f x)\right )-12 A c d \sin \left (\frac {5}{2} (e+f x)\right )+80 A d^2 \sin \left (\frac {1}{2} (e+f x)\right )+30 A d^2 \sin \left (\frac {3}{2} (e+f x)\right )-14 A d^2 \sin \left (\frac {5}{2} (e+f x)\right )+30 B c^2 \sin \left (\frac {1}{2} (e+f x)\right )-6 B c^2 \sin \left (\frac {5}{2} (e+f x)\right )+160 B c d \sin \left (\frac {1}{2} (e+f x)\right )+60 B c d \sin \left (\frac {3}{2} (e+f x)\right )-28 B c d \sin \left (\frac {5}{2} (e+f x)\right )-370 B d^2 \sin \left (\frac {1}{2} (e+f x)\right )+150 B d^2 e \sin \left (\frac {1}{2} (e+f x)\right )+150 B d^2 f x \sin \left (\frac {1}{2} (e+f x)\right )-90 B d^2 \sin \left (\frac {3}{2} (e+f x)\right )+75 B d^2 e \sin \left (\frac {3}{2} (e+f x)\right )+75 B d^2 f x \sin \left (\frac {3}{2} (e+f x)\right )+64 B d^2 \sin \left (\frac {5}{2} (e+f x)\right )-15 B d^2 e \sin \left (\frac {5}{2} (e+f x)\right )-15 B d^2 f x \sin \left (\frac {5}{2} (e+f x)\right )-15 B d^2 e \cos \left (\frac {5}{2} (e+f x)\right )-15 B d^2 f x \cos \left (\frac {5}{2} (e+f x)\right )\right )}{60 a^3 f (\sin (e+f x)+1)^3} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[((A + B*Sin[e + f*x])*(c + d*Sin[e + f*x])^2)/(a + a*Sin[e + f*x])^3,x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(30*(2*A*d*(c + d) + B*(c^2 + 4*c*d + d^2*(-9 + 5*e + 5*f*x)))*Cos[(e +

f*x)/2] - 5*(4*A*(c^2 + 3*c*d + 2*d^2) + B*(6*c^2 + 16*c*d + d^2*(-46 + 15*e + 15*f*x)))*Cos[(3*(e + f*x))/2]

- 15*B*d^2*e*Cos[(5*(e + f*x))/2] - 15*B*d^2*f*x*Cos[(5*(e + f*x))/2] + 40*A*c^2*Sin[(e + f*x)/2] + 30*B*c^2*

Sin[(e + f*x)/2] + 60*A*c*d*Sin[(e + f*x)/2] + 160*B*c*d*Sin[(e + f*x)/2] + 80*A*d^2*Sin[(e + f*x)/2] - 370*B*

d^2*Sin[(e + f*x)/2] + 150*B*d^2*e*Sin[(e + f*x)/2] + 150*B*d^2*f*x*Sin[(e + f*x)/2] + 60*B*c*d*Sin[(3*(e + f*

x))/2] + 30*A*d^2*Sin[(3*(e + f*x))/2] - 90*B*d^2*Sin[(3*(e + f*x))/2] + 75*B*d^2*e*Sin[(3*(e + f*x))/2] + 75*

B*d^2*f*x*Sin[(3*(e + f*x))/2] - 4*A*c^2*Sin[(5*(e + f*x))/2] - 6*B*c^2*Sin[(5*(e + f*x))/2] - 12*A*c*d*Sin[(5

*(e + f*x))/2] - 28*B*c*d*Sin[(5*(e + f*x))/2] - 14*A*d^2*Sin[(5*(e + f*x))/2] + 64*B*d^2*Sin[(5*(e + f*x))/2]

- 15*B*d^2*e*Sin[(5*(e + f*x))/2] - 15*B*d^2*f*x*Sin[(5*(e + f*x))/2]))/(60*a^3*f*(1 + Sin[e + f*x])^3)

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fricas [B] time = 0.46, size = 432, normalized size = 2.63 \[ -\frac {60 \, B d^{2} f x - {\left (15 \, B d^{2} f x - {\left (2 \, A + 3 \, B\right )} c^{2} - 2 \, {\left (3 \, A + 7 \, B\right )} c d - {\left (7 \, A - 32 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (A - B\right )} c^{2} + 6 \, {\left (A - B\right )} c d - 3 \, {\left (A - B\right )} d^{2} - {\left (45 \, B d^{2} f x + 2 \, {\left (2 \, A + 3 \, B\right )} c^{2} + 2 \, {\left (6 \, A - B\right )} c d - {\left (A + 19 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (10 \, B d^{2} f x - {\left (3 \, A + 2 \, B\right )} c^{2} - 2 \, {\left (2 \, A + 3 \, B\right )} c d - 3 \, {\left (A - 6 \, B\right )} d^{2}\right )} \cos \left (f x + e\right ) + {\left (60 \, B d^{2} f x + 3 \, {\left (A - B\right )} c^{2} - 6 \, {\left (A - B\right )} c d + 3 \, {\left (A - B\right )} d^{2} - {\left (15 \, B d^{2} f x + {\left (2 \, A + 3 \, B\right )} c^{2} + 2 \, {\left (3 \, A + 7 \, B\right )} c d + {\left (7 \, A - 32 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (10 \, B d^{2} f x - {\left (2 \, A + 3 \, B\right )} c^{2} - 2 \, {\left (3 \, A + 2 \, B\right )} c d - {\left (2 \, A - 17 \, B\right )} d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/15*(60*B*d^2*f*x - (15*B*d^2*f*x - (2*A + 3*B)*c^2 - 2*(3*A + 7*B)*c*d - (7*A - 32*B)*d^2)*cos(f*x + e)^3 -

3*(A - B)*c^2 + 6*(A - B)*c*d - 3*(A - B)*d^2 - (45*B*d^2*f*x + 2*(2*A + 3*B)*c^2 + 2*(6*A - B)*c*d - (A + 19

*B)*d^2)*cos(f*x + e)^2 + 3*(10*B*d^2*f*x - (3*A + 2*B)*c^2 - 2*(2*A + 3*B)*c*d - 3*(A - 6*B)*d^2)*cos(f*x + e

) + (60*B*d^2*f*x + 3*(A - B)*c^2 - 6*(A - B)*c*d + 3*(A - B)*d^2 - (15*B*d^2*f*x + (2*A + 3*B)*c^2 + 2*(3*A +

7*B)*c*d + (7*A - 32*B)*d^2)*cos(f*x + e)^2 + 3*(10*B*d^2*f*x - (2*A + 3*B)*c^2 - 2*(3*A + 2*B)*c*d - (2*A -

17*B)*d^2)*cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) -

4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(f*x + e))

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giac [B] time = 0.19, size = 382, normalized size = 2.33 \[ \frac {\frac {15 \, {\left (f x + e\right )} B d^{2}}{a^{3}} - \frac {2 \, {\left (15 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, A c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 75 \, B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 30 \, A c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 40 \, B c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, A d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 145 \, B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 30 \, A c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 20 \, B c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 10 \, A d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 95 \, B d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7 \, A c^{2} + 3 \, B c^{2} + 6 \, A c d + 4 \, B c d + 2 \, A d^{2} - 22 \, B d^{2}\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{15 \, f} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

1/15*(15*(f*x + e)*B*d^2/a^3 - 2*(15*A*c^2*tan(1/2*f*x + 1/2*e)^4 - 15*B*d^2*tan(1/2*f*x + 1/2*e)^4 + 30*A*c^2

*tan(1/2*f*x + 1/2*e)^3 + 15*B*c^2*tan(1/2*f*x + 1/2*e)^3 + 30*A*c*d*tan(1/2*f*x + 1/2*e)^3 - 75*B*d^2*tan(1/2

*f*x + 1/2*e)^3 + 40*A*c^2*tan(1/2*f*x + 1/2*e)^2 + 15*B*c^2*tan(1/2*f*x + 1/2*e)^2 + 30*A*c*d*tan(1/2*f*x + 1

/2*e)^2 + 40*B*c*d*tan(1/2*f*x + 1/2*e)^2 + 20*A*d^2*tan(1/2*f*x + 1/2*e)^2 - 145*B*d^2*tan(1/2*f*x + 1/2*e)^2

+ 20*A*c^2*tan(1/2*f*x + 1/2*e) + 15*B*c^2*tan(1/2*f*x + 1/2*e) + 30*A*c*d*tan(1/2*f*x + 1/2*e) + 20*B*c*d*ta

n(1/2*f*x + 1/2*e) + 10*A*d^2*tan(1/2*f*x + 1/2*e) - 95*B*d^2*tan(1/2*f*x + 1/2*e) + 7*A*c^2 + 3*B*c^2 + 6*A*c

*d + 4*B*c*d + 2*A*d^2 - 22*B*d^2)/(a^3*(tan(1/2*f*x + 1/2*e) + 1)^5))/f

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maple [B] time = 0.42, size = 617, normalized size = 3.76 \[ -\frac {2 B \,c^{2}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {2 B \,d^{2}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {2 B \,d^{2} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a^{3} f}-\frac {2 A \,c^{2}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {2 B \,d^{2}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {4 A \,c^{2}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {4 B \,d^{2}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {8 A \,c^{2}}{5 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {8 A \,d^{2}}{5 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {8 B \,c^{2}}{5 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {8 B \,d^{2}}{5 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {16 A \,c^{2}}{3 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {8 A \,d^{2}}{3 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4 B \,c^{2}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4 B \,d^{2}}{3 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4 A \,c^{2}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}+\frac {4 A \,d^{2}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {16 B c d}{5 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {4 B \,c^{2}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}+\frac {8 A c d}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {16 B c d}{3 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {4 A c d}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {8 A c d}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}+\frac {8 B c d}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}+\frac {16 A c d}{5 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x)

[Out]

-2/a^3/f/(tan(1/2*f*x+1/2*e)+1)^2*B*c^2+2/a^3/f/(tan(1/2*f*x+1/2*e)+1)^2*B*d^2+2/a^3/f*B*d^2*arctan(tan(1/2*f*

x+1/2*e))-2/a^3/f/(tan(1/2*f*x+1/2*e)+1)*A*c^2+2/a^3/f/(tan(1/2*f*x+1/2*e)+1)*B*d^2+4/a^3/f/(tan(1/2*f*x+1/2*e

)+1)^2*A*c^2-4/a^3/f/(tan(1/2*f*x+1/2*e)+1)^4*B*d^2-8/5/a^3/f/(tan(1/2*f*x+1/2*e)+1)^5*A*c^2-8/5/a^3/f/(tan(1/

2*f*x+1/2*e)+1)^5*A*d^2+8/5/a^3/f/(tan(1/2*f*x+1/2*e)+1)^5*B*c^2+8/5/a^3/f/(tan(1/2*f*x+1/2*e)+1)^5*B*d^2-16/3

/a^3/f/(tan(1/2*f*x+1/2*e)+1)^3*A*c^2-8/3/a^3/f/(tan(1/2*f*x+1/2*e)+1)^3*A*d^2+4/a^3/f/(tan(1/2*f*x+1/2*e)+1)^

3*B*c^2+4/3/a^3/f/(tan(1/2*f*x+1/2*e)+1)^3*B*d^2+4/a^3/f/(tan(1/2*f*x+1/2*e)+1)^4*A*c^2+4/a^3/f/(tan(1/2*f*x+1

/2*e)+1)^4*A*d^2-16/5/a^3/f/(tan(1/2*f*x+1/2*e)+1)^5*B*c*d-4/a^3/f/(tan(1/2*f*x+1/2*e)+1)^4*B*c^2+8/a^3/f/(tan

(1/2*f*x+1/2*e)+1)^3*A*c*d-16/3/a^3/f/(tan(1/2*f*x+1/2*e)+1)^3*B*c*d-4/a^3/f/(tan(1/2*f*x+1/2*e)+1)^2*A*c*d-8/

a^3/f/(tan(1/2*f*x+1/2*e)+1)^4*A*c*d+8/a^3/f/(tan(1/2*f*x+1/2*e)+1)^4*B*c*d+16/5/a^3/f/(tan(1/2*f*x+1/2*e)+1)^

5*A*c*d

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maxima [B] time = 0.55, size = 1132, normalized size = 6.90 \[ \text {result too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

2/15*(B*d^2*((95*sin(f*x + e)/(cos(f*x + e) + 1) + 145*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 75*sin(f*x + e)^3

/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 22)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) +

1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x

+ e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan(sin(f*x + e)/(cos(f*x + e)

+ 1))/a^3) - A*c^2*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x +

e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 7)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x +

e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(

f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 4*B*c*d*(5*sin(f*x + e)/(cos(f*x

+ e) + 1) + 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*

sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f

*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 2*A*d^2*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 10*si

n(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(c

os(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 +

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

f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10

*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(

cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 6*A*c*d*(5*sin(f*x + e)/(cos(f*x + e) + 1) +

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

cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 +

5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

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mupad [B] time = 16.54, size = 286, normalized size = 1.74 \[ \frac {B\,d^2\,x}{a^3}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {16\,A\,c^2}{3}+\frac {8\,A\,d^2}{3}+2\,B\,c^2-\frac {58\,B\,d^2}{3}+4\,A\,c\,d+\frac {16\,B\,c\,d}{3}\right )+\frac {14\,A\,c^2}{15}+\frac {4\,A\,d^2}{15}+\frac {2\,B\,c^2}{5}-\frac {44\,B\,d^2}{15}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (4\,A\,c^2+2\,B\,c^2-10\,B\,d^2+4\,A\,c\,d\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (2\,A\,c^2-2\,B\,d^2\right )+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {8\,A\,c^2}{3}+\frac {4\,A\,d^2}{3}+2\,B\,c^2-\frac {38\,B\,d^2}{3}+4\,A\,c\,d+\frac {8\,B\,c\,d}{3}\right )+\frac {4\,A\,c\,d}{5}+\frac {8\,B\,c\,d}{15}}{f\,\left (a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+5\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^3\right )} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(c + d*sin(e + f*x))^2)/(a + a*sin(e + f*x))^3,x)

[Out]

(B*d^2*x)/a^3 - (tan(e/2 + (f*x)/2)^2*((16*A*c^2)/3 + (8*A*d^2)/3 + 2*B*c^2 - (58*B*d^2)/3 + 4*A*c*d + (16*B*c

*d)/3) + (14*A*c^2)/15 + (4*A*d^2)/15 + (2*B*c^2)/5 - (44*B*d^2)/15 + tan(e/2 + (f*x)/2)^3*(4*A*c^2 + 2*B*c^2

- 10*B*d^2 + 4*A*c*d) + tan(e/2 + (f*x)/2)^4*(2*A*c^2 - 2*B*d^2) + tan(e/2 + (f*x)/2)*((8*A*c^2)/3 + (4*A*d^2)

/3 + 2*B*c^2 - (38*B*d^2)/3 + 4*A*c*d + (8*B*c*d)/3) + (4*A*c*d)/5 + (8*B*c*d)/15)/(f*(10*a^3*tan(e/2 + (f*x)/

2)^2 + 10*a^3*tan(e/2 + (f*x)/2)^3 + 5*a^3*tan(e/2 + (f*x)/2)^4 + a^3*tan(e/2 + (f*x)/2)^5 + a^3 + 5*a^3*tan(e

/2 + (f*x)/2)))

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sympy [A] time = 33.40, size = 3468, normalized size = 21.15 \[ \text {result too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c+d*sin(f*x+e))**2/(a+a*sin(f*x+e))**3,x)

[Out]

Piecewise((-30*A*c**2*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150

*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*A*

c**2*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 +

f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 80*A*c**2*tan(e/2 + f*

x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*

a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 40*A*c**2*tan(e/2 + f*x/2)/(15*a**3*f*t

an(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x

/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 14*A*c**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/

2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) +

15*a**3*f) - 60*A*c*d*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 15

0*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*A

*c*d*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 +

f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*A*c*d*tan(e/2 + f*x

/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3

*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 12*A*c*d/(15*a**3*f*tan(e/2 + f*x/2)**5 + 7

5*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan

(e/2 + f*x/2) + 15*a**3*f) - 40*A*d**2*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2

+ f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 1

5*a**3*f) - 20*A*d**2*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a*

*3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 4*A*d**2

/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*

tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30*B*c**2*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e

/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)*

*2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30*B*c**2*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 +

75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*ta

n(e/2 + f*x/2) + 15*a**3*f) - 30*B*c**2*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 +

f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*

a**3*f) - 6*B*c**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2

)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 80*B*c*d*tan(e/2 + f*x/2)**2

/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*

tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 40*B*c*d*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 +

f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 +

75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 8*B*c*d/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)

**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f

) + 15*B*d**2*f*x*tan(e/2 + f*x/2)**5/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**

3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 75*B*d**2

*f*x*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 +

f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 150*B*d**2*f*x*tan(e/2

+ f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 +

150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 150*B*d**2*f*x*tan(e/2 + f*x/2)**2

/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*

tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 75*B*d**2*f*x*tan(e/2 + f*x/2)/(15*a**3*f*tan(

e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)

**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 15*B*d**2*f*x/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e

/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2)

+ 15*a**3*f) + 30*B*d**2*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 +

150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 15

0*B*d**2*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e

/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 290*B*d**2*tan(e/2

+ f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 +

150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 190*B*d**2*tan(e/2 + f*x/2)/(15*a*

*3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2

+ f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 44*B*d**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*

tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*

x/2) + 15*a**3*f), Ne(f, 0)), (x*(A + B*sin(e))*(c + d*sin(e))**2/(a*sin(e) + a)**3, True))

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