3.1.75 \(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))} \, dx\) [75]

Optimal. Leaf size=102 \[ -\frac {(A-B) \sec (e+f x)}{5 a c f (a+a \sin (e+f x))^2}-\frac {(3 A+2 B) \sec (e+f x)}{15 c f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {2 (3 A+2 B) \tan (e+f x)}{15 a^3 c f} \]

[Out]

-1/5*(A-B)*sec(f*x+e)/a/c/f/(a+a*sin(f*x+e))^2-1/15*(3*A+2*B)*sec(f*x+e)/c/f/(a^3+a^3*sin(f*x+e))+2/15*(3*A+2*
B)*tan(f*x+e)/a^3/c/f

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Rubi [A]
time = 0.18, antiderivative size = 102, 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 = {3046, 2938, 2751, 3852, 8} \begin {gather*} \frac {2 (3 A+2 B) \tan (e+f x)}{15 a^3 c f}-\frac {(3 A+2 B) \sec (e+f x)}{15 c f \left (a^3 \sin (e+f x)+a^3\right )}-\frac {(A-B) \sec (e+f x)}{5 a c f (a \sin (e+f x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*((A - B)*Sec[e + f*x])/(a*c*f*(a + a*Sin[e + f*x])^2) - ((3*A + 2*B)*Sec[e + f*x])/(15*c*f*(a^3 + a^3*Sin
[e + f*x])) + (2*(3*A + 2*B)*Tan[e + f*x])/(15*a^3*c*f)

Rule 8

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

Rule 2751

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2938

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 3046

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 3852

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))^3 (c-c \sin (e+f x))} \, dx &=\frac {\int \frac {\sec ^2(e+f x) (A+B \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx}{a c}\\ &=-\frac {(A-B) \sec (e+f x)}{5 a c f (a+a \sin (e+f x))^2}+\frac {(3 A+2 B) \int \frac {\sec ^2(e+f x)}{a+a \sin (e+f x)} \, dx}{5 a^2 c}\\ &=-\frac {(A-B) \sec (e+f x)}{5 a c f (a+a \sin (e+f x))^2}-\frac {(3 A+2 B) \sec (e+f x)}{15 c f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {(2 (3 A+2 B)) \int \sec ^2(e+f x) \, dx}{15 a^3 c}\\ &=-\frac {(A-B) \sec (e+f x)}{5 a c f (a+a \sin (e+f x))^2}-\frac {(3 A+2 B) \sec (e+f x)}{15 c f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {(2 (3 A+2 B)) \text {Subst}(\int 1 \, dx,x,-\tan (e+f x))}{15 a^3 c f}\\ &=-\frac {(A-B) \sec (e+f x)}{5 a c f (a+a \sin (e+f x))^2}-\frac {(3 A+2 B) \sec (e+f x)}{15 c f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {2 (3 A+2 B) \tan (e+f x)}{15 a^3 c f}\\ \end {align*}

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Mathematica [A]
time = 0.56, size = 156, normalized size = 1.53 \begin {gather*} \frac {\cos (e+f x) (-80 B-5 (9 A+B) \cos (e+f x)+32 (3 A+2 B) \cos (2 (e+f x))+9 A \cos (3 (e+f x))+B \cos (3 (e+f x))-120 A \sin (e+f x)-80 B \sin (e+f x)-36 A \sin (2 (e+f x))-4 B \sin (2 (e+f x))+24 A \sin (3 (e+f x))+16 B \sin (3 (e+f x)))}{240 a^3 c f (-1+\sin (e+f x)) (1+\sin (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(-80*B - 5*(9*A + B)*Cos[e + f*x] + 32*(3*A + 2*B)*Cos[2*(e + f*x)] + 9*A*Cos[3*(e + f*x)] + B*C
os[3*(e + f*x)] - 120*A*Sin[e + f*x] - 80*B*Sin[e + f*x] - 36*A*Sin[2*(e + f*x)] - 4*B*Sin[2*(e + f*x)] + 24*A
*Sin[3*(e + f*x)] + 16*B*Sin[3*(e + f*x)]))/(240*a^3*c*f*(-1 + Sin[e + f*x])*(1 + Sin[e + f*x])^3)

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Maple [A]
time = 0.34, size = 145, normalized size = 1.42

method	result	size
risch	\(-\frac {4 \left (15 i A \,{\mathrm e}^{2 i \left (f x +e \right )}-12 A \,{\mathrm e}^{i \left (f x +e \right )}+10 B \,{\mathrm e}^{3 i \left (f x +e \right )}-3 i A -8 B \,{\mathrm e}^{i \left (f x +e \right )}+10 i B \,{\mathrm e}^{2 i \left (f x +e \right )}-2 i B \right )}{15 \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) c f \,a^{3}}\)	\(111\)
derivativedivides	\(\frac {-\frac {-4 A +4 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (2 A -2 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-\frac {5 A}{2}+\frac {3 B}{2}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {7 A}{8}-\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (\frac {9 A}{2}-\frac {7 B}{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {A}{8}+\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}}{c f \,a^{3}}\)	\(145\)
default	\(\frac {-\frac {-4 A +4 B}{2 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (2 A -2 B \right )}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-\frac {5 A}{2}+\frac {3 B}{2}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (\frac {7 A}{8}-\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {2 \left (\frac {9 A}{2}-\frac {7 B}{2}\right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (\frac {A}{8}+\frac {B}{8}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}}{c f \,a^{3}}\)	\(145\)
norman	\(\frac {\frac {12 A -2 B}{15 c f a}-\frac {2 \left (6 A +7 B \right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f a}-\frac {2 A \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a c f}+\frac {2 \left (9 A -4 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{15 c f a}+\frac {2 \left (2 A -7 B \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 c f a}-\frac {2 \left (9 A +4 B \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f a}-\frac {2 \left (2 A +B \right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f a}-\frac {2 \left (7 A +8 B \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 c f a}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}\)	\(258\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

2/f/a^3/c*(-1/4*(-4*A+4*B)/(tan(1/2*f*x+1/2*e)+1)^4-1/5*(2*A-2*B)/(tan(1/2*f*x+1/2*e)+1)^5-1/2*(-5/2*A+3/2*B)/
(tan(1/2*f*x+1/2*e)+1)^2-(7/8*A-1/8*B)/(tan(1/2*f*x+1/2*e)+1)-1/3*(9/2*A-7/2*B)/(tan(1/2*f*x+1/2*e)+1)^3-(1/8*
A+1/8*B)/(tan(1/2*f*x+1/2*e)-1))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 459 vs. \(2 (101) = 202\).
time = 0.31, size = 459, normalized size = 4.50 \begin {gather*} \frac {2 \, {\left (\frac {B {\left (\frac {4 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {20 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 1\right )}}{a^{3} c + \frac {4 \, a^{3} c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, a^{3} c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {5 \, a^{3} c \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {4 \, a^{3} c \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {a^{3} c \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}} - \frac {3 \, A {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {10 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {5 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + 2\right )}}{a^{3} c + \frac {4 \, a^{3} c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, a^{3} c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {5 \, a^{3} c \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {4 \, a^{3} c \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {a^{3} c \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}}\right )}}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(B*(4*sin(f*x + e)/(cos(f*x + e) + 1) + 20*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 20*sin(f*x + e)^3/(cos(f
*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 1)/(a^3*c + 4*a^3*c*sin(f*x + e)/(cos(f*x + e) + 1)
+ 5*a^3*c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 5*a^3*c*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 4*a^3*c*sin(f*x
+ e)^5/(cos(f*x + e) + 1)^5 - a^3*c*sin(f*x + e)^6/(cos(f*x + e) + 1)^6) - 3*A*(3*sin(f*x + e)/(cos(f*x + e) +
1) - 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 10*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 5*sin(f*x + e)^5/(cos(
f*x + e) + 1)^5 + 2)/(a^3*c + 4*a^3*c*sin(f*x + e)/(cos(f*x + e) + 1) + 5*a^3*c*sin(f*x + e)^2/(cos(f*x + e) +
1)^2 - 5*a^3*c*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 4*a^3*c*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - a^3*c*sin(
f*x + e)^6/(cos(f*x + e) + 1)^6))/f

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Fricas [A]
time = 0.34, size = 113, normalized size = 1.11 \begin {gather*} \frac {4 \, {\left (3 \, A + 2 \, B\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, {\left (3 \, A + 2 \, B\right )} \cos \left (f x + e\right )^{2} - 9 \, A - 6 \, B\right )} \sin \left (f x + e\right ) - 6 \, A - 9 \, B}{15 \, {\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \left (f x + e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(4*(3*A + 2*B)*cos(f*x + e)^2 + (2*(3*A + 2*B)*cos(f*x + e)^2 - 9*A - 6*B)*sin(f*x + e) - 6*A - 9*B)/(a^3
*c*f*cos(f*x + e)^3 - 2*a^3*c*f*cos(f*x + e)*sin(f*x + e) - 2*a^3*c*f*cos(f*x + e))

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 1236 vs. \(2 (85) = 170\).
time = 5.53, size = 1236, normalized size = 12.12 \begin {gather*} \begin {cases} - \frac {30 A \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} c f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 60 a^{3} c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 a^{3} c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 60 a^{3} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 a^{3} c f} - \frac {60 A \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} c f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 60 a^{3} c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 a^{3} c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 60 a^{3} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 a^{3} c f} - \frac {60 A \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} c f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 60 a^{3} c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 a^{3} c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 60 a^{3} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 a^{3} c f} + \frac {18 A \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} c f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 60 a^{3} c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 a^{3} c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 60 a^{3} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 a^{3} c f} + \frac {12 A}{15 a^{3} c f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 60 a^{3} c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 a^{3} c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 60 a^{3} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 a^{3} c f} - \frac {30 B \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} c f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 60 a^{3} c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 a^{3} c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 60 a^{3} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 a^{3} c f} - \frac {40 B \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} c f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 60 a^{3} c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 a^{3} c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 60 a^{3} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 a^{3} c f} - \frac {40 B \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} c f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 60 a^{3} c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 a^{3} c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 60 a^{3} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 a^{3} c f} - \frac {8 B \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} c f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 60 a^{3} c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 a^{3} c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 60 a^{3} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 a^{3} c f} - \frac {2 B}{15 a^{3} c f \tan ^{6}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 60 a^{3} c f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} c f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 a^{3} c f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 60 a^{3} c f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 a^{3} c f} & \text {for}\: f \neq 0 \\\frac {x \left (A + B \sin {\left (e \right )}\right )}{\left (a \sin {\left (e \right )} + a\right )^{3} \left (- c \sin {\left (e \right )} + c\right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-30*A*tan(e/2 + f*x/2)**5/(15*a**3*c*f*tan(e/2 + f*x/2)**6 + 60*a**3*c*f*tan(e/2 + f*x/2)**5 + 75*a
**3*c*f*tan(e/2 + f*x/2)**4 - 75*a**3*c*f*tan(e/2 + f*x/2)**2 - 60*a**3*c*f*tan(e/2 + f*x/2) - 15*a**3*c*f) -
60*A*tan(e/2 + f*x/2)**4/(15*a**3*c*f*tan(e/2 + f*x/2)**6 + 60*a**3*c*f*tan(e/2 + f*x/2)**5 + 75*a**3*c*f*tan(
e/2 + f*x/2)**4 - 75*a**3*c*f*tan(e/2 + f*x/2)**2 - 60*a**3*c*f*tan(e/2 + f*x/2) - 15*a**3*c*f) - 60*A*tan(e/2
+ f*x/2)**3/(15*a**3*c*f*tan(e/2 + f*x/2)**6 + 60*a**3*c*f*tan(e/2 + f*x/2)**5 + 75*a**3*c*f*tan(e/2 + f*x/2)
**4 - 75*a**3*c*f*tan(e/2 + f*x/2)**2 - 60*a**3*c*f*tan(e/2 + f*x/2) - 15*a**3*c*f) + 18*A*tan(e/2 + f*x/2)/(1
5*a**3*c*f*tan(e/2 + f*x/2)**6 + 60*a**3*c*f*tan(e/2 + f*x/2)**5 + 75*a**3*c*f*tan(e/2 + f*x/2)**4 - 75*a**3*c
*f*tan(e/2 + f*x/2)**2 - 60*a**3*c*f*tan(e/2 + f*x/2) - 15*a**3*c*f) + 12*A/(15*a**3*c*f*tan(e/2 + f*x/2)**6 +
60*a**3*c*f*tan(e/2 + f*x/2)**5 + 75*a**3*c*f*tan(e/2 + f*x/2)**4 - 75*a**3*c*f*tan(e/2 + f*x/2)**2 - 60*a**3
*c*f*tan(e/2 + f*x/2) - 15*a**3*c*f) - 30*B*tan(e/2 + f*x/2)**4/(15*a**3*c*f*tan(e/2 + f*x/2)**6 + 60*a**3*c*f
*tan(e/2 + f*x/2)**5 + 75*a**3*c*f*tan(e/2 + f*x/2)**4 - 75*a**3*c*f*tan(e/2 + f*x/2)**2 - 60*a**3*c*f*tan(e/2
+ f*x/2) - 15*a**3*c*f) - 40*B*tan(e/2 + f*x/2)**3/(15*a**3*c*f*tan(e/2 + f*x/2)**6 + 60*a**3*c*f*tan(e/2 + f
*x/2)**5 + 75*a**3*c*f*tan(e/2 + f*x/2)**4 - 75*a**3*c*f*tan(e/2 + f*x/2)**2 - 60*a**3*c*f*tan(e/2 + f*x/2) -
15*a**3*c*f) - 40*B*tan(e/2 + f*x/2)**2/(15*a**3*c*f*tan(e/2 + f*x/2)**6 + 60*a**3*c*f*tan(e/2 + f*x/2)**5 + 7
5*a**3*c*f*tan(e/2 + f*x/2)**4 - 75*a**3*c*f*tan(e/2 + f*x/2)**2 - 60*a**3*c*f*tan(e/2 + f*x/2) - 15*a**3*c*f)
- 8*B*tan(e/2 + f*x/2)/(15*a**3*c*f*tan(e/2 + f*x/2)**6 + 60*a**3*c*f*tan(e/2 + f*x/2)**5 + 75*a**3*c*f*tan(e
/2 + f*x/2)**4 - 75*a**3*c*f*tan(e/2 + f*x/2)**2 - 60*a**3*c*f*tan(e/2 + f*x/2) - 15*a**3*c*f) - 2*B/(15*a**3*
c*f*tan(e/2 + f*x/2)**6 + 60*a**3*c*f*tan(e/2 + f*x/2)**5 + 75*a**3*c*f*tan(e/2 + f*x/2)**4 - 75*a**3*c*f*tan(
e/2 + f*x/2)**2 - 60*a**3*c*f*tan(e/2 + f*x/2) - 15*a**3*c*f), Ne(f, 0)), (x*(A + B*sin(e))/((a*sin(e) + a)**3
*(-c*sin(e) + c)), True))

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Giac [A]
time = 0.44, size = 175, normalized size = 1.72 \begin {gather*} -\frac {\frac {15 \, {\left (A + B\right )}}{a^{3} c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}} + \frac {105 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 270 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 360 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 40 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 210 \, A \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 50 \, B \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 63 \, A + 7 \, B}{a^{3} c {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{60 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*(A + B)/(a^3*c*(tan(1/2*f*x + 1/2*e) - 1)) + (105*A*tan(1/2*f*x + 1/2*e)^4 - 15*B*tan(1/2*f*x + 1/2*
e)^4 + 270*A*tan(1/2*f*x + 1/2*e)^3 + 30*B*tan(1/2*f*x + 1/2*e)^3 + 360*A*tan(1/2*f*x + 1/2*e)^2 + 40*B*tan(1/
2*f*x + 1/2*e)^2 + 210*A*tan(1/2*f*x + 1/2*e) + 50*B*tan(1/2*f*x + 1/2*e) + 63*A + 7*B)/(a^3*c*(tan(1/2*f*x +
1/2*e) + 1)^5))/f

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Mupad [B]
time = 12.43, size = 178, normalized size = 1.75 \begin {gather*} -\frac {2\,\left (\frac {15\,A\,\cos \left (e+f\,x\right )}{4}-\frac {5\,B}{2}-\frac {5\,B\,\cos \left (e+f\,x\right )}{8}-\frac {15\,A\,\sin \left (e+f\,x\right )}{4}-\frac {5\,B\,\sin \left (e+f\,x\right )}{2}+3\,A\,\cos \left (2\,e+2\,f\,x\right )-\frac {3\,A\,\cos \left (3\,e+3\,f\,x\right )}{4}+2\,B\,\cos \left (2\,e+2\,f\,x\right )+\frac {B\,\cos \left (3\,e+3\,f\,x\right )}{8}+3\,A\,\sin \left (2\,e+2\,f\,x\right )+\frac {3\,A\,\sin \left (3\,e+3\,f\,x\right )}{4}-\frac {B\,\sin \left (2\,e+2\,f\,x\right )}{2}+\frac {B\,\sin \left (3\,e+3\,f\,x\right )}{2}\right )}{15\,a^3\,c\,f\,\left (\frac {5\,\cos \left (e+f\,x\right )}{4}-\frac {\cos \left (3\,e+3\,f\,x\right )}{4}+\sin \left (2\,e+2\,f\,x\right )\right )} \end {gather*}

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))),x)

[Out]

-(2*((15*A*cos(e + f*x))/4 - (5*B)/2 - (5*B*cos(e + f*x))/8 - (15*A*sin(e + f*x))/4 - (5*B*sin(e + f*x))/2 + 3
*A*cos(2*e + 2*f*x) - (3*A*cos(3*e + 3*f*x))/4 + 2*B*cos(2*e + 2*f*x) + (B*cos(3*e + 3*f*x))/8 + 3*A*sin(2*e +
2*f*x) + (3*A*sin(3*e + 3*f*x))/4 - (B*sin(2*e + 2*f*x))/2 + (B*sin(3*e + 3*f*x))/2))/(15*a^3*c*f*((5*cos(e +
f*x))/4 - cos(3*e + 3*f*x)/4 + sin(2*e + 2*f*x)))

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