3.2.28 \(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx\) [128]

Optimal. Leaf size=174 \[ \frac {(A+B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} a^3 \sqrt {c} f}-\frac {(A+B) \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f}-\frac {(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f} \]

[Out]

-1/6*(A+B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^3/c^2/f-1/5*(A-B)*sec(f*x+e)^5*(c-c*sin(f*x+e))^(5/2)/a^3/c^3
/f+1/8*(A+B)*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/a^3/f*2^(1/2)/c^(1/2)-1/4*(A+B)*se
c(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a^3/c/f

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Rubi [A]
time = 0.29, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3046, 2934, 2754, 2728, 212} \begin {gather*} -\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}-\frac {(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {(A+B) \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f}+\frac {(A+B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} a^3 \sqrt {c} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A + B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(4*Sqrt[2]*a^3*Sqrt[c]*f) - ((A +
B)*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(4*a^3*c*f) - ((A + B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(
6*a^3*c^2*f) - ((A - B)*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(5/2))/(5*a^3*c^3*f)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2754

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*(p + 1))), x] + Dist[a*((m + p + 1)/(g^2*(p + 1))), 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] && GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ[m + 1/2, 2*p]

Rule 2934

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*(p +
1))), x] + Dist[b*((a*d*m + b*c*(m + p + 1))/(a*g^2*(p + 1))), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*
x])^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

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]))

Rubi steps

\begin {align*} \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx &=\frac {\int \sec ^6(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx}{a^3 c^3}\\ &=-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac {(A+B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{2 a^3 c^2}\\ &=-\frac {(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac {(A+B) \int \sec ^2(e+f x) \sqrt {c-c \sin (e+f x)} \, dx}{4 a^3 c}\\ &=-\frac {(A+B) \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f}-\frac {(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac {(A+B) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{8 a^3}\\ &=-\frac {(A+B) \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f}-\frac {(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}-\frac {(A+B) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{4 a^3 f}\\ &=\frac {(A+B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} a^3 \sqrt {c} f}-\frac {(A+B) \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f}-\frac {(A+B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.54, size = 204, normalized size = 1.17 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (12 (-A+B)-10 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-15 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-(15+15 i) \sqrt [4]{-1} (A+B) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5\right )}{60 a^3 f (1+\sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(12*(-A + B) - 10*(A + B)*(Cos[(e
+ f*x)/2] + Sin[(e + f*x)/2])^2 - 15*(A + B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - (15 + 15*I)*(-1)^(1/4)
*(A + B)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5))/(60*a
^3*f*(1 + Sin[e + f*x])^3*Sqrt[c - c*Sin[e + f*x]])

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Maple [A]
time = 7.18, size = 200, normalized size = 1.15

method	result	size
default	\(-\frac {\left (\sin \left (f x +e \right )-1\right ) \left (-30 A \,c^{\frac {9}{2}} \left (\sin ^{2}\left (f x +e \right )\right )-30 B \,c^{\frac {9}{2}} \left (\sin ^{2}\left (f x +e \right )\right )-80 A \,c^{\frac {9}{2}} \sin \left (f x +e \right )-80 B \,c^{\frac {9}{2}} \sin \left (f x +e \right )+15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} c^{2} A -74 c^{\frac {9}{2}} A +15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} c^{2} B -26 c^{\frac {9}{2}} B \right )}{120 a^{3} c^{\frac {9}{2}} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(200\)

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))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/120*(sin(f*x+e)-1)*(-30*A*c^(9/2)*sin(f*x+e)^2-30*B*c^(9/2)*sin(f*x+e)^2-80*A*c^(9/2)*sin(f*x+e)-80*B*c^(9/
2)*sin(f*x+e)+15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*(c*(1+sin(f*x+e)))^(5/2)*c^2*A-
74*c^(9/2)*A+15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*(c*(1+sin(f*x+e)))^(5/2)*c^2*B-2
6*c^(9/2)*B)/a^3/c^(9/2)/(1+sin(f*x+e))^2/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 284, normalized size = 1.63 \begin {gather*} \frac {15 \, \sqrt {2} {\left ({\left (A + B\right )} \cos \left (f x + e\right )^{3} - 2 \, {\left (A + B\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, {\left (A + B\right )} \cos \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (15 \, {\left (A + B\right )} \cos \left (f x + e\right )^{2} - 40 \, {\left (A + B\right )} \sin \left (f x + e\right ) - 52 \, A - 28 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{240 \, {\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))^(1/2),x, algorithm="fricas")

[Out]

1/240*(15*sqrt(2)*((A + B)*cos(f*x + e)^3 - 2*(A + B)*cos(f*x + e)*sin(f*x + e) - 2*(A + B)*cos(f*x + e))*sqrt
(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3
*c*cos(f*x + e) + (c*cos(f*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e)
- cos(f*x + e) - 2)) - 4*(15*(A + B)*cos(f*x + e)^2 - 40*(A + B)*sin(f*x + e) - 52*A - 28*B)*sqrt(-c*sin(f*x
+ e) + c))/(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 [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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))**(1/2),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 472 vs. \(2 (159) = 318\).
time = 0.57, size = 472, normalized size = 2.71 \begin {gather*} \frac {\frac {15 \, \sqrt {2} {\left (A \sqrt {c} + B \sqrt {c}\right )} \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )}{a^{3} c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {4 \, \sqrt {2} {\left (23 \, A \sqrt {c} + 17 \, B \sqrt {c} + \frac {70 \, A \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {70 \, B \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {140 \, A \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \frac {80 \, B \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \frac {90 \, A \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} + \frac {90 \, B \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} + \frac {45 \, A \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{4}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{4}} + \frac {15 \, B \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{4}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{4}}\right )}}{a^{3} c {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )}^{5} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{240 \, 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))^(1/2),x, algorithm="giac")

[Out]

1/240*(15*sqrt(2)*(A*sqrt(c) + B*sqrt(c))*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1
/2*e) + 1))/(a^3*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 4*sqrt(2)*(23*A*sqrt(c) + 17*B*sqrt(c) + 70*A*sqrt(c
)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 70*B*sqrt(c)*(cos(-1/4*pi + 1/2*
f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 140*A*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/
(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + 80*B*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/
2*f*x + 1/2*e) + 1)^2 + 90*A*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3/(cos(-1/4*pi + 1/2*f*x + 1/2*e) +
1)^3 + 90*B*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3 + 45*A*sqrt(
c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4 + 15*B*sqrt(c)*(cos(-1/4*pi +
1/2*f*x + 1/2*e) - 1)^4/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4)/(a^3*c*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/
(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 1)^5*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \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))^(1/2)),x)

[Out]

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

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