3.130 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=258 \[ -\frac{(A-B) \sec ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{5 a^3 c^3 f}-\frac{(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{7 (9 A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{128 \sqrt{2} a^3 c^{5/2} f}+\frac{7 (9 A+B) \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac{7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}} \]

[Out]

(7*(9*A + B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(128*Sqrt[2]*a^3*c^(5/2)*f) +
(7*(9*A + B)*Cos[e + f*x])/(128*a^3*c*f*(c - c*Sin[e + f*x])^(3/2)) + (7*(9*A + B)*Sec[e + f*x])/(240*a^3*c*f
*(c - c*Sin[e + f*x])^(3/2)) - (7*(9*A + B)*Sec[e + f*x])/(96*a^3*c^2*f*Sqrt[c - c*Sin[e + f*x]]) - ((9*A + B)
*Sec[e + f*x]^3)/(30*a^3*c^2*f*Sqrt[c - c*Sin[e + f*x]]) - ((A - B)*Sec[e + f*x]^5*Sqrt[c - c*Sin[e + f*x]])/(
5*a^3*c^3*f)

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Rubi [A]  time = 0.55501, antiderivative size = 258, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.184, Rules used = {2967, 2855, 2687, 2681, 2650, 2649, 206} \[ -\frac{(A-B) \sec ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{5 a^3 c^3 f}-\frac{(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{7 (9 A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{128 \sqrt{2} a^3 c^{5/2} f}+\frac{7 (9 A+B) \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac{7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}} \]

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])^(5/2)),x]

[Out]

(7*(9*A + B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(128*Sqrt[2]*a^3*c^(5/2)*f) +
(7*(9*A + B)*Cos[e + f*x])/(128*a^3*c*f*(c - c*Sin[e + f*x])^(3/2)) + (7*(9*A + B)*Sec[e + f*x])/(240*a^3*c*f
*(c - c*Sin[e + f*x])^(3/2)) - (7*(9*A + B)*Sec[e + f*x])/(96*a^3*c^2*f*Sqrt[c - c*Sin[e + f*x]]) - ((9*A + B)
*Sec[e + f*x]^3)/(30*a^3*c^2*f*Sqrt[c - c*Sin[e + f*x]]) - ((A - B)*Sec[e + f*x]^5*Sqrt[c - c*Sin[e + f*x]])/(
5*a^3*c^3*f)

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 2855

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 2687

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Simp[(b*(g*
Cos[e + f*x])^(p + 1))/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(a*(2*p + 1))/(2*g^2*(p + 1)), Int[
(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[p, -1] && IntegerQ[2*p]

Rule 2681

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*(2*m + p + 1)), x] + Dist[(m + p + 1)/(a*(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] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && IntegersQ[2*m, 2*p]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2649

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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))^{5/2}} \, dx &=\frac{\int \sec ^6(e+f x) (A+B \sin (e+f x)) \sqrt{c-c \sin (e+f x)} \, dx}{a^3 c^3}\\ &=-\frac{(A-B) \sec ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac{(9 A+B) \int \frac{\sec ^4(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx}{10 a^3 c^2}\\ &=-\frac{(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac{(7 (9 A+B)) \int \frac{\sec ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{60 a^3 c}\\ &=\frac{7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac{(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac{(7 (9 A+B)) \int \frac{\sec ^2(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx}{96 a^3 c^2}\\ &=\frac{7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac{7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac{(7 (9 A+B)) \int \frac{1}{(c-c \sin (e+f x))^{3/2}} \, dx}{64 a^3 c}\\ &=\frac{7 (9 A+B) \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac{7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac{7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{5 a^3 c^3 f}+\frac{(7 (9 A+B)) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{256 a^3 c^2}\\ &=\frac{7 (9 A+B) \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac{7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac{7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{5 a^3 c^3 f}-\frac{(7 (9 A+B)) \operatorname{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 )}{128 a^3 c^2 f}\\ &=\frac{7 (9 A+B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{128 \sqrt{2} a^3 c^{5/2} f}+\frac{7 (9 A+B) \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac{7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac{7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec ^5(e+f x) \sqrt{c-c \sin (e+f x)}}{5 a^3 c^3 f}\\ \end{align*}

Mathematica [C]  time = 2.3776, size = 479, normalized size = 1.86 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (15 (15 A+7 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5+60 (A+B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5+30 (15 A+7 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5+120 (A+B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5+80 (B-3 A) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2+96 (B-A) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4+(-105-105 i) \sqrt [4]{-1} (9 A+B) \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5-720 A \cos ^4(e+f x)\right )}{1920 a^3 f (\sin (e+f x)+1)^3 (c-c \sin (e+f x))^{5/2}} \]

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])^(5/2)),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-720*A*Cos[e + f*x]^4 + 96*(-A +
B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 + 80*(-3*A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2])^2 + 60*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e +
f*x)/2])^5 + 15*(15*A + 7*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 -
(105 + 105*I)*(-1)^(1/4)*(9*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])^4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 + 120*(A + B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^5 + 30*(15*A + 7*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2]*(Cos[(e + f*x)
/2] + Sin[(e + f*x)/2])^5))/(1920*a^3*f*(1 + Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))

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Maple [A]  time = 1.475, size = 410, normalized size = 1.6 \begin{align*} -{\frac{1}{3840\,{a}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{2} \left ( -1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( \left ( 1260\,{c}^{9/2}A+140\,{c}^{9/2}B \right ) \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+ \left ( -1890\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( c+c\sin \left ( fx+e \right ) \right ) ^{5/2}{c}^{2}A+864\,{c}^{9/2}A-210\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( c+c\sin \left ( fx+e \right ) \right ) ^{5/2}{c}^{2}B+96\,{c}^{9/2}B \right ) \sin \left ( fx+e \right ) + \left ( -1890\,{c}^{9/2}A-210\,{c}^{9/2}B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( -945\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( c+c\sin \left ( fx+e \right ) \right ) ^{5/2}{c}^{2}A+252\,{c}^{9/2}A-105\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( c+c\sin \left ( fx+e \right ) \right ) ^{5/2}{c}^{2}B+28\,{c}^{9/2}B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+1890\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( c+c\sin \left ( fx+e \right ) \right ) ^{5/2}{c}^{2}A+96\,{c}^{9/2}A+210\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c+c\sin \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( c+c\sin \left ( fx+e \right ) \right ) ^{5/2}{c}^{2}B+864\,{c}^{9/2}B \right ){c}^{-{\frac{13}{2}}}{\frac{1}{\sqrt{c-c\sin \left ( fx+e \right ) }}}} \end{align*}

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))^(5/2),x)

[Out]

-1/3840/c^(13/2)/a^3*((1260*c^(9/2)*A+140*c^(9/2)*B)*sin(f*x+e)*cos(f*x+e)^2+(-1890*2^(1/2)*arctanh(1/2*(c+c*s
in(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))^(5/2)*c^2*A+864*c^(9/2)*A-210*2^(1/2)*arctanh(1/2*(c+c*sin(
f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))^(5/2)*c^2*B+96*c^(9/2)*B)*sin(f*x+e)+(-1890*c^(9/2)*A-210*c^(9
/2)*B)*cos(f*x+e)^4+(-945*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))^(5/2)*c
^2*A+252*c^(9/2)*A-105*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))^(5/2)*c^2*
B+28*c^(9/2)*B)*cos(f*x+e)^2+1890*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))
^(5/2)*c^2*A+96*c^(9/2)*A+210*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))^(5/
2)*c^2*B+864*c^(9/2)*B)/(1+sin(f*x+e))^2/(-1+sin(f*x+e))/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.8395, size = 659, normalized size = 2.55 \begin{align*} \frac{105 \, \sqrt{2}{\left (9 \, A + B\right )} \sqrt{c} \cos \left (f x + e\right )^{5} \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 (105 \,{\left (9 \, A + B\right )} \cos \left (f x + e\right )^{4} - 14 \,{\left (9 \, A + B\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (35 \,{\left (9 \, A + B\right )} \cos \left (f x + e\right )^{2} + 216 \, A + 24 \, B\right )} \sin \left (f x + e\right ) - 48 \, A - 432 \, B\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{7680 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}} \end{align*}

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))^(5/2),x, algorithm="fricas")

[Out]

1/7680*(105*sqrt(2)*(9*A + B)*sqrt(c)*cos(f*x + e)^5*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*(105*(9*A + B)*cos(f*x + e)^4 - 14*
(9*A + B)*cos(f*x + e)^2 - 2*(35*(9*A + B)*cos(f*x + e)^2 + 216*A + 24*B)*sin(f*x + e) - 48*A - 432*B)*sqrt(-c
*sin(f*x + e) + c))/(a^3*c^3*f*cos(f*x + e)^5)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

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))^(5/2),x, algorithm="giac")

[Out]

sage2