\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx\) [130]

Optimal result

Integrand size = 38, antiderivative size = 258 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {7 (9 A+B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} a^3 c^{5/2} f}+\frac {7 (9 A+B) \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac {7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac {7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f} \]

[Out]

7/128*(9*A+B)*cos(f*x+e)/a^3/c/f/(c-c*sin(f*x+e))^(3/2)+7/240*(9*A+B)*sec(f*x+e)/a^3/c/f/(c-c*sin(f*x+e))^(3/2
)+7/256*(9*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/c^(5/2)/f*2^(1/2)-7/96*(9*A
+B)*sec(f*x+e)/a^3/c^2/f/(c-c*sin(f*x+e))^(1/2)-1/30*(9*A+B)*sec(f*x+e)^3/a^3/c^2/f/(c-c*sin(f*x+e))^(1/2)-1/5
*(A-B)*sec(f*x+e)^5*(c-c*sin(f*x+e))^(1/2)/a^3/c^3/f

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {3046, 2934, 2766, 2760, 2729, 2728, 212} \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {7 (9 A+B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} a^3 c^{5/2} f}-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f}-\frac {(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {7 (9 A+B) \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac {7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}} \]

[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 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 2729

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 2760

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*(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 2766

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))), In
t[(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 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*} \text {integral}& = \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)) \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 )}{128 a^3 c^2 f} \\ & = \frac {7 (9 A+B) \text {arctanh}\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] (verified)

Result contains complex when optimal does not.

Time = 3.57 (sec) , antiderivative size = 479, normalized size of antiderivative = 1.86 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-720 A \cos ^4(e+f x)+96 (-A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+80 (-3 A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+60 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+15 (15 A+7 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5-(105+105 i) \sqrt [4]{-1} (9 A+B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+120 (A+B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+30 (15 A+7 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5\right )}{1920 a^3 f (1+\sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \]

[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))

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.59

method	result	size
default	\(-\frac {\left (-1890 c^{\frac {9}{2}} A -210 c^{\frac {9}{2}} B \right ) \left (\cos ^{4}\left (f x +e \right )\right )+\left (1260 c^{\frac {9}{2}} A +140 c^{\frac {9}{2}} B \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (-945 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} A +252 c^{\frac {9}{2}} A -105 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} B +28 c^{\frac {9}{2}} B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (-1890 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} A +864 c^{\frac {9}{2}} A -210 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} B +96 c^{\frac {9}{2}} B \right ) \sin \left (f x +e \right )+1890 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} A +96 c^{\frac {9}{2}} A +210 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} B +864 c^{\frac {9}{2}} B}{3840 c^{\frac {13}{2}} a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(410\)

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3840/c^(13/2)/a^3*((-1890*c^(9/2)*A-210*c^(9/2)*B)*cos(f*x+e)^4+(1260*c^(9/2)*A+140*c^(9/2)*B)*cos(f*x+e)^2
*sin(f*x+e)+(-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+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*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/(sin(f*x+e)-1)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.92 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {105 \, \sqrt {2} {\left (9 \, A + B\right )} \sqrt {c} \cos \left (f x + e\right )^{5} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (105 \, {\left (9 \, A + B\right )} \cos \left (f x + e\right )^{4} - 14 \, {\left (9 \, A + B\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (35 \, {\left (9 \, A + B\right )} \cos \left (f x + e\right )^{2} + 216 \, A + 24 \, B\right )} \sin \left (f x + e\right ) - 48 \, A - 432 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{7680 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[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

Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 901 vs. \(2 (227) = 454\).

Time = 0.56 (sec) , antiderivative size = 901, normalized size of antiderivative = 3.49 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \]

[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]

1/30720*(420*sqrt(2)*(9*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^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 15*sqrt(2)*(A*sqrt(c) + B*sqrt(c) - 32*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) - 16*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) + 378*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 + 42*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)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2/(a^3*c^3*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^
2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 256*sqrt(2)*(54*A*sqrt(c) - 4*B*sqrt(c) + 195*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) - 5*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) + 315*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 - 25*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 + 225*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 - 15*B*sqr
t(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 + 75*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^3*((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))) - 15*(32*sqrt(2)*A*a^3*c^(7/2)*(cos(-1/4*p
i + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 16*sqrt(2
)*B*a^3*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*
x + 1/2*e) + 1) - sqrt(2)*A*a^3*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2
*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - sqrt(2)*B*a^3*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn
(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)/(a^6*c^6))/f

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

[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]

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^(5/2)), x)