\(\int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx\) [156]

Optimal result

Integrand size = 40, antiderivative size = 196 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 B \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {a^3 B \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]

[Out]

1/6*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(7/2)-1/2*a*B*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)
/c/f/(c-c*sin(f*x+e))^(5/2)+a^2*B*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/c^2/f/(c-c*sin(f*x+e))^(3/2)+a^3*B*cos(f*x
+e)*ln(1-sin(f*x+e))/c^3/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3051, 2818, 2816, 2746, 31} \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {a^3 B \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {a^2 B \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {a B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}} \]

[In]

Int[((a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(6*f*(c - c*Sin[e + f*x])^(7/2)) - (a*B*Cos[e + f*x]*(a + a*
Sin[e + f*x])^(3/2))/(2*c*f*(c - c*Sin[e + f*x])^(5/2)) + (a^2*B*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(c^2*f
*(c - c*Sin[e + f*x])^(3/2)) + (a^3*B*Cos[e + f*x]*Log[1 - Sin[e + f*x]])/(c^3*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt
[c - c*Sin[e + f*x]])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2816

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

Rule 2818

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^n/(f*(2*n + 1))), x] - Dist[b*((2*m - 1)
/(d*(2*n + 1))), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e
, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] &&  !(ILtQ[m + n, 0] && G
tQ[2*m + n + 1, 0])

Rule 3051

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[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2
- b^2, 0] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] &&  !SumSimplerQ[n, 1])) && NeQ[2*m + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {B \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx}{c} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac {(a B) \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{3/2}} \, dx}{c^2} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 B \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}-\frac {\left (a^2 B\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^3} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 B \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}-\frac {\left (a^3 B \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{c^2 \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 B \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {\left (a^3 B \cos (e+f x)\right ) \text {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c^3 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{6 f (c-c \sin (e+f x))^{7/2}}-\frac {a B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 B \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {a^3 B \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.38 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.04 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {\left (4 (A+B)-6 (A+2 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+3 (A+5 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+6 B \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{5/2}}{3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 (c-c \sin (e+f x))^{7/2}} \]

[In]

Integrate[((a + a*Sin[e + f*x])^(5/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

((4*(A + B) - 6*(A + 2*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2 + 3*(A + 5*B)*(Cos[(e + f*x)/2] - Sin[(e + f
*x)/2])^4 + 6*B*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6)*(Cos[(e + f*
x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(5/2))/(3*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*(c - c*Si
n[e + f*x])^(7/2))

Maple [A] (verified)

Time = 4.02 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.63

method	result	size
default	\(\frac {a^{2} \sec \left (f x +e \right ) \left (-6 B \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )+3 B \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+A \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-7 B \left (\sin ^{3}\left (f x +e \right )\right )+18 B \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-9 B \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+6 B \left (\sin ^{2}\left (f x +e \right )\right )+24 B \sin \left (f x +e \right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-12 B \sin \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-4 A \sin \left (f x +e \right )-3 B \sin \left (f x +e \right )-24 B \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )+12 B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{3 c^{3} f \left (\cos ^{2}\left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(320\)
parts	\(\frac {A \tan \left (f x +e \right ) a^{2} \left (\cos ^{2}\left (f x +e \right )-4\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{3 f \left (\cos ^{2}\left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{3}}-\frac {B \sec \left (f x +e \right ) \left (6 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-3 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-18 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )+9 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-7 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-24 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \sin \left (f x +e \right )+12 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )+6 \left (\cos ^{2}\left (f x +e \right )\right )+24 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-12 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+10 \sin \left (f x +e \right )-6\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}{3 f \left (\cos ^{2}\left (f x +e \right )+2 \sin \left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{3}}\)	\(366\)

[In]

int((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/3*a^2/c^3/f*sec(f*x+e)*(-6*B*sin(f*x+e)*cos(f*x+e)^2*ln(csc(f*x+e)-cot(f*x+e)-1)+3*B*sin(f*x+e)*cos(f*x+e)^2
*ln(2/(1+cos(f*x+e)))+A*sin(f*x+e)*cos(f*x+e)^2-7*B*sin(f*x+e)^3+18*B*cos(f*x+e)^2*ln(csc(f*x+e)-cot(f*x+e)-1)
-9*B*cos(f*x+e)^2*ln(2/(1+cos(f*x+e)))+6*B*sin(f*x+e)^2+24*B*sin(f*x+e)*ln(csc(f*x+e)-cot(f*x+e)-1)-12*B*sin(f
*x+e)*ln(2/(1+cos(f*x+e)))-4*A*sin(f*x+e)-3*B*sin(f*x+e)-24*B*ln(csc(f*x+e)-cot(f*x+e)-1)+12*B*ln(2/(1+cos(f*x
+e))))*(a*(1+sin(f*x+e)))^(1/2)/(cos(f*x+e)^2+2*sin(f*x+e)-2)/(-c*(sin(f*x+e)-1))^(1/2)

Fricas [F]

\[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

integral(-((A + 2*B)*a^2*cos(f*x + e)^2 - 2*(A + B)*a^2 + (B*a^2*cos(f*x + e)^2 - 2*(A + B)*a^2)*sin(f*x + e))
*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^4*cos(f*x + e)^4 - 8*c^4*cos(f*x + e)^2 + 8*c^4 + 4*(c^
4*cos(f*x + e)^2 - 2*c^4)*sin(f*x + e)), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(7/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(5/2)/(-c*sin(f*x + e) + c)^(7/2), x)

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.41 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {\sqrt {2} \sqrt {a} {\left (\frac {6 \, \sqrt {2} B a^{2} \log \left (-2 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{c^{\frac {7}{2}} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} {\left (3 \, {\left (A a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, B a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + A a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 10 \, B a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, {\left (A a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 8 \, B a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{3} c^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )}}{12 \, f} \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

-1/12*sqrt(2)*sqrt(a)*(6*sqrt(2)*B*a^2*log(-2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 2)*sgn(cos(-1/4*pi + 1/2*f*x
+ 1/2*e))/(c^(7/2)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - sqrt(2)*(3*(A*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x
+ 1/2*e)) + 5*B*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4 + A*a^2*sqrt
(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 10*B*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*(A*a^2*sqrt
(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 8*B*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-1/4*pi + 1
/2*f*x + 1/2*e)^2)/((cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^3*c^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{7/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \]

[In]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2))/(c - c*sin(e + f*x))^(7/2),x)

[Out]

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(5/2))/(c - c*sin(e + f*x))^(7/2), x)