\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{5/2}} \, dx\) [188]

Optimal result

Integrand size = 40, antiderivative size = 263 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {6 (A-3 B) c^4 \cos (e+f x) \log (1+\sin (e+f x))}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {3 (A-3 B) c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {3 (A-3 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(A-3 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 f (a+a \sin (e+f x))^{5/2}} \]

[Out]

1/2*(A-3*B)*c*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/a/f/(a+a*sin(f*x+e))^(3/2)-1/4*(A-B)*cos(f*x+e)*(c-c*sin(f*x+e
))^(7/2)/f/(a+a*sin(f*x+e))^(5/2)+3/4*(A-3*B)*c^2*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a^2/f/(a+a*sin(f*x+e))^(1/
2)+6*(A-3*B)*c^4*cos(f*x+e)*ln(1+sin(f*x+e))/a^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)+3*(A-3*B)*c^3
*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a^2/f/(a+a*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3051, 2818, 2819, 2816, 2746, 31} \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {6 c^4 (A-3 B) \cos (e+f x) \log (\sin (e+f x)+1)}{a^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {3 c^3 (A-3 B) \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)+a}}+\frac {3 c^2 (A-3 B) \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a^2 f \sqrt {a \sin (e+f x)+a}}+\frac {c (A-3 B) \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 a f (a \sin (e+f x)+a)^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 f (a \sin (e+f x)+a)^{5/2}} \]

[In]

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

[Out]

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

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 2819

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Dist[a*((2*m - 1)/(
m + n)), 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, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m
]) &&  !(ILtQ[m + n, 0] && GtQ[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) (c-c \sin (e+f x))^{7/2}}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {(A-3 B) \int \frac {(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx}{2 a} \\ & = \frac {(A-3 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {(3 (A-3 B) c) \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{2 a^2} \\ & = \frac {3 (A-3 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(A-3 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (3 (A-3 B) c^2\right ) \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2} \\ & = \frac {3 (A-3 B) c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {3 (A-3 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(A-3 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (6 (A-3 B) c^3\right ) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2} \\ & = \frac {3 (A-3 B) c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {3 (A-3 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(A-3 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (6 (A-3 B) c^4 \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{a+a \sin (e+f x)} \, dx}{a \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {3 (A-3 B) c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {3 (A-3 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(A-3 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (6 (A-3 B) c^4 \cos (e+f x)\right ) \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,a \sin (e+f x)\right )}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {6 (A-3 B) c^4 \cos (e+f x) \log (1+\sin (e+f x))}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {3 (A-3 B) c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {3 (A-3 B) c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{4 a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {(A-3 B) c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 a f (a+a \sin (e+f x))^{3/2}}-\frac {(A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 f (a+a \sin (e+f x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 12.19 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c-c \sin (e+f x))^{7/2} \left (-16 (A-B)+16 (3 A-5 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+B \cos (2 (e+f x)) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+48 (A-3 B) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-4 (A-6 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin (e+f x)\right )}{4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 (a (1+\sin (e+f x)))^{5/2}} \]

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c - c*Sin[e + f*x])^(7/2)*(-16*(A - B) + 16*(3*A - 5*B)*(Cos[(e + f*x)
/2] + Sin[(e + f*x)/2])^2 + B*Cos[2*(e + f*x)]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 + 48*(A - 3*B)*Log[Cos[
(e + f*x)/2] + Sin[(e + f*x)/2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - 4*(A - 6*B)*(Cos[(e + f*x)/2] + Sin
[(e + f*x)/2])^4*Sin[e + f*x]))/(4*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(a*(1 + Sin[e + f*x]))^(5/2))

Maple [A] (verified)

Time = 3.13 (sec) , antiderivative size = 407, normalized size of antiderivative = 1.55

method	result	size
default	\(-\frac {c^{3} \sec \left (f x +e \right ) \left (-B \left (\sin ^{2}\left (f x +e \right )\right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \left (\sin ^{3}\left (f x +e \right )\right ) A -12 A \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+24 A \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-10 B \left (\sin ^{3}\left (f x +e \right )\right )+36 B \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-72 B \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+20 \left (\sin ^{2}\left (f x +e \right )\right ) A +24 A \sin \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-48 A \sin \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-54 B \left (\sin ^{2}\left (f x +e \right )\right )-72 B \sin \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+144 B \sin \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+10 A \sin \left (f x +e \right )+24 A \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-48 A \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-36 B \sin \left (f x +e \right )-72 B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+144 B \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}{2 a^{2} f \left (1+\sin \left (f x +e \right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}\)	\(407\)
parts	\(\frac {A \sec \left (f x +e \right ) \left (6 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-12 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-12 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )+24 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \sin \left (f x +e \right )+10 \left (\cos ^{2}\left (f x +e \right )\right )-12 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+24 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-6 \sin \left (f x +e \right )-10\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{3}}{f \left (1+\sin \left (f x +e \right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}-\frac {B \sec \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )+10 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-72 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+36 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+53 \left (\cos ^{2}\left (f x +e \right )\right )+144 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \sin \left (f x +e \right )-72 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )-46 \sin \left (f x +e \right )+144 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-72 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-54\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{3}}{2 f \left (1+\sin \left (f x +e \right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}\)	\(446\)

[In]

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

[Out]

-1/2*c^3/a^2/f*sec(f*x+e)*(-B*sin(f*x+e)^2*cos(f*x+e)^2+2*sin(f*x+e)^3*A-12*A*cos(f*x+e)^2*ln(2/(1+cos(f*x+e))
)+24*A*cos(f*x+e)^2*ln(-cot(f*x+e)+csc(f*x+e)+1)-10*B*sin(f*x+e)^3+36*B*cos(f*x+e)^2*ln(2/(1+cos(f*x+e)))-72*B
*cos(f*x+e)^2*ln(-cot(f*x+e)+csc(f*x+e)+1)+20*sin(f*x+e)^2*A+24*A*sin(f*x+e)*ln(2/(1+cos(f*x+e)))-48*A*sin(f*x
+e)*ln(-cot(f*x+e)+csc(f*x+e)+1)-54*B*sin(f*x+e)^2-72*B*sin(f*x+e)*ln(2/(1+cos(f*x+e)))+144*B*sin(f*x+e)*ln(-c
ot(f*x+e)+csc(f*x+e)+1)+10*A*sin(f*x+e)+24*A*ln(2/(1+cos(f*x+e)))-48*A*ln(-cot(f*x+e)+csc(f*x+e)+1)-36*B*sin(f
*x+e)-72*B*ln(2/(1+cos(f*x+e)))+144*B*ln(-cot(f*x+e)+csc(f*x+e)+1))*(-c*(sin(f*x+e)-1))^(1/2)/(1+sin(f*x+e))/(
a*(1+sin(f*x+e)))^(1/2)

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.57 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {2} \sqrt {c} {\left (\frac {6 \, \sqrt {2} {\left (A \sqrt {a} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, B \sqrt {a} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {2 \, {\left (\sqrt {2} B a^{\frac {7}{2}} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - \sqrt {2} A a^{\frac {7}{2}} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, \sqrt {2} B a^{\frac {7}{2}} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}}{a^{6}} + \frac {5 \, \sqrt {2} A \sqrt {a} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 9 \, \sqrt {2} B \sqrt {a} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, {\left (3 \, \sqrt {2} A \sqrt {a} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 5 \, \sqrt {2} B \sqrt {a} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )}}{2 \, f} \]

[In]

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

[Out]

-1/2*sqrt(2)*sqrt(c)*(6*sqrt(2)*(A*sqrt(a)*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*B*sqrt(a)*c^3*sgn(sin(-
1/4*pi + 1/2*f*x + 1/2*e)))*log(-sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)
)) - 2*(sqrt(2)*B*a^(7/2)*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4
*pi + 1/2*f*x + 1/2*e)^4 - sqrt(2)*A*a^(7/2)*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x
+ 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 5*sqrt(2)*B*a^(7/2)*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(
sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2)/a^6 + (5*sqrt(2)*A*sqrt(a)*c^3*sgn(sin(-1/4*
pi + 1/2*f*x + 1/2*e)) - 9*sqrt(2)*B*sqrt(a)*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*(3*sqrt(2)*A*sqrt(a)*
c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 5*sqrt(2)*B*sqrt(a)*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-1/
4*pi + 1/2*f*x + 1/2*e)^2)/((sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^2*a^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))))
/f

Mupad [F(-1)]

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

[In]

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

[Out]

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