\(\int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\) [155]

Optimal result

Integrand size = 40, antiderivative size = 212 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}-\frac {a^3 (A+5 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {a^2 (A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c^2 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3051, 2818, 2819, 2816, 2746, 31} \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {a^3 (A+5 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a^2 (A+5 B) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {a (A+5 B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}} \]

[In]

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

[Out]

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(4*f*(c - c*Sin[e + f*x])^(5/2)) - (a*(A + 5*B)*Cos[e + f*x]
*(a + a*Sin[e + f*x])^(3/2))/(4*c*f*(c - c*Sin[e + f*x])^(3/2)) - (a^3*(A + 5*B)*Cos[e + f*x]*Log[1 - Sin[e +
f*x]])/(c^2*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) - (a^2*(A + 5*B)*Cos[e + f*x]*Sqrt[a + a*Sin[
e + f*x]])/(2*c^2*f*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 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) (a+a \sin (e+f x))^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {(A+5 B) \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx}{4 c} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}+\frac {(a (A+5 B)) \int \frac {(a+a \sin (e+f x))^{3/2}}{\sqrt {c-c \sin (e+f x)}} \, dx}{2 c^2} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 (A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (a^2 (A+5 B)\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^2} \\ & = \frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 (A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (a^3 (A+5 B) \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{c \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}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 (A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (a^3 (A+5 B) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c^2 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}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a (A+5 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 c f (c-c \sin (e+f x))^{3/2}}-\frac {a^3 (A+5 B) \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {a^2 (A+5 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{2 c^2 f \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.41 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.98 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{5/2} \left (2 (A+B)-4 (A+2 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-2 (A+5 B) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-B \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin (e+f x)\right )}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 (c-c \sin (e+f x))^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.54 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.74

method	result	size
default	\(-\frac {a^{2} \sec \left (f x +e \right ) \left (2 A \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-A \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-B \left (\sin ^{3}\left (f x +e \right )\right )+10 B \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-5 B \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+2 \left (\sin ^{2}\left (f x +e \right )\right ) A +4 A \sin \left (f x +e \right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-2 A \sin \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+8 B \left (\sin ^{2}\left (f x +e \right )\right )+20 B \sin \left (f x +e \right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-10 B \sin \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-4 A \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )+2 A \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-5 B \sin \left (f x +e \right )-20 B \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )+10 B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{c^{2} f \left (\sin \left (f x +e \right )-1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(368\)
parts	\(\frac {A \sec \left (f x +e \right ) \left (\left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-2 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )+2 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )-4 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \sin \left (f x +e \right )+2 \left (\cos ^{2}\left (f x +e \right )\right )-2 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+4 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-2\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}{f \left (\sin \left (f x +e \right )-1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{2}}-\frac {B \sec \left (f x +e \right ) \left (10 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-5 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+\left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+20 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \sin \left (f x +e \right )-10 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )-8 \left (\cos ^{2}\left (f x +e \right )\right )-20 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )+10 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-6 \sin \left (f x +e \right )+8\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}{f \left (\sin \left (f x +e \right )-1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{2}}\)	\(413\)

[In]

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

[Out]

-a^2/c^2/f*sec(f*x+e)*(2*A*cos(f*x+e)^2*ln(csc(f*x+e)-cot(f*x+e)-1)-A*cos(f*x+e)^2*ln(2/(1+cos(f*x+e)))-B*sin(
f*x+e)^3+10*B*cos(f*x+e)^2*ln(csc(f*x+e)-cot(f*x+e)-1)-5*B*cos(f*x+e)^2*ln(2/(1+cos(f*x+e)))+2*sin(f*x+e)^2*A+
4*A*sin(f*x+e)*ln(csc(f*x+e)-cot(f*x+e)-1)-2*A*sin(f*x+e)*ln(2/(1+cos(f*x+e)))+8*B*sin(f*x+e)^2+20*B*sin(f*x+e
)*ln(csc(f*x+e)-cot(f*x+e)-1)-10*B*sin(f*x+e)*ln(2/(1+cos(f*x+e)))-4*A*ln(csc(f*x+e)-cot(f*x+e)-1)+2*A*ln(2/(1
+cos(f*x+e)))-5*B*sin(f*x+e)-20*B*ln(csc(f*x+e)-cot(f*x+e)-1)+10*B*ln(2/(1+cos(f*x+e))))*(a*(1+sin(f*x+e)))^(1
/2)/(sin(f*x+e)-1)/(-c*(sin(f*x+e)-1))^(1/2)

Fricas [F]

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

[In]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 506 vs. \(2 (190) = 380\).

Time = 0.32 (sec) , antiderivative size = 506, normalized size of antiderivative = 2.39 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {{\left (\frac {8 \, a^{\frac {5}{2}} \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (c^{3} - \frac {4 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {6 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {4 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, a^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac {5}{2}}} + \frac {a^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac {5}{2}}}\right )} A - B {\left (\frac {10 \, a^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac {5}{2}}} - \frac {5 \, a^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac {5}{2}}} + \frac {2 \, {\left (\frac {5 \, a^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {16 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {14 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {16 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {5 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{c^{\frac {5}{2}} - \frac {4 \, c^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {7 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {8 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {7 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {4 \, c^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {c^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}}\right )}}{f} \]

[In]

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

[Out]

-((8*a^(5/2)*sqrt(c)*sin(f*x + e)^2/((c^3 - 4*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 6*c^3*sin(f*x + e)^2/(cos(
f*x + e) + 1)^2 - 4*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)*(cos(f*
x + e) + 1)^2) - 2*a^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^(5/2) + a^(5/2)*log(sin(f*x + e)^2/(cos(
f*x + e) + 1)^2 + 1)/c^(5/2))*A - B*(10*a^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^(5/2) - 5*a^(5/2)*l
og(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/c^(5/2) + 2*(5*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) - 16*a^(5/2
)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 14*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 16*a^(5/2)*sin(f*x +
e)^4/(cos(f*x + e) + 1)^4 + 5*a^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(c^(5/2) - 4*c^(5/2)*sin(f*x + e)/(
cos(f*x + e) + 1) + 7*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 8*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1
)^3 + 7*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 4*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + c^(5/2)*
sin(f*x + e)^6/(cos(f*x + e) + 1)^6)))/f

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.40 \[ \int \frac {(a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {2} {\left (\frac {4 \, \sqrt {2} B a^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{c^{\frac {5}{2}} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {2 \, \sqrt {2} {\left (A a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, B a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \log \left (-32 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 32\right )}{c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {\sqrt {2} {\left (3 \, A a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 7 \, B a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 4 \, {\left (A a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, B a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \sqrt {a}}{4 \, f} \]

[In]

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

[Out]

1/4*sqrt(2)*(4*sqrt(2)*B*a^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))/(c^(5/2)*sgn
(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 2*sqrt(2)*(A*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 5*B*a^2*sqr
t(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*log(-32*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 32)/(c^3*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)) + 7*B*a^2*sqrt(c)*sgn(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)) + 2*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)^2*c^3*sg
n(sin(-1/4*pi + 1/2*f*x + 1/2*e))))*sqrt(a)/f

Mupad [F(-1)]

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

[In]

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

[Out]

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