\(\int \frac {(3+3 \sin (e+f x))^3}{(c-c \sin (e+f x))^{9/2}} \, dx\) [315]

Optimal result

Integrand size = 28, antiderivative size = 176 \[ \int \frac {(3+3 \sin (e+f x))^3}{(c-c \sin (e+f x))^{9/2}} \, dx=-\frac {135 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} c^{9/2} f}+\frac {27 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac {45 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{9/2}}+\frac {135 \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac {135 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}} \]

[Out]

1/4*a^3*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(13/2)-5/24*a^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(9/2)+5/32*a^3*cos
(f*x+e)/c^2/f/(c-c*sin(f*x+e))^(5/2)-5/128*a^3*cos(f*x+e)/c^3/f/(c-c*sin(f*x+e))^(3/2)-5/256*a^3*arctanh(1/2*c
os(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/c^(9/2)/f*2^(1/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2815, 2759, 2729, 2728, 212} \[ \int \frac {(3+3 \sin (e+f x))^3}{(c-c \sin (e+f x))^{9/2}} \, dx=-\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} c^{9/2} f}-\frac {5 a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}}+\frac {a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}+\frac {5 a^3 \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac {5 a^3 \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}} \]

[In]

Int[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(-5*a^3*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(128*Sqrt[2]*c^(9/2)*f) + (a^3*c^2
*Cos[e + f*x]^5)/(4*f*(c - c*Sin[e + f*x])^(13/2)) - (5*a^3*Cos[e + f*x]^3)/(24*f*(c - c*Sin[e + f*x])^(9/2))
+ (5*a^3*Cos[e + f*x])/(32*c^2*f*(c - c*Sin[e + f*x])^(5/2)) - (5*a^3*Cos[e + f*x])/(128*c^3*f*(c - c*Sin[e +
f*x])^(3/2))

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 2759

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[2*g*(g
*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(2*m +
p + 1))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2815

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 0]))

Rubi steps \begin{align*} \text {integral}& = \left (a^3 c^3\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{15/2}} \, dx \\ & = \frac {a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac {1}{8} \left (5 a^3 c\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{11/2}} \, dx \\ & = \frac {a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac {5 a^3 \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac {\left (5 a^3\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{7/2}} \, dx}{16 c} \\ & = \frac {a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac {5 a^3 \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac {5 a^3 \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac {\left (5 a^3\right ) \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{64 c^3} \\ & = \frac {a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac {5 a^3 \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac {5 a^3 \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac {5 a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}}-\frac {\left (5 a^3\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{256 c^4} \\ & = \frac {a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac {5 a^3 \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac {5 a^3 \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac {5 a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}}+\frac {\left (5 a^3\right ) \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 c^4 f} \\ & = -\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} c^{9/2} f}+\frac {a^3 c^2 \cos ^5(e+f x)}{4 f (c-c \sin (e+f x))^{13/2}}-\frac {5 a^3 \cos ^3(e+f x)}{24 f (c-c \sin (e+f x))^{9/2}}+\frac {5 a^3 \cos (e+f x)}{32 c^2 f (c-c \sin (e+f x))^{5/2}}-\frac {5 a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{3/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.34 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.90 \[ \int \frac {(3+3 \sin (e+f x))^3}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {9 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (384 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-544 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+236 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5-15 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7+(15+15 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^8+768 \sin \left (\frac {1}{2} (e+f x)\right )-1088 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )+472 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )-30 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 \sin \left (\frac {1}{2} (e+f x)\right )\right )}{128 f (c-c \sin (e+f x))^{9/2}} \]

[In]

Integrate[(3 + 3*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(9*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(384*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 544*(Cos[(e + f*x)/2] -
Sin[(e + f*x)/2])^3 + 236*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 - 15*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7
+ (15 + 15*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f
*x)/2])^8 + 768*Sin[(e + f*x)/2] - 1088*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 472*(Cos[(e
+ f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2] - 30*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6*Sin[(e + f*x)/2
]))/(128*f*(c - c*Sin[e + f*x])^(9/2))

Maple [A] (verified)

Time = 3.83 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.70

method	result	size
default	\(\frac {a^{3} \left (15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{5} \left (\sin ^{4}\left (f x +e \right )\right )-60 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{3}\left (f x +e \right )\right ) c^{5}-240 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {9}{2}}+440 \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {3}{2}} c^{\frac {7}{2}}-292 \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {5}{2}} c^{\frac {5}{2}}-30 \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {7}{2}} c^{\frac {3}{2}}+90 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{5}-60 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{5}+15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{5}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{768 c^{\frac {19}{2}} \left (\sin \left (f x +e \right )-1\right )^{3} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(299\)
parts	\(\text {Expression too large to display}\)	\(1216\)

[In]

int((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(9/2),x,method=_RETURNVERBOSE)

[Out]

1/768*a^3*(15*2^(1/2)*arctanh(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*c^5*sin(f*x+e)^4-60*2^(1/2)*arctan
h(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^3*c^5-240*(c*(sin(f*x+e)+1))^(1/2)*c^(9/2)+440*(c*(
sin(f*x+e)+1))^(3/2)*c^(7/2)-292*(c*(sin(f*x+e)+1))^(5/2)*c^(5/2)-30*(c*(sin(f*x+e)+1))^(7/2)*c^(3/2)+90*2^(1/
2)*arctanh(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*c^5-60*2^(1/2)*arctanh(1/2*(c*(sin(f*x+e
)+1))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*c^5+15*2^(1/2)*arctanh(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*c
^5)*(c*(sin(f*x+e)+1))^(1/2)/c^(19/2)/(sin(f*x+e)-1)^3/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 523 vs. \(2 (164) = 328\).

Time = 0.29 (sec) , antiderivative size = 523, normalized size of antiderivative = 2.97 \[ \int \frac {(3+3 \sin (e+f x))^3}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {15 \, \sqrt {2} {\left (a^{3} \cos \left (f x + e\right )^{5} + 5 \, a^{3} \cos \left (f x + e\right )^{4} - 8 \, a^{3} \cos \left (f x + e\right )^{3} - 20 \, a^{3} \cos \left (f x + e\right )^{2} + 8 \, a^{3} \cos \left (f x + e\right ) + 16 \, a^{3} - {\left (a^{3} \cos \left (f x + e\right )^{4} - 4 \, a^{3} \cos \left (f x + e\right )^{3} - 12 \, a^{3} \cos \left (f x + e\right )^{2} + 8 \, a^{3} \cos \left (f x + e\right ) + 16 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {c} \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 (15 \, a^{3} \cos \left (f x + e\right )^{4} - 191 \, a^{3} \cos \left (f x + e\right )^{3} - 338 \, a^{3} \cos \left (f x + e\right )^{2} + 252 \, a^{3} \cos \left (f x + e\right ) + 384 \, a^{3} - {\left (15 \, a^{3} \cos \left (f x + e\right )^{3} + 206 \, a^{3} \cos \left (f x + e\right )^{2} - 132 \, a^{3} \cos \left (f x + e\right ) - 384 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{1536 \, {\left (c^{5} f \cos \left (f x + e\right )^{5} + 5 \, c^{5} f \cos \left (f x + e\right )^{4} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} - 20 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f - {\left (c^{5} f \cos \left (f x + e\right )^{4} - 4 \, c^{5} f \cos \left (f x + e\right )^{3} - 12 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

1/1536*(15*sqrt(2)*(a^3*cos(f*x + e)^5 + 5*a^3*cos(f*x + e)^4 - 8*a^3*cos(f*x + e)^3 - 20*a^3*cos(f*x + e)^2 +
8*a^3*cos(f*x + e) + 16*a^3 - (a^3*cos(f*x + e)^4 - 4*a^3*cos(f*x + e)^3 - 12*a^3*cos(f*x + e)^2 + 8*a^3*cos(
f*x + e) + 16*a^3)*sin(f*x + e))*sqrt(c)*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*(15*a^3*cos(f*x + e)^4 - 191*a^3*cos(f*x + e)^3
- 338*a^3*cos(f*x + e)^2 + 252*a^3*cos(f*x + e) + 384*a^3 - (15*a^3*cos(f*x + e)^3 + 206*a^3*cos(f*x + e)^2 -
132*a^3*cos(f*x + e) - 384*a^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^5*f*cos(f*x + e)^5 + 5*c^5*f*cos(
f*x + e)^4 - 8*c^5*f*cos(f*x + e)^3 - 20*c^5*f*cos(f*x + e)^2 + 8*c^5*f*cos(f*x + e) + 16*c^5*f - (c^5*f*cos(f
*x + e)^4 - 4*c^5*f*cos(f*x + e)^3 - 12*c^5*f*cos(f*x + e)^2 + 8*c^5*f*cos(f*x + e) + 16*c^5*f)*sin(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^3}{(c-c \sin (e+f x))^{9/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(3+3 \sin (e+f x))^3}{(c-c \sin (e+f x))^{9/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^3/(-c*sin(f*x + e) + c)^(9/2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 513 vs. \(2 (164) = 328\).

Time = 0.45 (sec) , antiderivative size = 513, normalized size of antiderivative = 2.91 \[ \int \frac {(3+3 \sin (e+f x))^3}{(c-c \sin (e+f x))^{9/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(9/2),x, algorithm="giac")

[Out]

-1/12288*(120*sqrt(2)*a^3*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))/(c^(
9/2)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + sqrt(2)*(3*a^3*sqrt(c) + 16*a^3*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) + 24*a^3*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 - 48*a^3*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 - 250*a^3*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)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4/(c^5*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4*sgn(sin(-1/4*pi + 1/2*
f*x + 1/2*e))) + (48*sqrt(2)*a^3*c^(31/2)*(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) - 24*sqrt(2)*a^3*c^(31/2)*(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 - 16*sqrt(2)*a^3*c^(31/2)*(cos(-1/4*pi
+ 1/2*f*x + 1/2*e) - 1)^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3 - 3*sqrt
(2)*a^3*c^(31/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2
*f*x + 1/2*e) + 1)^4)/c^20)/f

Mupad [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^3}{(c-c \sin (e+f x))^{9/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{9/2}} \,d x \]

[In]

int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^(9/2),x)

[Out]

int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^(9/2), x)