\(\int \frac {(3+3 \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx\) [304]

Optimal result

Integrand size = 28, antiderivative size = 113 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {27 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} c^{5/2} f}+\frac {9 c \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}-\frac {27 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{3/2}} \]

[Out]

1/2*a^2*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(7/2)-3/4*a^2*cos(f*x+e)/c/f/(c-c*sin(f*x+e))^(3/2)+3/8*a^2*arctanh(
1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/c^(5/2)/f*2^(1/2)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2815, 2759, 2728, 212} \[ \int \frac {(3+3 \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} c^{5/2} f}+\frac {a^2 c \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}-\frac {3 a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{3/2}} \]

[In]

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

[Out]

(3*a^2*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(4*Sqrt[2]*c^(5/2)*f) + (a^2*c*Cos[
e + f*x]^3)/(2*f*(c - c*Sin[e + f*x])^(7/2)) - (3*a^2*Cos[e + f*x])/(4*c*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 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^2 c^2\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{9/2}} \, dx \\ & = \frac {a^2 c \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}-\frac {1}{4} \left (3 a^2\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx \\ & = \frac {a^2 c \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}-\frac {3 a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{3/2}}+\frac {\left (3 a^2\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{8 c^2} \\ & = \frac {a^2 c \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}-\frac {3 a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{3/2}}-\frac {\left (3 a^2\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 )}{4 c^2 f} \\ & = \frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} c^{5/2} f}+\frac {a^2 c \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}-\frac {3 a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{3/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.96 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.42 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {9 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (3 \cos \left (\frac {1}{2} (e+f x)\right )-5 \cos \left (\frac {3}{2} (e+f x)\right )+3 \sin \left (\frac {1}{2} (e+f x)\right )+(3+3 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) (-3+\cos (2 (e+f x))+4 \sin (e+f x))+5 \sin \left (\frac {3}{2} (e+f x)\right )\right )}{8 c^2 f (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[(3 + 3*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^(5/2),x]

[Out]

(9*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(3*Cos[(e + f*x)/2] - 5*Cos[(3*(e + f*x))/2] + 3*Sin[(e + f*x)/2] + (
3 + 3*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(-3 + Cos[2*(e + f*x)] + 4*Sin[e + f
*x]) + 5*Sin[(3*(e + f*x))/2]))/(8*c^2*f*(-1 + Sin[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 2.71 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.69

method	result	size
default	\(-\frac {a^{2} \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{2}-6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \sin \left (f x +e \right )+10 \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {3}{2}} \sqrt {c}+3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-12 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {3}{2}}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{8 c^{\frac {9}{2}} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(191\)
parts	\(-\frac {a^{2} \left (3 \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^{2}-6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \sin \left (f x +e \right )-6 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {3}{2}} \sin \left (f x +e \right )+3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+14 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {3}{2}}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{32 c^{\frac {9}{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}-\frac {a^{2} \left (19 \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^{2}-38 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \sin \left (f x +e \right )+26 \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {3}{2}} \sqrt {c}+19 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-44 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {3}{2}}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{32 c^{\frac {9}{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}-\frac {a^{2} \left (-5 \,\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 ) \sqrt {2}\, c^{3}+10 \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}+10 \,\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 ) \sqrt {2}\, c^{3}-12 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {5}{2}}-5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{16 c^{\frac {11}{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}\)	\(578\)

[In]

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

[Out]

-1/8/c^(9/2)*a^2*(3*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^2-6*2^(1/2)*a
rctanh(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*c^2*sin(f*x+e)+10*(c*(sin(f*x+e)+1))^(3/2)*c^(1/2)+3*2^(1
/2)*arctanh(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*c^2-12*(c*(sin(f*x+e)+1))^(1/2)*c^(3/2))*(c*(sin(f*x
+e)+1))^(1/2)/(sin(f*x+e)-1)/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. 362 vs. \(2 (103) = 206\).

Time = 0.27 (sec) , antiderivative size = 362, normalized size of antiderivative = 3.20 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {3 \, \sqrt {2} {\left (a^{2} \cos \left (f x + e\right )^{3} + 3 \, a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) - 4 \, a^{2} - {\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) - 4 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left (5 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2} \cos \left (f x + e\right ) - 4 \, a^{2} - {\left (5 \, a^{2} \cos \left (f x + e\right ) + 4 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{16 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

[In]

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

[Out]

1/16*(3*sqrt(2)*(a^2*cos(f*x + e)^3 + 3*a^2*cos(f*x + e)^2 - 2*a^2*cos(f*x + e) - 4*a^2 - (a^2*cos(f*x + e)^2
- 2*a^2*cos(f*x + e) - 4*a^2)*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*(5*a^2*cos(f*x + e)^2 + a^2*cos(f*x
+ e) - 4*a^2 - (5*a^2*cos(f*x + e) + 4*a^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^3*f*cos(f*x + e)^3 + 3
*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*
f)*sin(f*x + e))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (103) = 206\).

Time = 0.39 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.83 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\frac {12 \, \sqrt {2} a^{2} \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\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 {\sqrt {2} {\left (a^{2} \sqrt {c} + \frac {8 \, a^{2} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {18 \, a^{2} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}{c^{3} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {\frac {8 \, \sqrt {2} a^{2} c^{\frac {7}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {\sqrt {2} a^{2} c^{\frac {7}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}}{c^{6}}}{64 \, f} \]

[In]

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

[Out]

1/64*(12*sqrt(2)*a^2*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^(5/2)*
sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - sqrt(2)*(a^2*sqrt(c) + 8*a^2*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) + 18*a^2*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/(c^3*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*s
gn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + (8*sqrt(2)*a^2*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^2*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)/c^6)/f

Mupad [F(-1)]

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

[In]

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

[Out]

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