3.4.6 \(\int \frac {(3+3 \sin (e+f x))^2}{(c-c \sin (e+f x))^{9/2}} \, dx\) [306]

3.4.6.1 Optimal result
3.4.6.2 Mathematica [C] (verified)
3.4.6.3 Rubi [A] (verified)
3.4.6.4 Maple [A] (verified)
3.4.6.5 Fricas [B] (verification not implemented)
3.4.6.6 Sympy [F(-1)]
3.4.6.7 Maxima [F]
3.4.6.8 Giac [B] (verification not implemented)
3.4.6.9 Mupad [F(-1)]

3.4.6.1 Optimal result

Integrand size = 28, antiderivative size = 175 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {27 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{256 \sqrt {2} c^{9/2} f}+\frac {9 c \cos ^3(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {9 \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{7/2}}+\frac {9 \cos (e+f x)}{64 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {27 \cos (e+f x)}{256 c^3 f (c-c \sin (e+f x))^{3/2}} \]

output
1/4*a^2*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(11/2)-1/8*a^2*cos(f*x+e)/c/f/(c 
-c*sin(f*x+e))^(7/2)+1/64*a^2*cos(f*x+e)/c^2/f/(c-c*sin(f*x+e))^(5/2)+3/25 
6*a^2*cos(f*x+e)/c^3/f/(c-c*sin(f*x+e))^(3/2)+3/512*a^2*arctanh(1/2*cos(f* 
x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/c^(9/2)/f*2^(1/2)
 
3.4.6.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.40 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.91 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {9 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (128 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-96 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+4 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7-(3+3 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^8+256 \sin \left (\frac {1}{2} (e+f x)\right )-192 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )+8 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )+6 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 \sin \left (\frac {1}{2} (e+f x)\right )\right )}{256 f (c-c \sin (e+f x))^{9/2}} \]

input
Integrate[(3 + 3*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^(9/2),x]
 
output
(9*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(128*(Cos[(e + f*x)/2] - Sin[(e + 
 f*x)/2]) - 96*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 + 4*(Cos[(e + f*x)/ 
2] - Sin[(e + f*x)/2])^5 + 3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7 - (3 
+ 3*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(C 
os[(e + f*x)/2] - Sin[(e + f*x)/2])^8 + 256*Sin[(e + f*x)/2] - 192*(Cos[(e 
 + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 8*(Cos[(e + f*x)/2] - 
Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2] + 6*(Cos[(e + f*x)/2] - Sin[(e + f*x) 
/2])^6*Sin[(e + f*x)/2]))/(256*f*(c - c*Sin[e + f*x])^(9/2))
 
3.4.6.3 Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.15, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.464, Rules used = {3042, 3215, 3042, 3159, 3042, 3159, 3042, 3129, 3042, 3129, 3042, 3128, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{(c-c \sin (e+f x))^{9/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{(c-c \sin (e+f x))^{9/2}}dx\)

\(\Big \downarrow \) 3215

\(\displaystyle a^2 c^2 \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{13/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{13/2}}dx\)

\(\Big \downarrow \) 3159

\(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{9/2}}dx}{8 c^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^{9/2}}dx}{8 c^2}\right )\)

\(\Big \downarrow \) 3159

\(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\int \frac {1}{(c-c \sin (e+f x))^{5/2}}dx}{6 c^2}\right )}{8 c^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\int \frac {1}{(c-c \sin (e+f x))^{5/2}}dx}{6 c^2}\right )}{8 c^2}\right )\)

\(\Big \downarrow \) 3129

\(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}\right )\)

\(\Big \downarrow \) 3129

\(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}\right )\)

\(\Big \downarrow \) 3128

\(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \left (\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{2 c f}\right )}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{4 c f (c-c \sin (e+f x))^{11/2}}-\frac {3 \left (\frac {\cos (e+f x)}{3 c f (c-c \sin (e+f x))^{7/2}}-\frac {\frac {3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{3/2} f}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )}{8 c}+\frac {\cos (e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}}{6 c^2}\right )}{8 c^2}\right )\)

input
Int[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^(9/2),x]
 
output
a^2*c^2*(Cos[e + f*x]^3/(4*c*f*(c - c*Sin[e + f*x])^(11/2)) - (3*(Cos[e + 
f*x]/(3*c*f*(c - c*Sin[e + f*x])^(7/2)) - (Cos[e + f*x]/(4*f*(c - c*Sin[e 
+ f*x])^(5/2)) + (3*(ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Si 
n[e + f*x]])]/(2*Sqrt[2]*c^(3/2)*f) + Cos[e + f*x]/(2*f*(c - c*Sin[e + f*x 
])^(3/2))))/(8*c))/(6*c^2)))/(8*c^2))
 

3.4.6.3.1 Defintions of rubi rules used

rule 219
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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3128
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3129
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] + Simp[(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 3159
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] + Simp[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] && EqQ[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 3215
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.), x_Symbol] :> Simp[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] && ((Lt 
Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 
3.4.6.4 Maple [A] (verified)

Time = 3.03 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.71

method result size
default \(\frac {a^{2} \left (6 \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {7}{2}} c^{\frac {5}{2}}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{6} \left (\sin ^{4}\left (f x +e \right )\right )-44 \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {5}{2}} c^{\frac {7}{2}}+12 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{3}\left (f x +e \right )\right ) c^{6}-88 \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {3}{2}} c^{\frac {9}{2}}-18 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {2}\, c^{6}+48 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, c^{\frac {11}{2}}+12 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) \sqrt {2}\, c^{6}-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{6}\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}}{512 c^{\frac {21}{2}} \left (\sin \left (f x +e \right )-1\right )^{3} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) \(299\)
parts \(\text {Expression too large to display}\) \(918\)

input
int((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(9/2),x,method=_RETURNVERBOSE)
 
output
1/512/c^(21/2)*a^2*(6*(c*(sin(f*x+e)+1))^(7/2)*c^(5/2)-3*arctanh(1/2*(c*(s 
in(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^6*sin(f*x+e)^4-44*(c*(sin(f 
*x+e)+1))^(5/2)*c^(7/2)+12*2^(1/2)*arctanh(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^ 
(1/2)/c^(1/2))*sin(f*x+e)^3*c^6-88*(c*(sin(f*x+e)+1))^(3/2)*c^(9/2)-18*arc 
tanh(1/2*(c*(sin(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*2^(1/2)*c^ 
6+48*(c*(sin(f*x+e)+1))^(1/2)*c^(11/2)+12*arctanh(1/2*(c*(sin(f*x+e)+1))^( 
1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*2^(1/2)*c^6-3*2^(1/2)*arctanh(1/2*(c*(sin 
(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*c^6)*(c*(sin(f*x+e)+1))^(1/2)/(sin(f*x+ 
e)-1)^3/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
 
3.4.6.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 523 vs. \(2 (163) = 326\).

Time = 0.27 (sec) , antiderivative size = 523, normalized size of antiderivative = 2.99 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {3 \, \sqrt {2} {\left (a^{2} \cos \left (f x + e\right )^{5} + 5 \, a^{2} \cos \left (f x + e\right )^{4} - 8 \, a^{2} \cos \left (f x + e\right )^{3} - 20 \, a^{2} \cos \left (f x + e\right )^{2} + 8 \, a^{2} \cos \left (f x + e\right ) + 16 \, a^{2} - {\left (a^{2} \cos \left (f x + e\right )^{4} - 4 \, a^{2} \cos \left (f x + e\right )^{3} - 12 \, a^{2} \cos \left (f x + e\right )^{2} + 8 \, a^{2} \cos \left (f x + e\right ) + 16 \, 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 (3 \, a^{2} \cos \left (f x + e\right )^{4} + 13 \, a^{2} \cos \left (f x + e\right )^{3} + 86 \, a^{2} \cos \left (f x + e\right )^{2} - 52 \, a^{2} \cos \left (f x + e\right ) - 128 \, a^{2} - {\left (3 \, a^{2} \cos \left (f x + e\right )^{3} - 10 \, a^{2} \cos \left (f x + e\right )^{2} + 76 \, a^{2} \cos \left (f x + e\right ) + 128 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{1024 \, {\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 )}} \]

input
integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(9/2),x, algorithm="fricas")
 
output
1/1024*(3*sqrt(2)*(a^2*cos(f*x + e)^5 + 5*a^2*cos(f*x + e)^4 - 8*a^2*cos(f 
*x + e)^3 - 20*a^2*cos(f*x + e)^2 + 8*a^2*cos(f*x + e) + 16*a^2 - (a^2*cos 
(f*x + e)^4 - 4*a^2*cos(f*x + e)^3 - 12*a^2*cos(f*x + e)^2 + 8*a^2*cos(f*x 
 + e) + 16*a^2)*sin(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*s 
qrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*c 
os(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*(3*a^2*cos(f*x + 
 e)^4 + 13*a^2*cos(f*x + e)^3 + 86*a^2*cos(f*x + e)^2 - 52*a^2*cos(f*x + e 
) - 128*a^2 - (3*a^2*cos(f*x + e)^3 - 10*a^2*cos(f*x + e)^2 + 76*a^2*cos(f 
*x + e) + 128*a^2)*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))
 
3.4.6.6 Sympy [F(-1)]

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

input
integrate((a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**(9/2),x)
 
output
Timed out
 
3.4.6.7 Maxima [F]

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

input
integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(9/2),x, algorithm="maxima")
 
output
integrate((a*sin(f*x + e) + a)^2/(-c*sin(f*x + e) + c)^(9/2), x)
 
3.4.6.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 327 vs. \(2 (163) = 326\).

Time = 0.46 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.87 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {\frac {12 \, \sqrt {2} a^{2} \log \left (\frac {{\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 )}{c^{\frac {9}{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 )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \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 )}^{4}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{4}}\right )} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{4}}{c^{5} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{4} \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 {11}{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}} - \frac {\sqrt {2} a^{2} c^{\frac {11}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{4} \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 )}^{4}}}{c^{10}}}{8192 \, f} \]

input
integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(9/2),x, algorithm="giac")
 
output
1/8192*(12*sqrt(2)*a^2*log((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^(9/2)*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)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + 18*a^2*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))) - (8*sqrt(2)*a^2*c^(11/2)*(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 - sqrt(2)*a^2*c^(11/2)*(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^10)/f
 
3.4.6.9 Mupad [F(-1)]

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

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