3.4.75 \(\int \frac {(3+3 \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx\) [375]

3.4.75.1 Optimal result
3.4.75.2 Mathematica [A] (verified)
3.4.75.3 Rubi [A] (verified)
3.4.75.4 Maple [A] (verified)
3.4.75.5 Fricas [F]
3.4.75.6 Sympy [F(-1)]
3.4.75.7 Maxima [F]
3.4.75.8 Giac [A] (verification not implemented)
3.4.75.9 Mupad [F(-1)]

3.4.75.1 Optimal result

Integrand size = 30, antiderivative size = 183 \[ \int \frac {(3+3 \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {3 \cos (e+f x) (3+3 \sin (e+f x))^{5/2}}{f (c-c \sin (e+f x))^{3/2}}+\frac {972 \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {3+3 \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {162 \cos (e+f x) \sqrt {3+3 \sin (e+f x)}}{c f \sqrt {c-c \sin (e+f x)}}+\frac {27 \cos (e+f x) (3+3 \sin (e+f x))^{3/2}}{2 c f \sqrt {c-c \sin (e+f x)}} \]

output
a*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(3/2)+3/2*a^2*cos(f 
*x+e)*(a+a*sin(f*x+e))^(3/2)/c/f/(c-c*sin(f*x+e))^(1/2)+12*a^4*cos(f*x+e)* 
ln(1-sin(f*x+e))/c/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)+6*a^3*c 
os(f*x+e)*(a+a*sin(f*x+e))^(1/2)/c/f/(c-c*sin(f*x+e))^(1/2)
 
3.4.75.2 Mathematica [A] (verified)

Time = 7.89 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.98 \[ \int \frac {(3+3 \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {27 \sqrt {3} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^{7/2} \left (44+18 \cos (2 (e+f x))+192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\left (39-192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)+\sin (3 (e+f x))\right )}{8 c f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 (-1+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \]

input
Integrate[(3 + 3*Sin[e + f*x])^(7/2)/(c - c*Sin[e + f*x])^(3/2),x]
 
output
(-27*Sqrt[3]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^(7/2 
)*(44 + 18*Cos[2*(e + f*x)] + 192*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] 
 + (39 - 192*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]])*Sin[e + f*x] + Sin[ 
3*(e + f*x)]))/(8*c*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(-1 + Sin[e 
+ f*x])*Sqrt[c - c*Sin[e + f*x]])
 
3.4.75.3 Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {3042, 3218, 3042, 3219, 3042, 3219, 3042, 3216, 3042, 3146, 16}

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.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{7/2}}{(c-c \sin (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3218

\(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {3 a \int \frac {(\sin (e+f x) a+a)^{5/2}}{\sqrt {c-c \sin (e+f x)}}dx}{c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {3 a \int \frac {(\sin (e+f x) a+a)^{5/2}}{\sqrt {c-c \sin (e+f x)}}dx}{c}\)

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 16

\(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {3 a \left (2 a \left (-\frac {2 a^2 \cos (e+f x) \log (c-c \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\)

input
Int[(a + a*Sin[e + f*x])^(7/2)/(c - c*Sin[e + f*x])^(3/2),x]
 
output
(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(f*(c - c*Sin[e + f*x])^(3/2)) 
 - (3*a*(-1/2*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(f*Sqrt[c - c*Si 
n[e + f*x]]) + 2*a*((-2*a^2*Cos[e + f*x]*Log[c - c*Sin[e + f*x]])/(f*Sqrt[ 
a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) - (a*Cos[e + f*x]*Sqrt[a + a 
*Sin[e + f*x]])/(f*Sqrt[c - c*Sin[e + f*x]]))))/c
 

3.4.75.3.1 Defintions of rubi rules used

rule 16
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

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

rule 3146
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[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] && I 
ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/ 
2])
 

rule 3216
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)]], x_Symbol] :> Simp[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 3218
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] - Simp[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] && GtQ[2*m + 
n + 1, 0])
 

rule 3219
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] + Simp[a*((2*m - 1)/(m + n 
))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; Fre 
eQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I 
GtQ[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])
 
3.4.75.4 Maple [A] (verified)

Time = 2.62 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.86

method result size
default \(\frac {\sec \left (f x +e \right ) \left (\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+9 \left (\cos ^{2}\left (f x +e \right )\right )-48 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right ) \sin \left (f x +e \right )+24 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \sin \left (f x +e \right )+25 \sin \left (f x +e \right )+48 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )-24 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-9\right ) \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, a^{3}}{2 f c \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\) \(158\)

input
int((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
 
output
1/2/f*sec(f*x+e)*(sin(f*x+e)*cos(f*x+e)^2+9*cos(f*x+e)^2-48*ln(-cot(f*x+e) 
+csc(f*x+e)-1)*sin(f*x+e)+24*ln(2/(cos(f*x+e)+1))*sin(f*x+e)+25*sin(f*x+e) 
+48*ln(-cot(f*x+e)+csc(f*x+e)-1)-24*ln(2/(cos(f*x+e)+1))-9)*(a*(sin(f*x+e) 
+1))^(1/2)*a^3/c/(-c*(sin(f*x+e)-1))^(1/2)
 
3.4.75.5 Fricas [F]

\[ \int \frac {(3+3 \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]

input
integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="fric 
as")
 
output
integral((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))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^2*cos(f*x 
+ e)^2 + 2*c^2*sin(f*x + e) - 2*c^2), x)
 
3.4.75.6 Sympy [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]

input
integrate((a+a*sin(f*x+e))**(7/2)/(c-c*sin(f*x+e))**(3/2),x)
 
output
Timed out
 
3.4.75.7 Maxima [F]

\[ \int \frac {(3+3 \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]

input
integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxi 
ma")
 
output
integrate((a*sin(f*x + e) + a)^(7/2)/(-c*sin(f*x + e) + c)^(3/2), x)
 
3.4.75.8 Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.93 \[ \int \frac {(3+3 \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {2 \, a^{\frac {7}{2}} \sqrt {c} {\left (\frac {6 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 4 \, c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{c^{4}} - \frac {2}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \]

input
integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac 
")
 
output
-2*a^(7/2)*sqrt(c)*(6*log(-cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(c^2*sgn( 
sin(-1/4*pi + 1/2*f*x + 1/2*e))) + (c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*s 
gn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 4*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^ 
2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))/c^4 - 2/((cos(-1/4*pi + 1/2*f*x + 1 
/2*e)^2 - 1)*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))*sgn(cos(-1/4*pi + 1 
/2*f*x + 1/2*e))/f
 
3.4.75.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]

input
int((a + a*sin(e + f*x))^(7/2)/(c - c*sin(e + f*x))^(3/2),x)
 
output
int((a + a*sin(e + f*x))^(7/2)/(c - c*sin(e + f*x))^(3/2), x)