\(\int \frac {(3+3 \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{7/2}} \, dx\) [377]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F]
   Sympy [F(-1)]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 30, antiderivative size = 182 \[ \int \frac {(3+3 \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {\cos (e+f x) (3+3 \sin (e+f x))^{5/2}}{f (c-c \sin (e+f x))^{7/2}}-\frac {9 \cos (e+f x) (3+3 \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac {27 \cos (e+f x) \sqrt {3+3 \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {81 \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt {3+3 \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2818, 2816, 2746, 31} \[ \int \frac {(3+3 \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {a^4 \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {a^3 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c^2 f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}} \]

[In]

Int[(a + a*Sin[e + f*x])^(7/2)/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(3*f*(c - c*Sin[e + f*x])^(7/2)) - (a^2*Cos[e + f*x]*(a + a*Sin[e
+ f*x])^(3/2))/(2*c*f*(c - c*Sin[e + f*x])^(5/2)) + (a^3*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(c^2*f*(c - c*
Sin[e + f*x])^(3/2)) + (a^4*Cos[e + f*x]*Log[1 - Sin[e + f*x]])/(c^3*f*Sqrt[a + a*Sin[e + f*x]]*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])

Rubi steps \begin{align*} \text {integral}& = \frac {a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}}-\frac {a \int \frac {(a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx}{c} \\ & = \frac {a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}}-\frac {a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 \int \frac {(a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{3/2}} \, dx}{c^2} \\ & = \frac {a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}}-\frac {a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}-\frac {a^3 \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx}{c^3} \\ & = \frac {a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}}-\frac {a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}-\frac {\left (a^4 \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{c^2 \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}}-\frac {a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {\left (a^4 \cos (e+f x)\right ) \text {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{c^3 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 f (c-c \sin (e+f x))^{7/2}}-\frac {a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f (c-c \sin (e+f x))^{5/2}}+\frac {a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c^2 f (c-c \sin (e+f x))^{3/2}}+\frac {a^4 \cos (e+f x) \log (1-\sin (e+f x))}{c^3 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 8.42 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.27 \[ \int \frac {(3+3 \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {9 \sqrt {3} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^{7/2} \left (-34-30 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+18 \cos (2 (e+f x)) \left (1+\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+9 \left (4+5 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)-3 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sin (3 (e+f x))\right )}{2 c^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 (-1+\sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[(3 + 3*Sin[e + f*x])^(7/2)/(c - c*Sin[e + f*x])^(7/2),x]

[Out]

(9*Sqrt[3]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^(7/2)*(-34 - 30*Log[Cos[(e + f*x)/2] - Sin
[(e + f*x)/2]] + 18*Cos[2*(e + f*x)]*(1 + Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]) + 9*(4 + 5*Log[Cos[(e + f*
x)/2] - Sin[(e + f*x)/2]])*Sin[e + f*x] - 3*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]]*Sin[3*(e + f*x)]))/(2*c^3
*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7*(-1 + Sin[e + f*x])^3*Sqrt[c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 2.95 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.60

method result size
default \(-\frac {\sec \left (f x +e \right ) \left (6 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )-3 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-8 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-18 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )+9 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-24 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right ) \sin \left (f x +e \right )+12 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \sin \left (f x +e \right )+6 \left (\cos ^{2}\left (f x +e \right )\right )+14 \sin \left (f x +e \right )+24 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )-12 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-6\right ) \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, a^{3}}{3 f \left (2 \sin \left (f x +e \right )+\cos ^{2}\left (f x +e \right )-2\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{3}}\) \(291\)

[In]

int((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/f*sec(f*x+e)*(6*cos(f*x+e)^2*sin(f*x+e)*ln(-cot(f*x+e)+csc(f*x+e)-1)-3*cos(f*x+e)^2*sin(f*x+e)*ln(2/(cos(
f*x+e)+1))-8*sin(f*x+e)*cos(f*x+e)^2-18*cos(f*x+e)^2*ln(-cot(f*x+e)+csc(f*x+e)-1)+9*cos(f*x+e)^2*ln(2/(cos(f*x
+e)+1))-24*ln(-cot(f*x+e)+csc(f*x+e)-1)*sin(f*x+e)+12*ln(2/(cos(f*x+e)+1))*sin(f*x+e)+6*cos(f*x+e)^2+14*sin(f*
x+e)+24*ln(-cot(f*x+e)+csc(f*x+e)-1)-12*ln(2/(cos(f*x+e)+1))-6)*(a*(sin(f*x+e)+1))^(1/2)*a^3/(2*sin(f*x+e)+cos
(f*x+e)^2-2)/(-c*(sin(f*x+e)-1))^(1/2)/c^3

Fricas [F]

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

[In]

integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

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^4*cos(f*x + e)^4 - 8*c^4*cos(f*x + e)^2 + 8*c^4 + 4*(c^4*cos(f*x + e)^2 - 2*c^4)*
sin(f*x + e)), x)

Sympy [F(-1)]

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

[In]

integrate((a+a*sin(f*x+e))**(7/2)/(c-c*sin(f*x+e))**(7/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.85 \[ \int \frac {(3+3 \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {\frac {6 \, a^{\frac {7}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac {7}{2}}} - \frac {3 \, a^{\frac {7}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac {7}{2}}} + \frac {4 \, {\left (\frac {3 \, a^{\frac {7}{2}} \sqrt {c} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {6 \, a^{\frac {7}{2}} \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {22 \, a^{\frac {7}{2}} \sqrt {c} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {6 \, a^{\frac {7}{2}} \sqrt {c} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {3 \, a^{\frac {7}{2}} \sqrt {c} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{c^{4} - \frac {6 \, c^{4} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {15 \, c^{4} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {20 \, c^{4} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, c^{4} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {6 \, c^{4} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {c^{4} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}}}{3 \, f} \]

[In]

integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

-1/3*(6*a^(7/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^(7/2) - 3*a^(7/2)*log(sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 + 1)/c^(7/2) + 4*(3*a^(7/2)*sqrt(c)*sin(f*x + e)/(cos(f*x + e) + 1) - 6*a^(7/2)*sqrt(c)*sin(f*x + e)^2/
(cos(f*x + e) + 1)^2 + 22*a^(7/2)*sqrt(c)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 6*a^(7/2)*sqrt(c)*sin(f*x + e)
^4/(cos(f*x + e) + 1)^4 + 3*a^(7/2)*sqrt(c)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(c^4 - 6*c^4*sin(f*x + e)/(co
s(f*x + e) + 1) + 15*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 20*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15
*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 6*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + c^4*sin(f*x + e)^6/(cos
(f*x + e) + 1)^6))/f

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.80 \[ \int \frac {(3+3 \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {\sqrt {2} a^{\frac {7}{2}} \sqrt {c} {\left (\frac {6 \, \sqrt {2} \log \left (-2 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2\right )}{c^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} {\left (18 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 27 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 11\right )}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{3} c^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{12 \, f} \]

[In]

integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

-1/12*sqrt(2)*a^(7/2)*sqrt(c)*(6*sqrt(2)*log(-2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 2)/(c^4*sgn(sin(-1/4*pi + 1
/2*f*x + 1/2*e))) - sqrt(2)*(18*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4 - 27*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 11)/(
(cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^3*c^4*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

Mupad [F(-1)]

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

[In]

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

[Out]

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