\(\int \frac {(c-c \sin (e+f x))^{7/2}}{3+3 \sin (e+f x)} \, dx\) [317]

Optimal result

Integrand size = 28, antiderivative size = 120 \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{3+3 \sin (e+f x)} \, dx=-\frac {256 c^3 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{15 f}+\frac {64 c^2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{15 f}+\frac {8 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{15 f}+\frac {2 \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{15 f} \]

[Out]

64/5*c^2*sec(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a/f+8/5*c*sec(f*x+e)*(c-c*sin(f*x+e))^(5/2)/a/f+2/5*sec(f*x+e)*(c-c
*sin(f*x+e))^(7/2)/a/f-256/5*c^3*sec(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/f

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2815, 2753, 2752} \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{3+3 \sin (e+f x)} \, dx=-\frac {256 c^3 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{5 a f}+\frac {64 c^2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}+\frac {2 \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 a f}+\frac {8 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{5 a f} \]

[In]

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

[Out]

(-256*c^3*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(5*a*f) + (64*c^2*Sec[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(5
*a*f) + (8*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(5*a*f) + (2*Sec[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(5
*a*f)

Rule 2752

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

Rule 2753

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

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}& = \frac {\int \sec ^2(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{a c} \\ & = \frac {2 \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 a f}+\frac {12 \int \sec ^2(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{5 a} \\ & = \frac {8 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{5 a f}+\frac {2 \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 a f}+\frac {(32 c) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{5 a} \\ & = \frac {64 c^2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}+\frac {8 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{5 a f}+\frac {2 \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 a f}+\frac {\left (128 c^2\right ) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{5 a} \\ & = -\frac {256 c^3 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{5 a f}+\frac {64 c^2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{5 a f}+\frac {8 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{5 a f}+\frac {2 \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{5 a f} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.46 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.91 \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{3+3 \sin (e+f x)} \, dx=\frac {c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} (-350-14 \cos (2 (e+f x))-175 \sin (e+f x)+\sin (3 (e+f x)))}{30 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))} \]

[In]

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

[Out]

(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(-350 - 14*Cos[2*(e + f*x)] - 175*Sin[e +
f*x] + Sin[3*(e + f*x)]))/(30*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x]))

Maple [A] (verified)

Time = 1.71 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.58

method	result	size
default	\(\frac {2 c^{4} \left (\sin \left (f x +e \right )-1\right ) \left (\sin ^{3}\left (f x +e \right )-7 \left (\sin ^{2}\left (f x +e \right )\right )+43 \sin \left (f x +e \right )+91\right )}{5 a \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(69\)

[In]

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

[Out]

2/5*c^4/a*(sin(f*x+e)-1)*(sin(f*x+e)^3-7*sin(f*x+e)^2+43*sin(f*x+e)+91)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.62 \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{3+3 \sin (e+f x)} \, dx=-\frac {2 \, {\left (7 \, c^{3} \cos \left (f x + e\right )^{2} + 84 \, c^{3} - {\left (c^{3} \cos \left (f x + e\right )^{2} - 44 \, c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{5 \, a f \cos \left (f x + e\right )} \]

[In]

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

[Out]

-2/5*(7*c^3*cos(f*x + e)^2 + 84*c^3 - (c^3*cos(f*x + e)^2 - 44*c^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(a
*f*cos(f*x + e))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (116) = 232\).

Time = 0.34 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.98 \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{3+3 \sin (e+f x)} \, dx=\frac {2 \, {\left (91 \, c^{\frac {7}{2}} + \frac {86 \, c^{\frac {7}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {336 \, c^{\frac {7}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {266 \, c^{\frac {7}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {490 \, c^{\frac {7}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {266 \, c^{\frac {7}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {336 \, c^{\frac {7}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {86 \, c^{\frac {7}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac {91 \, c^{\frac {7}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\right )}}{5 \, {\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} f {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {7}{2}}} \]

[In]

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

[Out]

2/5*(91*c^(7/2) + 86*c^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 336*c^(7/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 266*c^(7/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 490*c^(7/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 266*c^(
7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 336*c^(7/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 86*c^(7/2)*sin(f*
x + e)^7/(cos(f*x + e) + 1)^7 + 91*c^(7/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)/((a + a*sin(f*x + e)/(cos(f*x
+ e) + 1))*f*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(7/2))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (116) = 232\).

Time = 0.36 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.71 \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{3+3 \sin (e+f x)} \, dx=\frac {16 \, \sqrt {2} {\left (\frac {5 \, c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{a {\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} + 1\right )}} - \frac {11 \, c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \frac {50 \, c^{3} {\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 {80 \, 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 )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} - \frac {30 \, c^{3} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3} \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 )}^{3}} + \frac {5 \, c^{3} {\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}}}{a {\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} - 1\right )}^{5}}\right )} \sqrt {c}}{5 \, f} \]

[In]

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

[Out]

16/5*sqrt(2)*(5*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(a*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi
+ 1/2*f*x + 1/2*e) + 1) + 1)) - (11*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 50*c^3*(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) + 80*c^3*(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 - 30*c^3*(co
s(-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
+ 5*c^3*(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)/(a*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 1)^5))*sqrt(c)/f

Mupad [F(-1)]

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

[In]

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

[Out]

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