\(\int \frac {(c-c \sin (e+f x))^{7/2}}{(3+3 \sin (e+f x))^{5/2}} \, dx\) [397]

Optimal result

Integrand size = 30, antiderivative size = 188 \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(3+3 \sin (e+f x))^{5/2}} \, dx=\frac {2 c^4 \cos (e+f x) \log (1+\sin (e+f x))}{3 f \sqrt {3+3 \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 f \sqrt {3+3 \sin (e+f x)}}+\frac {c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f (3+3 \sin (e+f x))^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (3+3 \sin (e+f x))^{5/2}} \]

[Out]

3/2*c^2*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a/f/(a+a*sin(f*x+e))^(3/2)-1/2*c*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/f
/(a+a*sin(f*x+e))^(5/2)+6*c^4*cos(f*x+e)*ln(1+sin(f*x+e))/a^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)+
3*c^3*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a^2/f/(a+a*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2818, 2819, 2816, 2746, 31} \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(3+3 \sin (e+f x))^{5/2}} \, dx=\frac {6 c^4 \cos (e+f x) \log (\sin (e+f x)+1)}{a^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {3 c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)+a}}+\frac {3 c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f (a \sin (e+f x)+a)^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a \sin (e+f x)+a)^{5/2}} \]

[In]

Int[(c - c*Sin[e + f*x])^(7/2)/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

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

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])

Rule 2819

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] + Dist[a*((2*m - 1)/(
m + n)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[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])

Rubi steps \begin{align*} \text {integral}& = -\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a+a \sin (e+f x))^{5/2}}-\frac {(3 c) \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx}{2 a} \\ & = \frac {3 c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f (a+a \sin (e+f x))^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (3 c^2\right ) \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2} \\ & = \frac {3 c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {3 c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f (a+a \sin (e+f x))^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (6 c^3\right ) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2} \\ & = \frac {3 c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {3 c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f (a+a \sin (e+f x))^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (6 c^4 \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{a+a \sin (e+f x)} \, dx}{a \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {3 c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {3 c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f (a+a \sin (e+f x))^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (6 c^4 \cos (e+f x)\right ) \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,a \sin (e+f x)\right )}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {6 c^4 \cos (e+f x) \log (1+\sin (e+f x))}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {3 c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}+\frac {3 c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f (a+a \sin (e+f x))^{3/2}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{2 f (a+a \sin (e+f x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 7.66 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.01 \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(3+3 \sin (e+f x))^{5/2}} \, dx=\frac {c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} \left (28+\cos (2 (e+f x)) \left (4-24 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+72 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\left (41+96 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)+\sin (3 (e+f x))\right )}{36 \sqrt {3} f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^{5/2}} \]

[In]

Integrate[(c - c*Sin[e + f*x])^(7/2)/(3 + 3*Sin[e + f*x])^(5/2),x]

[Out]

(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(28 + Cos[2*(e + f*x)]*(4 - 24*Log[Cos[(e
+ f*x)/2] + Sin[(e + f*x)/2]]) + 72*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (41 + 96*Log[Cos[(e + f*x)/2] +
Sin[(e + f*x)/2]])*Sin[e + f*x] + Sin[3*(e + f*x)]))/(36*Sqrt[3]*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 +
Sin[e + f*x])^(5/2))

Maple [A] (verified)

Time = 2.99 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.15

method	result	size
default	\(\frac {\sec \left (f x +e \right ) \left (\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+6 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-12 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+10 \left (\cos ^{2}\left (f x +e \right )\right )-12 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \sin \left (f x +e \right )+24 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \sin \left (f x +e \right )-6 \sin \left (f x +e \right )-12 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+24 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-10\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{3}}{f \left (\sin \left (f x +e \right )+1\right ) \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, a^{2}}\)	\(217\)

[In]

int((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/f*sec(f*x+e)*(sin(f*x+e)*cos(f*x+e)^2+6*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))-12*cos(f*x+e)^2*ln(-cot(f*x+e)+csc
(f*x+e)+1)+10*cos(f*x+e)^2-12*ln(2/(cos(f*x+e)+1))*sin(f*x+e)+24*ln(-cot(f*x+e)+csc(f*x+e)+1)*sin(f*x+e)-6*sin
(f*x+e)-12*ln(2/(cos(f*x+e)+1))+24*ln(-cot(f*x+e)+csc(f*x+e)+1)-10)*(-c*(sin(f*x+e)-1))^(1/2)*c^3/(sin(f*x+e)+
1)/(a*(sin(f*x+e)+1))^(1/2)/a^2

Fricas [F]

\[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(3+3 \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral((3*c^3*cos(f*x + e)^2 - 4*c^3 - (c^3*cos(f*x + e)^2 - 4*c^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*s
qrt(-c*sin(f*x + e) + c)/(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)), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(3+3 \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((c-c*sin(f*x+e))**(7/2)/(a+a*sin(f*x+e))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(3+3 \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((-c*sin(f*x + e) + c)^(7/2)/(a*sin(f*x + e) + a)^(5/2), x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.82 \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(3+3 \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {a} c^{\frac {7}{2}} {\left (\frac {2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {6 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {6 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 5}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \]

[In]

integrate((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(3+3 \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

[In]

int((c - c*sin(e + f*x))^(7/2)/(a + a*sin(e + f*x))^(5/2),x)

[Out]

int((c - c*sin(e + f*x))^(7/2)/(a + a*sin(e + f*x))^(5/2), x)