\(\int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {3+3 \sin (e+f x)}} \, dx\) [383]

Optimal result

Integrand size = 30, antiderivative size = 139 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\frac {4 c^3 \cos (e+f x) \log (1+\sin (e+f x))}{f \sqrt {3+3 \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {2 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {3+3 \sin (e+f x)}}+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {3+3 \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2819, 2816, 2746, 31} \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\frac {4 c^3 \cos (e+f x) \log (\sin (e+f x)+1)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {2 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}} \]

[In]

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

[Out]

(4*c^3*Cos[e + f*x]*Log[1 + Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) + (2*c^2*Cos[
e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]]) + (c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/
(2*f*Sqrt[a + a*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 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))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}+(2 c) \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx \\ & = \frac {2 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}+\left (4 c^2\right ) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx \\ & = \frac {2 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}+\frac {\left (4 a c^3 \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{a+a \sin (e+f x)} \, dx}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {2 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}}+\frac {\left (4 c^3 \cos (e+f x)\right ) \text {Subst}\left (\int \frac {1}{a+x} \, dx,x,a \sin (e+f x)\right )}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = \frac {4 c^3 \cos (e+f x) \log (1+\sin (e+f x))}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {2 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a+a \sin (e+f x)}}+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a+a \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 12.22 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {3+3 \sin (e+f x)}} \, dx=-\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^2 \left (\cos (2 (e+f x))-32 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+12 \sin (e+f x)\right ) \sqrt {c-c \sin (e+f x)}}{4 \sqrt {3} f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \sqrt {1+\sin (e+f x)}} \]

[In]

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

[Out]

-1/4*(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^2*(Cos[2*(e + f*x)] - 32*Log[Cos[(e + f*x)
/2] + Sin[(e + f*x)/2]] + 12*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(Sqrt[3]*f*(Cos[(e + f*x)/2] - Sin[(e + f
*x)/2])^5*Sqrt[1 + Sin[e + f*x]])

Maple [A] (verified)

Time = 3.14 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.75

method	result	size
default	\(-\frac {\left (\cos ^{3}\left (f x +e \right )+\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-5 \left (\cos ^{2}\left (f x +e \right )\right )+6 \sin \left (f x +e \right ) \cos \left (f x +e \right )+8 \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-16 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \cos \left (f x +e \right )+8 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \sin \left (f x +e \right )-16 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \sin \left (f x +e \right )-\cos \left (f x +e \right )+5 \sin \left (f x +e \right )+8 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-16 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )+5\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{2}}{2 f \left (1+\cos \left (f x +e \right )-\sin \left (f x +e \right )\right ) \sqrt {a \left (\sin \left (f x +e \right )+1\right )}}\)	\(243\)

[In]

int((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2/f*(cos(f*x+e)^3+sin(f*x+e)*cos(f*x+e)^2-5*cos(f*x+e)^2+6*sin(f*x+e)*cos(f*x+e)+8*cos(f*x+e)*ln(2/(cos(f*x
+e)+1))-16*ln(-cot(f*x+e)+csc(f*x+e)+1)*cos(f*x+e)+8*ln(2/(cos(f*x+e)+1))*sin(f*x+e)-16*ln(-cot(f*x+e)+csc(f*x
+e)+1)*sin(f*x+e)-cos(f*x+e)+5*sin(f*x+e)+8*ln(2/(cos(f*x+e)+1))-16*ln(-cot(f*x+e)+csc(f*x+e)+1)+5)*(-c*(sin(f
*x+e)-1))^(1/2)*c^2/(1+cos(f*x+e)-sin(f*x+e))/(a*(sin(f*x+e)+1))^(1/2)

Fricas [F]

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

[In]

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

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

[In]

integrate((c-c*sin(f*x+e))**(5/2)/(a+a*sin(f*x+e))**(1/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.92 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {3+3 \sin (e+f x)}} \, dx=-\frac {2 \, \sqrt {a} c^{\frac {5}{2}} {\left (\frac {a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{2}} + \frac {2 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \]

[In]

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

-2*sqrt(a)*c^(5/2)*((a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^4 + 2*a*sgn(cos(-1/4
*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2)/a^2 + 2*log(-sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(
a*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. \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\int \frac {{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}}{\sqrt {a+a\,\sin \left (e+f\,x\right )}} \,d x \]

[In]

int((c - c*sin(e + f*x))^(5/2)/(a + a*sin(e + f*x))^(1/2),x)

[Out]

int((c - c*sin(e + f*x))^(5/2)/(a + a*sin(e + f*x))^(1/2), x)