\(\int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx\) [363]

Optimal result

Integrand size = 30, antiderivative size = 138 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {3 \cos (e+f x) (3+3 \sin (e+f x))^{3/2}}{f (c-c \sin (e+f x))^{3/2}}+\frac {108 \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {3+3 \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {18 \cos (e+f x) \sqrt {3+3 \sin (e+f x)}}{c f \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2818, 2819, 2816, 2746, 31} \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {4 a^3 \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {2 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{c f \sqrt {c-c \sin (e+f x)}}+\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{f (c-c \sin (e+f x))^{3/2}} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 6.85 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.22 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {9 \sqrt {3} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^{5/2} \left (7+\cos (2 (e+f x))+16 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\left (2-16 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right )}{2 c f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 (-1+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

(-9*Sqrt[3]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^(5/2)*(7 + Cos[2*(e + f*x)] + 16*Log[Cos[
(e + f*x)/2] - Sin[(e + f*x)/2]] + (2 - 16*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]])*Sin[e + f*x]))/(2*c*f*(Co
s[(e + f*x)/2] + Sin[(e + f*x)/2])^5*(-1 + Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 3.02 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.01

method	result	size
default	\(\frac {\sec \left (f x +e \right ) \left (4 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \sin \left (f x +e \right )-8 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right ) \sin \left (f x +e \right )+\cos ^{2}\left (f x +e \right )-4 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+8 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )+5 \sin \left (f x +e \right )-1\right ) \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, a^{2}}{f \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c}\)	\(140\)

[In]

int((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/f*sec(f*x+e)*(4*ln(2/(cos(f*x+e)+1))*sin(f*x+e)-8*ln(-cot(f*x+e)+csc(f*x+e)-1)*sin(f*x+e)+cos(f*x+e)^2-4*ln(
2/(cos(f*x+e)+1))+8*ln(-cot(f*x+e)+csc(f*x+e)-1)+5*sin(f*x+e)-1)*(a*(sin(f*x+e)+1))^(1/2)*a^2/(-c*(sin(f*x+e)-
1))^(1/2)/c

Fricas [F]

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

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral((a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2)*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)

Sympy [F(-1)]

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

[In]

integrate((a+a*sin(f*x+e))**(5/2)/(c-c*sin(f*x+e))**(3/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)/(-c*sin(f*x + e) + c)^(3/2), x)

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.98 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {2 \, a^{\frac {5}{2}} \sqrt {c} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {2 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {1}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \]

[In]

integrate((a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + a*sin(e + f*x))^(5/2)/(c - c*sin(e + f*x))^(3/2),x)

[Out]

int((a + a*sin(e + f*x))^(5/2)/(c - c*sin(e + f*x))^(3/2), x)