\(\int \frac {1}{(3+3 \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx\) [337]

Optimal result

Integrand size = 28, antiderivative size = 148 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{108 \sqrt {2} \sqrt {c} f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{108 c f}-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{162 c^2 f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{135 c^3 f} \]

[Out]

-1/6*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^3/c^2/f-1/5*sec(f*x+e)^5*(c-c*sin(f*x+e))^(5/2)/a^3/c^3/f+1/8*arcta
nh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/a^3/f*2^(1/2)/c^(1/2)-1/4*sec(f*x+e)*(c-c*sin(f*x+e)
)^(1/2)/a^3/c/f

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2815, 2754, 2728, 212} \[ \int \frac {1}{(3+3 \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} a^3 \sqrt {c} f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f} \]

[In]

Int[1/((a + a*Sin[e + f*x])^3*Sqrt[c - c*Sin[e + f*x]]),x]

[Out]

ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])]/(4*Sqrt[2]*a^3*Sqrt[c]*f) - (Sec[e + f*x]*S
qrt[c - c*Sin[e + f*x]])/(4*a^3*c*f) - (Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(6*a^3*c^2*f) - (Sec[e + f*
x]^5*(c - c*Sin[e + f*x])^(5/2))/(5*a^3*c^3*f)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2754

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/(a*f*g*(p + 1))), x] + Dist[a*((m + p + 1)/(g^2*(p + 1))), Int
[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2,
0] && GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ[m + 1/2, 2*p]

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 ^6(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{a^3 c^3} \\ & = -\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac {\int \sec ^4(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{2 a^3 c^2} \\ & = -\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac {\int \sec ^2(e+f x) \sqrt {c-c \sin (e+f x)} \, dx}{4 a^3 c} \\ & = -\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f}-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}+\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{8 a^3} \\ & = -\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f}-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f}-\frac {\text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{4 a^3 f} \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} a^3 \sqrt {c} f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{4 a^3 c f}-\frac {\sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{6 a^3 c^2 f}-\frac {\sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^3 f} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.62 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.26 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-12-10 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-15 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-(15+15 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5\right )}{1620 f (1+\sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \]

[In]

Integrate[1/((3 + 3*Sin[e + f*x])^3*Sqrt[c - c*Sin[e + f*x]]),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-12 - 10*(Cos[(e + f*x)/2] + Sin
[(e + f*x)/2])^2 - 15*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - (15 + 15*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)
^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5))/(1620*f*(1 + Sin[e + f*x])^3*Sqrt[c -
c*Sin[e + f*x]])

Maple [A] (verified)

Time = 0.91 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {\left (\sin \left (f x +e \right )-1\right ) \left (30 \left (\sin ^{2}\left (f x +e \right )\right ) c^{\frac {11}{2}}+80 \sin \left (f x +e \right ) c^{\frac {11}{2}}+74 c^{\frac {11}{2}}-15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} \left (c \left (\sin \left (f x +e \right )+1\right )\right )^{\frac {5}{2}}\right )}{120 a^{3} c^{\frac {11}{2}} \left (\sin \left (f x +e \right )+1\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(122\)

[In]

int(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/120*(sin(f*x+e)-1)*(30*sin(f*x+e)^2*c^(11/2)+80*sin(f*x+e)*c^(11/2)+74*c^(11/2)-15*2^(1/2)*arctanh(1/2*(c*(s
in(f*x+e)+1))^(1/2)*2^(1/2)/c^(1/2))*c^3*(c*(sin(f*x+e)+1))^(5/2))/a^3/c^(11/2)/(sin(f*x+e)+1)^2/cos(f*x+e)/(c
-c*sin(f*x+e))^(1/2)/f

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.63 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=\frac {15 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{3} - 2 \, \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (15 \, \cos \left (f x + e\right )^{2} - 40 \, \sin \left (f x + e\right ) - 52\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{240 \, {\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \left (f x + e\right )\right )}} \]

[In]

integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

1/240*(15*sqrt(2)*(cos(f*x + e)^3 - 2*cos(f*x + e)*sin(f*x + e) - 2*cos(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)
^2 + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos
(f*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) -
4*(15*cos(f*x + e)^2 - 40*sin(f*x + e) - 52)*sqrt(-c*sin(f*x + e) + c))/(a^3*c*f*cos(f*x + e)^3 - 2*a^3*c*f*co
s(f*x + e)*sin(f*x + e) - 2*a^3*c*f*cos(f*x + e))

Sympy [F]

\[ \int \frac {1}{(3+3 \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\int \frac {1}{\sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )} + 3 \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )} + 3 \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx}{a^{3}} \]

[In]

integrate(1/(a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**(1/2),x)

[Out]

Integral(1/(sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**3 + 3*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2 + 3*sqrt(-
c*sin(e + f*x) + c)*sin(e + f*x) + sqrt(-c*sin(e + f*x) + c)), x)/a**3

Maxima [F]

\[ \int \frac {1}{(3+3 \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{3} \sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \]

[In]

integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((a*sin(f*x + e) + a)^3*sqrt(-c*sin(f*x + e) + c)), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 278 vs. \(2 (137) = 274\).

Time = 0.40 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.88 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\frac {15 \, \sqrt {2} \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )}{a^{3} \sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {4 \, \sqrt {2} {\left (23 \, \sqrt {c} + \frac {70 \, \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {140 \, \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \frac {90 \, \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} + \frac {45 \, \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{4}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{4}}\right )}}{a^{3} c {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )}^{5} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{240 \, f} \]

[In]

integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

1/240*(15*sqrt(2)*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1))/(a^3*sqrt(c)
*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 4*sqrt(2)*(23*sqrt(c) + 70*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1
)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 140*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2
*f*x + 1/2*e) + 1)^2 + 90*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^
3 + 45*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4)/(a^3*c*((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*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))
/f

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(3+3 \sin (e+f x))^3 \sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]

[In]

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

[Out]

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