\(\int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {3+3 \sin (e+f x)}} \, dx\) [385]

Optimal result

Integrand size = 30, antiderivative size = 49 \[ \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\frac {c \cos (e+f x) \log (1+\sin (e+f x))}{f \sqrt {3+3 \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]

[Out]

c*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)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2816, 2746, 31} \[ \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\frac {c \cos (e+f x) \log (\sin (e+f x)+1)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}} \]

[In]

Int[Sqrt[c - c*Sin[e + f*x]]/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

(c*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]])

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]

Rubi steps \begin{align*} \text {integral}& = \frac {(a c \cos (e+f x)) \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 {(c \cos (e+f x)) \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 {c \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)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.19 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.43 \[ \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {3+3 \sin (e+f x)}} \, dx=-\frac {\sqrt {\frac {2}{3}} \left (i+e^{i (e+f x)}\right ) \left (f x+2 i \log \left (i+e^{i (e+f x)}\right )\right ) \sqrt {c-c \sin (e+f x)}}{\left (-i+e^{i (e+f x)}\right ) \sqrt {-i e^{-i (e+f x)} \left (i+e^{i (e+f x)}\right )^2} f} \]

[In]

Integrate[Sqrt[c - c*Sin[e + f*x]]/Sqrt[3 + 3*Sin[e + f*x]],x]

[Out]

-((Sqrt[2/3]*(I + E^(I*(e + f*x)))*(f*x + (2*I)*Log[I + E^(I*(e + f*x))])*Sqrt[c - c*Sin[e + f*x]])/((-I + E^(
I*(e + f*x)))*Sqrt[((-I)*(I + E^(I*(e + f*x)))^2)/E^(I*(e + f*x))]*f))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(94\) vs. \(2(45)=90\).

Time = 2.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.94

method	result	size
default	\(\frac {\sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \left (\ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )\right ) \left (1+\cos \left (f x +e \right )+\sin \left (f x +e \right )\right )}{f \left (-\cos \left (f x +e \right )+\sin \left (f x +e \right )-1\right ) \sqrt {a \left (\sin \left (f x +e \right )+1\right )}}\)	\(95\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.31 \[ \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {3+3 \sin (e+f x)}} \, dx=-\frac {\frac {2 \, \sqrt {c} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{\sqrt {a}} - \frac {\sqrt {c} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt {a}}}{f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (45) = 90\).

Time = 0.37 (sec) , antiderivative size = 219, normalized size of antiderivative = 4.47 \[ \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {3+3 \sin (e+f x)}} \, dx=\frac {\sqrt {2} \sqrt {c} {\left (\frac {\sqrt {2} \log \left ({\left | -\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1} - \frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 2 \right |}\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {\sqrt {2} \log \left ({\left | -\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1} - \frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - 2 \right |}\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )}}{2 \, f} \]

[In]

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

[Out]

1/2*sqrt(2)*sqrt(c)*(sqrt(2)*log(abs(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1
) - (cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(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))/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - sqrt(2)*log(abs(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1
)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1) - (cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(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))/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))))/f

Mupad [F(-1)]

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

[In]

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

[Out]

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