\(\int \frac {a+a \sin (e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx\) [294]

Optimal result

Integrand size = 26, antiderivative size = 77 \[ \int \frac {a+a \sin (e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {2 \sqrt {2} a \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2815, 2758, 2728, 212} \[ \int \frac {a+a \sin (e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {2 \sqrt {2} a \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

(2*Sqrt[2]*a*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[c]*f) - (2*a*Cos[e + f*
x])/(f*Sqrt[c - c*Sin[e + f*x]])

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 2758

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + p))), x] + Dist[g^2*((p - 1)/(a*(m + p))), 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]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 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}& = (a c) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx \\ & = -\frac {2 a \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}+(2 a) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx \\ & = -\frac {2 a \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}-\frac {(4 a) \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 )}{f} \\ & = \frac {2 \sqrt {2} a \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.46 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.75 \[ \int \frac {a+a \sin (e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {2 a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sqrt {c} (1+\sin (e+f x))+\sqrt {2} \arctan \left (\frac {\sqrt {-c (1+\sin (e+f x))}}{\sqrt {2} \sqrt {c}}\right ) \sqrt {-c (1+\sin (e+f x))}\right )}{\sqrt {c} f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

(-2*a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Sqrt[c]*(1 + Sin[e + f*x]) + Sqrt[2]*ArcTan[Sqrt[-(c*(1 + Sin[e +
f*x]))]/(Sqrt[2]*Sqrt[c])]*Sqrt[-(c*(1 + Sin[e + f*x]))]))/(Sqrt[c]*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*S
qrt[c - c*Sin[e + f*x]])

Maple [A] (verified)

Time = 3.84 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.21

method	result	size
default	\(\frac {2 \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, a \left (\sqrt {c \left (\sin \left (f x +e \right )+1\right )}-\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{c \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(93\)
parts	\(-\frac {a \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{\sqrt {c}\, \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {a \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \left (\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-2 \sqrt {c \left (\sin \left (f x +e \right )+1\right )}\right )}{c \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(169\)
risch	\(-\frac {\left ({\mathrm e}^{i \left (f x +e \right )}-3 i\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) a \sqrt {2}\, \sqrt {-i c \left (-{\mathrm e}^{2 i \left (f x +e \right )}+1+2 i {\mathrm e}^{i \left (f x +e \right )}\right ) {\mathrm e}^{i \left (f x +e \right )}}\, {\mathrm e}^{-i \left (f x +e \right )}}{f \sqrt {i c \left ({\mathrm e}^{2 i \left (f x +e \right )}-1-2 i {\mathrm e}^{i \left (f x +e \right )}\right ) {\mathrm e}^{i \left (f x +e \right )}}\, \sqrt {c \left (i {\mathrm e}^{2 i \left (f x +e \right )}-i+2 \,{\mathrm e}^{i \left (f x +e \right )}\right ) {\mathrm e}^{-i \left (f x +e \right )}}}+\frac {4 i \left (-{\mathrm e}^{i \left (f x +e \right )}+i\right ) \left (c^{\frac {3}{2}}+\arctan \left (\frac {\sqrt {i {\mathrm e}^{i \left (f x +e \right )} c}}{\sqrt {c}}\right ) c \sqrt {i {\mathrm e}^{i \left (f x +e \right )} c}\right ) a \sqrt {2}\, {\mathrm e}^{-i \left (f x +e \right )}}{f \,c^{\frac {3}{2}} \sqrt {c \left (i {\mathrm e}^{2 i \left (f x +e \right )}-i+2 \,{\mathrm e}^{i \left (f x +e \right )}\right ) {\mathrm e}^{-i \left (f x +e \right )}}}\)	\(293\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (66) = 132\).

Time = 0.27 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.55 \[ \int \frac {a+a \sin (e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\frac {\sqrt {2} {\left (a c \cos \left (f x + e\right ) - a c \sin \left (f x + e\right ) + a c\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} - 2 \, {\left (a \cos \left (f x + e\right ) + a \sin \left (f x + e\right ) + a\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.56 \[ \int \frac {a+a \sin (e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\frac {\sqrt {2} a \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )}{\sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {4 \, \sqrt {2} a}{\sqrt {c} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{f} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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