\(\int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx\) [135]

Optimal result

Integrand size = 40, antiderivative size = 100 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {a (A+B) \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {a B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{c f \sqrt {a+a \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3050, 2817, 2816, 2746, 31} \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {a B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{c f \sqrt {a \sin (e+f x)+a}}-\frac {a (A+B) \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

-((a*(A + B)*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]])) + (a*B
*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(c*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 2817

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[
-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]

Rule 3050

Int[Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[B/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x
] - Dist[(B*c - A*d)/d, Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f
, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = (A+B) \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx-\frac {B \int \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)} \, dx}{c} \\ & = \frac {a B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{c f \sqrt {a+a \sin (e+f x)}}+\frac {(a (A+B) c \cos (e+f x)) \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 B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{c f \sqrt {a+a \sin (e+f x)}}-\frac {(a (A+B) \cos (e+f x)) \text {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = -\frac {a (A+B) \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 {a B \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{c f \sqrt {a+a \sin (e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.36 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} \left ((A+B) \left (-i f x+2 \log \left (i-e^{i (e+f x)}\right )\right )+B \sin (e+f x)\right )}{f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)}} \]

[In]

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

[Out]

-(((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*((A + B)*((-I)*f*x + 2*Log[I - E^(I*(e + f
*x))]) + B*Sin[e + f*x]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(299\) vs. \(2(92)=184\).

Time = 2.53 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.00

method	result	size
parts	\(\frac {A \left (2 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-\ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \left (-\cos \left (f x +e \right )+\sin \left (f x +e \right )-1\right )}{f \left (\cos \left (f x +e \right )+\sin \left (f x +e \right )+1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}-\frac {B \left (2 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \cos \left (f x +e \right )-2 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \sin \left (f x +e \right )-\ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \cos \left (f x +e \right )+\ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )+\cos ^{2}\left (f x +e \right )+\cos \left (f x +e \right ) \sin \left (f x +e \right )+2 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-\ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+\sin \left (f x +e \right )-1\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{f \left (\cos \left (f x +e \right )+\sin \left (f x +e \right )+1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(300\)
default	\(-\frac {\left (2 A \cos \left (f x +e \right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-2 A \sin \left (f x +e \right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-A \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+A \sin \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+2 B \cos \left (f x +e \right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-2 B \sin \left (f x +e \right ) \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-B \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+B \sin \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+B \left (\cos ^{2}\left (f x +e \right )\right )+B \cos \left (f x +e \right ) \sin \left (f x +e \right )+2 A \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-A \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+2 B \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-B \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )+B \sin \left (f x +e \right )-B \right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{f \left (\cos \left (f x +e \right )+\sin \left (f x +e \right )+1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}\)	\(345\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.75 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {B {\left (\frac {2 \, \sqrt {a} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{\sqrt {c}} - \frac {\sqrt {a} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt {c}} + \frac {2 \, \sqrt {a} \sqrt {c} \sin \left (f x + e\right )}{{\left (c + \frac {c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + A {\left (\frac {2 \, \sqrt {a} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{\sqrt {c}} - \frac {\sqrt {a} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt {c}}\right )}}{f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.41 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\sqrt {2} {\left (\frac {2 \, \sqrt {2} B \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\sqrt {c} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {\sqrt {2} {\left (A \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + B \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \log \left (-2 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2\right )}{c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \sqrt {a}}{2 \, f} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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