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

Optimal. Leaf size=113 \[ \frac {(A-B) \cos (e+f x) \log (\sin (e+f x)+1)}{2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {(A+B) \cos (e+f x) \log (1-\sin (e+f x))}{2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}} \]

[Out]

-1/2*(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)+1/2*(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)

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Rubi [A]  time = 0.36, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2969, 2737, 2667, 31} \[ \frac {(A-B) \cos (e+f x) \log (\sin (e+f x)+1)}{2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {(A+B) \cos (e+f x) \log (1-\sin (e+f x))}{2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((A + B)*Cos[e + f*x]*Log[1 - Sin[e + f*x]])/(2*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) + ((A -
B)*Cos[e + f*x]*Log[1 + Sin[e + f*x]])/(2*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 2667

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 2737

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 2969

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

Rubi steps

\begin {align*} \int \frac {A+B \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \, dx &=\frac {(A+B) \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c-c \sin (e+f x)}} \, dx}{2 a}+\frac {(A-B) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx}{2 c}\\ &=\frac {(a (A-B) \cos (e+f x)) \int \frac {\cos (e+f x)}{a+a \sin (e+f x)} \, dx}{2 \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {((A+B) c \cos (e+f x)) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)} \, dx}{2 \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=\frac {((A-B) \cos (e+f x)) \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,a \sin (e+f x)\right )}{2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {((A+B) \cos (e+f x)) \operatorname {Subst}\left (\int \frac {1}{c+x} \, dx,x,-c \sin (e+f x)\right )}{2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {(A+B) \cos (e+f x) \log (1-\sin (e+f x))}{2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {(A-B) \cos (e+f x) \log (1+\sin (e+f x))}{2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 97, normalized size = 0.86 \[ -\frac {\cos (e+f x) \left ((A+B) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+(B-A) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{f \sqrt {a (\sin (e+f x)+1)} \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sin \left (f x + e\right ) + A\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{a c \cos \left (f x + e\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sin \left (f x + e\right ) + A}{\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.59, size = 165, normalized size = 1.46 \[ -\frac {\left (A \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-A \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+B \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-B \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+B \ln \left (\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )\right ) \cos \left (f x +e \right )}{f \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sin \left (f x + e\right ) + A}{\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,\sin \left (e+f\,x\right )}{\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \sin {\left (e + f x \right )}}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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