3.6.26 \(\int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx\) [526]

Optimal. Leaf size=61 \[ -\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {d} \sqrt {c+d} f} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2852, 214} \begin {gather*} -\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {d} f \sqrt {c+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 214

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

Rule 2852

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

Rubi steps

\begin {align*} \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx &=-\frac {(2 a) \text {Subst}\left (\int \frac {1}{a c+a d-d x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{f}\\ &=-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {d} \sqrt {c+d} f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 3.84, size = 657, normalized size = 10.77 \begin {gather*} \frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (\text {RootSum}\left [-d+2 i c e^{i e} \text {$\#$1}^2+d e^{2 i e} \text {$\#$1}^4\&,\frac {(1+i) d \sqrt {e^{-i e}} f x-(2-2 i) d \sqrt {e^{-i e}} \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right )-i \sqrt {d} \sqrt {c+d} f x \text {$\#$1}+2 \sqrt {d} \sqrt {c+d} \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right ) \text {$\#$1}+\frac {(1-i) c f x \text {$\#$1}^2}{\sqrt {e^{-i e}}}+\frac {(2+2 i) c \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right ) \text {$\#$1}^2}{\sqrt {e^{-i e}}}-\sqrt {d} \sqrt {c+d} e^{i e} f x \text {$\#$1}^3-2 i \sqrt {d} \sqrt {c+d} e^{i e} \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right ) \text {$\#$1}^3}{d-i c e^{i e} \text {$\#$1}^2}\&\right ]-i \text {RootSum}\left [-d+2 i c e^{i e} \text {$\#$1}^2+d e^{2 i e} \text {$\#$1}^4\&,\frac {(1-i) d \sqrt {e^{-i e}} f x+(2+2 i) d \sqrt {e^{-i e}} \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right )+\sqrt {d} \sqrt {c+d} f x \text {$\#$1}+2 i \sqrt {d} \sqrt {c+d} \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right ) \text {$\#$1}-\frac {(1+i) c f x \text {$\#$1}^2}{\sqrt {e^{-i e}}}+\frac {(2-2 i) c \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right ) \text {$\#$1}^2}{\sqrt {e^{-i e}}}-i \sqrt {d} \sqrt {c+d} e^{i e} f x \text {$\#$1}^3+2 \sqrt {d} \sqrt {c+d} e^{i e} \log \left (e^{\frac {i f x}{2}}-\text {$\#$1}\right ) \text {$\#$1}^3}{d-i c e^{i e} \text {$\#$1}^2}\&\right ]\right ) \left (\cos \left (\frac {e}{2}\right )+i \sin \left (\frac {e}{2}\right )\right ) \sqrt {a (1+\sin (e+f x))}}{\sqrt {d} \sqrt {c+d} f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1/8 + I/8)*(RootSum[-d + (2*I)*c*E^(I*e)*#1^2 + d*E^((2*I)*e)*#1^4 & , ((1 + I)*d*Sqrt[E^((-I)*e)]*f*x - (2
- 2*I)*d*Sqrt[E^((-I)*e)]*Log[E^((I/2)*f*x) - #1] - I*Sqrt[d]*Sqrt[c + d]*f*x*#1 + 2*Sqrt[d]*Sqrt[c + d]*Log[E
^((I/2)*f*x) - #1]*#1 + ((1 - I)*c*f*x*#1^2)/Sqrt[E^((-I)*e)] + ((2 + 2*I)*c*Log[E^((I/2)*f*x) - #1]*#1^2)/Sqr
t[E^((-I)*e)] - Sqrt[d]*Sqrt[c + d]*E^(I*e)*f*x*#1^3 - (2*I)*Sqrt[d]*Sqrt[c + d]*E^(I*e)*Log[E^((I/2)*f*x) - #
1]*#1^3)/(d - I*c*E^(I*e)*#1^2) & ] - I*RootSum[-d + (2*I)*c*E^(I*e)*#1^2 + d*E^((2*I)*e)*#1^4 & , ((1 - I)*d*
Sqrt[E^((-I)*e)]*f*x + (2 + 2*I)*d*Sqrt[E^((-I)*e)]*Log[E^((I/2)*f*x) - #1] + Sqrt[d]*Sqrt[c + d]*f*x*#1 + (2*
I)*Sqrt[d]*Sqrt[c + d]*Log[E^((I/2)*f*x) - #1]*#1 - ((1 + I)*c*f*x*#1^2)/Sqrt[E^((-I)*e)] + ((2 - 2*I)*c*Log[E
^((I/2)*f*x) - #1]*#1^2)/Sqrt[E^((-I)*e)] - I*Sqrt[d]*Sqrt[c + d]*E^(I*e)*f*x*#1^3 + 2*Sqrt[d]*Sqrt[c + d]*E^(
I*e)*Log[E^((I/2)*f*x) - #1]*#1^3)/(d - I*c*E^(I*e)*#1^2) & ])*(Cos[e/2] + I*Sin[e/2])*Sqrt[a*(1 + Sin[e + f*x
])])/(Sqrt[d]*Sqrt[c + d]*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

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Maple [A]
time = 2.80, size = 80, normalized size = 1.31

method	result	size
default	\(-\frac {2 a \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, d}{\sqrt {a \left (c +d \right ) d}}\right )}{\sqrt {a \left (c +d \right ) d}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.43, size = 484, normalized size = 7.93 \begin {gather*} \left [\frac {\sqrt {\frac {a}{c d + d^{2}}} \log \left (\frac {a d^{2} \cos \left (f x + e\right )^{3} - a c^{2} - 2 \, a c d - a d^{2} - {\left (6 \, a c d + 7 \, a d^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left (c^{2} d + 4 \, c d^{2} + 3 \, d^{3} - {\left (c d^{2} + d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (c^{2} d + 3 \, c d^{2} + 2 \, d^{3}\right )} \cos \left (f x + e\right ) - {\left (c^{2} d + 4 \, c d^{2} + 3 \, d^{3} + {\left (c d^{2} + d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {\frac {a}{c d + d^{2}}} - {\left (a c^{2} + 8 \, a c d + 9 \, a d^{2}\right )} \cos \left (f x + e\right ) + {\left (a d^{2} \cos \left (f x + e\right )^{2} - a c^{2} - 2 \, a c d - a d^{2} + 2 \, {\left (3 \, a c d + 4 \, a d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{d^{2} \cos \left (f x + e\right )^{3} + {\left (2 \, c d + d^{2}\right )} \cos \left (f x + e\right )^{2} - c^{2} - 2 \, c d - d^{2} - {\left (c^{2} + d^{2}\right )} \cos \left (f x + e\right ) + {\left (d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \cos \left (f x + e\right ) - c^{2} - 2 \, c d - d^{2}\right )} \sin \left (f x + e\right )}\right )}{2 \, f}, -\frac {\sqrt {-\frac {a}{c d + d^{2}}} \arctan \left (\frac {\sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) - c - 2 \, d\right )} \sqrt {-\frac {a}{c d + d^{2}}}}{2 \, a \cos \left (f x + e\right )}\right )}{f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*sqrt(a/(c*d + d^2))*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e
)^2 + 4*(c^2*d + 4*c*d^2 + 3*d^3 - (c*d^2 + d^3)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + e) - (c^
2*d + 4*c*d^2 + 3*d^3 + (c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(a/(c*d + d^2))
- (a*c^2 + 8*a*c*d + 9*a*d^2)*cos(f*x + e) + (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4
*a*d^2)*cos(f*x + e))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (
c^2 + d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e)))/f, -sqr
t(-a/(c*d + d^2))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) - c - 2*d)*sqrt(-a/(c*d + d^2))/(a*cos(f
*x + e)))/f]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )}}{c + d \sin {\left (e + f x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.47, size = 65, normalized size = 1.07 \begin {gather*} -\frac {2 \, \sqrt {a} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\sqrt {-c d - d^{2}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {a+a\,\sin \left (e+f\,x\right )}}{c+d\,\sin \left (e+f\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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