3.1.64 \(\int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx\) [64]

Optimal. Leaf size=47 \[ -\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{\sqrt {a} d} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2728, 212} \begin {gather*} -\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{\sqrt {a} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

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

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]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.04, size = 73, normalized size = 1.55 \begin {gather*} \frac {(2+2 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d \sqrt {a (1+\sin (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

((2 + 2*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(Cos[(c + d*x)/2] + Sin[(c + d*x
)/2]))/(d*Sqrt[a*(1 + Sin[c + d*x])])

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Maple [A]
time = 0.01, size = 75, normalized size = 1.60

method result size
default \(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}\, \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) \(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(1/sqrt(a*sin(d*x + c) + a), x)

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Fricas [A]
time = 0.35, size = 167, normalized size = 3.55 \begin {gather*} \left [\frac {\sqrt {2} \log \left (-\frac {\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (d x + c\right ) + 2}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right )}{2 \, \sqrt {a} d}, \frac {\sqrt {2} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {-\frac {1}{a}}}{\cos \left (d x + c\right )}\right )}{d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*sqrt(2)*log(-(cos(d*x + c)^2 - (cos(d*x + c) - 2)*sin(d*x + c) - 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*(cos(
d*x + c) - sin(d*x + c) + 1)/sqrt(a) + 3*cos(d*x + c) + 2)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) -
 cos(d*x + c) - 2))/(sqrt(a)*d), sqrt(2)*sqrt(-1/a)*arctan(sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(-1/a)/cos(d*x
 + c))/d]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 111 vs. \(2 (38) = 76\).
time = 0.52, size = 111, normalized size = 2.36 \begin {gather*} \frac {\frac {\sqrt {2} \log \left ({\left | \frac {1}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} \log \left ({\left | \frac {1}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 6.44, size = 49, normalized size = 1.04 \begin {gather*} -\frac {\mathrm {F}\left (\frac {\pi }{4}-\frac {c}{2}-\frac {d\,x}{2}\middle |1\right )\,\sqrt {\frac {2\,\left (a+a\,\sin \left (c+d\,x\right )\right )}{a}}}{d\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a*sin(c + d*x))^(1/2),x)

[Out]

-(ellipticF(pi/4 - c/2 - (d*x)/2, 1)*((2*(a + a*sin(c + d*x)))/a)^(1/2))/(d*(a + a*sin(c + d*x))^(1/2))

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