### 3.5 $$\int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0248641, antiderivative size = 47, normalized size of antiderivative = 1., 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 = {2649, 206} $-\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{\sqrt{a} d}$

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 2649

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx &=-\frac{2 \operatorname{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*}

Mathematica [C]  time = 0.0686954, size = 73, normalized size = 1.55 $\frac{(2+2 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )}{d \sqrt{a (\sin (c+d x)+1)}}$

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 = 0.072, size = 75, normalized size = 1.6 \begin{align*} -{\frac{ \left ( 1+\sin \left ( dx+c \right ) \right ) \sqrt{2}}{d\cos \left ( dx+c \right ) }\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{\frac{1}{\sqrt{a}}}} \right ){\frac{1}{\sqrt{a}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \sin \left (d x + c\right ) + a}}\,{d x} \end{align*}

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 = 1.6461, size = 463, normalized size = 9.85 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a \sin{\left (c + d x \right )} + a}}\, dx \end{align*}

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

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]

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