3.65 \(\int \frac{\csc (c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=84 \[ \frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{\sqrt{a} d} \]

[Out]

(-2*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(Sqrt[a]*d) + (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 = 0.113939, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2780, 2649, 206, 2773} \[ \frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{\sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

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

Rule 2780

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[
b/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] - Dist[d/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c +
d*Sin[e + f*x]), 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]

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])

Rule 2773

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{\csc (c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx &=\frac{\int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{a}-\int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=-\frac{2 \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{d}+\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{2 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{\sqrt{a} 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 = 0.0987107, size = 128, normalized size = 1.52 \[ -\frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left ((2+2 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )+\log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )\right )}{d \sqrt{a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]/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])] + Log[1 + Cos[(c + d*x)/2] -
Sin[(c + d*x)/2]] - Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))/(d*Sq
rt[a*(1 + Sin[c + d*x])]))

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Maple [A]  time = 0.506, size = 96, normalized size = 1.1 \begin{align*}{\frac{1+\sin \left ( dx+c \right ) }{d\cos \left ( dx+c \right ) }\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) } \left ( \sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{a}}}} \right ) -2\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( dx+c \right ) }}{\sqrt{a}}} \right ) \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(csc(d*x+c)/(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)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-2*arctan
h((a-a*sin(d*x+c))^(1/2)/a^(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{\csc \left (d x + c\right )}{\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(csc(d*x+c)/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.90382, size = 795, normalized size = 9.46 \begin{align*} \frac{\sqrt{2} \sqrt{a} \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 ) + \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \,{\left (\cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt{a \sin \left (d x + c\right ) + a} \sqrt{a} - 9 \, a \cos \left (d x + c\right ) +{\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right )}{2 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt(2)*sqrt(a)*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)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 + (cos(d*
x + c) + 3)*sin(d*x + c) - 2*cos(d*x + c) - 3)*sqrt(a*sin(d*x + c) + a)*sqrt(a) - 9*a*cos(d*x + c) + (a*cos(d*
x + c)^2 + 8*a*cos(d*x + c) - a)*sin(d*x + c) - a)/(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin
(d*x + c) - cos(d*x + c) - 1)))/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError