3.227 \(\int \frac{\csc (a+b x)}{\sqrt{d \cos (a+b x)}} \, dx\)

Optimal. Leaf size=59 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b \sqrt{d}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b \sqrt{d}} \]

[Out]

-(ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]]/(b*Sqrt[d])) - ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]]/(b*Sqrt[d])

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Rubi [A]  time = 0.0510998, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {2565, 329, 212, 206, 203} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b \sqrt{d}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b \sqrt{d}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]/Sqrt[d*Cos[a + b*x]],x]

[Out]

-(ArcTan[Sqrt[d*Cos[a + b*x]]/Sqrt[d]]/(b*Sqrt[d])) - ArcTanh[Sqrt[d*Cos[a + b*x]]/Sqrt[d]]/(b*Sqrt[d])

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 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 203

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

Rubi steps

\begin{align*} \int \frac{\csc (a+b x)}{\sqrt{d \cos (a+b x)}} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1-\frac{x^2}{d^2}\right )} \, dx,x,d \cos (a+b x)\right )}{b d}\\ &=-\frac{2 \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^4}{d^2}} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{b d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{d-x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{b}-\frac{\operatorname{Subst}\left (\int \frac{1}{d+x^2} \, dx,x,\sqrt{d \cos (a+b x)}\right )}{b}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b \sqrt{d}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{d \cos (a+b x)}}{\sqrt{d}}\right )}{b \sqrt{d}}\\ \end{align*}

Mathematica [A]  time = 0.0399, size = 50, normalized size = 0.85 \[ -\frac{\sqrt{\cos (a+b x)} \left (\tan ^{-1}\left (\sqrt{\cos (a+b x)}\right )+\tanh ^{-1}\left (\sqrt{\cos (a+b x)}\right )\right )}{b \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]/Sqrt[d*Cos[a + b*x]],x]

[Out]

-(((ArcTan[Sqrt[Cos[a + b*x]]] + ArcTanh[Sqrt[Cos[a + b*x]]])*Sqrt[Cos[a + b*x]])/(b*Sqrt[d*Cos[a + b*x]]))

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Maple [B]  time = 0.104, size = 177, normalized size = 3. \begin{align*} -{\frac{1}{2\,b}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}+2\,d\cos \left ( 1/2\,bx+a/2 \right ) -d}{\cos \left ( 1/2\,bx+a/2 \right ) -1}} \right ){\frac{1}{\sqrt{d}}}}-{\frac{1}{2\,b}\ln \left ( 2\,{\frac{\sqrt{d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}-2\,d\cos \left ( 1/2\,bx+a/2 \right ) -d}{\cos \left ( 1/2\,bx+a/2 \right ) +1}} \right ){\frac{1}{\sqrt{d}}}}+{\frac{1}{b}\ln \left ( 2\,{\frac{\sqrt{-d}\sqrt{-2\, \left ( \sin \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}d+d}-d}{\cos \left ( 1/2\,bx+a/2 \right ) }} \right ){\frac{1}{\sqrt{-d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)/(d*cos(b*x+a))^(1/2),x)

[Out]

-1/2/d^(1/2)/b*ln(2/(cos(1/2*b*x+1/2*a)-1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)+2*d*cos(1/2*b*x+1/2*a)
-d))-1/2/d^(1/2)/b*ln(2/(cos(1/2*b*x+1/2*a)+1)*(d^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-2*d*cos(1/2*b*x+1/
2*a)-d))+1/(-d)^(1/2)/b*ln(2/cos(1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*sin(1/2*b*x+1/2*a)^2*d+d)^(1/2)-d))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.5388, size = 676, normalized size = 11.46 \begin{align*} \left [\frac{2 \, \sqrt{-d} \arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )} \sqrt{-d}{\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) - \sqrt{-d} \log \left (\frac{d \cos \left (b x + a\right )^{2} + 4 \, \sqrt{d \cos \left (b x + a\right )} \sqrt{-d}{\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right )}{4 \, b d}, -\frac{2 \, \sqrt{d} \arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )}{\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt{d} \cos \left (b x + a\right )}\right ) - \sqrt{d} \log \left (\frac{d \cos \left (b x + a\right )^{2} - 4 \, \sqrt{d \cos \left (b x + a\right )} \sqrt{d}{\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right )}{4 \, b d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*sqrt(-d)*arctan(1/2*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) + 1)/(d*cos(b*x + a))) - sqrt(-d)*log(
(d*cos(b*x + a)^2 + 4*sqrt(d*cos(b*x + a))*sqrt(-d)*(cos(b*x + a) - 1) - 6*d*cos(b*x + a) + d)/(cos(b*x + a)^2
+ 2*cos(b*x + a) + 1)))/(b*d), -1/4*(2*sqrt(d)*arctan(1/2*sqrt(d*cos(b*x + a))*(cos(b*x + a) - 1)/(sqrt(d)*co
s(b*x + a))) - sqrt(d)*log((d*cos(b*x + a)^2 - 4*sqrt(d*cos(b*x + a))*sqrt(d)*(cos(b*x + a) + 1) + 6*d*cos(b*x
+ a) + d)/(cos(b*x + a)^2 - 2*cos(b*x + a) + 1)))/(b*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*cos(b*x+a))**(1/2),x)

[Out]

Integral(csc(a + b*x)/sqrt(d*cos(a + b*x)), x)

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Giac [A]  time = 1.12228, size = 70, normalized size = 1.19 \begin{align*} \frac{d{\left (\frac{\arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )}}{\sqrt{-d}}\right )}{\sqrt{-d} d} - \frac{\arctan \left (\frac{\sqrt{d \cos \left (b x + a\right )}}{\sqrt{d}}\right )}{d^{\frac{3}{2}}}\right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*(arctan(sqrt(d*cos(b*x + a))/sqrt(-d))/(sqrt(-d)*d) - arctan(sqrt(d*cos(b*x + a))/sqrt(d))/d^(3/2))/b