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

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 59, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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])

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 = 0.04, 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]]))

________________________________________________________________________________________

fricas [B]  time = 0.52, size = 246, normalized size = 4.17 \[ \left [\frac {2 \, \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) - \sqrt {-d} \log \left (\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right )}{4 \, b d}, -\frac {2 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) - \sqrt {d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right )}{4 \, b d}\right ] \]

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)]

________________________________________________________________________________________

giac [B]  time = 1.59, size = 105, normalized size = 1.78 \[ \frac {\frac {2 \, \arctan \left (-\frac {\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} - \frac {\log \left ({\left | -\sqrt {-d} \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} + \sqrt {-d \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{4} + d} \right |}\right )}{\sqrt {-d}}}{2 \, b} \]

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]

1/2*(2*arctan(-(sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 - sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d))/sqrt(-d))/sqrt(-d) - lo
g(abs(-sqrt(-d)*tan(1/2*b*x + 1/2*a)^2 + sqrt(-d*tan(1/2*b*x + 1/2*a)^4 + d)))/sqrt(-d))/b

________________________________________________________________________________________

maple [B]  time = 0.20, size = 177, normalized size = 3.00 \[ -\frac {\ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}+4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right )}{2 \sqrt {d}\, b}-\frac {\ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right )}{2 \sqrt {d}\, b}+\frac {\ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right )}{\sqrt {-d}\, b} \]

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))

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 68, normalized size = 1.15 \[ -\frac {2 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) - \sqrt {d} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{2 \, b d} \]

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]

-1/2*(2*sqrt(d)*arctan(sqrt(d*cos(b*x + a))/sqrt(d)) - sqrt(d)*log((sqrt(d*cos(b*x + a)) - sqrt(d))/(sqrt(d*co
s(b*x + a)) + sqrt(d))))/(b*d)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sin \left (a+b\,x\right )\,\sqrt {d\,\cos \left (a+b\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc {\left (a + b x \right )}}{\sqrt {d \cos {\left (a + b x \right )}}}\, dx \]

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)

________________________________________________________________________________________