3.3.27 \(\int \frac {\csc (a+b x)}{\sqrt {d \cos (a+b x)}} \, dx\) [227]

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)

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Rubi [A]
time = 0.03, 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 = {2645, 335, 218, 212, 209} \begin {gather*} -\frac {\text {ArcTan}\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 {gather*}

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 209

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

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 218

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] &&  !Gt
Q[a/b, 0]

Rule 335

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 2645

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 {\text {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 \text {Subst}\left (\int \frac {1}{1-\frac {x^4}{d^2}} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b d}\\ &=-\frac {\text {Subst}\left (\int \frac {1}{d-x^2} \, dx,x,\sqrt {d \cos (a+b x)}\right )}{b}-\frac {\text {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*}

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Mathematica [A]
time = 0.03, size = 50, normalized size = 0.85 \begin {gather*} -\frac {\left (\tan ^{-1}\left (\sqrt {\cos (a+b x)}\right )+\tanh ^{-1}\left (\sqrt {\cos (a+b x)}\right )\right ) \sqrt {\cos (a+b x)}}{b \sqrt {d \cos (a+b x)}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(181\) vs. \(2(47)=94\).
time = 0.43, size = 182, normalized size = 3.08

method	result	size
default	\(-\frac {\ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}+\ln \left (\frac {2 \sqrt {d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}+4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}-2 \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d +d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) \sqrt {d}}{2 \sqrt {d}\, \sqrt {-d}\, b}\)	\(182\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d^(1/2)/(-d)^(1/2)*(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))*(-d)^(1/2)+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))*(-d)^(1/2)-2*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))*d^(1/
2))/b

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Maxima [A]
time = 0.51, size = 68, normalized size = 1.15 \begin {gather*} -\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} \end {gather*}

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)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (47) = 94\).
time = 0.40, size = 246, normalized size = 4.17 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc {\left (a + b x \right )}}{\sqrt {d \cos {\left (a + b x \right )}}}\, dx \end {gather*}

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 = 4.15, size = 52, normalized size = 0.88 \begin {gather*} \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 {gather*}

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

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

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)

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