Integrand size = 19, antiderivative size = 59 \[ \int \frac {\csc (a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b \sqrt {d}}-\frac {\text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b \sqrt {d}} \]
-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)
Time = 0.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.85 \[ \int \frac {\csc (a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=-\frac {\left (\arctan \left (\sqrt {\cos (a+b x)}\right )+\text {arctanh}\left (\sqrt {\cos (a+b x)}\right )\right ) \sqrt {\cos (a+b x)}}{b \sqrt {d \cos (a+b x)}} \]
-(((ArcTan[Sqrt[Cos[a + b*x]]] + ArcTanh[Sqrt[Cos[a + b*x]]])*Sqrt[Cos[a + b*x]])/(b*Sqrt[d*Cos[a + b*x]]))
Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.86, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 3045, 27, 266, 756, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\csc (a+b x)}{\sqrt {d \cos (a+b x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x) \sqrt {d \cos (a+b x)}}dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\int \frac {d^2}{\sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {d \int \frac {1}{\sqrt {d \cos (a+b x)} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))}{b}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {2 d \int \frac {1}{d^2-d^4 \cos ^4(a+b x)}d\sqrt {d \cos (a+b x)}}{b}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle -\frac {2 d \left (\frac {\int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d}+\frac {\int \frac {1}{d^2 \cos ^2(a+b x)+d}d\sqrt {d \cos (a+b x)}}{2 d}\right )}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {2 d \left (\frac {\int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}}{2 d}+\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2 d \left (\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}+\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 d^{3/2}}\right )}{b}\) |
(-2*d*(ArcTan[Sqrt[d]*Cos[a + b*x]]/(2*d^(3/2)) + ArcTanh[Sqrt[d]*Cos[a + b*x]]/(2*d^(3/2))))/b
3.3.27.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ Symbol] :> Simp[-(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])
Leaf count of result is larger than twice the leaf count of optimal. \(180\) vs. \(2(47)=94\).
Time = 0.10 (sec) , antiderivative size = 181, normalized size of antiderivative = 3.07
method | result | size |
default | \(-\frac {\ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}-2 \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) \sqrt {d}+\ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}}{2 \sqrt {-d}\, \sqrt {d}\, b}\) | \(181\) |
-1/2/(-d)^(1/2)/d^(1/2)*(ln(-2/(cos(1/2*b*x+1/2*a)+1)*(2*d*cos(1/2*b*x+1/2 *a)-d^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)+d))*(-d)^(1/2)-2*ln(2/cos( 1/2*b*x+1/2*a)*((-d)^(1/2)*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)-d))*d^(1/2) +ln(2/(cos(1/2*b*x+1/2*a)-1)*(2*d*cos(1/2*b*x+1/2*a)+d^(1/2)*(-2*d*sin(1/2 *b*x+1/2*a)^2+d)^(1/2)-d))*(-d)^(1/2))/b
Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (47) = 94\).
Time = 0.34 (sec) , antiderivative size = 246, normalized size of antiderivative = 4.17 \[ \int \frac {\csc (a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\left [\frac {2 \, \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) - \sqrt {-d} \log \left (\frac {d \cos \left (b x + a\right )^{2} + 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right )}{4 \, b d}, -\frac {2 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) - \sqrt {d} \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right )}{4 \, b d}\right ] \]
[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)*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)]
\[ \int \frac {\csc (a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\int \frac {\csc {\left (a + b x \right )}}{\sqrt {d \cos {\left (a + b x \right )}}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.15 \[ \int \frac {\csc (a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=-\frac {2 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right ) - \sqrt {d} \log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{2 \, b d} \]
-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*cos(b*x + a)) + sqrt(d))))/(b*d)
Time = 0.35 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.88 \[ \int \frac {\csc (a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\frac {d {\left (\frac {\arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {-d}}\right )}{\sqrt {-d} d} - \frac {\arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{d^{\frac {3}{2}}}\right )}}{b} \]
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
Timed out. \[ \int \frac {\csc (a+b x)}{\sqrt {d \cos (a+b x)}} \, dx=\int \frac {1}{\sin \left (a+b\,x\right )\,\sqrt {d\,\cos \left (a+b\,x\right )}} \,d x \]