Integrand size = 19, antiderivative size = 78 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d \cos (a+b x)}}{\sqrt {d}}\right )}{b d^{3/2}}+\frac {2}{b d \sqrt {d \cos (a+b x)}} \]
arctan((d*cos(b*x+a))^(1/2)/d^(1/2))/b/d^(3/2)-arctanh((d*cos(b*x+a))^(1/2 )/d^(1/2))/b/d^(3/2)+2/b/d/(d*cos(b*x+a))^(1/2)
Time = 0.17 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.85 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {2+\arctan \left (\sqrt {\cos (a+b x)}\right ) \sqrt {\cos (a+b x)}-\text {arctanh}\left (\sqrt {\cos (a+b x)}\right ) \sqrt {\cos (a+b x)}}{b d \sqrt {d \cos (a+b x)}} \]
(2 + ArcTan[Sqrt[Cos[a + b*x]]]*Sqrt[Cos[a + b*x]] - ArcTanh[Sqrt[Cos[a + b*x]]]*Sqrt[Cos[a + b*x]])/(b*d*Sqrt[d*Cos[a + b*x]])
Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 3045, 27, 264, 266, 827, 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.
\(\displaystyle \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x) (d \cos (a+b x))^{3/2}}dx\) |
\(\Big \downarrow \) 3045 |
\(\displaystyle -\frac {\int \frac {d^2}{(d \cos (a+b x))^{3/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))}{b d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {d \int \frac {1}{(d \cos (a+b x))^{3/2} \left (d^2-d^2 \cos ^2(a+b x)\right )}d(d \cos (a+b x))}{b}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {d \left (\frac {\int \frac {\sqrt {d \cos (a+b x)}}{d^2-d^2 \cos ^2(a+b x)}d(d \cos (a+b x))}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}\right )}{b}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {d \left (\frac {2 \int \frac {d^2 \cos ^2(a+b x)}{d^2-d^4 \cos ^4(a+b x)}d\sqrt {d \cos (a+b x)}}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}\right )}{b}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle -\frac {d \left (\frac {2 \left (\frac {1}{2} \int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}-\frac {1}{2} \int \frac {1}{d^2 \cos ^2(a+b x)+d}d\sqrt {d \cos (a+b x)}\right )}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}\right )}{b}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {d \left (\frac {2 \left (\frac {1}{2} \int \frac {1}{d-d^2 \cos ^2(a+b x)}d\sqrt {d \cos (a+b x)}-\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}\right )}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {d \left (\frac {2 \left (\frac {\text {arctanh}\left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}-\frac {\arctan \left (\sqrt {d} \cos (a+b x)\right )}{2 \sqrt {d}}\right )}{d^2}-\frac {2}{d^2 \sqrt {d \cos (a+b x)}}\right )}{b}\) |
-((d*((2*(-1/2*ArcTan[Sqrt[d]*Cos[a + b*x]]/Sqrt[d] + ArcTanh[Sqrt[d]*Cos[ a + b*x]]/(2*Sqrt[d])))/d^2 - 2/(d^2*Sqrt[d*Cos[a + b*x]])))/b)
3.3.28.3.1 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] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) 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. \(440\) vs. \(2(64)=128\).
Time = 0.10 (sec) , antiderivative size = 441, normalized size of antiderivative = 5.65
method | result | size |
default | \(-\frac {4 d^{\frac {5}{2}} \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+2 \sqrt {-d}\, \ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d^{2}+2 \sqrt {-d}\, \ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) d^{2}-2 \ln \left (\frac {2 \sqrt {-d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )}\right ) d^{\frac {5}{2}}+4 \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}\, \sqrt {-d}\, d^{\frac {3}{2}}-\ln \left (-\frac {2 \left (2 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-\sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}+d \right )}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\right ) \sqrt {-d}\, d^{2}-\ln \left (\frac {4 d \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+2 \sqrt {d}\, \sqrt {-2 d \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+d}-2 d}{\cos \left (\frac {b x}{2}+\frac {a}{2}\right )-1}\right ) \sqrt {-d}\, d^{2}}{2 \sqrt {-d}\, d^{\frac {7}{2}} \left (2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) b}\) | \(441\) |
-1/2/(-d)^(1/2)/d^(7/2)/(2*sin(1/2*b*x+1/2*a)^2-1)*(4*d^(5/2)*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))*sin(1/2*b* x+1/2*a)^2+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))*sin(1/2*b*x+1/2*a)^2*d^ 2+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))*sin(1/2*b*x+1/2*a)^2*d^2-2*ln(2/c os(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^(5 /2)+4*(-2*d*sin(1/2*b*x+1/2*a)^2+d)^(1/2)*(-d)^(1/2)*d^(3/2)-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)*d^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)*d^2)/b
Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (64) = 128\).
Time = 0.32 (sec) , antiderivative size = 309, normalized size of antiderivative = 3.96 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\left [\frac {2 \, \sqrt {-d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) + 1\right )}}{2 \, d \cos \left (b x + a\right )}\right ) \cos \left (b x + a\right ) - \sqrt {-d} \cos \left (b x + a\right ) \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {-d} {\left (\cos \left (b x + a\right ) - 1\right )} - 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )}}{4 \, b d^{2} \cos \left (b x + a\right )}, \frac {2 \, \sqrt {d} \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - 1\right )}}{2 \, \sqrt {d} \cos \left (b x + a\right )}\right ) \cos \left (b x + a\right ) + \sqrt {d} \cos \left (b x + a\right ) \log \left (\frac {d \cos \left (b x + a\right )^{2} - 4 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {d} {\left (\cos \left (b x + a\right ) + 1\right )} + 6 \, d \cos \left (b x + a\right ) + d}{\cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1}\right ) + 8 \, \sqrt {d \cos \left (b x + a\right )}}{4 \, b d^{2} \cos \left (b x + a\right )}\right ] \]
[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)))*cos(b*x + a) - sqrt(-d)*cos(b*x + a)*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)) + 8*sqrt(d*cos(b*x + a)))/ (b*d^2*cos(b*x + a)), 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)))*cos(b*x + a) + sqrt(d)*cos(b*x + a)* 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)) + 8*sqrt(d* cos(b*x + a)))/(b*d^2*cos(b*x + a))]
\[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int \frac {\csc {\left (a + b x \right )}}{\left (d \cos {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.01 \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\frac {\frac {2 \, \arctan \left (\frac {\sqrt {d \cos \left (b x + a\right )}}{\sqrt {d}}\right )}{\sqrt {d}} + \frac {\log \left (\frac {\sqrt {d \cos \left (b x + a\right )} - \sqrt {d}}{\sqrt {d \cos \left (b x + a\right )} + \sqrt {d}}\right )}{\sqrt {d}} + \frac {4}{\sqrt {d \cos \left (b x + a\right )}}}{2 \, b d} \]
1/2*(2*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)))/sqrt(d) + 4/sqrt(d*cos( b*x + a)))/(b*d)
\[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int { \frac {\csc \left (b x + a\right )}{\left (d \cos \left (b x + a\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\csc (a+b x)}{(d \cos (a+b x))^{3/2}} \, dx=\int \frac {1}{\sin \left (a+b\,x\right )\,{\left (d\,\cos \left (a+b\,x\right )\right )}^{3/2}} \,d x \]