Integrand size = 19, antiderivative size = 78 \[ \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f}-\frac {\text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f}+\frac {2}{b f \sqrt {b \sec (e+f x)}} \]
arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/b^(3/2)/f-arctanh((b*sec(f*x+e))^(1/2 )/b^(1/2))/b^(3/2)/f+2/b/f/(b*sec(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.46 \[ \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},\sec ^2(e+f x)\right )}{b f \sqrt {b \sec (e+f x)}} \]
Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {3042, 3102, 25, 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.
\(\displaystyle \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int -\frac {b^2}{(b \sec (e+f x))^{3/2} \left (b^2-b^2 \sec ^2(e+f x)\right )}d(b \sec (e+f x))}{b f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {b^2}{(b \sec (e+f x))^{3/2} \left (b^2-b^2 \sec ^2(e+f x)\right )}d(b \sec (e+f x))}{b f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b \int \frac {1}{(b \sec (e+f x))^{3/2} \left (b^2-b^2 \sec ^2(e+f x)\right )}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {b \left (\frac {\int \frac {\sqrt {b \sec (e+f x)}}{b^2-b^2 \sec ^2(e+f x)}d(b \sec (e+f x))}{b^2}-\frac {2}{b^2 \sqrt {b \sec (e+f x)}}\right )}{f}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {b \left (\frac {2 \int \frac {b^2 \sec ^2(e+f x)}{b^2-b^4 \sec ^4(e+f x)}d\sqrt {b \sec (e+f x)}}{b^2}-\frac {2}{b^2 \sqrt {b \sec (e+f x)}}\right )}{f}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle -\frac {b \left (\frac {2 \left (\frac {1}{2} \int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {1}{2} \int \frac {1}{b^2 \sec ^2(e+f x)+b}d\sqrt {b \sec (e+f x)}\right )}{b^2}-\frac {2}{b^2 \sqrt {b \sec (e+f x)}}\right )}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {b \left (\frac {2 \left (\frac {1}{2} \int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}\right )}{b^2}-\frac {2}{b^2 \sqrt {b \sec (e+f x)}}\right )}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {b \left (\frac {2 \left (\frac {\text {arctanh}\left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}\right )}{b^2}-\frac {2}{b^2 \sqrt {b \sec (e+f x)}}\right )}{f}\) |
-((b*((2*(-1/2*ArcTan[Sqrt[b]*Sec[e + f*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*Sec[ e + f*x]]/(2*Sqrt[b])))/b^2 - 2/(b^2*Sqrt[b*Sec[e + f*x]])))/f)
3.5.28.3.1 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[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S ymbol] :> Simp[1/(f*a^n) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/ 2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1 )/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Leaf count of result is larger than twice the leaf count of optimal. \(196\) vs. \(2(64)=128\).
Time = 0.18 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.53
method | result | size |
default | \(\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right )}{2 f \left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {b \sec \left (f x +e \right )}\, b}\) | \(197\) |
1/2/f*(4*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+arctan(1/2/(-cos( f*x+e)/(cos(f*x+e)+1)^2)^(1/2))+4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-ln( (2*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+2*(-cos(f*x+e)/(cos(f*x +e)+1)^2)^(1/2)-cos(f*x+e)+1)/(cos(f*x+e)+1)))/(cos(f*x+e)+1)/(-cos(f*x+e) /(cos(f*x+e)+1)^2)^(1/2)/(b*sec(f*x+e))^(1/2)/b
Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (64) = 128\).
Time = 0.43 (sec) , antiderivative size = 314, normalized size of antiderivative = 4.03 \[ \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\left [\frac {2 \, \sqrt {-b} \arctan \left (\frac {2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) + b}\right ) + 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) - \sqrt {-b} \log \left (-\frac {b \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, b^{2} f}, \frac {2 \, \sqrt {b} \arctan \left (\frac {2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) - b}\right ) + 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) + \sqrt {b} \log \left (-\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, b^{2} f}\right ] \]
[1/4*(2*sqrt(-b)*arctan(2*sqrt(-b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/(b*co s(f*x + e) + b)) + 8*sqrt(b/cos(f*x + e))*cos(f*x + e) - sqrt(-b)*log(-(b* cos(f*x + e)^2 + 4*(cos(f*x + e)^2 - cos(f*x + e))*sqrt(-b)*sqrt(b/cos(f*x + e)) - 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)))/(b^ 2*f), 1/4*(2*sqrt(b)*arctan(2*sqrt(b)*sqrt(b/cos(f*x + e))*cos(f*x + e)/(b *cos(f*x + e) - b)) + 8*sqrt(b/cos(f*x + e))*cos(f*x + e) + sqrt(b)*log(-( b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(b)*sqrt(b/cos(f* x + e)) + 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)))/(b ^2*f)]
\[ \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int \frac {\csc {\left (e + f x \right )}}{\left (b \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.34 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.14 \[ \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {b {\left (\frac {2 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{b^{\frac {5}{2}}} + \frac {\log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{b^{\frac {5}{2}}} + \frac {4}{b^{2} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}}{2 \, f} \]
1/2*b*(2*arctan(sqrt(b/cos(f*x + e))/sqrt(b))/b^(5/2) + log(-(sqrt(b) - sq rt(b/cos(f*x + e)))/(sqrt(b) + sqrt(b/cos(f*x + e))))/b^(5/2) + 4/(b^2*sqr t(b/cos(f*x + e))))/f
Time = 0.29 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94 \[ \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\frac {\frac {b \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \sqrt {b} \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right ) + 2 \, \sqrt {b \cos \left (f x + e\right )}}{b^{2} f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \]
(b*arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/sqrt(-b) - sqrt(b)*arctan(sqrt(b* cos(f*x + e))/sqrt(b)) + 2*sqrt(b*cos(f*x + e)))/(b^2*f*sgn(cos(f*x + e)))
Timed out. \[ \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]