Integrand size = 21, antiderivative size = 93 \[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {3 \arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 b^{5/2} f}+\frac {3 \text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 b^{5/2} f}-\frac {\cot ^2(e+f x) \sqrt {b \sec (e+f x)}}{2 b^3 f} \]
3/4*arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/b^(5/2)/f+3/4*arctanh((b*sec(f*x+ e))^(1/2)/b^(1/2))/b^(5/2)/f-1/2*cot(f*x+e)^2*(b*sec(f*x+e))^(1/2)/b^3/f
Time = 0.68 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.05 \[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {\left (6 \arctan \left (\sqrt {\sec (e+f x)}\right )-3 \log \left (1-\sqrt {\sec (e+f x)}\right )+3 \log \left (1+\sqrt {\sec (e+f x)}\right )-\frac {4 \csc ^2(e+f x)}{\sec ^{\frac {3}{2}}(e+f x)}\right ) \sqrt {\sec (e+f x)}}{8 b^2 f \sqrt {b \sec (e+f x)}} \]
((6*ArcTan[Sqrt[Sec[e + f*x]]] - 3*Log[1 - Sqrt[Sec[e + f*x]]] + 3*Log[1 + Sqrt[Sec[e + f*x]]] - (4*Csc[e + f*x]^2)/Sec[e + f*x]^(3/2))*Sqrt[Sec[e + f*x]])/(8*b^2*f*Sqrt[b*Sec[e + f*x]])
Time = 0.27 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 3102, 27, 253, 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.
| \(\displaystyle \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc (e+f x)^3}{(b \sec (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3102 |
| \(\displaystyle \frac {\int \frac {b^4}{\sqrt {b \sec (e+f x)} \left (b^2-b^2 \sec ^2(e+f x)\right )^2}d(b \sec (e+f x))}{b^3 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \int \frac {1}{\sqrt {b \sec (e+f x)} \left (b^2-b^2 \sec ^2(e+f x)\right )^2}d(b \sec (e+f x))}{f}\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle \frac {b \left (\frac {3 \int \frac {1}{\sqrt {b \sec (e+f x)} \left (b^2-b^2 \sec ^2(e+f x)\right )}d(b \sec (e+f x))}{4 b^2}+\frac {\sqrt {b \sec (e+f x)}}{2 b^2 \left (b^2-b^2 \sec ^2(e+f x)\right )}\right )}{f}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {b \left (\frac {3 \int \frac {1}{b^2-b^4 \sec ^4(e+f x)}d\sqrt {b \sec (e+f x)}}{2 b^2}+\frac {\sqrt {b \sec (e+f x)}}{2 b^2 \left (b^2-b^2 \sec ^2(e+f x)\right )}\right )}{f}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {b \left (\frac {3 \left (\frac {\int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}}{2 b}+\frac {\int \frac {1}{b^2 \sec ^2(e+f x)+b}d\sqrt {b \sec (e+f x)}}{2 b}\right )}{2 b^2}+\frac {\sqrt {b \sec (e+f x)}}{2 b^2 \left (b^2-b^2 \sec ^2(e+f x)\right )}\right )}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {b \left (\frac {3 \left (\frac {\int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}}{2 b}+\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}\right )}{2 b^2}+\frac {\sqrt {b \sec (e+f x)}}{2 b^2 \left (b^2-b^2 \sec ^2(e+f x)\right )}\right )}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {b \left (\frac {3 \left (\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}\right )}{2 b^2}+\frac {\sqrt {b \sec (e+f x)}}{2 b^2 \left (b^2-b^2 \sec ^2(e+f x)\right )}\right )}{f}\) |
(b*((3*(ArcTan[Sqrt[b]*Sec[e + f*x]]/(2*b^(3/2)) + ArcTanh[Sqrt[b]*Sec[e + f*x]]/(2*b^(3/2))))/(2*b^2) + Sqrt[b*Sec[e + f*x]]/(2*b^2*(b^2 - b^2*Sec[ e + f*x]^2))))/f
3.5.42.3.1 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)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -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[((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[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. \(289\) vs. \(2(73)=146\).
Time = 0.19 (sec) , antiderivative size = 290, normalized size of antiderivative = 3.12
| method | result | size |
| default | \(\frac {4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-3 \cos \left (f x +e \right ) \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )-3 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \cos \left (f x +e \right )+3 \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )+3 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right )}{8 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {b \sec \left (f x +e \right )}\, b^{2} \left (\cos ^{2}\left (f x +e \right )-1\right )}\) | \(290\) |
1/8/f*(4*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-3*cos(f*x+e)*arct an(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))-3*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)+3*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^ 2)^(1/2))+3*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) /(cos(f*x+e)+1)^2)^(1/2)/(b*sec(f*x+e))^(1/2)/b^2/(cos(f*x+e)^2-1)
Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (73) = 146\).
Time = 0.36 (sec) , antiderivative size = 370, normalized size of antiderivative = 3.98 \[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\left [-\frac {6 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) - 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + 3 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {-b} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right )}{16 \, {\left (b^{3} f \cos \left (f x + e\right )^{2} - b^{3} f\right )}}, -\frac {6 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {b}}\right ) - 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} - 3 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt {b} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right )}{16 \, {\left (b^{3} f \cos \left (f x + e\right )^{2} - b^{3} f\right )}}\right ] \]
[-1/16*(6*(cos(f*x + e)^2 - 1)*sqrt(-b)*arctan(1/2*sqrt(-b)*sqrt(b/cos(f*x + e))*(cos(f*x + e) + 1)/b) - 8*sqrt(b/cos(f*x + e))*cos(f*x + e)^2 + 3*( cos(f*x + e)^2 - 1)*sqrt(-b)*log((b*cos(f*x + e)^2 + 4*(cos(f*x + e)^2 - c os(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^3*f*cos(f*x + e)^2 - b^3*f), -1/16*(6* (cos(f*x + e)^2 - 1)*sqrt(b)*arctan(1/2*sqrt(b/cos(f*x + e))*(cos(f*x + e) - 1)/sqrt(b)) - 8*sqrt(b/cos(f*x + e))*cos(f*x + e)^2 - 3*(cos(f*x + e)^2 - 1)*sqrt(b)*log((b*cos(f*x + e)^2 + 4*(cos(f*x + e)^2 + cos(f*x + e))*sq rt(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^3*f*cos(f*x + e)^2 - b^3*f)]
\[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\int \frac {\csc ^{3}{\left (e + f x \right )}}{\left (b \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.14 \[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {b {\left (\frac {4 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{b^{4} - \frac {b^{4}}{\cos \left (f x + e\right )^{2}}} + \frac {6 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{b^{\frac {7}{2}}} - \frac {3 \, \log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{b^{\frac {7}{2}}}\right )}}{8 \, f} \]
1/8*b*(4*sqrt(b/cos(f*x + e))/(b^4 - b^4/cos(f*x + e)^2) + 6*arctan(sqrt(b /cos(f*x + e))/sqrt(b))/b^(7/2) - 3*log(-(sqrt(b) - sqrt(b/cos(f*x + e)))/ (sqrt(b) + sqrt(b/cos(f*x + e))))/b^(7/2))/f
Time = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.09 \[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {\frac {2 \, \sqrt {b \cos \left (f x + e\right )} b \cos \left (f x + e\right )}{b^{2} \cos \left (f x + e\right )^{2} - b^{2}} - \frac {3 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {3 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{\sqrt {b}}}{4 \, b^{2} f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \]
1/4*(2*sqrt(b*cos(f*x + e))*b*cos(f*x + e)/(b^2*cos(f*x + e)^2 - b^2) - 3* arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/sqrt(-b) - 3*arctan(sqrt(b*cos(f*x + e))/sqrt(b))/sqrt(b))/(b^2*f*sgn(cos(f*x + e)))
Timed out. \[ \int \frac {\csc ^3(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^3\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]