Integrand size = 21, antiderivative size = 123 \[ \int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {3 \arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 b^{5/2} f}+\frac {3 \text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 b^{5/2} f}-\frac {\cot ^2(e+f x) \sqrt {b \sec (e+f x)}}{16 b^3 f}-\frac {\cot ^4(e+f x) \sqrt {b \sec (e+f x)}}{4 b^3 f} \]
3/32*arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/b^(5/2)/f+3/32*arctanh((b*sec(f* x+e))^(1/2)/b^(1/2))/b^(5/2)/f-1/16*cot(f*x+e)^2*(b*sec(f*x+e))^(1/2)/b^3/ f-1/4*cot(f*x+e)^4*(b*sec(f*x+e))^(1/2)/b^3/f
Time = 1.22 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.89 \[ \int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\frac {\left (6 \arctan \left (\sqrt {\sec (e+f x)}\right )-3 \log \left (1-\sqrt {\sec (e+f x)}\right )+3 \log \left (1+\sqrt {\sec (e+f x)}\right )-\frac {2 (5+3 \cos (2 (e+f x))) \csc ^4(e+f x)}{\sec ^{\frac {3}{2}}(e+f x)}\right ) \sqrt {\sec (e+f x)}}{64 b^2 f \sqrt {b \sec (e+f x)}} \]
((6*ArcTan[Sqrt[Sec[e + f*x]]] - 3*Log[1 - Sqrt[Sec[e + f*x]]] + 3*Log[1 + Sqrt[Sec[e + f*x]]] - (2*(5 + 3*Cos[2*(e + f*x)])*Csc[e + f*x]^4)/Sec[e + f*x]^(3/2))*Sqrt[Sec[e + f*x]])/(64*b^2*f*Sqrt[b*Sec[e + f*x]])
Time = 0.30 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3102, 25, 27, 252, 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.
| \(\displaystyle \int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc (e+f x)^5}{(b \sec (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3102 |
| \(\displaystyle \frac {\int -\frac {b^6 (b \sec (e+f x))^{3/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^3}d(b \sec (e+f x))}{b^5 f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {b^6 (b \sec (e+f x))^{3/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^3}d(b \sec (e+f x))}{b^5 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {b \int \frac {(b \sec (e+f x))^{3/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^3}d(b \sec (e+f x))}{f}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {b \left (\frac {\sqrt {b \sec (e+f x)}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {1}{8} \int \frac {1}{\sqrt {b \sec (e+f x)} \left (b^2-b^2 \sec ^2(e+f x)\right )^2}d(b \sec (e+f x))\right )}{f}\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle -\frac {b \left (\frac {1}{8} \left (-\frac {3 \int \frac {1}{\sqrt {b \sec (e+f x)} \left (b^2-b^2 \sec ^2(e+f x)\right )}d(b \sec (e+f x))}{4 b^2}-\frac {\sqrt {b \sec (e+f x)}}{2 b^2 \left (b^2-b^2 \sec ^2(e+f x)\right )}\right )+\frac {\sqrt {b \sec (e+f x)}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}\right )}{f}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {b \left (\frac {1}{8} \left (-\frac {3 \int \frac {1}{b^2-b^4 \sec ^4(e+f x)}d\sqrt {b \sec (e+f x)}}{2 b^2}-\frac {\sqrt {b \sec (e+f x)}}{2 b^2 \left (b^2-b^2 \sec ^2(e+f x)\right )}\right )+\frac {\sqrt {b \sec (e+f x)}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}\right )}{f}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle -\frac {b \left (\frac {1}{8} \left (-\frac {3 \left (\frac {\int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}}{2 b}+\frac {\int \frac {1}{b^2 \sec ^2(e+f x)+b}d\sqrt {b \sec (e+f x)}}{2 b}\right )}{2 b^2}-\frac {\sqrt {b \sec (e+f x)}}{2 b^2 \left (b^2-b^2 \sec ^2(e+f x)\right )}\right )+\frac {\sqrt {b \sec (e+f x)}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}\right )}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {b \left (\frac {1}{8} \left (-\frac {3 \left (\frac {\int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}}{2 b}+\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}\right )}{2 b^2}-\frac {\sqrt {b \sec (e+f x)}}{2 b^2 \left (b^2-b^2 \sec ^2(e+f x)\right )}\right )+\frac {\sqrt {b \sec (e+f x)}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}\right )}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {b \left (\frac {1}{8} \left (-\frac {3 \left (\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}\right )}{2 b^2}-\frac {\sqrt {b \sec (e+f x)}}{2 b^2 \left (b^2-b^2 \sec ^2(e+f x)\right )}\right )+\frac {\sqrt {b \sec (e+f x)}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}\right )}{f}\) |
-((b*(Sqrt[b*Sec[e + f*x]]/(4*(b^2 - b^2*Sec[e + f*x]^2)^2) + ((-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)))/ 8))/f)
3.5.43.3.1 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*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
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. \(347\) vs. \(2(99)=198\).
Time = 0.19 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.83
| method | result | size |
| default | \(-\frac {\left (12 \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-3 \left (\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right ) \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )-3 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )+3 \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right ) \left (\sin ^{2}\left (f x +e \right )\right )+3 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \left (\sin ^{2}\left (f x +e \right )\right )+4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right ) \left (\csc ^{4}\left (f x +e \right )\right )}{64 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sqrt {b \sec \left (f x +e \right )}\, b^{2}}\) | \(348\) |
-1/64/f*(12*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-3*sin(f*x+e) ^2*cos(f*x+e)*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))*sin(f*x+e)^2*cos(f*x+e)+3*arctan(1 /2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))*sin(f*x+e)^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))*sin(f*x+e)^2+4*cos(f*x+e)*(-cos(f*x+e)/(cos (f*x+e)+1)^2)^(1/2))/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(b*sec(f*x+e))^( 1/2)/b^2*csc(f*x+e)^4
Leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (99) = 198\).
Time = 0.39 (sec) , antiderivative size = 458, normalized size of antiderivative = 3.72 \[ \int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\left [-\frac {6 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) + 3 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {-b} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, {\left (3 \, \cos \left (f x + e\right )^{4} + \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{128 \, {\left (b^{3} f \cos \left (f x + e\right )^{4} - 2 \, b^{3} f \cos \left (f x + e\right )^{2} + b^{3} f\right )}}, -\frac {6 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {b}}\right ) - 3 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {b} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, {\left (3 \, \cos \left (f x + e\right )^{4} + \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{128 \, {\left (b^{3} f \cos \left (f x + e\right )^{4} - 2 \, b^{3} f \cos \left (f x + e\right )^{2} + b^{3} f\right )}}\right ] \]
[-1/128*(6*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(-b)*arctan(1/2*sqr t(-b)*sqrt(b/cos(f*x + e))*(cos(f*x + e) + 1)/b) + 3*(cos(f*x + e)^4 - 2*c os(f*x + e)^2 + 1)*sqrt(-b)*log((b*cos(f*x + e)^2 + 4*(cos(f*x + e)^2 - co s(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)) + 8*(3*cos(f*x + e)^4 + cos(f*x + e)^2)*sqr t(b/cos(f*x + e)))/(b^3*f*cos(f*x + e)^4 - 2*b^3*f*cos(f*x + e)^2 + b^3*f) , -1/128*(6*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(b)*arctan(1/2*sqr t(b/cos(f*x + e))*(cos(f*x + e) - 1)/sqrt(b)) - 3*(cos(f*x + e)^4 - 2*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))*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)) + 8*(3*cos(f*x + e)^4 + cos(f*x + e)^2)*sqrt(b/c os(f*x + e)))/(b^3*f*cos(f*x + e)^4 - 2*b^3*f*cos(f*x + e)^2 + b^3*f)]
\[ \int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\int \frac {\csc ^{5}{\left (e + f x \right )}}{\left (b \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.26 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.11 \[ \int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=-\frac {b {\left (\frac {4 \, {\left (3 \, b^{2} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {5}{2}}\right )}}{b^{6} - \frac {2 \, b^{6}}{\cos \left (f x + e\right )^{2}} + \frac {b^{6}}{\cos \left (f x + e\right )^{4}}} - \frac {6 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{b^{\frac {7}{2}}} + \frac {3 \, \log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{b^{\frac {7}{2}}}\right )}}{64 \, f} \]
-1/64*b*(4*(3*b^2*sqrt(b/cos(f*x + e)) + (b/cos(f*x + e))^(5/2))/(b^6 - 2* b^6/cos(f*x + e)^2 + b^6/cos(f*x + e)^4) - 6*arctan(sqrt(b/cos(f*x + e))/s qrt(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 = 0.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.07 \[ \int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=-\frac {\frac {3 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {3 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, \sqrt {b \cos \left (f x + e\right )} b^{3} \cos \left (f x + e\right )^{3} + \sqrt {b \cos \left (f x + e\right )} b^{3} \cos \left (f x + e\right )\right )}}{{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )}^{2} b^{2}}}{32 \, f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \]
-1/32*(3*arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b^2) + 3*arctan(s qrt(b*cos(f*x + e))/sqrt(b))/b^(5/2) + 2*(3*sqrt(b*cos(f*x + e))*b^3*cos(f *x + e)^3 + sqrt(b*cos(f*x + e))*b^3*cos(f*x + e))/((b^2*cos(f*x + e)^2 - b^2)^2*b^2))/(f*sgn(cos(f*x + e)))
Timed out. \[ \int \frac {\csc ^5(e+f x)}{(b \sec (e+f x))^{5/2}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^5\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]