Integrand size = 21, antiderivative size = 123 \[ \int \frac {\csc ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=-\frac {5 \arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 \sqrt {b} f}-\frac {5 \text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{32 \sqrt {b} f}-\frac {5 \cot ^2(e+f x) \sqrt {b \sec (e+f x)}}{16 b f}-\frac {\cot ^4(e+f x) (b \sec (e+f x))^{5/2}}{4 b^3 f} \]
-1/4*cot(f*x+e)^4*(b*sec(f*x+e))^(5/2)/b^3/f-5/32*arctan((b*sec(f*x+e))^(1 /2)/b^(1/2))/f/b^(1/2)-5/32*arctanh((b*sec(f*x+e))^(1/2)/b^(1/2))/f/b^(1/2 )-5/16*cot(f*x+e)^2*(b*sec(f*x+e))^(1/2)/b/f
Time = 0.63 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.87 \[ \int \frac {\csc ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=-\frac {\left (10 \arctan \left (\sqrt {\sec (e+f x)}\right )-5 \log \left (1-\sqrt {\sec (e+f x)}\right )+5 \log \left (1+\sqrt {\sec (e+f x)}\right )+4 \left (-5+\csc ^2(e+f x)+4 \csc ^4(e+f x)\right ) \sqrt {\sec (e+f x)}\right ) \sqrt {\sec (e+f x)}}{64 f \sqrt {b \sec (e+f x)}} \]
-1/64*((10*ArcTan[Sqrt[Sec[e + f*x]]] - 5*Log[1 - Sqrt[Sec[e + f*x]]] + 5* Log[1 + Sqrt[Sec[e + f*x]]] + 4*(-5 + Csc[e + f*x]^2 + 4*Csc[e + f*x]^4)*S qrt[Sec[e + f*x]])*Sqrt[Sec[e + f*x]])/(f*Sqrt[b*Sec[e + f*x]])
Time = 0.29 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.07, 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, 252, 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.
\(\displaystyle \int \frac {\csc ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc (e+f x)^5}{\sqrt {b \sec (e+f x)}}dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int -\frac {b^6 (b \sec (e+f x))^{7/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^3}d(b \sec (e+f x))}{b^5 f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {b^6 (b \sec (e+f x))^{7/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^3}d(b \sec (e+f x))}{b^5 f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b \int \frac {(b \sec (e+f x))^{7/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^3}d(b \sec (e+f x))}{f}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{5/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {5}{8} \int \frac {(b \sec (e+f x))^{3/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^2}d(b \sec (e+f x))\right )}{f}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{5/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {5}{8} \left (\frac {\sqrt {b \sec (e+f x)}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {1}{4} \int \frac {1}{\sqrt {b \sec (e+f x)} \left (b^2-b^2 \sec ^2(e+f x)\right )}d(b \sec (e+f x))\right )\right )}{f}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{5/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {5}{8} \left (\frac {\sqrt {b \sec (e+f x)}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {1}{2} \int \frac {1}{b^2-b^4 \sec ^4(e+f x)}d\sqrt {b \sec (e+f x)}\right )\right )}{f}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{5/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {5}{8} \left (\frac {1}{2} \left (-\frac {\int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}}{2 b}-\frac {\int \frac {1}{b^2 \sec ^2(e+f x)+b}d\sqrt {b \sec (e+f x)}}{2 b}\right )+\frac {\sqrt {b \sec (e+f x)}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}\right )\right )}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{5/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {5}{8} \left (\frac {1}{2} \left (-\frac {\int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}}{2 b}-\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}\right )+\frac {\sqrt {b \sec (e+f x)}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}\right )\right )}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {b \left (\frac {(b \sec (e+f x))^{5/2}}{4 \left (b^2-b^2 \sec ^2(e+f x)\right )^2}-\frac {5}{8} \left (\frac {1}{2} \left (-\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}-\frac {\text {arctanh}\left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}\right )+\frac {\sqrt {b \sec (e+f x)}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}\right )\right )}{f}\) |
-((b*((b*Sec[e + f*x])^(5/2)/(4*(b^2 - b^2*Sec[e + f*x]^2)^2) - (5*((-1/2* ArcTan[Sqrt[b]*Sec[e + f*x]]/b^(3/2) - ArcTanh[Sqrt[b]*Sec[e + f*x]]/(2*b^ (3/2)))/2 + Sqrt[b*Sec[e + f*x]]/(2*(b^2 - b^2*Sec[e + f*x]^2))))/8))/f)
3.5.16.3.1 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] :> 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. \(344\) vs. \(2(99)=198\).
Time = 0.22 (sec) , antiderivative size = 345, normalized size of antiderivative = 2.80
method | result | size |
default | \(\frac {\left (20 \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-5 \left (\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right ) \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )-5 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) \cos \left (f x +e \right )+5 \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right ) \left (\sin ^{2}\left (f x +e \right )\right )+5 \ln \left (\frac {2 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\cos \left (f x +e \right )+1}{\cos \left (f x +e \right )+1}\right ) \left (\sin ^{2}\left (f x +e \right )\right )-36 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right ) \left (\csc ^{4}\left (f x +e \right )\right )}{64 f \sqrt {b \sec \left (f x +e \right )}\, \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\) | \(345\) |
1/64/f*(20*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-5*sin(f*x+e)^ 2*cos(f*x+e)*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))-5*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)+5*arctan(1/ 2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))*sin(f*x+e)^2+5*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)^2-36*cos(f*x+e)*(-cos(f*x+e)/(cos (f*x+e)+1)^2)^(1/2))/(b*sec(f*x+e))^(1/2)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^( 1/2)*csc(f*x+e)^4
Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (99) = 198\).
Time = 0.34 (sec) , antiderivative size = 450, normalized size of antiderivative = 3.66 \[ \int \frac {\csc ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\left [\frac {10 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) - 5 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {-b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, {\left (5 \, \cos \left (f x + e\right )^{4} - 9 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{128 \, {\left (b f \cos \left (f x + e\right )^{4} - 2 \, b f \cos \left (f x + e\right )^{2} + b f\right )}}, \frac {10 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {b}}\right ) + 5 \, {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sqrt {b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, {\left (5 \, \cos \left (f x + e\right )^{4} - 9 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{128 \, {\left (b f \cos \left (f x + e\right )^{4} - 2 \, b f \cos \left (f x + e\right )^{2} + b f\right )}}\right ] \]
[1/128*(10*(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) - 5*(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*(5*cos(f*x + e)^4 - 9*cos(f*x + e)^2)*s qrt(b/cos(f*x + e)))/(b*f*cos(f*x + e)^4 - 2*b*f*cos(f*x + e)^2 + b*f), 1/ 128*(10*(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(b)*arctan(1/2*sqrt(b/ cos(f*x + e))*(cos(f*x + e) - 1)/sqrt(b)) + 5*(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*(5*cos(f*x + e)^4 - 9*cos(f*x + e)^2)*sqrt(b/cos (f*x + e)))/(b*f*cos(f*x + e)^4 - 2*b*f*cos(f*x + e)^2 + b*f)]
\[ \int \frac {\csc ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\int \frac {\csc ^{5}{\left (e + f x \right )}}{\sqrt {b \sec {\left (e + f x \right )}}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.12 \[ \int \frac {\csc ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {b {\left (\frac {4 \, {\left (5 \, b^{2} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 9 \, \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {5}{2}}\right )}}{b^{4} - \frac {2 \, b^{4}}{\cos \left (f x + e\right )^{2}} + \frac {b^{4}}{\cos \left (f x + e\right )^{4}}} - \frac {10 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{b^{\frac {3}{2}}} + \frac {5 \, \log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{b^{\frac {3}{2}}}\right )}}{64 \, f} \]
1/64*b*(4*(5*b^2*sqrt(b/cos(f*x + e)) - 9*(b/cos(f*x + e))^(5/2))/(b^4 - 2 *b^4/cos(f*x + e)^2 + b^4/cos(f*x + e)^4) - 10*arctan(sqrt(b/cos(f*x + e)) /sqrt(b))/b^(3/2) + 5*log(-(sqrt(b) - sqrt(b/cos(f*x + e)))/(sqrt(b) + sqr t(b/cos(f*x + e))))/b^(3/2))/f
Time = 0.31 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.10 \[ \int \frac {\csc ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\frac {b^{4} {\left (\frac {5 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{4}} + \frac {5 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {9}{2}}} + \frac {2 \, {\left (5 \, \sqrt {b \cos \left (f x + e\right )} b^{3} \cos \left (f x + e\right )^{3} - 9 \, \sqrt {b \cos \left (f x + e\right )} b^{3} \cos \left (f x + e\right )\right )}}{{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )}^{2} b^{4}}\right )}}{32 \, f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \]
1/32*b^4*(5*arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b^4) + 5*arcta n(sqrt(b*cos(f*x + e))/sqrt(b))/b^(9/2) + 2*(5*sqrt(b*cos(f*x + e))*b^3*co s(f*x + e)^3 - 9*sqrt(b*cos(f*x + e))*b^3*cos(f*x + e))/((b^2*cos(f*x + e) ^2 - b^2)^2*b^4))/(f*sgn(cos(f*x + e)))
Timed out. \[ \int \frac {\csc ^5(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^5\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}} \,d x \]