Integrand size = 21, antiderivative size = 113 \[ \int \csc ^3(e+f x) (b \sec (e+f x))^{3/2} \, dx=-\frac {5 b^{3/2} \arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 f}-\frac {5 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 f}+\frac {5 b \sqrt {b \sec (e+f x)}}{2 f}-\frac {\cot ^2(e+f x) (b \sec (e+f x))^{5/2}}{2 b f} \]
-5/4*b^(3/2)*arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/f-5/4*b^(3/2)*arctanh((b *sec(f*x+e))^(1/2)/b^(1/2))/f-1/2*cot(f*x+e)^2*(b*sec(f*x+e))^(5/2)/b/f+5/ 2*b*(b*sec(f*x+e))^(1/2)/f
Time = 0.67 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.86 \[ \int \csc ^3(e+f x) (b \sec (e+f x))^{3/2} \, dx=-\frac {\left (10 \arctan \left (\sqrt {\sec (e+f x)}\right )-5 \log \left (1-\sqrt {\sec (e+f x)}\right )+5 \log \left (1+\sqrt {\sec (e+f x)}\right )+4 \left (-5+\csc ^2(e+f x)\right ) \sqrt {\sec (e+f x)}\right ) (b \sec (e+f x))^{3/2}}{8 f \sec ^{\frac {3}{2}}(e+f x)} \]
-1/8*((10*ArcTan[Sqrt[Sec[e + f*x]]] - 5*Log[1 - Sqrt[Sec[e + f*x]]] + 5*L og[1 + Sqrt[Sec[e + f*x]]] + 4*(-5 + Csc[e + f*x]^2)*Sqrt[Sec[e + f*x]])*( b*Sec[e + f*x])^(3/2))/(f*Sec[e + f*x]^(3/2))
Time = 0.29 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3102, 27, 252, 262, 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.
| \(\displaystyle \int \csc ^3(e+f x) (b \sec (e+f x))^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc (e+f x)^3 (b \sec (e+f x))^{3/2}dx\) |
| \(\Big \downarrow \) 3102 |
| \(\displaystyle \frac {\int \frac {b^4 (b \sec (e+f x))^{7/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^2}d(b \sec (e+f x))}{b^3 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \int \frac {(b \sec (e+f x))^{7/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^2}d(b \sec (e+f x))}{f}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {b \left (\frac {(b \sec (e+f x))^{5/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {5}{4} \int \frac {(b \sec (e+f x))^{3/2}}{b^2-b^2 \sec ^2(e+f x)}d(b \sec (e+f x))\right )}{f}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {b \left (\frac {(b \sec (e+f x))^{5/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {5}{4} \left (b^2 \int \frac {1}{\sqrt {b \sec (e+f x)} \left (b^2-b^2 \sec ^2(e+f x)\right )}d(b \sec (e+f x))-2 \sqrt {b \sec (e+f x)}\right )\right )}{f}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {b \left (\frac {(b \sec (e+f x))^{5/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {5}{4} \left (2 b^2 \int \frac {1}{b^2-b^4 \sec ^4(e+f x)}d\sqrt {b \sec (e+f x)}-2 \sqrt {b \sec (e+f x)}\right )\right )}{f}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {b \left (\frac {(b \sec (e+f x))^{5/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {5}{4} \left (2 b^2 \left (\frac {\int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}}{2 b}+\frac {\int \frac {1}{b^2 \sec ^2(e+f x)+b}d\sqrt {b \sec (e+f x)}}{2 b}\right )-2 \sqrt {b \sec (e+f x)}\right )\right )}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {b \left (\frac {(b \sec (e+f x))^{5/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {5}{4} \left (2 b^2 \left (\frac {\int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}}{2 b}+\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}\right )-2 \sqrt {b \sec (e+f x)}\right )\right )}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {b \left (\frac {(b \sec (e+f x))^{5/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {5}{4} \left (2 b^2 \left (\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \sec (e+f x)\right )}{2 b^{3/2}}\right )-2 \sqrt {b \sec (e+f x)}\right )\right )}{f}\) |
(b*((b*Sec[e + f*x])^(5/2)/(2*(b^2 - b^2*Sec[e + f*x]^2)) - (5*(2*b^2*(Arc Tan[Sqrt[b]*Sec[e + f*x]]/(2*b^(3/2)) + ArcTanh[Sqrt[b]*Sec[e + f*x]]/(2*b ^(3/2))) - 2*Sqrt[b*Sec[e + f*x]]))/4))/f
3.4.90.3.1 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*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && 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. \(600\) vs. \(2(89)=178\).
Time = 0.24 (sec) , antiderivative size = 601, normalized size of antiderivative = 5.32
| method | result | size |
| default | \(\frac {\sqrt {b \sec \left (f x +e \right )}\, b \left (4 \left (\cos ^{4}\left (f x +e \right )\right ) \left (-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}\right )^{\frac {3}{2}}+8 \left (\cos ^{3}\left (f x +e \right )\right ) \left (-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}\right )^{\frac {3}{2}}+4 \left (\cos ^{2}\left (f x +e \right )\right ) \left (-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}\right )^{\frac {3}{2}}-16 \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 (\cos ^{3}\left (f x +e \right )\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 )-4 \left (\cos ^{3}\left (f x +e \right )\right ) \ln \left (\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}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right )-\left (\cos ^{3}\left (f x +e \right )\right ) \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 )+5 \left (\cos ^{2}\left (f x +e \right )\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 )+4 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\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}}}+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-2 \cos \left (f x +e \right )+2}{\cos \left (f x +e \right )+1}\right )+\left (\cos ^{2}\left (f x +e \right )\right ) \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 )+16 \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 )}{8 f \left (\cos \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right )+1\right )^{3} \left (-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}\right )^{\frac {3}{2}}}\) | \(601\) |
1/8/f*(b*sec(f*x+e))^(1/2)*b/(cos(f*x+e)-1)/(cos(f*x+e)+1)^3/(-cos(f*x+e)/ (cos(f*x+e)+1)^2)^(3/2)*(4*cos(f*x+e)^4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/ 2)+8*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)+4*cos(f*x+e)^2*(-co s(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)-16*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+ 1)^2)^(1/2)-5*cos(f*x+e)^3*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2) )-4*cos(f*x+e)^3*ln(2*(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*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))+5*cos(f*x+e)^2*arc tan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))+4*cos(f*x+e)^2*ln(2*(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)^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))+16*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/ 2))
Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (89) = 178\).
Time = 0.35 (sec) , antiderivative size = 388, normalized size of antiderivative = 3.43 \[ \int \csc ^3(e+f x) (b \sec (e+f x))^{3/2} \, dx=\left [\frac {10 \, {\left (b \cos \left (f x + e\right )^{2} - b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) + 5 \, {\left (b \cos \left (f x + e\right )^{2} - b\right )} \sqrt {-b} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, {\left (5 \, b \cos \left (f x + e\right )^{2} - 4 \, b\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{16 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}}, \frac {10 \, {\left (b \cos \left (f x + e\right )^{2} - b\right )} \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {b}}\right ) + 5 \, {\left (b \cos \left (f x + e\right )^{2} - b\right )} \sqrt {b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, {\left (5 \, b \cos \left (f x + e\right )^{2} - 4 \, b\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{16 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}}\right ] \]
[1/16*(10*(b*cos(f*x + e)^2 - b)*sqrt(-b)*arctan(1/2*sqrt(-b)*sqrt(b/cos(f *x + e))*(cos(f*x + e) + 1)/b) + 5*(b*cos(f*x + e)^2 - b)*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*b*cos(f*x + e)^2 - 4*b)*sqrt(b/cos(f*x + e)))/(f*cos(f*x + e)^2 - f), 1 /16*(10*(b*cos(f*x + e)^2 - b)*sqrt(b)*arctan(1/2*sqrt(b/cos(f*x + e))*(co s(f*x + e) - 1)/sqrt(b)) + 5*(b*cos(f*x + e)^2 - b)*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*b*cos (f*x + e)^2 - 4*b)*sqrt(b/cos(f*x + e)))/(f*cos(f*x + e)^2 - f)]
Timed out. \[ \int \csc ^3(e+f x) (b \sec (e+f x))^{3/2} \, dx=\text {Timed out} \]
Time = 0.30 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09 \[ \int \csc ^3(e+f x) (b \sec (e+f x))^{3/2} \, dx=\frac {{\left (\frac {4 \, b^{2} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{b^{2} - \frac {b^{2}}{\cos \left (f x + e\right )^{2}}} - 10 \, \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right ) + 5 \, \sqrt {b} \log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right ) + 16 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}}\right )} b}{8 \, f} \]
1/8*(4*b^2*sqrt(b/cos(f*x + e))/(b^2 - b^2/cos(f*x + e)^2) - 10*sqrt(b)*ar ctan(sqrt(b/cos(f*x + e))/sqrt(b)) + 5*sqrt(b)*log(-(sqrt(b) - sqrt(b/cos( f*x + e)))/(sqrt(b) + sqrt(b/cos(f*x + e)))) + 16*sqrt(b/cos(f*x + e)))*b/ f
Time = 0.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.12 \[ \int \csc ^3(e+f x) (b \sec (e+f x))^{3/2} \, dx=\frac {b^{6} {\left (\frac {5 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{4}} + \frac {5 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {9}{2}}} + \frac {2 \, {\left (5 \, b^{2} \cos \left (f x + e\right )^{2} - 4 \, b^{2}\right )}}{{\left (\sqrt {b \cos \left (f x + e\right )} b^{2} \cos \left (f x + e\right )^{2} - \sqrt {b \cos \left (f x + e\right )} b^{2}\right )} b^{4}}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{4 \, f} \]
1/4*b^6*(5*arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b^4) + 5*arctan (sqrt(b*cos(f*x + e))/sqrt(b))/b^(9/2) + 2*(5*b^2*cos(f*x + e)^2 - 4*b^2)/ ((sqrt(b*cos(f*x + e))*b^2*cos(f*x + e)^2 - sqrt(b*cos(f*x + e))*b^2)*b^4) )*sgn(cos(f*x + e))/f
Timed out. \[ \int \csc ^3(e+f x) (b \sec (e+f x))^{3/2} \, dx=\int \frac {{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{{\sin \left (e+f\,x\right )}^3} \,d x \]