Integrand size = 21, antiderivative size = 113 \[ \int \csc ^3(e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {7 b^{5/2} \arctan \left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 f}-\frac {7 b^{5/2} \text {arctanh}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 f}+\frac {7 b (b \sec (e+f x))^{3/2}}{6 f}-\frac {\cot ^2(e+f x) (b \sec (e+f x))^{7/2}}{2 b f} \]
7/4*b^(5/2)*arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/f-7/4*b^(5/2)*arctanh((b* sec(f*x+e))^(1/2)/b^(1/2))/f+7/6*b*(b*sec(f*x+e))^(3/2)/f-1/2*cot(f*x+e)^2 *(b*sec(f*x+e))^(7/2)/b/f
Time = 1.46 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.96 \[ \int \csc ^3(e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {b^3 \left (-12 \csc ^2(e+f x)+42 \arctan \left (\sqrt {\sec (e+f x)}\right ) \sqrt {\sec (e+f x)}+21 \left (\log \left (1-\sqrt {\sec (e+f x)}\right )-\log \left (1+\sqrt {\sec (e+f x)}\right )\right ) \sqrt {\sec (e+f x)}+16 \sec ^2(e+f x)\right )}{24 f \sqrt {b \sec (e+f x)}} \]
(b^3*(-12*Csc[e + f*x]^2 + 42*ArcTan[Sqrt[Sec[e + f*x]]]*Sqrt[Sec[e + f*x] ] + 21*(Log[1 - Sqrt[Sec[e + f*x]]] - Log[1 + Sqrt[Sec[e + f*x]]])*Sqrt[Se c[e + f*x]] + 16*Sec[e + f*x]^2))/(24*f*Sqrt[b*Sec[e + f*x]])
Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.99, 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, 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.
| \(\displaystyle \int \csc ^3(e+f x) (b \sec (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc (e+f x)^3 (b \sec (e+f x))^{5/2}dx\) |
| \(\Big \downarrow \) 3102 |
| \(\displaystyle \frac {\int \frac {b^4 (b \sec (e+f x))^{9/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^2}d(b \sec (e+f x))}{b^3 f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \int \frac {(b \sec (e+f x))^{9/2}}{\left (b^2-b^2 \sec ^2(e+f x)\right )^2}d(b \sec (e+f x))}{f}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {b \left (\frac {(b \sec (e+f x))^{7/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {7}{4} \int \frac {(b \sec (e+f x))^{5/2}}{b^2-b^2 \sec ^2(e+f x)}d(b \sec (e+f x))\right )}{f}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {b \left (\frac {(b \sec (e+f x))^{7/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {7}{4} \left (b^2 \int \frac {\sqrt {b \sec (e+f x)}}{b^2-b^2 \sec ^2(e+f x)}d(b \sec (e+f x))-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )\right )}{f}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {b \left (\frac {(b \sec (e+f x))^{7/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {7}{4} \left (2 b^2 \int \frac {b^2 \sec ^2(e+f x)}{b^2-b^4 \sec ^4(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )\right )}{f}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle \frac {b \left (\frac {(b \sec (e+f x))^{7/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {7}{4} \left (2 b^2 \left (\frac {1}{2} \int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {1}{2} \int \frac {1}{b^2 \sec ^2(e+f x)+b}d\sqrt {b \sec (e+f x)}\right )-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )\right )}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {b \left (\frac {(b \sec (e+f x))^{7/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {7}{4} \left (2 b^2 \left (\frac {1}{2} \int \frac {1}{b-b^2 \sec ^2(e+f x)}d\sqrt {b \sec (e+f x)}-\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}\right )-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )\right )}{f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {b \left (\frac {(b \sec (e+f x))^{7/2}}{2 \left (b^2-b^2 \sec ^2(e+f x)\right )}-\frac {7}{4} \left (2 b^2 \left (\frac {\text {arctanh}\left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}-\frac {\arctan \left (\sqrt {b} \sec (e+f x)\right )}{2 \sqrt {b}}\right )-\frac {2}{3} (b \sec (e+f x))^{3/2}\right )\right )}{f}\) |
(b*((b*Sec[e + f*x])^(7/2)/(2*(b^2 - b^2*Sec[e + f*x]^2)) - (7*(2*b^2*(-1/ 2*ArcTan[Sqrt[b]*Sec[e + f*x]]/Sqrt[b] + ArcTanh[Sqrt[b]*Sec[e + f*x]]/(2* Sqrt[b])) - (2*(b*Sec[e + f*x])^(3/2))/3))/4))/f
3.5.2.3.1 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[(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. \(501\) vs. \(2(89)=178\).
Time = 31.80 (sec) , antiderivative size = 502, normalized size of antiderivative = 4.44
| method | result | size |
| default | \(-\frac {\sqrt {b \sec \left (f x +e \right )}\, b^{2} \left (21 \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 )+3 \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 )-24 \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 )+28 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \left (\cos ^{2}\left (f x +e \right )\right )-21 \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 )-3 \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 )+24 \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 )-16 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right ) \sec \left (f x +e \right ) \left (\csc ^{2}\left (f x +e \right )\right )}{24 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\) | \(502\) |
-1/24/f*(b*sec(f*x+e))^(1/2)*b^2*(21*cos(f*x+e)^3*arctan(1/2/(-cos(f*x+e)/ (cos(f*x+e)+1)^2)^(1/2))+3*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))-24*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))+28*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^2-21*cos(f*x+e) ^2*arctan(1/2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2))-3*cos(f*x+e)^2*ln((2*c os(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))+24*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))-16*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2) )/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*sec(f*x+e)*csc(f*x+e)^2
Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (89) = 178\).
Time = 0.34 (sec) , antiderivative size = 448, normalized size of antiderivative = 3.96 \[ \int \csc ^3(e+f x) (b \sec (e+f x))^{5/2} \, dx=\left [\frac {42 \, {\left (b^{2} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) + 21 \, {\left (b^{2} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt {-b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, {\left (7 \, b^{2} \cos \left (f x + e\right )^{2} - 4 \, b^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{48 \, {\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )}}, -\frac {42 \, {\left (b^{2} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {b}}\right ) - 21 \, {\left (b^{2} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt {b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) - 8 \, {\left (7 \, b^{2} \cos \left (f x + e\right )^{2} - 4 \, b^{2}\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{48 \, {\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )}}\right ] \]
[1/48*(42*(b^2*cos(f*x + e)^3 - b^2*cos(f*x + e))*sqrt(-b)*arctan(1/2*sqrt (-b)*sqrt(b/cos(f*x + e))*(cos(f*x + e) + 1)/b) + 21*(b^2*cos(f*x + e)^3 - b^2*cos(f*x + e))*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*(7*b^2*cos(f*x + e)^2 - 4*b^2)*sqrt(b/c os(f*x + e)))/(f*cos(f*x + e)^3 - f*cos(f*x + e)), -1/48*(42*(b^2*cos(f*x + e)^3 - b^2*cos(f*x + e))*sqrt(b)*arctan(1/2*sqrt(b/cos(f*x + e))*(cos(f* x + e) - 1)/sqrt(b)) - 21*(b^2*cos(f*x + e)^3 - b^2*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/c os(f*x + e)) + 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 - 2*cos(f*x + e) + 1) ) - 8*(7*b^2*cos(f*x + e)^2 - 4*b^2)*sqrt(b/cos(f*x + e)))/(f*cos(f*x + e) ^3 - f*cos(f*x + e))]
Timed out. \[ \int \csc ^3(e+f x) (b \sec (e+f x))^{5/2} \, dx=\text {Timed out} \]
Time = 0.30 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09 \[ \int \csc ^3(e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {{\left (\frac {12 \, b^{2} \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}}}{b^{2} - \frac {b^{2}}{\cos \left (f x + e\right )^{2}}} + 42 \, b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right ) + 21 \, b^{\frac {3}{2}} \log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right ) + 16 \, \left (\frac {b}{\cos \left (f x + e\right )}\right )^{\frac {3}{2}}\right )} b}{24 \, f} \]
1/24*(12*b^2*(b/cos(f*x + e))^(3/2)/(b^2 - b^2/cos(f*x + e)^2) + 42*b^(3/2 )*arctan(sqrt(b/cos(f*x + e))/sqrt(b)) + 21*b^(3/2)*log(-(sqrt(b) - sqrt(b /cos(f*x + e)))/(sqrt(b) + sqrt(b/cos(f*x + e)))) + 16*(b/cos(f*x + e))^(3 /2))*b/f
Time = 0.31 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.07 \[ \int \csc ^3(e+f x) (b \sec (e+f x))^{5/2} \, dx=\frac {b^{8} {\left (\frac {6 \, \sqrt {b \cos \left (f x + e\right )}}{{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} b^{4}} + \frac {21 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{5}} - \frac {21 \, \arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {11}{2}}} + \frac {8}{\sqrt {b \cos \left (f x + e\right )} b^{5} \cos \left (f x + e\right )}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{12 \, f} \]
1/12*b^8*(6*sqrt(b*cos(f*x + e))/((b^2*cos(f*x + e)^2 - b^2)*b^4) + 21*arc tan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sqrt(-b)*b^5) - 21*arctan(sqrt(b*cos(f *x + e))/sqrt(b))/b^(11/2) + 8/(sqrt(b*cos(f*x + e))*b^5*cos(f*x + e)))*sg n(cos(f*x + e))/f
Timed out. \[ \int \csc ^3(e+f x) (b \sec (e+f x))^{5/2} \, dx=\int \frac {{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\sin \left (e+f\,x\right )}^3} \,d x \]