Integrand size = 23, antiderivative size = 169 \[ \int \sqrt {a+b \sec (c+d x)} \tan ^5(c+d x) \, dx=-\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 \sqrt {a+b \sec (c+d x)}}{d}-\frac {2 a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{3/2}}{3 b^4 d}+\frac {2 \left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{5/2}}{5 b^4 d}-\frac {6 a (a+b \sec (c+d x))^{7/2}}{7 b^4 d}+\frac {2 (a+b \sec (c+d x))^{9/2}}{9 b^4 d} \]
-2/3*a*(a^2-2*b^2)*(a+b*sec(d*x+c))^(3/2)/b^4/d+2/5*(3*a^2-2*b^2)*(a+b*sec (d*x+c))^(5/2)/b^4/d-6/7*a*(a+b*sec(d*x+c))^(7/2)/b^4/d+2/9*(a+b*sec(d*x+c ))^(9/2)/b^4/d-2*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))*a^(1/2)/d+2*(a+b* sec(d*x+c))^(1/2)/d
Time = 1.91 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.90 \[ \int \sqrt {a+b \sec (c+d x)} \tan ^5(c+d x) \, dx=\frac {-2 \sqrt {a} b^4 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )+2 b^4 \sqrt {a+b \sec (c+d x)}-\frac {2}{3} a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{3/2}+\frac {2}{5} \left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{5/2}-\frac {6}{7} a (a+b \sec (c+d x))^{7/2}+\frac {2}{9} (a+b \sec (c+d x))^{9/2}}{b^4 d} \]
(-2*Sqrt[a]*b^4*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]] + 2*b^4*Sqrt[a + b*Sec[c + d*x]] - (2*a*(a^2 - 2*b^2)*(a + b*Sec[c + d*x])^(3/2))/3 + (2*( 3*a^2 - 2*b^2)*(a + b*Sec[c + d*x])^(5/2))/5 - (6*a*(a + b*Sec[c + d*x])^( 7/2))/7 + (2*(a + b*Sec[c + d*x])^(9/2))/9)/(b^4*d)
Time = 0.36 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.78, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 25, 4373, 517, 25, 1584, 2009}
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.
| \(\displaystyle \int \tan ^5(c+d x) \sqrt {a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^5 \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \sqrt {a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx\) |
| \(\Big \downarrow \) 4373 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) \sqrt {a+b \sec (c+d x)} \left (b^2-b^2 \sec ^2(c+d x)\right )^2}{b}d(b \sec (c+d x))}{b^4 d}\) |
| \(\Big \downarrow \) 517 |
| \(\displaystyle \frac {2 \int -\frac {b^2 \sec ^2(c+d x) \left (b^4 \sec ^4(c+d x)-2 a b^2 \sec ^2(c+d x)+a^2-b^2\right )^2}{a-b^2 \sec ^2(c+d x)}d\sqrt {a+b \sec (c+d x)}}{b^4 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {2 \int \frac {b^2 \sec ^2(c+d x) \left (b^4 \sec ^4(c+d x)-2 a b^2 \sec ^2(c+d x)+a^2-b^2\right )^2}{a-b^2 \sec ^2(c+d x)}d\sqrt {a+b \sec (c+d x)}}{b^4 d}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle -\frac {2 \int \left (-b^8 \sec ^8(c+d x)+3 a b^6 \sec ^6(c+d x)-b^4 \left (3 a^2-2 b^2\right ) \sec ^4(c+d x)+a b^2 \left (a^2-2 b^2\right ) \sec ^2(c+d x)-b^4+\frac {a b^4}{a-b^2 \sec ^2(c+d x)}\right )d\sqrt {a+b \sec (c+d x)}}{b^4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \left (\frac {1}{5} b^5 \left (3 a^2-2 b^2\right ) \sec ^5(c+d x)-\frac {1}{3} a b^3 \left (a^2-2 b^2\right ) \sec ^3(c+d x)-\sqrt {a} b^4 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )-\frac {3}{7} a b^7 \sec ^7(c+d x)+\frac {1}{9} b^9 \sec ^9(c+d x)+b^5 \sec (c+d x)\right )}{b^4 d}\) |
(2*(-(Sqrt[a]*b^4*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]]) + b^5*Sec[c + d*x] - (a*b^3*(a^2 - 2*b^2)*Sec[c + d*x]^3)/3 + (b^5*(3*a^2 - 2*b^2)*Sec[ c + d*x]^5)/5 - (3*a*b^7*Sec[c + d*x]^7)/7 + (b^9*Sec[c + d*x]^9)/9))/(b^4 *d)
3.4.18.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1)) Subst[Int[x^(2*n + 1)*(-c + x^ 2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1)) Subst[Int[(b^2 - x^ 2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(882\) vs. \(2(145)=290\).
Time = 16.53 (sec) , antiderivative size = 883, normalized size of antiderivative = 5.22
-1/315/d/b^4*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)+1)/((b+a*cos(d*x+c))*cos(d *x+c)/(cos(d*x+c)+1)^2)^(1/2)*(315*cos(d*x+c)*ln(4*cos(d*x+c)*((b+a*cos(d* x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2)* ((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*a^(1/2)*b^4+32*( (b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^4*cos(d*x+c)-168*((b +a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^2*b^2*cos(d*x+c)-630*c os(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*b^4+32*((b+ a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^4-16*((b+a*cos(d*x+c))* cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^3*b-168*((b+a*cos(d*x+c))*cos(d*x+c)/ (cos(d*x+c)+1)^2)^(1/2)*a^2*b^2+84*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c )+1)^2)^(1/2)*a*b^3-630*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/ 2)*b^4-16*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^3*b*sec(d *x+c)+12*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^2*b^2*sec( d*x+c)+84*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a*b^3*sec(d *x+c)+252*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*b^4*sec(d*x +c)+12*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^2*b^2*sec(d* x+c)^2-10*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a*b^3*sec(d *x+c)^2+252*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*b^4*sec(d *x+c)^2-10*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a*b^3*sec( d*x+c)^3-70*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*b^4*se...
Time = 0.44 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.51 \[ \int \sqrt {a+b \sec (c+d x)} \tan ^5(c+d x) \, dx=\left [\frac {315 \, \sqrt {a} b^{4} \cos \left (d x + c\right )^{4} \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} + 4 \, {\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right ) + 4 \, {\left (5 \, a b^{3} \cos \left (d x + c\right ) - {\left (16 \, a^{4} - 84 \, a^{2} b^{2} - 315 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 35 \, b^{4} + 2 \, {\left (4 \, a^{3} b - 21 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (a^{2} b^{2} + 21 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{630 \, b^{4} d \cos \left (d x + c\right )^{4}}, \frac {315 \, \sqrt {-a} b^{4} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) \cos \left (d x + c\right )^{4} + 2 \, {\left (5 \, a b^{3} \cos \left (d x + c\right ) - {\left (16 \, a^{4} - 84 \, a^{2} b^{2} - 315 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 35 \, b^{4} + 2 \, {\left (4 \, a^{3} b - 21 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (a^{2} b^{2} + 21 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{315 \, b^{4} d \cos \left (d x + c\right )^{4}}\right ] \]
[1/630*(315*sqrt(a)*b^4*cos(d*x + c)^4*log(-8*a^2*cos(d*x + c)^2 - 8*a*b*c os(d*x + c) - b^2 + 4*(2*a*cos(d*x + c)^2 + b*cos(d*x + c))*sqrt(a)*sqrt(( a*cos(d*x + c) + b)/cos(d*x + c))) + 4*(5*a*b^3*cos(d*x + c) - (16*a^4 - 8 4*a^2*b^2 - 315*b^4)*cos(d*x + c)^4 + 35*b^4 + 2*(4*a^3*b - 21*a*b^3)*cos( d*x + c)^3 - 6*(a^2*b^2 + 21*b^4)*cos(d*x + c)^2)*sqrt((a*cos(d*x + c) + b )/cos(d*x + c)))/(b^4*d*cos(d*x + c)^4), 1/315*(315*sqrt(-a)*b^4*arctan(2* sqrt(-a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b))*cos(d*x + c)^4 + 2*(5*a*b^3*cos(d*x + c) - (16*a^4 - 84*a^2*b^ 2 - 315*b^4)*cos(d*x + c)^4 + 35*b^4 + 2*(4*a^3*b - 21*a*b^3)*cos(d*x + c) ^3 - 6*(a^2*b^2 + 21*b^4)*cos(d*x + c)^2)*sqrt((a*cos(d*x + c) + b)/cos(d* x + c)))/(b^4*d*cos(d*x + c)^4)]
\[ \int \sqrt {a+b \sec (c+d x)} \tan ^5(c+d x) \, dx=\int \sqrt {a + b \sec {\left (c + d x \right )}} \tan ^{5}{\left (c + d x \right )}\, dx \]
Time = 0.28 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.13 \[ \int \sqrt {a+b \sec (c+d x)} \tan ^5(c+d x) \, dx=\frac {315 \, \sqrt {a} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 630 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \frac {70 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {9}{2}}}{b^{4}} - \frac {270 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {7}{2}} a}{b^{4}} + \frac {378 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}} a^{2}}{b^{4}} - \frac {210 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a^{3}}{b^{4}} - \frac {252 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{b^{2}} + \frac {420 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a}{b^{2}}}{315 \, d} \]
1/315*(315*sqrt(a)*log((sqrt(a + b/cos(d*x + c)) - sqrt(a))/(sqrt(a + b/co s(d*x + c)) + sqrt(a))) + 630*sqrt(a + b/cos(d*x + c)) + 70*(a + b/cos(d*x + c))^(9/2)/b^4 - 270*(a + b/cos(d*x + c))^(7/2)*a/b^4 + 378*(a + b/cos(d *x + c))^(5/2)*a^2/b^4 - 210*(a + b/cos(d*x + c))^(3/2)*a^3/b^4 - 252*(a + b/cos(d*x + c))^(5/2)/b^2 + 420*(a + b/cos(d*x + c))^(3/2)*a/b^2)/d
Leaf count of result is larger than twice the leaf count of optimal. 966 vs. \(2 (145) = 290\).
Time = 1.76 (sec) , antiderivative size = 966, normalized size of antiderivative = 5.72 \[ \int \sqrt {a+b \sec (c+d x)} \tan ^5(c+d x) \, dx=\text {Too large to display} \]
2/315*(315*a*arctan(-1/2*(sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan( 1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b) + sqrt(a - b))/sqrt(-a))/sqrt(-a) - 2*(315*(sqrt(a - b)*tan(1/2* d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b))^8*a - 3150*(sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2 *a*tan(1/2*d*x + 1/2*c)^2 + a + b))^7*sqrt(a - b)*a + 210*(sqrt(a - b)*tan (1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2* c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b))^6*(39*a^2 - 5*a*b - 32*b^2) - 630*(sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b))^5*(9*a^2 + 15*a*b - 16*b^2)*sqrt(a - b) - 252*(25*a^3 - 37*a^2*b + 80*a*b^2 - 72*b^3 )*(sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b* tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b))^4 + 945*a^5 + 3864*a^4*b + 2562*a^3*b^2 + 2448*a^2*b^3 - 1083*a*b^4 + 224*b^5 + 42*(25 5*a^3 + 2*a^2*b + 655*a*b^2 - 288*b^3)*(sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2* d*x + 1/2*c)^2 + a + b))^3*sqrt(a - b) - 18*(175*a^4 - 483*a^3*b + 1113*a^ 2*b^2 - 773*a*b^3 + 448*b^4)*(sqrt(a - b)*tan(1/2*d*x + 1/2*c)^2 - sqrt(a* tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1...
Timed out. \[ \int \sqrt {a+b \sec (c+d x)} \tan ^5(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^5\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \]