Integrand size = 23, antiderivative size = 169 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 a \sqrt {a+a \sec (c+d x)}}{d}+\frac {2 (a+a \sec (c+d x))^{3/2}}{3 d}+\frac {2 (a+a \sec (c+d x))^{5/2}}{5 a d}+\frac {2 (a+a \sec (c+d x))^{7/2}}{7 a^2 d}-\frac {2 (a+a \sec (c+d x))^{9/2}}{3 a^3 d}+\frac {2 (a+a \sec (c+d x))^{11/2}}{11 a^4 d} \] Output:
-2*a^(3/2)*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/d+2*a*(a+a*sec(d*x+c))^ (1/2)/d+2/3*(a+a*sec(d*x+c))^(3/2)/d+2/5*(a+a*sec(d*x+c))^(5/2)/a/d+2/7*(a +a*sec(d*x+c))^(7/2)/a^2/d-2/3*(a+a*sec(d*x+c))^(9/2)/a^3/d+2/11*(a+a*sec( d*x+c))^(11/2)/a^4/d
Time = 0.33 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.66 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=\frac {2 (a (1+\sec (c+d x)))^{3/2} \left (-1155 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right )+\sqrt {1+\sec (c+d x)} \left (1656+327 \sec (c+d x)-534 \sec ^2(c+d x)-325 \sec ^3(c+d x)+140 \sec ^4(c+d x)+105 \sec ^5(c+d x)\right )\right )}{1155 d (1+\sec (c+d x))^{3/2}} \] Input:
Integrate[(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x]^5,x]
Output:
(2*(a*(1 + Sec[c + d*x]))^(3/2)*(-1155*ArcTanh[Sqrt[1 + Sec[c + d*x]]] + S qrt[1 + Sec[c + d*x]]*(1656 + 327*Sec[c + d*x] - 534*Sec[c + d*x]^2 - 325* Sec[c + d*x]^3 + 140*Sec[c + d*x]^4 + 105*Sec[c + d*x]^5)))/(1155*d*(1 + S ec[c + d*x])^(3/2))
Time = 0.34 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 25, 4368, 27, 99, 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.
| \(\displaystyle \int \tan ^5(c+d x) (a \sec (c+d x)+a)^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\cot \left (c+d x+\frac {\pi }{2}\right )^5 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^{3/2}dx\) |
| \(\Big \downarrow \) 4368 |
| \(\displaystyle \frac {\int a^2 \cos (c+d x) (1-\sec (c+d x))^2 (\sec (c+d x) a+a)^{7/2}d\sec (c+d x)}{a^4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \cos (c+d x) (1-\sec (c+d x))^2 (\sec (c+d x) a+a)^{7/2}d\sec (c+d x)}{a^2 d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\frac {(\sec (c+d x) a+a)^{9/2}}{a}+\cos (c+d x) (\sec (c+d x) a+a)^{7/2}-3 (\sec (c+d x) a+a)^{7/2}\right )d\sec (c+d x)}{a^2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-2 a^{7/2} \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )+2 a^3 \sqrt {a \sec (c+d x)+a}+\frac {2 (a \sec (c+d x)+a)^{11/2}}{11 a^2}+\frac {2}{3} a^2 (a \sec (c+d x)+a)^{3/2}-\frac {2 (a \sec (c+d x)+a)^{9/2}}{3 a}+\frac {2}{7} (a \sec (c+d x)+a)^{7/2}+\frac {2}{5} a (a \sec (c+d x)+a)^{5/2}}{a^2 d}\) |
Input:
Int[(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x]^5,x]
Output:
(-2*a^(7/2)*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]] + 2*a^3*Sqrt[a + a*S ec[c + d*x]] + (2*a^2*(a + a*Sec[c + d*x])^(3/2))/3 + (2*a*(a + a*Sec[c + d*x])^(5/2))/5 + (2*(a + a*Sec[c + d*x])^(7/2))/7 - (2*(a + a*Sec[c + d*x] )^(9/2))/(3*a) + (2*(a + a*Sec[c + d*x])^(11/2))/(11*a^2))/(a^2*d)
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_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(d*b^(m - 1))^(-1) Subst[Int[(-a + b*x)^((m - 1)/2 )*((a + b*x)^((m - 1)/2 + n)/x), x], x, Csc[c + d*x]], x] /; FreeQ[{a, b, c , d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && !IntegerQ[n]
Time = 1.18 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.69
| method | result | size |
| default | \(\frac {\left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\frac {1104}{385}+\frac {218 \sec \left (d x +c \right )}{385}-\frac {356 \sec \left (d x +c \right )^{2}}{385}-\frac {130 \sec \left (d x +c \right )^{3}}{231}+\frac {8 \sec \left (d x +c \right )^{4}}{33}+\frac {2 \sec \left (d x +c \right )^{5}}{11}\right ) a \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d}\) | \(116\) |
Input:
int((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^5,x,method=_RETURNVERBOSE)
Output:
1/d*(2*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*(-cos(d*x+ c)/(1+cos(d*x+c)))^(1/2)+1104/385+218/385*sec(d*x+c)-356/385*sec(d*x+c)^2- 130/231*sec(d*x+c)^3+8/33*sec(d*x+c)^4+2/11*sec(d*x+c)^5)*a*(a*(1+sec(d*x+ c)))^(1/2)
Time = 0.13 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.98 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=\left [\frac {1155 \, a^{\frac {3}{2}} \cos \left (d x + c\right )^{5} \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) + 4 \, {\left (1656 \, a \cos \left (d x + c\right )^{5} + 327 \, a \cos \left (d x + c\right )^{4} - 534 \, a \cos \left (d x + c\right )^{3} - 325 \, a \cos \left (d x + c\right )^{2} + 140 \, a \cos \left (d x + c\right ) + 105 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{2310 \, d \cos \left (d x + c\right )^{5}}, \frac {1155 \, \sqrt {-a} a \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right )^{5} + 2 \, {\left (1656 \, a \cos \left (d x + c\right )^{5} + 327 \, a \cos \left (d x + c\right )^{4} - 534 \, a \cos \left (d x + c\right )^{3} - 325 \, a \cos \left (d x + c\right )^{2} + 140 \, a \cos \left (d x + c\right ) + 105 \, a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}}{1155 \, d \cos \left (d x + c\right )^{5}}\right ] \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^5,x, algorithm="fricas")
Output:
[1/2310*(1155*a^(3/2)*cos(d*x + c)^5*log(-8*a*cos(d*x + c)^2 + 4*(2*cos(d* x + c)^2 + cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)) - 8*a*cos(d*x + c) - a) + 4*(1656*a*cos(d*x + c)^5 + 327*a*cos(d*x + c)^4 - 534*a*cos(d*x + c)^3 - 325*a*cos(d*x + c)^2 + 140*a*cos(d*x + c) + 105*a) *sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos(d*x + c)^5), 1/1155*(1155 *sqrt(-a)*a*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos( d*x + c)/(2*a*cos(d*x + c) + a))*cos(d*x + c)^5 + 2*(1656*a*cos(d*x + c)^5 + 327*a*cos(d*x + c)^4 - 534*a*cos(d*x + c)^3 - 325*a*cos(d*x + c)^2 + 14 0*a*cos(d*x + c) + 105*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(d*cos( d*x + c)^5)]
\[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}} \tan ^{5}{\left (c + d x \right )}\, dx \] Input:
integrate((a+a*sec(d*x+c))**(3/2)*tan(d*x+c)**5,x)
Output:
Integral((a*(sec(c + d*x) + 1))**(3/2)*tan(c + d*x)**5, x)
Time = 0.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.96 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=\frac {1155 \, a^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 770 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} + \frac {210 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {11}{2}}}{a^{4}} - \frac {770 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {9}{2}}}{a^{3}} + \frac {330 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {7}{2}}}{a^{2}} + \frac {462 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{a} + 2310 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a}{1155 \, d} \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^5,x, algorithm="maxima")
Output:
1/1155*(1155*a^(3/2)*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/ cos(d*x + c)) + sqrt(a))) + 770*(a + a/cos(d*x + c))^(3/2) + 210*(a + a/co s(d*x + c))^(11/2)/a^4 - 770*(a + a/cos(d*x + c))^(9/2)/a^3 + 330*(a + a/c os(d*x + c))^(7/2)/a^2 + 462*(a + a/cos(d*x + c))^(5/2)/a + 2310*sqrt(a + a/cos(d*x + c))*a)/d
Time = 0.96 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.29 \[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=\frac {\sqrt {2} {\left (\frac {1155 \, \sqrt {2} a \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} + \frac {2 \, {\left (1155 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} a - 770 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} a^{2} + 924 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{3} a^{3} - 1320 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} a^{4} - 6160 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a^{5} - 3360 \, a^{6}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{5} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{1155 \, d} \] Input:
integrate((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^5,x, algorithm="giac")
Output:
1/1155*sqrt(2)*(1155*sqrt(2)*a*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/ 2*c)^2 + a)/sqrt(-a))/sqrt(-a) + 2*(1155*(a*tan(1/2*d*x + 1/2*c)^2 - a)^5* a - 770*(a*tan(1/2*d*x + 1/2*c)^2 - a)^4*a^2 + 924*(a*tan(1/2*d*x + 1/2*c) ^2 - a)^3*a^3 - 1320*(a*tan(1/2*d*x + 1/2*c)^2 - a)^2*a^4 - 6160*(a*tan(1/ 2*d*x + 1/2*c)^2 - a)*a^5 - 3360*a^6)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^5*sq rt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))*a*sgn(cos(d*x + c))/d
Timed out. \[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^5\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \] Input:
int(tan(c + d*x)^5*(a + a/cos(c + d*x))^(3/2),x)
Output:
int(tan(c + d*x)^5*(a + a/cos(c + d*x))^(3/2), x)
\[ \int (a+a \sec (c+d x))^{3/2} \tan ^5(c+d x) \, dx=\frac {\sqrt {a}\, a \left (210 \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )^{4}-240 \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}+320 \sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )+280 \sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{4}-496 \sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{2}+2304 \sqrt {\sec \left (d x +c \right )+1}+35 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{5}}{\sec \left (d x +c \right )+1}d x \right ) d -128 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{3}}{\sec \left (d x +c \right )+1}d x \right ) d +992 \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )+1}d x \right ) d \right )}{1155 d} \] Input:
int((a+a*sec(d*x+c))^(3/2)*tan(d*x+c)^5,x)
Output:
(sqrt(a)*a*(210*sqrt(sec(c + d*x) + 1)*sec(c + d*x)*tan(c + d*x)**4 - 240* sqrt(sec(c + d*x) + 1)*sec(c + d*x)*tan(c + d*x)**2 + 320*sqrt(sec(c + d*x ) + 1)*sec(c + d*x) + 280*sqrt(sec(c + d*x) + 1)*tan(c + d*x)**4 - 496*sqr t(sec(c + d*x) + 1)*tan(c + d*x)**2 + 2304*sqrt(sec(c + d*x) + 1) + 35*int ((sqrt(sec(c + d*x) + 1)*tan(c + d*x)**5)/(sec(c + d*x) + 1),x)*d - 128*in t((sqrt(sec(c + d*x) + 1)*tan(c + d*x)**3)/(sec(c + d*x) + 1),x)*d + 992*i nt((sqrt(sec(c + d*x) + 1)*tan(c + d*x))/(sec(c + d*x) + 1),x)*d))/(1155*d )