Integrand size = 23, antiderivative size = 78 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}-\frac {6 \sqrt {a+a \sec (c+d x)}}{a^3 d}+\frac {2 (a+a \sec (c+d x))^{3/2}}{3 a^4 d} \] Output:
-2*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/a^(5/2)/d-6*(a+a*sec(d*x+c))^(1 /2)/a^3/d+2/3*(a+a*sec(d*x+c))^(3/2)/a^4/d
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.88 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {2 \left (-8-7 \sec (c+d x)+\sec ^2(c+d x)-3 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right ) \sqrt {1+\sec (c+d x)}\right )}{3 a^2 d \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[Tan[c + d*x]^5/(a + a*Sec[c + d*x])^(5/2),x]
Output:
(2*(-8 - 7*Sec[c + d*x] + Sec[c + d*x]^2 - 3*ArcTanh[Sqrt[1 + Sec[c + d*x] ]]*Sqrt[1 + Sec[c + d*x]]))/(3*a^2*d*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.97, 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.
\(\displaystyle \int \frac {\tan ^5(c+d x)}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\cot \left (c+d x+\frac {\pi }{2}\right )^5}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^5}{\left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4368 |
\(\displaystyle \frac {\int \frac {a^2 \cos (c+d x) (1-\sec (c+d x))^2}{\sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{a^4 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\cos (c+d x) (1-\sec (c+d x))^2}{\sqrt {\sec (c+d x) a+a}}d\sec (c+d x)}{a^2 d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle \frac {\int \left (\frac {\cos (c+d x)}{\sqrt {\sec (c+d x) a+a}}+\frac {\sqrt {\sec (c+d x) a+a}}{a}-\frac {3}{\sqrt {\sec (c+d x) a+a}}\right )d\sec (c+d x)}{a^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2 (a \sec (c+d x)+a)^{3/2}}{3 a^2}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {6 \sqrt {a \sec (c+d x)+a}}{a}}{a^2 d}\) |
Input:
Int[Tan[c + d*x]^5/(a + a*Sec[c + d*x])^(5/2),x]
Output:
((-2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/Sqrt[a] - (6*Sqrt[a + a*Se c[c + d*x]])/a + (2*(a + a*Sec[c + d*x])^(3/2))/(3*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 = 1.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {\left (2 \arctan \left (\frac {\sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\frac {16}{3}+\frac {2 \sec \left (d x +c \right )}{3}\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}}{d \,a^{3}}\) | \(78\) |
Input:
int(tan(d*x+c)^5/(a+a*sec(d*x+c))^(5/2),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)-16/3+2/3*sec(d*x+c))/a^3*(a*(1+sec(d*x+c)))^(1/2)
Time = 0.12 (sec) , antiderivative size = 241, normalized size of antiderivative = 3.09 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [\frac {3 \, \sqrt {a} \cos \left (d x + c\right ) \log \left (-8 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) - 4 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (8 \, \cos \left (d x + c\right ) - 1\right )}}{6 \, a^{3} d \cos \left (d x + c\right )}, \frac {3 \, \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) \cos \left (d x + c\right ) - 2 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (8 \, \cos \left (d x + c\right ) - 1\right )}}{3 \, a^{3} d \cos \left (d x + c\right )}\right ] \] Input:
integrate(tan(d*x+c)^5/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
Output:
[1/6*(3*sqrt(a)*cos(d*x + c)*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*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(8*cos(d*x + c) - 1))/(a^3*d*cos(d*x + c)), 1/3*(3*sqrt(-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) - 2*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(8*cos(d*x + c) - 1))/(a^ 3*d*cos(d*x + c))]
\[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\tan ^{5}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(tan(d*x+c)**5/(a+a*sec(d*x+c))**(5/2),x)
Output:
Integral(tan(c + d*x)**5/(a*(sec(c + d*x) + 1))**(5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (66) = 132\).
Time = 0.12 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.09 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\frac {3 \, \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} + \frac {2 \, {\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}}}{a^{4}} - \frac {18 \, \sqrt {a + \frac {a}{\cos \left (d x + c\right )}}}{a^{3}} + \frac {2 \, {\left (4 \, a + \frac {3 \, a}{\cos \left (d x + c\right )}\right )}}{{\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a^{2}} - \frac {6}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a^{2}} - \frac {2}{{\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a}}{3 \, d} \] Input:
integrate(tan(d*x+c)^5/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
Output:
1/3*(3*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos(d*x + c)) + sqrt(a)))/a^(5/2) + 2*(a + a/cos(d*x + c))^(3/2)/a^4 - 18*sqrt(a + a/cos (d*x + c))/a^3 + 2*(4*a + 3*a/cos(d*x + c))/((a + a/cos(d*x + c))^(3/2)*a^ 2) - 6/(sqrt(a + a/cos(d*x + c))*a^2) - 2/((a + a/cos(d*x + c))^(3/2)*a))/ d
Time = 1.21 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.62 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {2 \, {\left (\frac {3 \, \arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\sqrt {2} {\left (9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 7 \, a\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )}}{3 \, d} \] Input:
integrate(tan(d*x+c)^5/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
Output:
2/3*(3*arctan(1/2*sqrt(2)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)/sqrt(-a))/(s qrt(-a)*a^2*sgn(cos(d*x + c))) - sqrt(2)*(9*a*tan(1/2*d*x + 1/2*c)^2 - 7*a )/((a*tan(1/2*d*x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*a^2* sgn(cos(d*x + c))))/d
Timed out. \[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^5}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:
int(tan(c + d*x)^5/(a + a/cos(c + d*x))^(5/2),x)
Output:
int(tan(c + d*x)^5/(a + a/cos(c + d*x))^(5/2), x)
\[ \int \frac {\tan ^5(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \tan \left (d x +c \right )^{5}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:
int(tan(d*x+c)^5/(a+a*sec(d*x+c))^(5/2),x)
Output:
(sqrt(a)*int((sqrt(sec(c + d*x) + 1)*tan(c + d*x)**5)/(sec(c + d*x)**3 + 3 *sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x))/a**3