Integrand size = 26, antiderivative size = 130 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {4 i a^2 \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d} \]
4*I*a^(5/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/ d-4*I*a^2*(a+I*a*tan(d*x+c))^(1/2)/d-2/3*I*a*(a+I*a*tan(d*x+c))^(3/2)/d-2/ 7*I*(a+I*a*tan(d*x+c))^(7/2)/a/d
Time = 1.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.82 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {84 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+2 a^2 \sqrt {a+i a \tan (c+d x)} \left (-52 i+16 \tan (c+d x)+9 i \tan ^2(c+d x)-3 \tan ^3(c+d x)\right )}{21 d} \]
((84*I)*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a ])] + 2*a^2*Sqrt[a + I*a*Tan[c + d*x]]*(-52*I + 16*Tan[c + d*x] + (9*I)*Ta n[c + d*x]^2 - 3*Tan[c + d*x]^3))/(21*d)
Time = 0.51 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3042, 4026, 25, 3042, 3959, 3042, 3959, 3042, 3961, 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.
\(\displaystyle \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^2 (a+i a \tan (c+d x))^{5/2}dx\) |
\(\Big \downarrow \) 4026 |
\(\displaystyle \int -(i \tan (c+d x) a+a)^{5/2}dx-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int (i \tan (c+d x) a+a)^{5/2}dx-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\int (i \tan (c+d x) a+a)^{5/2}dx-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle -2 a \int (i \tan (c+d x) a+a)^{3/2}dx-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -2 a \int (i \tan (c+d x) a+a)^{3/2}dx-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle -2 a \left (2 a \int \sqrt {i \tan (c+d x) a+a}dx+\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -2 a \left (2 a \int \sqrt {i \tan (c+d x) a+a}dx+\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}\right )-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle -2 a \left (\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}-\frac {4 i a^2 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}\right )-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -2 a \left (\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}\right )-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}-\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\) |
(((-2*I)/3)*a*(a + I*a*Tan[c + d*x])^(3/2))/d - (((2*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a*d) - 2*a*(((-2*I)*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*T an[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + ((2*I)*a*Sqrt[a + I*a*Tan[c + d*x]])/ d)
3.1.100.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[2*a Int[(a + b*Tan[c + d* x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && GtQ[n , 1]
Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a , b, c, d}, x] && EqQ[a^2 + b^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[d^2*((a + b*Tan[e + f*x])^(m + 1)/(b*f*( m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e + f* x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && !LeQ [m, -1] && !(EqQ[m, 2] && EqQ[a, 0])
Time = 0.90 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.74
method | result | size |
derivativedivides | \(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}+2 a^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d a}\) | \(96\) |
default | \(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a^{3} \sqrt {a +i a \tan \left (d x +c \right )}+2 a^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d a}\) | \(96\) |
2*I/d/a*(-1/7*(a+I*a*tan(d*x+c))^(7/2)-1/3*a^2*(a+I*a*tan(d*x+c))^(3/2)-2* a^3*(a+I*a*tan(d*x+c))^(1/2)+2*a^(7/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+ c))^(1/2)*2^(1/2)/a^(1/2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 371 vs. \(2 (95) = 190\).
Time = 0.25 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.85 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {2 \, {\left (21 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{5}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 21 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{5}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) + 2 \, \sqrt {2} {\left (40 i \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} + 77 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 70 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 21 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )}}{21 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
-2/21*(21*sqrt(2)*sqrt(-a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(4*(a^3*e^(I*d*x + I*c) + sqrt(- a^5/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) )*e^(-I*d*x - I*c)/a^2) - 21*sqrt(2)*sqrt(-a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(4*(a^3*e^(I* d*x + I*c) + sqrt(-a^5/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e^(2* I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) + 2*sqrt(2)*(40*I*a^2*e^(7*I*d *x + 7*I*c) + 77*I*a^2*e^(5*I*d*x + 5*I*c) + 70*I*a^2*e^(3*I*d*x + 3*I*c) + 21*I*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(6*I*d *x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \tan ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.27 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.92 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {2 i \, {\left (21 \, \sqrt {2} a^{\frac {11}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} + 7 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{4} + 42 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{5}\right )}}{21 \, a^{3} d} \]
-2/21*I*(21*sqrt(2)*a^(11/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) + 3*(I*a*tan(d*x + c) + a)^(7/2)*a^2 + 7*(I*a*tan(d*x + c) + a)^(3/2)*a^4 + 42*sqrt(I*a*tan(d *x + c) + a)*a^5)/(a^3*d)
Timed out. \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]
Time = 5.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.82 \[ \int \tan ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{d}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}\,2{}\mathrm {i}}{7\,a\,d}-\frac {a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}+\frac {\sqrt {2}\,{\left (-a\right )}^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,4{}\mathrm {i}}{d} \]