Integrand size = 26, antiderivative size = 133 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {i}{2 a d (a+i a \tan (c+d x))^{3/2}}-\frac {i}{4 a^2 d \sqrt {a+i a \tan (c+d x)}} \]
1/8*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^(1 /2)-1/4*I/a^2/d/(a+I*a*tan(d*x+c))^(1/2)-1/5*I/d/(a+I*a*tan(d*x+c))^(5/2)+ 1/2*I/a/d/(a+I*a*tan(d*x+c))^(3/2)
Time = 0.97 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {i \sec ^2(c+d x) \left (2 \sqrt {a} (-3+2 \cos (2 (c+d x)))-5 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}\right )}{40 a^{5/2} d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]
((I/40)*Sec[c + d*x]^2*(2*Sqrt[a]*(-3 + 2*Cos[2*(c + d*x)]) - 5*Sqrt[2]*Ar cTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]*(Cos[2*(c + d*x)] + I* Sin[2*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x]]))/(a^(5/2)*d*(-I + Tan[c + d* x])^2*Sqrt[a + I*a*Tan[c + d*x]])
Time = 0.59 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 4023, 3042, 4009, 3042, 3960, 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.
\(\displaystyle \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^2}{(a+i a \tan (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 4023 |
\(\displaystyle \frac {\int \frac {a-2 i a \tan (c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx}{2 a^2}-\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a-2 i a \tan (c+d x)}{(i \tan (c+d x) a+a)^{3/2}}dx}{2 a^2}-\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 4009 |
\(\displaystyle \frac {\frac {i a}{d (a+i a \tan (c+d x))^{3/2}}-\frac {1}{2} \int \frac {1}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}-\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {i a}{d (a+i a \tan (c+d x))^{3/2}}-\frac {1}{2} \int \frac {1}{\sqrt {i \tan (c+d x) a+a}}dx}{2 a^2}-\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {1}{2} \left (-\frac {\int \sqrt {i \tan (c+d x) a+a}dx}{2 a}-\frac {i}{d \sqrt {a+i a \tan (c+d x)}}\right )+\frac {i a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}-\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} \left (-\frac {\int \sqrt {i \tan (c+d x) a+a}dx}{2 a}-\frac {i}{d \sqrt {a+i a \tan (c+d x)}}\right )+\frac {i a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}-\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle \frac {\frac {1}{2} \left (\frac {i \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}-\frac {i}{d \sqrt {a+i a \tan (c+d x)}}\right )+\frac {i a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}-\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {1}{2} \left (\frac {i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}-\frac {i}{d \sqrt {a+i a \tan (c+d x)}}\right )+\frac {i a}{d (a+i a \tan (c+d x))^{3/2}}}{2 a^2}-\frac {i}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
(-1/5*I)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + ((I*a)/(d*(a + I*a*Tan[c + d*x ])^(3/2)) + ((I*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sq rt[2]*Sqrt[a]*d) - I/(d*Sqrt[a + I*a*Tan[c + d*x]]))/2)/(2*a^2)
3.2.29.3.1 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[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[1/(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] && LtQ[n, 0]
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_)]), x_Symbol] :> Simp[(-(b*c - a*d))*((a + b*Tan[e + f*x])^m/(2*a *f*m)), x] + Simp[(b*c + a*d)/(2*a*b) Int[(a + b*Tan[e + f*x])^(m + 1), x ], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2 , 0] && LtQ[m, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(-b)*(a*c + b*d)^2*((a + b*Tan[e + f*x])^ m/(2*a^3*f*m)), x] + Simp[1/(2*a^2) Int[(a + b*Tan[e + f*x])^(m + 1)*Simp [a*c^2 - 2*b*c*d + a*d^2 - 2*b*d^2*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LeQ[m, -1] && EqQ[a^2 + b^2, 0]
Time = 0.88 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {2 i \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {3}{2}}}+\frac {1}{4 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {1}{8 a \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d a}\) | \(94\) |
default | \(\frac {2 i \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {3}{2}}}+\frac {1}{4 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {1}{8 a \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d a}\) | \(94\) |
2*I/d/a*(1/16/a^(3/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2) /a^(1/2))+1/4/(a+I*a*tan(d*x+c))^(3/2)-1/8/a/(a+I*a*tan(d*x+c))^(1/2)-1/10 *a/(a+I*a*tan(d*x+c))^(5/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (94) = 188\).
Time = 0.24 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.13 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (5 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 5 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 2 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{40 \, a^{3} d} \]
1/40*(5*I*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(4*(sqr t(2)*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2* I*c) + 1))*sqrt(1/(a^5*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 5*I* sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt (1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1) )*sqrt(1/(a^5*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sqrt(2)*sqrt( a/(e^(2*I*d*x + 2*I*c) + 1))*(-I*e^(6*I*d*x + 6*I*c) + 2*I*e^(4*I*d*x + 4* I*c) + 2*I*e^(2*I*d*x + 2*I*c) - I))*e^(-5*I*d*x - 5*I*c)/(a^3*d)
\[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\tan ^{2}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.89 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \, {\left (5 \, \sqrt {2} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a - 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} + 4 \, a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\right )}}{80 \, a^{3} d} \]
-1/80*I*(5*sqrt(2)*sqrt(a)*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))) + 4*(5*(I*a*tan(d*x + c) + a)^2*a - 10*(I*a*tan(d*x + c) + a)*a^2 + 4*a^3)/(I*a*tan(d*x + c) + a)^(5/2))/(a^3*d)
\[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Time = 0.24 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.66 \[ \int \frac {\tan ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {1{}\mathrm {i}}{20\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{4\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{5/2}\,d} \]