Integrand size = 26, antiderivative size = 110 \[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{7/2}} \, dx=-\frac {2 (-1)^{3/4} a \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{7/2} f}-\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac {2 i a}{3 d^2 f (d \tan (e+f x))^{3/2}}+\frac {2 a}{d^3 f \sqrt {d \tan (e+f x)}} \]
-2*(-1)^(3/4)*a*arctanh((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/d^(7/2)/f +2*a/d^3/f/(d*tan(f*x+e))^(1/2)-2/5*a/d/f/(d*tan(f*x+e))^(5/2)+2/3*I*a/d^2 /f/(d*tan(f*x+e))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.14 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.37 \[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{7/2}} \, dx=-\frac {2 a \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-i \tan (e+f x)\right )}{5 d f (d \tan (e+f x))^{5/2}} \]
Time = 0.59 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {3042, 4012, 25, 3042, 4012, 3042, 4012, 25, 3042, 4016, 221}
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.
\(\displaystyle \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{7/2}}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle -\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}+\frac {\int -\frac {i a d+a \tan (e+f x) d}{(d \tan (e+f x))^{5/2}}dx}{d^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac {\int \frac {i a d+a \tan (e+f x) d}{(d \tan (e+f x))^{5/2}}dx}{d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac {\int \frac {i a d+a \tan (e+f x) d}{(d \tan (e+f x))^{5/2}}dx}{d^2}\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle -\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac {\frac {\int \frac {a d^2-i a d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}}dx}{d^2}-\frac {2 i a}{3 f (d \tan (e+f x))^{3/2}}}{d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac {\frac {\int \frac {a d^2-i a d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}}dx}{d^2}-\frac {2 i a}{3 f (d \tan (e+f x))^{3/2}}}{d^2}\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle -\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac {\frac {-\frac {2 a d}{f \sqrt {d \tan (e+f x)}}+\frac {\int -\frac {i a d^3+a \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)}}dx}{d^2}}{d^2}-\frac {2 i a}{3 f (d \tan (e+f x))^{3/2}}}{d^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac {\frac {-\frac {2 a d}{f \sqrt {d \tan (e+f x)}}-\frac {\int \frac {i a d^3+a \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)}}dx}{d^2}}{d^2}-\frac {2 i a}{3 f (d \tan (e+f x))^{3/2}}}{d^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac {\frac {-\frac {2 a d}{f \sqrt {d \tan (e+f x)}}-\frac {\int \frac {i a d^3+a \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)}}dx}{d^2}}{d^2}-\frac {2 i a}{3 f (d \tan (e+f x))^{3/2}}}{d^2}\) |
\(\Big \downarrow \) 4016 |
\(\displaystyle -\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac {\frac {-\frac {2 a d}{f \sqrt {d \tan (e+f x)}}+\frac {2 a^2 d^4 \int \frac {1}{i a d^4-a d^4 \tan (e+f x)}d\sqrt {d \tan (e+f x)}}{f}}{d^2}-\frac {2 i a}{3 f (d \tan (e+f x))^{3/2}}}{d^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 a}{5 d f (d \tan (e+f x))^{5/2}}-\frac {\frac {\frac {2 (-1)^{3/4} a \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{f}-\frac {2 a d}{f \sqrt {d \tan (e+f x)}}}{d^2}-\frac {2 i a}{3 f (d \tan (e+f x))^{3/2}}}{d^2}\) |
(-2*a)/(5*d*f*(d*Tan[e + f*x])^(5/2)) - ((((-2*I)/3)*a)/(f*(d*Tan[e + f*x] )^(3/2)) + ((2*(-1)^(3/4)*a*Sqrt[d]*ArcTanh[((-1)^(3/4)*Sqrt[d*Tan[e + f*x ]])/Sqrt[d]])/f - (2*a*d)/(f*Sqrt[d*Tan[e + f*x]]))/d^2)/d^2
3.2.48.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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 + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2*(c^2/f) Subst[Int[1/(b*c - d*x^2), x], x, Sqrt[b *Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (89 ) = 178\).
Time = 0.81 (sec) , antiderivative size = 326, normalized size of antiderivative = 2.96
method | result | size |
derivativedivides | \(-\frac {a \left (\frac {-\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{3}}-\frac {2}{d^{3} \sqrt {d \tan \left (f x +e \right )}}-\frac {2 i}{3 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {2}{5 d \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f}\) | \(326\) |
default | \(-\frac {a \left (\frac {-\frac {i \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d^{3}}-\frac {2}{d^{3} \sqrt {d \tan \left (f x +e \right )}}-\frac {2 i}{3 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {2}{5 d \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\right )}{f}\) | \(326\) |
parts | \(\frac {2 a d \left (-\frac {1}{5 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}+\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d^{4} \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}-\frac {i a \left (-\frac {2}{3 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d^{4}}\right )}{f}\) | \(330\) |
-1/f*a*(2/d^3*(-1/8*I/d*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)+(d^2)^(1/4)* (d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)-(d^2)^(1/4)*(d*tan (f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f *x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))-1/8 /(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^ (1/2)+(d^2)^(1/2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2)+ (d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arcta n(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)))-2/d^3/(d*tan(f*x+e))^(1/2 )-2/3*I/d^2/(d*tan(f*x+e))^(3/2)+2/5/d/(d*tan(f*x+e))^(5/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 437 vs. \(2 (88) = 176\).
Time = 0.25 (sec) , antiderivative size = 437, normalized size of antiderivative = 3.97 \[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{7/2}} \, dx=-\frac {15 \, {\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )} \sqrt {-\frac {4 i \, a^{2}}{d^{7} f^{2}}} \log \left (-\frac {{\left ({\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 i \, a^{2}}{d^{7} f^{2}}} + 2 \, a\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3} f}\right ) - 15 \, {\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )} \sqrt {-\frac {4 i \, a^{2}}{d^{7} f^{2}}} \log \left (\frac {{\left ({\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {4 i \, a^{2}}{d^{7} f^{2}}} - 2 \, a\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3} f}\right ) + 8 \, {\left (-13 i \, a e^{\left (6 i \, f x + 6 i \, e\right )} + 11 i \, a e^{\left (4 i \, f x + 4 i \, e\right )} + i \, a e^{\left (2 i \, f x + 2 i \, e\right )} - 23 i \, a\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, {\left (d^{4} f e^{\left (6 i \, f x + 6 i \, e\right )} - 3 \, d^{4} f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, d^{4} f e^{\left (2 i \, f x + 2 i \, e\right )} - d^{4} f\right )}} \]
-1/60*(15*(d^4*f*e^(6*I*f*x + 6*I*e) - 3*d^4*f*e^(4*I*f*x + 4*I*e) + 3*d^4 *f*e^(2*I*f*x + 2*I*e) - d^4*f)*sqrt(-4*I*a^2/(d^7*f^2))*log(-((d^3*f*e^(2 *I*f*x + 2*I*e) + d^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-4*I*a^2/(d^7*f^2)) + 2*a)*e^(-2*I*f*x - 2*I*e)/(d^3* f)) - 15*(d^4*f*e^(6*I*f*x + 6*I*e) - 3*d^4*f*e^(4*I*f*x + 4*I*e) + 3*d^4* f*e^(2*I*f*x + 2*I*e) - d^4*f)*sqrt(-4*I*a^2/(d^7*f^2))*log(((d^3*f*e^(2*I *f*x + 2*I*e) + d^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-4*I*a^2/(d^7*f^2)) - 2*a)*e^(-2*I*f*x - 2*I*e)/(d^3*f) ) + 8*(-13*I*a*e^(6*I*f*x + 6*I*e) + 11*I*a*e^(4*I*f*x + 4*I*e) + I*a*e^(2 *I*f*x + 2*I*e) - 23*I*a)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f* x + 2*I*e) + 1)))/(d^4*f*e^(6*I*f*x + 6*I*e) - 3*d^4*f*e^(4*I*f*x + 4*I*e) + 3*d^4*f*e^(2*I*f*x + 2*I*e) - d^4*f)
\[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{7/2}} \, dx=- i a \left (\int \frac {i}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx + \int \frac {\tan {\left (e + f x \right )}}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}\, dx\right ) \]
-I*a*(Integral(I/(d*tan(e + f*x))**(7/2), x) + Integral(tan(e + f*x)/(d*ta n(e + f*x))**(7/2), x))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (88) = 176\).
Time = 0.29 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.92 \[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{7/2}} \, dx=-\frac {\frac {15 \, a {\left (-\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d^{2}} - \frac {8 \, {\left (15 \, a d^{2} \tan \left (f x + e\right )^{2} + 5 i \, a d^{2} \tan \left (f x + e\right ) - 3 \, a d^{2}\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {5}{2}} d^{2}}}{60 \, d f} \]
-1/60*(15*a*(-(2*I + 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sq rt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - (2*I + 2)*sqrt(2)*arctan(-1/2*sqrt( 2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - (I - 1)*s qrt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt (d) + (I - 1)*sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sq rt(d) + d)/sqrt(d))/d^2 - 8*(15*a*d^2*tan(f*x + e)^2 + 5*I*a*d^2*tan(f*x + e) - 3*a*d^2)/((d*tan(f*x + e))^(5/2)*d^2))/(d*f)
Time = 0.81 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.13 \[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{7/2}} \, dx=\frac {2}{15} \, a {\left (\frac {15 \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{d^{\frac {7}{2}} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {15 \, d^{2} \tan \left (f x + e\right )^{2} + 5 i \, d^{2} \tan \left (f x + e\right ) - 3 \, d^{2}}{\sqrt {d \tan \left (f x + e\right )} d^{5} f \tan \left (f x + e\right )^{2}}\right )} \]
2/15*a*(15*sqrt(2)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-4*I*sqrt(2)*d ^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(d^(7/2)*f*(-I*d/sqrt(d^2) + 1)) + (15*d^2*tan(f*x + e)^2 + 5*I*d^2*tan(f*x + e) - 3*d^2)/(sqrt(d*tan(f*x + e ))*d^5*f*tan(f*x + e)^2))
Time = 5.73 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.79 \[ \int \frac {a-i a \tan (e+f x)}{(d \tan (e+f x))^{7/2}} \, dx=\frac {2\,{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {d}}\right )}{d^{7/2}\,f}-\frac {\frac {2\,a}{5\,d}-\frac {2\,a\,{\mathrm {tan}\left (e+f\,x\right )}^2}{d}}{f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}+\frac {a\,2{}\mathrm {i}}{3\,d^2\,f\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \]