Integrand size = 28, antiderivative size = 326 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\left (\frac {25}{16}-\frac {21 i}{16}\right ) d^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {25}{16}-\frac {21 i}{16}\right ) d^{7/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {25}{32}+\frac {21 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {25}{32}+\frac {21 i}{32}\right ) d^{7/2} \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a^2 f}-\frac {25 d^3 \sqrt {d \tan (e+f x)}}{8 a^2 f}+\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2} \]
(-25/32+21/32*I)*d^(7/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a^ 2/f*2^(1/2)+(25/32-21/32*I)*d^(7/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/ d^(1/2))/a^2/f*2^(1/2)-(25/64+21/64*I)*d^(7/2)*ln(d^(1/2)-2^(1/2)*(d*tan(f *x+e))^(1/2)+d^(1/2)*tan(f*x+e))/a^2/f*2^(1/2)+(25/64+21/64*I)*d^(7/2)*ln( d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/a^2/f*2^(1/2)-25/ 8*d^3*(d*tan(f*x+e))^(1/2)/a^2/f+7/8*I*d^2*(d*tan(f*x+e))^(3/2)/a^2/f/(1+I *tan(f*x+e))-1/4*d*(d*tan(f*x+e))^(5/2)/f/(a+I*a*tan(f*x+e))^2
Time = 1.21 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.57 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {d^3 \sec ^2(e+f x) \left (4 \sqrt [4]{-1} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+46 \sqrt [4]{-1} \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+(9+41 \cos (2 (e+f x))+43 i \sin (2 (e+f x))) \sqrt {d \tan (e+f x)}\right )}{16 a^2 f (-i+\tan (e+f x))^2} \]
(d^3*Sec[e + f*x]^2*(4*(-1)^(1/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]) + 46*(-1)^(1/4)* Sqrt[d]*ArcTanh[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]*(Cos[2*(e + f*x )] + I*Sin[2*(e + f*x)]) + (9 + 41*Cos[2*(e + f*x)] + (43*I)*Sin[2*(e + f* x)])*Sqrt[d*Tan[e + f*x]]))/(16*a^2*f*(-I + Tan[e + f*x])^2)
Time = 1.01 (sec) , antiderivative size = 314, normalized size of antiderivative = 0.96, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.679, Rules used = {3042, 4041, 27, 3042, 4078, 3042, 4011, 3042, 4017, 25, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2}dx\) |
| \(\Big \downarrow \) 4041 |
| \(\displaystyle -\frac {\int -\frac {(d \tan (e+f x))^{3/2} \left (5 a d^2-9 i a d^2 \tan (e+f x)\right )}{2 (i \tan (e+f x) a+a)}dx}{4 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{3/2} \left (5 a d^2-9 i a d^2 \tan (e+f x)\right )}{i \tan (e+f x) a+a}dx}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(d \tan (e+f x))^{3/2} \left (5 a d^2-9 i a d^2 \tan (e+f x)\right )}{i \tan (e+f x) a+a}dx}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\int \sqrt {d \tan (e+f x)} \left (21 i a^2 d^3+25 a^2 \tan (e+f x) d^3\right )dx}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\int \sqrt {d \tan (e+f x)} \left (21 i a^2 d^3+25 a^2 \tan (e+f x) d^3\right )dx}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}+\int \frac {21 i a^2 d^4 \tan (e+f x)-25 a^2 d^4}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}+\int \frac {21 i a^2 d^4 \tan (e+f x)-25 a^2 d^4}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}+\frac {2 \int -\frac {a^2 d^4 (25 d-21 i d \tan (e+f x))}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 \int \frac {a^2 d^4 (25 d-21 i d \tan (e+f x))}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \int \frac {25 d-21 i d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {25}{2}-\frac {21 i}{2}\right ) \int \frac {\tan (e+f x) d+d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}\right )\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}+\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )+\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )+\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {7 i d^2 (d \tan (e+f x))^{3/2}}{f (1+i \tan (e+f x))}-\frac {\frac {50 a^2 d^3 \sqrt {d \tan (e+f x)}}{f}-\frac {2 a^2 d^4 \left (\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )+\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{5/2}}{4 f (a+i a \tan (e+f x))^2}\) |
-1/4*(d*(d*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])^2) + (((7*I)*d^2 *(d*Tan[e + f*x])^(3/2))/(f*(1 + I*Tan[e + f*x])) - ((-2*a^2*d^4*((25/2 - (21*I)/2)*(-(ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*S qrt[d])) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*Sqr t[d])) + (25/2 + (21*I)/2)*(-1/2*Log[d + d*Tan[e + f*x] - Sqrt[2]*Sqrt[d]* Sqrt[d*Tan[e + f*x]]]/(Sqrt[2]*Sqrt[d]) + Log[d + d*Tan[e + f*x] + Sqrt[2] *Sqrt[d]*Sqrt[d*Tan[e + f*x]]]/(2*Sqrt[2]*Sqrt[d]))))/f + (50*a^2*d^3*Sqrt [d*Tan[e + f*x]])/f)/(2*a^2))/(8*a^2)
3.2.72.3.1 Defintions of rubi rules used
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_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a*c, 2]}, Simp[(d*q + a*e)/(2*a*c) Int[(q + c*x^2)/(a + c*x^4), x], x] + Simp[(d*q - a*e)/(2*a*c) Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a , c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- a)*c]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2/f Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & & NeQ[c^2 + d^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* ((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && GtQ[n, 1] && (In tegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-(A*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f* x])^(n - 1)*Simp[A*(a*c*m + b*d*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a *A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]
Time = 0.88 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.39
| method | result | size |
| derivativedivides | \(\frac {2 d^{3} \left (-\sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {\frac {11 i \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}+\frac {9 d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {23 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}\right )}{8}+\frac {i d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 \sqrt {i d}}\right )}{f \,a^{2}}\) | \(126\) |
| default | \(\frac {2 d^{3} \left (-\sqrt {d \tan \left (f x +e \right )}-\frac {d \left (\frac {\frac {11 i \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}+\frac {9 d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {23 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}\right )}{8}+\frac {i d \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 \sqrt {i d}}\right )}{f \,a^{2}}\) | \(126\) |
2/f/a^2*d^3*(-(d*tan(f*x+e))^(1/2)-1/8*d*((11/2*I*(d*tan(f*x+e))^(3/2)+9/2 *d*(d*tan(f*x+e))^(1/2))/(I*d*tan(f*x+e)+d)^2+23/2*I/(-I*d)^(1/2)*arctan(( d*tan(f*x+e))^(1/2)/(-I*d)^(1/2)))+1/8*I*d/(I*d)^(1/2)*arctan((d*tan(f*x+e ))^(1/2)/(I*d)^(1/2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 573 vs. \(2 (240) = 480\).
Time = 0.27 (sec) , antiderivative size = 573, normalized size of antiderivative = 1.76 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {{\left (4 \, a^{2} \sqrt {-\frac {i \, d^{7}}{16 \, a^{4} f^{2}}} f e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} + 4 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {-\frac {i \, d^{7}}{16 \, a^{4} f^{2}}} \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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3}}\right ) - 4 \, a^{2} \sqrt {-\frac {i \, d^{7}}{16 \, a^{4} f^{2}}} f e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, d^{4} e^{\left (2 i \, f x + 2 i \, e\right )} - 4 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {-\frac {i \, d^{7}}{16 \, a^{4} f^{2}}} \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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{d^{3}}\right ) - 4 \, a^{2} \sqrt {\frac {529 i \, d^{7}}{64 \, a^{4} f^{2}}} f e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (23 i \, d^{4} + 8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {529 i \, d^{7}}{64 \, a^{4} f^{2}}} \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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} f}\right ) + 4 \, a^{2} \sqrt {\frac {529 i \, d^{7}}{64 \, a^{4} f^{2}}} f e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (23 i \, d^{4} - 8 \, {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {529 i \, d^{7}}{64 \, a^{4} f^{2}}} \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}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{8 \, a^{2} f}\right ) + {\left (42 \, d^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 9 \, d^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - d^{3}\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}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{16 \, a^{2} f} \]
-1/16*(4*a^2*sqrt(-1/16*I*d^7/(a^4*f^2))*f*e^(4*I*f*x + 4*I*e)*log(-2*(I*d ^4*e^(2*I*f*x + 2*I*e) + 4*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(-1/16* I*d^7/(a^4*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e ) + 1)))*e^(-2*I*f*x - 2*I*e)/d^3) - 4*a^2*sqrt(-1/16*I*d^7/(a^4*f^2))*f*e ^(4*I*f*x + 4*I*e)*log(-2*(I*d^4*e^(2*I*f*x + 2*I*e) - 4*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(-1/16*I*d^7/(a^4*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I *e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/d^3) - 4*a^2*s qrt(529/64*I*d^7/(a^4*f^2))*f*e^(4*I*f*x + 4*I*e)*log(1/8*(23*I*d^4 + 8*(a ^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(529/64*I*d^7/(a^4*f^2))*sqrt((-I*d* e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e )/(a^2*f)) + 4*a^2*sqrt(529/64*I*d^7/(a^4*f^2))*f*e^(4*I*f*x + 4*I*e)*log( 1/8*(23*I*d^4 - 8*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(529/64*I*d^7/(a ^4*f^2))*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))) *e^(-2*I*f*x - 2*I*e)/(a^2*f)) + (42*d^3*e^(4*I*f*x + 4*I*e) + 9*d^3*e^(2* I*f*x + 2*I*e) - d^3)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-4*I*f*x - 4*I*e)/(a^2*f)
\[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {\left (d \tan {\left (e + f x \right )}\right )^{\frac {7}{2}}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \]
Exception generated. \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \]
Time = 0.61 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.67 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {1}{8} \, d^{3} {\left (-\frac {2 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {23 i \, \sqrt {2} \sqrt {d} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a^{2} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} + \frac {16 \, \sqrt {d \tan \left (f x + e\right )}}{a^{2} f} + \frac {-11 i \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - 9 \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2} a^{2} f}\right )} \]
-1/8*d^3*(-2*I*sqrt(2)*sqrt(d)*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)))/(a^2*f*(I*d/sqrt(d^2) + 1)) + 23*I*sqrt(2)*sqrt(d)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-4*I*s qrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a^2*f*(-I*d/sqrt(d^2) + 1) ) + 16*sqrt(d*tan(f*x + e))/(a^2*f) + (-11*I*sqrt(d*tan(f*x + e))*d^2*tan( f*x + e) - 9*sqrt(d*tan(f*x + e))*d^2)/((d*tan(f*x + e) - I*d)^2*a^2*f))
Time = 6.56 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.62 \[ \int \frac {(d \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\frac {9\,d^5\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{8\,a^2\,f}+\frac {d^4\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,11{}\mathrm {i}}{8\,a^2\,f}}{-d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+d^2\,\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}+d^2}-\frac {2\,d^3\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{a^2\,f}+\mathrm {atan}\left (\frac {8\,a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{64\,a^4\,f^2}}}{d^4}\right )\,\sqrt {-\frac {d^7\,1{}\mathrm {i}}{64\,a^4\,f^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {16\,a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^7\,529{}\mathrm {i}}{256\,a^4\,f^2}}}{23\,d^4}\right )\,\sqrt {\frac {d^7\,529{}\mathrm {i}}{256\,a^4\,f^2}}\,2{}\mathrm {i} \]
atan((8*a^2*f*(d*tan(e + f*x))^(1/2)*(-(d^7*1i)/(64*a^4*f^2))^(1/2))/d^4)* (-(d^7*1i)/(64*a^4*f^2))^(1/2)*2i - ((9*d^5*(d*tan(e + f*x))^(1/2))/(8*a^2 *f) + (d^4*(d*tan(e + f*x))^(3/2)*11i)/(8*a^2*f))/(d^2*tan(e + f*x)*2i + d ^2 - d^2*tan(e + f*x)^2) - atan((16*a^2*f*(d*tan(e + f*x))^(1/2)*((d^7*529 i)/(256*a^4*f^2))^(1/2))/(23*d^4))*((d^7*529i)/(256*a^4*f^2))^(1/2)*2i - ( 2*d^3*(d*tan(e + f*x))^(1/2))/(a^2*f)