Integrand size = 28, antiderivative size = 266 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \, dx=\frac {\left (\frac {25}{16}+\frac {21 i}{16}\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 d^{3/2} f}-\frac {\left (\frac {25}{16}+\frac {21 i}{16}\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 d^{3/2} f}+\frac {\left (\frac {25}{16}-\frac {21 i}{16}\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{\sqrt {2} a^2 d^{3/2} f}-\frac {25}{8 a^2 d f \sqrt {d \tan (e+f x)}}+\frac {7}{8 a^2 d f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}+\frac {1}{4 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))^2} \] Output:
(25/32+21/32*I)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/2)/a^2 /d^(3/2)/f-(25/32+21/32*I)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))* 2^(1/2)/a^2/d^(3/2)/f+(25/32-21/32*I)*arctanh(2^(1/2)*(d*tan(f*x+e))^(1/2) /(d^(1/2)+d^(1/2)*tan(f*x+e)))*2^(1/2)/a^2/d^(3/2)/f-25/8/a^2/d/f/(d*tan(f *x+e))^(1/2)+7/8/a^2/d/f/(1+I*tan(f*x+e))/(d*tan(f*x+e))^(1/2)+1/4/d/f/(d* tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^2
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.40 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.57 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \, dx=\frac {\sec ^2(e+f x) \left (-9-9 \cos (2 (e+f x))+46 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-i \tan (e+f x)\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},i \tan (e+f x)\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))-7 i \sin (2 (e+f x))\right )}{16 a^2 d f \sqrt {d \tan (e+f x)} (-i+\tan (e+f x))^2} \] Input:
Integrate[1/((d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x])^2),x]
Output:
(Sec[e + f*x]^2*(-9 - 9*Cos[2*(e + f*x)] + 46*Hypergeometric2F1[-1/2, 1, 1 /2, (-I)*Tan[e + f*x]]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]) + 4*Hyperge ometric2F1[-1/2, 1, 1/2, I*Tan[e + f*x]]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]) - (7*I)*Sin[2*(e + f*x)]))/(16*a^2*d*f*Sqrt[d*Tan[e + f*x]]*(-I + T an[e + f*x])^2)
Time = 1.06 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.18, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.679, Rules used = {3042, 4042, 27, 3042, 4079, 3042, 4012, 25, 3042, 4017, 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.
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^2 (d \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4042 |
| \(\displaystyle \frac {\int \frac {9 a d-5 i a d \tan (e+f x)}{2 (d \tan (e+f x))^{3/2} (i \tan (e+f x) a+a)}dx}{4 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {9 a d-5 i a d \tan (e+f x)}{(d \tan (e+f x))^{3/2} (i \tan (e+f x) a+a)}dx}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {9 a d-5 i a d \tan (e+f x)}{(d \tan (e+f x))^{3/2} (i \tan (e+f x) a+a)}dx}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {\int \frac {25 a^2 d^2-21 i a^2 d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}}dx}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {25 a^2 d^2-21 i a^2 d^2 \tan (e+f x)}{(d \tan (e+f x))^{3/2}}dx}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {\frac {-\frac {50 a^2 d}{f \sqrt {d \tan (e+f x)}}+\frac {\int -\frac {21 i a^2 d^3+25 a^2 \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)}}dx}{d^2}}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {-\frac {50 a^2 d}{f \sqrt {d \tan (e+f x)}}-\frac {\int \frac {21 i a^2 d^3+25 a^2 \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)}}dx}{d^2}}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {-\frac {50 a^2 d}{f \sqrt {d \tan (e+f x)}}-\frac {\int \frac {21 i a^2 d^3+25 a^2 \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)}}dx}{d^2}}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {\frac {-\frac {50 a^2 d}{f \sqrt {d \tan (e+f x)}}-\frac {2 \int \frac {a^2 d^3 (25 \tan (e+f x) d+21 i d)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{d^2 f}}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {-\frac {50 a^2 d}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^2 d \int \frac {25 \tan (e+f x) d+21 i d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {\frac {-\frac {50 a^2 d}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^2 d \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \int \frac {\tan (e+f x) d+d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}-\left (\frac {25}{2}-\frac {21 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {\frac {-\frac {50 a^2 d}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^2 d \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}\right )-\left (\frac {25}{2}-\frac {21 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {-\frac {50 a^2 d}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^2 d \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {25}{2}-\frac {21 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {-\frac {50 a^2 d}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^2 d \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {25}{2}-\frac {21 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {\frac {-\frac {50 a^2 d}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^2 d \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {-\frac {50 a^2 d}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^2 d \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {-\frac {50 a^2 d}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^2 d \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )\right )}{f}}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {-\frac {50 a^2 d}{f \sqrt {d \tan (e+f x)}}-\frac {2 a^2 d \left (\left (\frac {25}{2}+\frac {21 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {25}{2}-\frac {21 i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}}{2 a^2 d}+\frac {7}{f (1+i \tan (e+f x)) \sqrt {d \tan (e+f x)}}}{8 a^2 d}+\frac {1}{4 d f (a+i a \tan (e+f x))^2 \sqrt {d \tan (e+f x)}}\) |
Input:
Int[1/((d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x])^2),x]
Output:
1/(4*d*f*Sqrt[d*Tan[e + f*x]]*(a + I*a*Tan[e + f*x])^2) + (7/(f*(1 + I*Tan [e + f*x])*Sqrt[d*Tan[e + f*x]]) + ((-2*a^2*d*((25/2 + (21*I)/2)*(-(ArcTan [1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*Sqrt[d])) - (25/2 - (2 1*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)/(f*Sqrt[d*Tan[e + f*x]]) )/(2*a^2*d))/(8*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_)^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[(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/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[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) In t[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && (IntegerQ[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*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n, 0]
Time = 1.42 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.49
| method | result | size |
| derivativedivides | \(\frac {2 d^{3} \left (-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 d^{4} \sqrt {i d}}-\frac {\frac {-\frac {9 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}+\frac {11 i d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {23 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}}{8 d^{4}}-\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}\right )}{f \,a^{2}}\) | \(131\) |
| default | \(\frac {2 d^{3} \left (-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 d^{4} \sqrt {i d}}-\frac {\frac {-\frac {9 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}+\frac {11 i d \sqrt {d \tan \left (f x +e \right )}}{2}}{\left (i d \tan \left (f x +e \right )+d \right )^{2}}+\frac {23 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{2 \sqrt {-i d}}}{8 d^{4}}-\frac {1}{d^{4} \sqrt {d \tan \left (f x +e \right )}}\right )}{f \,a^{2}}\) | \(131\) |
Input:
int(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
Output:
2/f/a^2*d^3*(-1/8/d^4/(I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(I*d)^(1/2)) -1/8/d^4*((-9/2*(d*tan(f*x+e))^(3/2)+11/2*I*d*(d*tan(f*x+e))^(1/2))/(I*d*t an(f*x+e)+d)^2+23/2/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2)) )-1/d^4/(d*tan(f*x+e))^(1/2))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 697 vs. \(2 (200) = 400\).
Time = 0.10 (sec) , antiderivative size = 697, normalized size of antiderivative = 2.62 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \, dx =\text {Too large to display} \] Input:
integrate(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas ")
Output:
1/16*(4*(a^2*d^2*f*e^(6*I*f*x + 6*I*e) - a^2*d^2*f*e^(4*I*f*x + 4*I*e))*sq rt(1/16*I/(a^4*d^3*f^2))*log(-2*(4*(I*a^2*d^2*f*e^(2*I*f*x + 2*I*e) + I*a^ 2*d^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))* sqrt(1/16*I/(a^4*d^3*f^2)) + I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e) ) - 4*(a^2*d^2*f*e^(6*I*f*x + 6*I*e) - a^2*d^2*f*e^(4*I*f*x + 4*I*e))*sqrt (1/16*I/(a^4*d^3*f^2))*log(-2*(4*(-I*a^2*d^2*f*e^(2*I*f*x + 2*I*e) - I*a^2 *d^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*s qrt(1/16*I/(a^4*d^3*f^2)) + I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) + 4*(a^2*d^2*f*e^(6*I*f*x + 6*I*e) - a^2*d^2*f*e^(4*I*f*x + 4*I*e))*sqrt( -529/64*I/(a^4*d^3*f^2))*log(1/8*(8*(a^2*d*f*e^(2*I*f*x + 2*I*e) + a^2*d*f )*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-5 29/64*I/(a^4*d^3*f^2)) + 23)*e^(-2*I*f*x - 2*I*e)/(a^2*d*f)) - 4*(a^2*d^2* f*e^(6*I*f*x + 6*I*e) - a^2*d^2*f*e^(4*I*f*x + 4*I*e))*sqrt(-529/64*I/(a^4 *d^3*f^2))*log(-1/8*(8*(a^2*d*f*e^(2*I*f*x + 2*I*e) + a^2*d*f)*sqrt((-I*d* e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-529/64*I/(a^4* d^3*f^2)) - 23)*e^(-2*I*f*x - 2*I*e)/(a^2*d*f)) + sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*(-42*I*e^(6*I*f*x + 6*I*e) - 33*I *e^(4*I*f*x + 4*I*e) + 10*I*e^(2*I*f*x + 2*I*e) + I))/(a^2*d^2*f*e^(6*I*f* x + 6*I*e) - a^2*d^2*f*e^(4*I*f*x + 4*I*e))
\[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (e + f x \right )} - 2 i \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )} - \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx}{a^{2}} \] Input:
integrate(1/(d*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e))**2,x)
Output:
-Integral(1/((d*tan(e + f*x))**(3/2)*tan(e + f*x)**2 - 2*I*(d*tan(e + f*x) )**(3/2)*tan(e + f*x) - (d*tan(e + f*x))**(3/2)), x)/a**2
Exception generated. \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima ")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.19 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \, dx=-\frac {\frac {2 \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\sqrt {d} {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )}} + \frac {23 \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{-i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\sqrt {d} {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )}} + \frac {16}{\sqrt {d \tan \left (f x + e\right )}} + \frac {9 \, \sqrt {d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) - 11 i \, \sqrt {d \tan \left (f x + e\right )} d}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2}}}{8 \, a^{2} d f} \] Input:
integrate(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")
Output:
-1/8*(2*sqrt(2)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(I*sqrt(2)*d^(3/2) + sqrt(2)*sqrt(d)*abs(d)))/(sqrt(d)*(I*d/abs(d) + 1)) + 23*sqrt(2)*arctan(2* sqrt(d*tan(f*x + e))*abs(d)/(-I*sqrt(2)*d^(3/2) + sqrt(2)*sqrt(d)*abs(d))) /(sqrt(d)*(-I*d/abs(d) + 1)) + 16/sqrt(d*tan(f*x + e)) + (9*sqrt(d*tan(f*x + e))*d*tan(f*x + e) - 11*I*sqrt(d*tan(f*x + e))*d)/(d*tan(f*x + e) - I*d )^2)/(a^2*d*f)
Time = 3.04 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \, dx=2\,\mathrm {atanh}\left (8\,a^2\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {1{}\mathrm {i}}{64\,a^4\,d^3\,f^2}}\right )\,\sqrt {\frac {1{}\mathrm {i}}{64\,a^4\,d^3\,f^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^2\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {529{}\mathrm {i}}{256\,a^4\,d^3\,f^2}}}{23}\right )\,\sqrt {-\frac {529{}\mathrm {i}}{256\,a^4\,d^3\,f^2}}-\frac {\frac {2\,d}{a^2\,f}-\frac {25\,d\,{\mathrm {tan}\left (e+f\,x\right )}^2}{8\,a^2\,f}+\frac {d\,\mathrm {tan}\left (e+f\,x\right )\,43{}\mathrm {i}}{8\,a^2\,f}}{d^2\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}-{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}+d\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,2{}\mathrm {i}} \] Input:
int(1/((d*tan(e + f*x))^(3/2)*(a + a*tan(e + f*x)*1i)^2),x)
Output:
2*atanh(8*a^2*d*f*(d*tan(e + f*x))^(1/2)*(1i/(64*a^4*d^3*f^2))^(1/2))*(1i/ (64*a^4*d^3*f^2))^(1/2) + 2*atanh((16*a^2*d*f*(d*tan(e + f*x))^(1/2)*(-529 i/(256*a^4*d^3*f^2))^(1/2))/23)*(-529i/(256*a^4*d^3*f^2))^(1/2) - ((2*d)/( a^2*f) - (25*d*tan(e + f*x)^2)/(8*a^2*f) + (d*tan(e + f*x)*43i)/(8*a^2*f)) /(d*(d*tan(e + f*x))^(3/2)*2i - (d*tan(e + f*x))^(5/2) + d^2*(d*tan(e + f* x))^(1/2))
\[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))^2} \, dx=-\frac {\int \frac {1}{\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{3}-2 \sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )^{2} i -\sqrt {\tan \left (f x +e \right )}\, \tan \left (f x +e \right )}d x}{\sqrt {d}\, a^{2} d} \] Input:
int(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^2,x)
Output:
( - int(1/(sqrt(tan(e + f*x))*tan(e + f*x)**3 - 2*sqrt(tan(e + f*x))*tan(e + f*x)**2*i - sqrt(tan(e + f*x))*tan(e + f*x)),x))/(sqrt(d)*a**2*d)