Integrand size = 32, antiderivative size = 354 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} (c-i d)^{5/2} f}-\frac {1}{3 (i c-d) f (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}+\frac {i c-5 d}{2 a (c+i d)^2 f \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}+\frac {d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{6 a^2 (c-i d) (c+i d)^3 f (c+d \tan (e+f x))^{3/2}}+\frac {(c-3 i d) d \left (3 c^2+22 i c d+13 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{6 a^2 (c-i d)^2 (c+i d)^4 f \sqrt {c+d \tan (e+f x)}} \] Output:
-1/4*I*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2)/(a+I*a *tan(f*x+e))^(1/2))*2^(1/2)/a^(3/2)/(c-I*d)^(5/2)/f-1/3/(I*c-d)/f/(a+I*a*t an(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(3/2)+1/2*(I*c-5*d)/a/(c+I*d)^2/f/(a+I*a *tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2)+1/6*d*(3*c^2+14*I*c*d+21*d^2)*(a +I*a*tan(f*x+e))^(1/2)/a^2/(c-I*d)/(c+I*d)^3/f/(c+d*tan(f*x+e))^(3/2)+1/6* (c-3*I*d)*d*(3*c^2+22*I*c*d+13*d^2)*(a+I*a*tan(f*x+e))^(1/2)/a^2/(c-I*d)^2 /(c+I*d)^4/f/(c+d*tan(f*x+e))^(1/2)
Time = 5.56 (sec) , antiderivative size = 326, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx=\frac {\frac {4 i (c+i d)}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}+\frac {6 i (c+5 i d)}{a \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac {\frac {3 i \sqrt {2} a (c+i d)^4 \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a (c-i d)}}-\frac {2 d \left (c^2+d^2\right ) \left (3 c^2+14 i c d+21 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}}-\frac {2 d \left (3 c^3+13 i c^2 d+79 c d^2-39 i d^3\right ) \sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}}{a^2 \left (c^2+d^2\right )^2}}{12 (c+i d)^2 f} \] Input:
Integrate[1/((a + I*a*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(5/2)),x]
Output:
(((4*I)*(c + I*d))/((a + I*a*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(3/2 )) + ((6*I)*(c + (5*I)*d))/(a*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f* x])^(3/2)) - (((3*I)*Sqrt[2]*a*(c + I*d)^4*ArcTan[(Sqrt[-(a*(c - I*d))]*Sq rt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*a*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[-(a* (c - I*d))] - (2*d*(c^2 + d^2)*(3*c^2 + (14*I)*c*d + 21*d^2)*Sqrt[a + I*a* Tan[e + f*x]])/(c + d*Tan[e + f*x])^(3/2) - (2*d*(3*c^3 + (13*I)*c^2*d + 7 9*c*d^2 - (39*I)*d^3)*Sqrt[a + I*a*Tan[e + f*x]])/Sqrt[c + d*Tan[e + f*x]] )/(a^2*(c^2 + d^2)^2))/(12*(c + I*d)^2*f)
Time = 1.95 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.14, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.469, Rules used = {3042, 4042, 27, 3042, 4079, 27, 3042, 4081, 27, 3042, 4081, 27, 3042, 4027, 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.
\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 4042 |
\(\displaystyle -\frac {\int -\frac {3 (a (i c-3 d)+2 i a d \tan (e+f x))}{2 \sqrt {i \tan (e+f x) a+a} (c+d \tan (e+f x))^{5/2}}dx}{3 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a (i c-3 d)+2 i a d \tan (e+f x)}{\sqrt {i \tan (e+f x) a+a} (c+d \tan (e+f x))^{5/2}}dx}{2 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a (i c-3 d)+2 i a d \tan (e+f x)}{\sqrt {i \tan (e+f x) a+a} (c+d \tan (e+f x))^{5/2}}dx}{2 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 4079 |
\(\displaystyle \frac {-\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (c^2+6 i d c-21 d^2\right ) a^2+4 (c+5 i d) d \tan (e+f x) a^2\right )}{2 (c+d \tan (e+f x))^{5/2}}dx}{a^2 (-d+i c)}-\frac {a (c+5 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (c^2+6 i d c-21 d^2\right ) a^2+4 (c+5 i d) d \tan (e+f x) a^2\right )}{(c+d \tan (e+f x))^{5/2}}dx}{2 a^2 (-d+i c)}-\frac {a (c+5 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (c^2+6 i d c-21 d^2\right ) a^2+4 (c+5 i d) d \tan (e+f x) a^2\right )}{(c+d \tan (e+f x))^{5/2}}dx}{2 a^2 (-d+i c)}-\frac {a (c+5 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {-\frac {\frac {2 \int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (3 c^3+15 i d c^2-37 d^2 c+39 i d^3\right ) a^3+2 d \left (3 c^2+14 i d c+21 d^2\right ) \tan (e+f x) a^3\right )}{2 (c+d \tan (e+f x))^{3/2}}dx}{3 a \left (c^2+d^2\right )}+\frac {2 a^2 d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {a (c+5 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-\frac {\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (3 c^3+15 i d c^2-37 d^2 c+39 i d^3\right ) a^3+2 d \left (3 c^2+14 i d c+21 d^2\right ) \tan (e+f x) a^3\right )}{(c+d \tan (e+f x))^{3/2}}dx}{3 a \left (c^2+d^2\right )}+\frac {2 a^2 d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {a (c+5 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (3 c^3+15 i d c^2-37 d^2 c+39 i d^3\right ) a^3+2 d \left (3 c^2+14 i d c+21 d^2\right ) \tan (e+f x) a^3\right )}{(c+d \tan (e+f x))^{3/2}}dx}{3 a \left (c^2+d^2\right )}+\frac {2 a^2 d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {a (c+5 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {-\frac {\frac {\frac {2 \int \frac {3 a^4 (c+i d)^4 \sqrt {i \tan (e+f x) a+a}}{2 \sqrt {c+d \tan (e+f x)}}dx}{a \left (c^2+d^2\right )}+\frac {2 a^3 d \left (3 c^3+13 i c^2 d+79 c d^2-39 i d^3\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{3 a \left (c^2+d^2\right )}+\frac {2 a^2 d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {a (c+5 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-\frac {\frac {\frac {3 a^3 (c+i d)^4 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^3 d \left (3 c^3+13 i c^2 d+79 c d^2-39 i d^3\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{3 a \left (c^2+d^2\right )}+\frac {2 a^2 d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {a (c+5 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {\frac {\frac {3 a^3 (c+i d)^4 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^3 d \left (3 c^3+13 i c^2 d+79 c d^2-39 i d^3\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{3 a \left (c^2+d^2\right )}+\frac {2 a^2 d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {a (c+5 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 4027 |
\(\displaystyle \frac {-\frac {\frac {\frac {2 a^3 d \left (3 c^3+13 i c^2 d+79 c d^2-39 i d^3\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {6 i a^5 (c+i d)^4 \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f \left (c^2+d^2\right )}}{3 a \left (c^2+d^2\right )}+\frac {2 a^2 d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {a (c+5 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {-\frac {\frac {2 a^2 d \left (3 c^2+14 i c d+21 d^2\right ) \sqrt {a+i a \tan (e+f x)}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac {\frac {2 a^3 d \left (3 c^3+13 i c^2 d+79 c d^2-39 i d^3\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}-\frac {3 i \sqrt {2} a^{7/2} (c+i d)^4 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f \sqrt {c-i d} \left (c^2+d^2\right )}}{3 a \left (c^2+d^2\right )}}{2 a^2 (-d+i c)}-\frac {a (c+5 i d)}{f (c+i d) \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}}}{2 a^2 (-d+i c)}-\frac {1}{3 f (-d+i c) (a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{3/2}}\) |
Input:
Int[1/((a + I*a*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(5/2)),x]
Output:
-1/3*1/((I*c - d)*f*(a + I*a*Tan[e + f*x])^(3/2)*(c + d*Tan[e + f*x])^(3/2 )) + (-((a*(c + (5*I)*d))/((c + I*d)*f*Sqrt[a + I*a*Tan[e + f*x]]*(c + d*T an[e + f*x])^(3/2))) - ((2*a^2*d*(3*c^2 + (14*I)*c*d + 21*d^2)*Sqrt[a + I* a*Tan[e + f*x]])/(3*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)) + (((-3*I)*S qrt[2]*a^(7/2)*(c + I*d)^4*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x ]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[c - I*d]*(c^2 + d^2 )*f) + (2*a^3*d*(3*c^3 + (13*I)*c^2*d + 79*c*d^2 - (39*I)*d^3)*Sqrt[a + I* a*Tan[e + f*x]])/((c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]))/(3*a*(c^2 + d^2 )))/(2*a^2*(I*c - d)))/(2*a^2*(I*c - 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/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f) Subst[Int[1/(a*c - b*d - 2* a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N eQ[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]
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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 7060 vs. \(2 (294 ) = 588\).
Time = 0.64 (sec) , antiderivative size = 7061, normalized size of antiderivative = 19.95
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(7061\) |
default | \(\text {Expression too large to display}\) | \(7061\) |
Input:
int(1/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x,method=_RETURNVERB OSE)
Output:
result too large to display
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1546 vs. \(2 (276) = 552\).
Time = 0.23 (sec) , antiderivative size = 1546, normalized size of antiderivative = 4.37 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm=" fricas")
Output:
1/12*(sqrt(2)*(I*c^5 - c^4*d + 2*I*c^3*d^2 - 2*c^2*d^3 + I*c*d^4 - d^5 - 4 *(-I*c^5 + c^4*d - 14*I*c^3*d^2 - 50*c^2*d^3 + 51*I*c*d^4 + 13*d^5)*e^(8*I *f*x + 8*I*e) + (13*I*c^5 - 25*c^4*d + 134*I*c^3*d^2 + 314*c^2*d^3 - 87*I* c*d^4 + 35*d^5)*e^(6*I*f*x + 6*I*e) - 3*(-5*I*c^5 + 13*c^4*d - 30*I*c^3*d^ 2 - 26*c^2*d^3 - 41*I*c*d^4 - 23*d^5)*e^(4*I*f*x + 4*I*e) + (7*I*c^5 - 19* c^4*d + 14*I*c^3*d^2 - 38*c^2*d^3 + 7*I*c*d^4 - 19*d^5)*e^(2*I*f*x + 2*I*e ))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1 ))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)) - 3*((a^2*c^8 + 4*a^2*c^6*d^2 + 6*a^2 *c^4*d^4 + 4*a^2*c^2*d^6 + a^2*d^8)*f*e^(7*I*f*x + 7*I*e) + 2*(a^2*c^8 + 2 *I*a^2*c^7*d + 2*a^2*c^6*d^2 + 6*I*a^2*c^5*d^3 + 6*I*a^2*c^3*d^5 - 2*a^2*c ^2*d^6 + 2*I*a^2*c*d^7 - a^2*d^8)*f*e^(5*I*f*x + 5*I*e) + (a^2*c^8 + 4*I*a ^2*c^7*d - 4*a^2*c^6*d^2 + 4*I*a^2*c^5*d^3 - 10*a^2*c^4*d^4 - 4*I*a^2*c^3* d^5 - 4*a^2*c^2*d^6 - 4*I*a^2*c*d^7 + a^2*d^8)*f*e^(3*I*f*x + 3*I*e))*sqrt (1/2*I/((-I*a^3*c^5 - 5*a^3*c^4*d + 10*I*a^3*c^3*d^2 + 10*a^3*c^2*d^3 - 5* I*a^3*c*d^4 - a^3*d^5)*f^2))*log(-2*(I*a^2*c^3 + 3*a^2*c^2*d - 3*I*a^2*c*d ^2 - a^2*d^3)*f*sqrt(1/2*I/((-I*a^3*c^5 - 5*a^3*c^4*d + 10*I*a^3*c^3*d^2 + 10*a^3*c^2*d^3 - 5*I*a^3*c*d^4 - a^3*d^5)*f^2))*e^(I*f*x + I*e) + sqrt(2) *sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)) *sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I*e) + 1)) + 3*((a^2*c^ 8 + 4*a^2*c^6*d^2 + 6*a^2*c^4*d^4 + 4*a^2*c^2*d^6 + a^2*d^8)*f*e^(7*I*f...
\[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(a+I*a*tan(f*x+e))**(3/2)/(c+d*tan(f*x+e))**(5/2),x)
Output:
Integral(1/((I*a*(tan(e + f*x) - I))**(3/2)*(c + d*tan(e + f*x))**(5/2)), x)
Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm=" maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm=" giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeRecursive ass umption s
Timed out. \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {1}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int(1/((a + a*tan(e + f*x)*1i)^(3/2)*(c + d*tan(e + f*x))^(5/2)),x)
Output:
int(1/((a + a*tan(e + f*x)*1i)^(3/2)*(c + d*tan(e + f*x))^(5/2)), x)
\[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {1}{\left (a +i a \tan \left (f x +e \right )\right )^{\frac {3}{2}} \left (d \tan \left (f x +e \right )+c \right )^{\frac {5}{2}}}d x \] Input:
int(1/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x)
Output:
int(1/(a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(5/2),x)