Integrand size = 30, antiderivative size = 185 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx=-\frac {i (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 a f}+\frac {(c+i d)^{3/2} (i c+4 d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 a f}-\frac {(c+5 i d) d \sqrt {c+d \tan (e+f x)}}{2 a f}+\frac {(i c-d) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))} \]
-1/2*I*(c-I*d)^(5/2)*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/a/f+1/2 *(c+I*d)^(3/2)*(I*c+4*d)*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/a/f -1/2*(c+5*I*d)*d*(c+d*tan(f*x+e))^(1/2)/a/f+1/2*(I*c-d)*(c+d*tan(f*x+e))^( 3/2)/f/(a+I*a*tan(f*x+e))
Time = 1.26 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.90 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx=\frac {-i (c-i d)^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\sqrt {c+i d} \left (i c^2+3 c d+4 i d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )+\frac {\sqrt {c+d \tan (e+f x)} \left (c^2+2 i c d-5 d^2-4 i d^2 \tan (e+f x)\right )}{-i+\tan (e+f x)}}{2 a f} \]
((-I)*(c - I*d)^(5/2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + Sq rt[c + I*d]*(I*c^2 + 3*c*d + (4*I)*d^2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/S qrt[c + I*d]] + (Sqrt[c + d*Tan[e + f*x]]*(c^2 + (2*I)*c*d - 5*d^2 - (4*I) *d^2*Tan[e + f*x]))/(-I + Tan[e + f*x]))/(2*a*f)
Time = 0.91 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.87, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4033, 27, 3042, 4011, 3042, 4022, 3042, 4020, 25, 73, 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.
| \(\displaystyle \int \frac {(c+d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c+d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)}dx\) |
| \(\Big \downarrow \) 4033 |
| \(\displaystyle \frac {\int \frac {1}{2} \sqrt {c+d \tan (e+f x)} (a (2 c+i d) (c-3 i d)-a (c+5 i d) d \tan (e+f x))dx}{2 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sqrt {c+d \tan (e+f x)} (a (2 c+i d) (c-3 i d)-a (c+5 i d) d \tan (e+f x))dx}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sqrt {c+d \tan (e+f x)} (a (2 c+i d) (c-3 i d)-a (c+5 i d) d \tan (e+f x))dx}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 4011 |
| \(\displaystyle \frac {\int \frac {a \left (2 c^3-5 i d c^2+4 d^2 c+5 i d^3\right )+a d \left (c^2-10 i d c+3 d^2\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 a d (c+5 i d) \sqrt {c+d \tan (e+f x)}}{f}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a \left (2 c^3-5 i d c^2+4 d^2 c+5 i d^3\right )+a d \left (c^2-10 i d c+3 d^2\right ) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 a d (c+5 i d) \sqrt {c+d \tan (e+f x)}}{f}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 4022 |
| \(\displaystyle \frac {a (c-i d)^3 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+a (c+i d)^2 (c-4 i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 a d (c+5 i d) \sqrt {c+d \tan (e+f x)}}{f}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (c-i d)^3 \int \frac {i \tan (e+f x)+1}{\sqrt {c+d \tan (e+f x)}}dx+a (c+i d)^2 (c-4 i d) \int \frac {1-i \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx-\frac {2 a d (c+5 i d) \sqrt {c+d \tan (e+f x)}}{f}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {\frac {i a (c-i d)^3 \int -\frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}-\frac {i a (c+i d)^2 (c-4 i d) \int -\frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}-\frac {2 a d (c+5 i d) \sqrt {c+d \tan (e+f x)}}{f}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {i a (c-i d)^3 \int \frac {1}{(1-i \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(i \tan (e+f x))}{f}+\frac {i a (c+i d)^2 (c-4 i d) \int \frac {1}{(i \tan (e+f x)+1) \sqrt {c+d \tan (e+f x)}}d(-i \tan (e+f x))}{f}-\frac {2 a d (c+5 i d) \sqrt {c+d \tan (e+f x)}}{f}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {2 a (c-i d)^3 \int \frac {1}{\frac {i \tan ^2(e+f x)}{d}+\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}+\frac {2 a (c+i d)^2 (c-4 i d) \int \frac {1}{-\frac {i \tan ^2(e+f x)}{d}-\frac {i c}{d}+1}d\sqrt {c+d \tan (e+f x)}}{d f}-\frac {2 a d (c+5 i d) \sqrt {c+d \tan (e+f x)}}{f}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {2 a (c-i d)^{5/2} \arctan \left (\frac {\tan (e+f x)}{\sqrt {c-i d}}\right )}{f}+\frac {2 a (c+i d)^{3/2} (c-4 i d) \arctan \left (\frac {\tan (e+f x)}{\sqrt {c+i d}}\right )}{f}-\frac {2 a d (c+5 i d) \sqrt {c+d \tan (e+f x)}}{f}}{4 a^2}+\frac {(-d+i c) (c+d \tan (e+f x))^{3/2}}{2 f (a+i a \tan (e+f x))}\) |
((I*c - d)*(c + d*Tan[e + f*x])^(3/2))/(2*f*(a + I*a*Tan[e + f*x])) + ((2* a*(c - I*d)^(5/2)*ArcTan[Tan[e + f*x]/Sqrt[c - I*d]])/f + (2*a*(c + I*d)^( 3/2)*(c - (4*I)*d)*ArcTan[Tan[e + f*x]/Sqrt[c + I*d]])/f - (2*a*(c + (5*I) *d)*d*Sqrt[c + d*Tan[e + f*x]])/f)/(4*a^2)
3.12.16.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, 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[((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[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f) Subst[Int[(a + (b/d)*x)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2 Int[(a + b*Tan[e + f*x])^m*( 1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2 Int[(a + b*Tan[e + f*x])^m *(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !IntegerQ[m]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((c + d*Tan[e + f*x])^(n - 1)/( 2*a*f*(a + b*Tan[e + f*x]))), x] + Simp[1/(2*a^2) Int[(c + d*Tan[e + f*x] )^(n - 2)*Simp[a*c^2 + a*d^2*(n - 1) - b*c*d*n - d*(a*c*(n - 2) + b*d*n)*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] && GtQ[n, 1]
Time = 0.57 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.28
| method | result | size |
| derivativedivides | \(\frac {2 d^{2} \left (-i \sqrt {c +d \tan \left (f x +e \right )}+\frac {\left (i c^{3}-3 i c \,d^{2}+3 c^{2} d -d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 d^{2} \sqrt {i d -c}}+\frac {-\frac {d \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{\left (i d +c \right ) \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {\left (i c^{4}+9 i c^{2} d^{2}-4 i d^{4}+c^{3} d -11 c \,d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\left (i d +c \right ) \sqrt {-i d -c}}}{4 d^{2}}\right )}{f a}\) | \(237\) |
| default | \(\frac {2 d^{2} \left (-i \sqrt {c +d \tan \left (f x +e \right )}+\frac {\left (i c^{3}-3 i c \,d^{2}+3 c^{2} d -d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 d^{2} \sqrt {i d -c}}+\frac {-\frac {d \left (3 i c^{2} d -i d^{3}+c^{3}-3 c \,d^{2}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{\left (i d +c \right ) \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {\left (i c^{4}+9 i c^{2} d^{2}-4 i d^{4}+c^{3} d -11 c \,d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\left (i d +c \right ) \sqrt {-i d -c}}}{4 d^{2}}\right )}{f a}\) | \(237\) |
2/f/a*d^2*(-I*(c+d*tan(f*x+e))^(1/2)+1/4*(I*c^3-3*I*c*d^2+3*c^2*d-d^3)/d^2 /(I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))+1/4/d^2*(-d*(3 *I*c^2*d-I*d^3+c^3-3*c*d^2)/(c+I*d)*(c+d*tan(f*x+e))^(1/2)/(-d*tan(f*x+e)+ I*d)-(c^3*d-11*c*d^3+I*c^4+9*I*c^2*d^2-4*I*d^4)/(c+I*d)/(-I*d-c)^(1/2)*arc tan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1025 vs. \(2 (141) = 282\).
Time = 0.40 (sec) , antiderivative size = 1025, normalized size of antiderivative = 5.54 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx=\text {Too large to display} \]
1/8*(a*f*sqrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I* d^5)/(a^2*f^2))*e^(2*I*f*x + 2*I*e)*log(2*(c^3 - 2*I*c^2*d - c*d^2 + (I*a* f*e^(2*I*f*x + 2*I*e) + I*a*f)*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(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I* c^2*d^3 + 5*c*d^4 - I*d^5)/(a^2*f^2)) + (c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3 )*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(c^2 - 2*I*c*d - d^2)) - a*f*s qrt(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^2* f^2))*e^(2*I*f*x + 2*I*e)*log(2*(c^3 - 2*I*c^2*d - c*d^2 + (-I*a*f*e^(2*I* f*x + 2*I*e) - I*a*f)*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(-(c^5 - 5*I*c^4*d - 10*c^3*d^2 + 10*I*c^2*d^3 + 5*c*d^4 - I*d^5)/(a^2*f^2)) + (c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)*e^(2*I* f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(c^2 - 2*I*c*d - d^2)) + a*f*sqrt(-(c^5 - 5*I*c^4*d + 5*c^3*d^2 - 25*I*c^2*d^3 + 40*c*d^4 + 16*I*d^5)/(a^2*f^2))* e^(2*I*f*x + 2*I*e)*log(1/2*(I*c^3 + 2*c^2*d + 7*I*c*d^2 - 4*d^3 + (a*f*e^ (2*I*f*x + 2*I*e) + a*f)*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(-(c^5 - 5*I*c^4*d + 5*c^3*d^2 - 25*I*c^2*d^3 + 40*c*d^4 + 16*I*d^5)/(a^2*f^2)) + (I*c^3 + 3*c^2*d + 4*I*c*d^2)*e^(2*I* f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/(a*f)) - a*f*sqrt(-(c^5 - 5*I*c^4*d + 5 *c^3*d^2 - 25*I*c^2*d^3 + 40*c*d^4 + 16*I*d^5)/(a^2*f^2))*e^(2*I*f*x + 2*I *e)*log(1/2*(I*c^3 + 2*c^2*d + 7*I*c*d^2 - 4*d^3 - (a*f*e^(2*I*f*x + 2*...
\[ \int \frac {(c+d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx=- \frac {i \left (\int \frac {c^{2} \sqrt {c + d \tan {\left (e + f x \right )}}}{\tan {\left (e + f x \right )} - i}\, dx + \int \frac {d^{2} \sqrt {c + d \tan {\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}}{\tan {\left (e + f x \right )} - i}\, dx + \int \frac {2 c d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}{\tan {\left (e + f x \right )} - i}\, dx\right )}{a} \]
-I*(Integral(c**2*sqrt(c + d*tan(e + f*x))/(tan(e + f*x) - I), x) + Integr al(d**2*sqrt(c + d*tan(e + f*x))*tan(e + f*x)**2/(tan(e + f*x) - I), x) + Integral(2*c*d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)/(tan(e + f*x) - I), x ))/a
Exception generated. \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx=\text {Exception raised: RuntimeError} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 467 vs. \(2 (141) = 282\).
Time = 0.60 (sec) , antiderivative size = 467, normalized size of antiderivative = 2.52 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx=-\frac {1}{2} \, d^{2} {\left (\frac {4 i \, \sqrt {d \tan \left (f x + e\right ) + c}}{a f} - \frac {\sqrt {d \tan \left (f x + e\right ) + c} c^{2} + 2 i \, \sqrt {d \tan \left (f x + e\right ) + c} c d - \sqrt {d \tan \left (f x + e\right ) + c} d^{2}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )} a d f} + \frac {2 \, {\left (i \, c^{3} + 2 \, c^{2} d + 7 i \, c d^{2} - 4 \, d^{3}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} + i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{a \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d^{2} f {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {2 \, {\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{a \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d^{2} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}}\right )} \]
-1/2*d^2*(4*I*sqrt(d*tan(f*x + e) + c)/(a*f) - (sqrt(d*tan(f*x + e) + c)*c ^2 + 2*I*sqrt(d*tan(f*x + e) + c)*c*d - sqrt(d*tan(f*x + e) + c)*d^2)/((d* tan(f*x + e) - I*d)*a*d*f) + 2*(I*c^3 + 2*c^2*d + 7*I*c*d^2 - 4*d^3)*arcta n(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c) )/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2)) + I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2))))/(a*sqrt(-2*c + 2*sqrt(c^ 2 + d^2))*d^2*f*(I*d/(c - sqrt(c^2 + d^2)) + 1)) + 2*(-I*c^3 - 3*c^2*d + 3 *I*c*d^2 + d^3)*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqr t(d*tan(f*x + e) + c))/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2)) - I*sqrt(-2*c + 2 *sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2))))/(a* sqrt(-2*c + 2*sqrt(c^2 + d^2))*d^2*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)))
Time = 10.47 (sec) , antiderivative size = 4857, normalized size of antiderivative = 26.25 \[ \int \frac {(c+d \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx=\text {Too large to display} \]
log((a*f*(12*d^11 - c*d^10*25i + 75*c^2*d^9 - c^3*d^8*35i + 113*c^4*d^7 + c^5*d^6*5i + 49*c^6*d^5 + c^7*d^4*15i - c^8*d^3))/2 - (((a*f*(a^2*d^6*f^2* 80i + 16*a^2*c*d^5*f^2 + a^2*c^2*d^4*f^2*80i + 16*a^2*c^3*d^3*f^2))/2 + 64 *a^4*c*d^2*f^4*(c + d*tan(e + f*x))^(1/2)*(-(720*c*d^10 + d^11*240i + 640* c^3*d^8 - c^4*d^7*400i - 48*c^5*d^6 - c^6*d^5*160i + 32*c^7*d^4 + a^2*f^2* ((((400*c^4*d^7 - 240*d^11 + 160*c^6*d^5)*1i)/(a^2*f^2) - (720*c*d^10 + 64 0*c^3*d^8 - 48*c^5*d^6 + 32*c^7*d^4)/(a^2*f^2))^2 - 4*(256*d^6 + 256*c^2*d ^4)*(((40*c*d^15 + 150*c^3*d^13 + 200*c^5*d^11 + 100*c^7*d^9 - 10*c^11*d^5 )*1i)/(a^4*f^4) - (35*c^4*d^12 - 31*c^2*d^14 - 16*d^16 + 130*c^6*d^10 + 11 0*c^8*d^8 + 29*c^10*d^6 - c^12*d^4)/(a^4*f^4)))^(1/2))/(512*a^2*f^2*(d^6 + c^2*d^4)))^(1/2))*(-(720*c*d^10 + d^11*240i + 640*c^3*d^8 - c^4*d^7*400i - 48*c^5*d^6 - c^6*d^5*160i + 32*c^7*d^4 + a^2*f^2*((((400*c^4*d^7 - 240*d ^11 + 160*c^6*d^5)*1i)/(a^2*f^2) - (720*c*d^10 + 640*c^3*d^8 - 48*c^5*d^6 + 32*c^7*d^4)/(a^2*f^2))^2 - 4*(256*d^6 + 256*c^2*d^4)*(((40*c*d^15 + 150* c^3*d^13 + 200*c^5*d^11 + 100*c^7*d^9 - 10*c^11*d^5)*1i)/(a^4*f^4) - (35*c ^4*d^12 - 31*c^2*d^14 - 16*d^16 + 130*c^6*d^10 + 110*c^8*d^8 + 29*c^10*d^6 - c^12*d^4)/(a^4*f^4)))^(1/2))/(512*a^2*f^2*(d^6 + c^2*d^4)))^(1/2) + 2*a ^2*f^2*(c + d*tan(e + f*x))^(1/2)*(c*d^7*50i - 17*d^8 + 80*c^2*d^6 - 5*c^4 *d^4 - c^5*d^3*10i + 2*c^6*d^2))*(-(720*c*d^10 + d^11*240i + 640*c^3*d^8 - c^4*d^7*400i - 48*c^5*d^6 - c^6*d^5*160i + 32*c^7*d^4 + a^2*f^2*((((40...