Integrand size = 43, antiderivative size = 223 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {(7 i A-3 B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a c^{5/2} f}-\frac {7 i A-3 B}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 i A-3 B}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 i A-3 B}{16 a c^2 f \sqrt {c-i c \tan (e+f x)}} \] Output:
1/32*(7*I*A-3*B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^( 1/2)/a/c^(5/2)/f-1/20*(7*I*A-3*B)/a/f/(c-I*c*tan(f*x+e))^(5/2)+1/2*(I*A-B) /a/f/(1+I*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2)-1/24*(7*I*A-3*B)/a/c/f/(c-I *c*tan(f*x+e))^(3/2)-1/16*(7*I*A-3*B)/a/c^2/f/(c-I*c*tan(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 3.92 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.57 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {5 (7 A+3 i B) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\frac {1}{2} i (i+\tan (e+f x))\right ) \sec ^2(e+f x)+6 i (3 i A-7 B+(7 A+3 i B) \tan (e+f x))}{120 a c^2 f (-i+\tan (e+f x)) (i+\tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \] Input:
Integrate[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f* x])^(5/2)),x]
Output:
(5*(7*A + (3*I)*B)*Hypergeometric2F1[-3/2, 1, -1/2, (-1/2*I)*(I + Tan[e + f*x])]*Sec[e + f*x]^2 + (6*I)*((3*I)*A - 7*B + (7*A + (3*I)*B)*Tan[e + f*x ]))/(120*a*c^2*f*(-I + Tan[e + f*x])*(I + Tan[e + f*x])^2*Sqrt[c - I*c*Tan [e + f*x]])
Time = 0.69 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.92, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {3042, 4071, 27, 87, 61, 61, 61, 73, 219}
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.
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{a^2 (i \tan (e+f x)+1)^2 (c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {c \int \frac {A+B \tan (e+f x)}{(i \tan (e+f x)+1)^2 (c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{a f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (7 A+3 i B) \int \frac {1}{(i \tan (e+f x)+1) (c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)+\frac {-B+i A}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}\right )}{a f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (7 A+3 i B) \left (\frac {\int \frac {1}{(i \tan (e+f x)+1) (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{2 c}-\frac {i}{5 c (c-i c \tan (e+f x))^{5/2}}\right )+\frac {-B+i A}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}\right )}{a f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (7 A+3 i B) \left (\frac {\frac {\int \frac {1}{(i \tan (e+f x)+1) (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{2 c}-\frac {i}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {i}{5 c (c-i c \tan (e+f x))^{5/2}}\right )+\frac {-B+i A}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}\right )}{a f}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (7 A+3 i B) \left (\frac {\frac {\frac {\int \frac {1}{(i \tan (e+f x)+1) \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)}{2 c}-\frac {i}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {i}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {i}{5 c (c-i c \tan (e+f x))^{5/2}}\right )+\frac {-B+i A}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}\right )}{a f}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (7 A+3 i B) \left (\frac {\frac {\frac {i \int \frac {1}{2-\frac {c-i c \tan (e+f x)}{c}}d\sqrt {c-i c \tan (e+f x)}}{c^2}-\frac {i}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {i}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {i}{5 c (c-i c \tan (e+f x))^{5/2}}\right )+\frac {-B+i A}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}\right )}{a f}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {c \left (\frac {1}{4} (7 A+3 i B) \left (\frac {\frac {\frac {i \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} c^{3/2}}-\frac {i}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {i}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {i}{5 c (c-i c \tan (e+f x))^{5/2}}\right )+\frac {-B+i A}{2 c (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}\right )}{a f}\) |
Input:
Int[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5 /2)),x]
Output:
(c*((I*A - B)/(2*c*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)) + (( 7*A + (3*I)*B)*((-1/5*I)/(c*(c - I*c*Tan[e + f*x])^(5/2)) + ((-1/3*I)/(c*( c - I*c*Tan[e + f*x])^(3/2)) + ((I*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqr t[2]*Sqrt[c])])/(Sqrt[2]*c^(3/2)) - I/(c*Sqrt[c - I*c*Tan[e + f*x]]))/(2*c ))/(2*c)))/4))/(a*f)
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] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x], x , Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
Time = 0.46 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.75
| method | result | size |
| derivativedivides | \(\frac {2 i c \left (-\frac {i B +3 A}{16 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {A}{12 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {-i B +A}{20 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {\frac {\left (\frac {i B}{4}+\frac {A}{4}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\frac {c}{2}+\frac {i c \tan \left (f x +e \right )}{2}}+\frac {\left (\frac {7 A}{2}+\frac {3 i B}{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}}{16 c^{3}}\right )}{f a}\) | \(168\) |
| default | \(\frac {2 i c \left (-\frac {i B +3 A}{16 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {A}{12 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {-i B +A}{20 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {\frac {\left (\frac {i B}{4}+\frac {A}{4}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\frac {c}{2}+\frac {i c \tan \left (f x +e \right )}{2}}+\frac {\left (\frac {7 A}{2}+\frac {3 i B}{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}}{16 c^{3}}\right )}{f a}\) | \(168\) |
Input:
int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x,method= _RETURNVERBOSE)
Output:
2*I/f/a*c*(-1/16/c^3*(3*A+I*B)/(c-I*c*tan(f*x+e))^(1/2)-1/12/c^2*A/(c-I*c* tan(f*x+e))^(3/2)-1/20/c*(A-I*B)/(c-I*c*tan(f*x+e))^(5/2)+1/16/c^3*((1/4*I *B+1/4*A)*(c-I*c*tan(f*x+e))^(1/2)/(1/2*c+1/2*I*c*tan(f*x+e))+1/2*(7/2*A+3 /2*I*B)*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/ 2))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (172) = 344\).
Time = 0.11 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.87 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {{\left (15 \, \sqrt {\frac {1}{2}} a c^{3} f \sqrt {-\frac {49 \, A^{2} + 42 i \, A B - 9 \, B^{2}}{a^{2} c^{5} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {49 \, A^{2} + 42 i \, A B - 9 \, B^{2}}{a^{2} c^{5} f^{2}}} + 7 i \, A - 3 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a c^{2} f}\right ) - 15 \, \sqrt {\frac {1}{2}} a c^{3} f \sqrt {-\frac {49 \, A^{2} + 42 i \, A B - 9 \, B^{2}}{a^{2} c^{5} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {49 \, A^{2} + 42 i \, A B - 9 \, B^{2}}{a^{2} c^{5} f^{2}}} - 7 i \, A + 3 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a c^{2} f}\right ) - \sqrt {2} {\left (6 \, {\left (i \, A + B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 2 \, {\left (19 i \, A + 9 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 4 \, {\left (37 i \, A - 3 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (-101 i \, A + 9 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 15 i \, A + 15 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{480 \, a c^{3} f} \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")
Output:
1/480*(15*sqrt(1/2)*a*c^3*f*sqrt(-(49*A^2 + 42*I*A*B - 9*B^2)/(a^2*c^5*f^2 ))*e^(2*I*f*x + 2*I*e)*log(1/8*(sqrt(2)*sqrt(1/2)*(a*c^2*f*e^(2*I*f*x + 2* I*e) + a*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(49*A^2 + 42*I*A*B - 9*B^2)/(a^2*c^5*f^2)) + 7*I*A - 3*B)*e^(-I*f*x - I*e)/(a*c^2*f)) - 15*s qrt(1/2)*a*c^3*f*sqrt(-(49*A^2 + 42*I*A*B - 9*B^2)/(a^2*c^5*f^2))*e^(2*I*f *x + 2*I*e)*log(-1/8*(sqrt(2)*sqrt(1/2)*(a*c^2*f*e^(2*I*f*x + 2*I*e) + a*c ^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(49*A^2 + 42*I*A*B - 9*B^2)/ (a^2*c^5*f^2)) - 7*I*A + 3*B)*e^(-I*f*x - I*e)/(a*c^2*f)) - sqrt(2)*(6*(I* A + B)*e^(8*I*f*x + 8*I*e) + 2*(19*I*A + 9*B)*e^(6*I*f*x + 6*I*e) + 4*(37* I*A - 3*B)*e^(4*I*f*x + 4*I*e) - (-101*I*A + 9*B)*e^(2*I*f*x + 2*I*e) - 15 *I*A + 15*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a*c^ 3*f)
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=- \frac {i \left (\int \frac {A}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} - i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan {\left (e + f x \right )}}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} - i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx\right )}{a} \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))**(5/2),x)
Output:
-I*(Integral(A/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - I*c**2 *sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) - I*c**2*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(B* tan(e + f*x)/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - I*c**2*s qrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) - I*c**2*sqrt(-I*c*tan(e + f*x) + c)), x))/a
Time = 0.12 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.87 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i \, {\left (\frac {4 \, {\left (15 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} {\left (7 \, A + 3 i \, B\right )} - 20 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (7 \, A + 3 i \, B\right )} c - 8 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (7 \, A + 3 i \, B\right )} c^{2} - 48 \, {\left (A - i \, B\right )} c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a c - 2 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a c^{2}} + \frac {15 \, \sqrt {2} {\left (7 \, A + 3 i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a c^{\frac {3}{2}}}\right )}}{960 \, c f} \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")
Output:
-1/960*I*(4*(15*(-I*c*tan(f*x + e) + c)^3*(7*A + 3*I*B) - 20*(-I*c*tan(f*x + e) + c)^2*(7*A + 3*I*B)*c - 8*(-I*c*tan(f*x + e) + c)*(7*A + 3*I*B)*c^2 - 48*(A - I*B)*c^3)/((-I*c*tan(f*x + e) + c)^(7/2)*a*c - 2*(-I*c*tan(f*x + e) + c)^(5/2)*a*c^2) + 15*sqrt(2)*(7*A + 3*I*B)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c )))/(a*c^(3/2)))/(c*f)
Exception generated. \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*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 TypeDone
Time = 7.09 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.38 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\frac {B\,c}{5}-\frac {B\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{10}-\frac {B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{4\,c}+\frac {3\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3}{16\,c^2}}{a\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}-2\,a\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\frac {A\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{30\,a\,f}+\frac {A\,c\,1{}\mathrm {i}}{5\,a\,f}+\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,7{}\mathrm {i}}{12\,a\,c\,f}-\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,7{}\mathrm {i}}{16\,a\,c^2\,f}}{2\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,7{}\mathrm {i}}{32\,a\,{\left (-c\right )}^{5/2}\,f}-\frac {3\,\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{32\,a\,c^{5/2}\,f} \] Input:
int((A + B*tan(e + f*x))/((a + a*tan(e + f*x)*1i)*(c - c*tan(e + f*x)*1i)^ (5/2)),x)
Output:
((B*c)/5 - (B*(c - c*tan(e + f*x)*1i))/10 - (B*(c - c*tan(e + f*x)*1i)^2)/ (4*c) + (3*B*(c - c*tan(e + f*x)*1i)^3)/(16*c^2))/(a*f*(c - c*tan(e + f*x) *1i)^(7/2) - 2*a*c*f*(c - c*tan(e + f*x)*1i)^(5/2)) - ((A*(c - c*tan(e + f *x)*1i)*7i)/(30*a*f) + (A*c*1i)/(5*a*f) + (A*(c - c*tan(e + f*x)*1i)^2*7i) /(12*a*c*f) - (A*(c - c*tan(e + f*x)*1i)^3*7i)/(16*a*c^2*f))/(2*c*(c - c*t an(e + f*x)*1i)^(5/2) - (c - c*tan(e + f*x)*1i)^(7/2)) - (2^(1/2)*A*atan(( 2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*7i)/(32*a*(-c)^(5/2 )*f) - (3*2^(1/2)*B*atanh((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/ 2))))/(32*a*c^(5/2)*f)
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx =\text {Too large to display} \] Input:
int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x)
Output:
( - sqrt( - tan(e + f*x)*i + 1)*a*i + 165888*int(( - sqrt( - tan(e + f*x)* i + 1))/(36864*tan(e + f*x)**5*i - 36864*tan(e + f*x)**4 + 73728*tan(e + f *x)**3*i - 73728*tan(e + f*x)**2 + 36864*tan(e + f*x)*i - 36864),x)*tan(e + f*x)**4*a*f + 331776*int(( - sqrt( - tan(e + f*x)*i + 1))/(36864*tan(e + f*x)**5*i - 36864*tan(e + f*x)**4 + 73728*tan(e + f*x)**3*i - 73728*tan(e + f*x)**2 + 36864*tan(e + f*x)*i - 36864),x)*tan(e + f*x)**2*a*f + 165888 *int(( - sqrt( - tan(e + f*x)*i + 1))/(36864*tan(e + f*x)**5*i - 36864*tan (e + f*x)**4 + 73728*tan(e + f*x)**3*i - 73728*tan(e + f*x)**2 + 36864*tan (e + f*x)*i - 36864),x)*a*f - 129024*int(( - sqrt( - tan(e + f*x)*i + 1)*t an(e + f*x)**2)/(36864*tan(e + f*x)**5*i - 36864*tan(e + f*x)**4 + 73728*t an(e + f*x)**3*i - 73728*tan(e + f*x)**2 + 36864*tan(e + f*x)*i - 36864),x )*tan(e + f*x)**4*a*f - 258048*int(( - sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(36864*tan(e + f*x)**5*i - 36864*tan(e + f*x)**4 + 73728*tan(e + f*x)**3*i - 73728*tan(e + f*x)**2 + 36864*tan(e + f*x)*i - 36864),x)*tan( e + f*x)**2*a*f - 129024*int(( - sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)* *2)/(36864*tan(e + f*x)**5*i - 36864*tan(e + f*x)**4 + 73728*tan(e + f*x)* *3*i - 73728*tan(e + f*x)**2 + 36864*tan(e + f*x)*i - 36864),x)*a*f - 4*in t(tan(e + f*x)/(sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**3*i - sqrt( - ta n(e + f*x)*i + 1)*tan(e + f*x)**2 + sqrt( - tan(e + f*x)*i + 1)*tan(e + f* x)*i - sqrt( - tan(e + f*x)*i + 1)),x)*tan(e + f*x)**4*b*f - 8*int(tan(...