Integrand size = 32, antiderivative size = 262 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} \sqrt {c-i d} f}-\frac {\sqrt {c+d \tan (e+f x)}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac {(5 i c-13 d) \sqrt {c+d \tan (e+f x)}}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2}}+\frac {\left (15 c^2+50 i c d-67 d^2\right ) \sqrt {c+d \tan (e+f x)}}{60 a^2 (i c-d)^3 f \sqrt {a+i a \tan (e+f x)}} \] Output:
-1/8*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^(5/2)/(c-I*d)^(1/2)/f-1/5*(c+d*tan(f*x+e))^( 1/2)/(I*c-d)/f/(a+I*a*tan(f*x+e))^(5/2)+1/30*(5*I*c-13*d)*(c+d*tan(f*x+e)) ^(1/2)/a/(c+I*d)^2/f/(a+I*a*tan(f*x+e))^(3/2)+1/60*(15*c^2+50*I*c*d-67*d^2 )*(c+d*tan(f*x+e))^(1/2)/a^2/(I*c-d)^3/f/(a+I*a*tan(f*x+e))^(1/2)
Time = 2.35 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {\frac {15 i \sqrt {2} a^2 (c+i d)^2 \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a (c-i d)}}+\frac {24 a^4 (-i c+d) \sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/2}}+\frac {2 a^2 \left (-15 i c^2+50 c d+67 i d^2\right ) \sqrt {c+d \tan (e+f x)}}{(c+i d) \sqrt {a+i a \tan (e+f x)}}-\frac {4 a^2 (5 c+13 i d) \sqrt {c+d \tan (e+f x)}}{(-i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}}}{120 a^4 (c+i d)^2 f} \] Input:
Integrate[1/((a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]]),x]
Output:
-1/120*(((15*I)*Sqrt[2]*a^2*(c + I*d)^2*ArcTan[(Sqrt[-(a*(c - I*d))]*Sqrt[ a + I*a*Tan[e + f*x]])/(Sqrt[2]*a*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[-(a*(c - I*d))] + (24*a^4*((-I)*c + d)*Sqrt[c + d*Tan[e + f*x]])/(a + I*a*Tan[e + f*x])^(5/2) + (2*a^2*((-15*I)*c^2 + 50*c*d + (67*I)*d^2)*Sqrt[c + d*Tan[e + f*x]])/((c + I*d)*Sqrt[a + I*a*Tan[e + f*x]]) - (4*a^2*(5*c + (13*I)*d) *Sqrt[c + d*Tan[e + f*x]])/((-I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]] ))/(a^4*(c + I*d)^2*f)
Time = 1.34 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.16, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 4042, 27, 3042, 4079, 27, 3042, 4079, 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.
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}dx\) |
| \(\Big \downarrow \) 4042 |
| \(\displaystyle -\frac {\int -\frac {a (5 i c-9 d)+4 i a d \tan (e+f x)}{2 (i \tan (e+f x) a+a)^{3/2} \sqrt {c+d \tan (e+f x)}}dx}{5 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (5 i c-9 d)+4 i a d \tan (e+f x)}{(i \tan (e+f x) a+a)^{3/2} \sqrt {c+d \tan (e+f x)}}dx}{10 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (5 i c-9 d)+4 i a d \tan (e+f x)}{(i \tan (e+f x) a+a)^{3/2} \sqrt {c+d \tan (e+f x)}}dx}{10 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {-\frac {\int \frac {\left (15 c^2+40 i d c-41 d^2\right ) a^2+2 (5 c+13 i d) d \tan (e+f x) a^2}{2 \sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)}}dx}{3 a^2 (-d+i c)}-\frac {a (5 c+13 i d) \sqrt {c+d \tan (e+f x)}}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2}}}{10 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \frac {\left (15 c^2+40 i d c-41 d^2\right ) a^2+2 (5 c+13 i d) d \tan (e+f x) a^2}{\sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)}}dx}{6 a^2 (-d+i c)}-\frac {a (5 c+13 i d) \sqrt {c+d \tan (e+f x)}}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2}}}{10 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\left (15 c^2+40 i d c-41 d^2\right ) a^2+2 (5 c+13 i d) d \tan (e+f x) a^2}{\sqrt {i \tan (e+f x) a+a} \sqrt {c+d \tan (e+f x)}}dx}{6 a^2 (-d+i c)}-\frac {a (5 c+13 i d) \sqrt {c+d \tan (e+f x)}}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2}}}{10 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {-\frac {\frac {a^2 \left (15 i c^2-50 c d-67 i d^2\right ) \sqrt {c+d \tan (e+f x)}}{f (c+i d) \sqrt {a+i a \tan (e+f x)}}-\frac {\int \frac {15 a^3 (i c-d)^3 \sqrt {i \tan (e+f x) a+a}}{2 \sqrt {c+d \tan (e+f x)}}dx}{a^2 (-d+i c)}}{6 a^2 (-d+i c)}-\frac {a (5 c+13 i d) \sqrt {c+d \tan (e+f x)}}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2}}}{10 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\frac {a^2 \left (15 i c^2-50 c d-67 i d^2\right ) \sqrt {c+d \tan (e+f x)}}{f (c+i d) \sqrt {a+i a \tan (e+f x)}}-\frac {15}{2} a (-d+i c)^2 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{6 a^2 (-d+i c)}-\frac {a (5 c+13 i d) \sqrt {c+d \tan (e+f x)}}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2}}}{10 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {a^2 \left (15 i c^2-50 c d-67 i d^2\right ) \sqrt {c+d \tan (e+f x)}}{f (c+i d) \sqrt {a+i a \tan (e+f x)}}-\frac {15}{2} a (-d+i c)^2 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{6 a^2 (-d+i c)}-\frac {a (5 c+13 i d) \sqrt {c+d \tan (e+f x)}}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2}}}{10 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \frac {-\frac {\frac {15 i a^3 (-d+i c)^2 \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f}+\frac {a^2 \left (15 i c^2-50 c d-67 i d^2\right ) \sqrt {c+d \tan (e+f x)}}{f (c+i d) \sqrt {a+i a \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {a (5 c+13 i d) \sqrt {c+d \tan (e+f x)}}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2}}}{10 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {-\frac {\frac {15 i a^{3/2} (-d+i c)^2 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} f \sqrt {c-i d}}+\frac {a^2 \left (15 i c^2-50 c d-67 i d^2\right ) \sqrt {c+d \tan (e+f x)}}{f (c+i d) \sqrt {a+i a \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {a (5 c+13 i d) \sqrt {c+d \tan (e+f x)}}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2}}}{10 a^2 (-d+i c)}-\frac {\sqrt {c+d \tan (e+f x)}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}\) |
Input:
Int[1/((a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]]),x]
Output:
-1/5*Sqrt[c + d*Tan[e + f*x]]/((I*c - d)*f*(a + I*a*Tan[e + f*x])^(5/2)) + (-1/3*(a*(5*c + (13*I)*d)*Sqrt[c + d*Tan[e + f*x]])/((c + I*d)*f*(a + I*a *Tan[e + f*x])^(3/2)) - (((15*I)*a^(3/2)*(I*c - d)^2*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[2]*Sqrt[c - I*d]*f) + (a^2*((15*I)*c^2 - 50*c*d - (67*I)*d^2)*Sqrt[ c + d*Tan[e + f*x]])/((c + I*d)*f*Sqrt[a + I*a*Tan[e + f*x]]))/(6*a^2*(I*c - d)))/(10*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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 5217 vs. \(2 (214 ) = 428\).
Time = 0.80 (sec) , antiderivative size = 5218, normalized size of antiderivative = 19.92
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(5218\) |
| default | \(\text {Expression too large to display}\) | \(5218\) |
Input:
int(1/(a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(1/2),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. 571 vs. \(2 (202) = 404\).
Time = 0.17 (sec) , antiderivative size = 571, normalized size of antiderivative = 2.18 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=\frac {{\left (30 \, {\left (i \, a^{3} c^{3} - 3 \, a^{3} c^{2} d - 3 i \, a^{3} c d^{2} + a^{3} d^{3}\right )} f \sqrt {\frac {i}{8 \, {\left (-i \, a^{5} c - a^{5} d\right )} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (-4 \, {\left (i \, a^{3} c + a^{3} d\right )} f \sqrt {\frac {i}{8 \, {\left (-i \, a^{5} c - a^{5} d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) + 30 \, {\left (-i \, a^{3} c^{3} + 3 \, a^{3} c^{2} d + 3 i \, a^{3} c d^{2} - a^{3} d^{3}\right )} f \sqrt {\frac {i}{8 \, {\left (-i \, a^{5} c - a^{5} d\right )} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (-4 \, {\left (-i \, a^{3} c - a^{3} d\right )} f \sqrt {\frac {i}{8 \, {\left (-i \, a^{5} c - a^{5} d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right ) + \sqrt {2} {\left (3 \, c^{2} + 6 i \, c d - 3 \, d^{2} + {\left (23 \, c^{2} + 74 i \, c d - 83 \, d^{2}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, {\left (17 \, c^{2} + 52 i \, c d - 51 \, d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (7 \, c^{2} + 18 i \, c d - 11 \, d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{120 \, {\left (-i \, a^{3} c^{3} + 3 \, a^{3} c^{2} d + 3 i \, a^{3} c d^{2} - a^{3} d^{3}\right )} f} \] Input:
integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm=" fricas")
Output:
1/120*(30*(I*a^3*c^3 - 3*a^3*c^2*d - 3*I*a^3*c*d^2 + a^3*d^3)*f*sqrt(1/8*I /((-I*a^5*c - a^5*d)*f^2))*e^(5*I*f*x + 5*I*e)*log(-4*(I*a^3*c + a^3*d)*f* sqrt(1/8*I/((-I*a^5*c - a^5*d)*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)) + 30*(-I*a^3*c^3 + 3*a^3 *c^2*d + 3*I*a^3*c*d^2 - a^3*d^3)*f*sqrt(1/8*I/((-I*a^5*c - a^5*d)*f^2))*e ^(5*I*f*x + 5*I*e)*log(-4*(-I*a^3*c - a^3*d)*f*sqrt(1/8*I/((-I*a^5*c - a^5 *d)*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)) + sqrt(2)*(3*c^2 + 6*I*c*d - 3*d^2 + (23*c^2 + 74*I *c*d - 83*d^2)*e^(6*I*f*x + 6*I*e) + 2*(17*c^2 + 52*I*c*d - 51*d^2)*e^(4*I *f*x + 4*I*e) + 2*(7*c^2 + 18*I*c*d - 11*d^2)*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)))*e^(-5*I*f*x - 5*I*e)/((-I*a^3*c^3 + 3*a^3*c^2* d + 3*I*a^3*c*d^2 - a^3*d^3)*f)
\[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \] Input:
integrate(1/(a+I*a*tan(f*x+e))**(5/2)/(c+d*tan(f*x+e))**(1/2),x)
Output:
Integral(1/((I*a*(tan(e + f*x) - I))**(5/2)*sqrt(c + d*tan(e + f*x))), x)
Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm=" maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(1/2),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. \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=\int \frac {1}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \] Input:
int(1/((a + a*tan(e + f*x)*1i)^(5/2)*(c + d*tan(e + f*x))^(1/2)),x)
Output:
int(1/((a + a*tan(e + f*x)*1i)^(5/2)*(c + d*tan(e + f*x))^(1/2)), x)
\[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx=-\frac {\int \frac {1}{\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}-2 \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right ) i -\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {d \tan \left (f x +e \right )+c}}d x}{\sqrt {a}\, a^{2}} \] Input:
int(1/(a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(1/2),x)
Output:
( - int(1/(sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x)* *2 - 2*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x)*i - sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)),x))/(sqrt(a)*a**2)