Integrand size = 28, antiderivative size = 76 \[ \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {2 i a \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}-\frac {2 a}{(i c+d) f \sqrt {c+d \tan (e+f x)}} \] Output:
-2*I*a*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(3/2)/f-2*a/( I*c+d)/f/(c+d*tan(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.78 \[ \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {2 i a \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )}{(c-i d) f \sqrt {c+d \tan (e+f x)}} \] Input:
Integrate[(a + I*a*Tan[e + f*x])/(c + d*Tan[e + f*x])^(3/2),x]
Output:
((2*I)*a*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*Tan[e + f*x])/(c - I*d)])/ ((c - I*d)*f*Sqrt[c + d*Tan[e + f*x]])
Time = 0.48 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.30, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4012, 3042, 4020, 25, 27, 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.
\(\displaystyle \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {\int \frac {a (c+i d)+a (i c-d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}-\frac {2 a}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a (c+i d)+a (i c-d) \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}-\frac {2 a}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 4020 |
\(\displaystyle \frac {i a^2 (c+i d)^2 \int -\frac {1}{a (c+i d) (a (c+i d)-a (i c-d) \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(a (i c-d) \tan (e+f x))}{f \left (c^2+d^2\right )}-\frac {2 a}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {i a^2 (c+i d)^2 \int \frac {1}{a (c+i d) (a (c+i d)-a (i c-d) \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(a (i c-d) \tan (e+f x))}{f \left (c^2+d^2\right )}-\frac {2 a}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {i a (c+i d) \int \frac {1}{(a (c+i d)-a (i c-d) \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}d(a (i c-d) \tan (e+f x))}{f \left (c^2+d^2\right )}-\frac {2 a}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {2 i a^2 (-d+i c) (c+i d) \int \frac {1}{\frac {i a \left (c^2+d^2\right )}{d}-\frac {a^3 (i c-d)^3 \tan ^2(e+f x)}{d}}d\sqrt {c+d \tan (e+f x)}}{d f \left (c^2+d^2\right )}-\frac {2 a}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 a (-d+i c) \sqrt {c-i d} (c+i d) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{f \left (c^2+d^2\right )^2}-\frac {2 a}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\) |
Input:
Int[(a + I*a*Tan[e + f*x])/(c + d*Tan[e + f*x])^(3/2),x]
Output:
(-2*a*(I*c - d)*Sqrt[c - I*d]*(c + I*d)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/S qrt[c - I*d]])/((c^2 + d^2)^2*f) - (2*a)/((I*c + d)*f*Sqrt[c + d*Tan[e + f *x]])
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[(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[((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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2439 vs. \(2 (64 ) = 128\).
Time = 0.40 (sec) , antiderivative size = 2440, normalized size of antiderivative = 32.11
method | result | size |
derivativedivides | \(\text {Expression too large to display}\) | \(2440\) |
default | \(\text {Expression too large to display}\) | \(2440\) |
parts | \(\text {Expression too large to display}\) | \(3682\) |
Input:
int((a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
Output:
2/f*a/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*ar ctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2) ^(1/2)-2*c)^(1/2))*c^2*d+I/f*a/(c^2+d^2)^(1/2)/((c^2+d^2)^(1/2)+c)/(2*(c^2 +d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^(1/2) +2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c+I/f*a/(c^2+d^2)^(1/2)/((c^2+ d^2)^(1/2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/ 2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c-1/2/f*a /(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)*ln(d*tan(f*x+e)+c-(c+d*tan(f*x+e))^(1 /2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)^(1/2)+2*c) ^(1/2)*c*d+I/f*a/(c^2+d^2)/((c^2+d^2)^(1/2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/ 2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2 +d^2)^(1/2)-2*c)^(1/2))*d^2+1/f*a/(c^2+d^2)/((c^2+d^2)^(1/2)+c)/(2*(c^2+d^ 2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*(c^2+d^2)^(1/2)+2* c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c*d+I/f*a/(c^2+d^2)/((c^2+d^2)^(1 /2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)-(2*( c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*d^2+1/f*a/(c^2+d ^2)/((c^2+d^2)^(1/2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f *x+e))^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)) *c*d-1/4*I/f*a/(c^2+d^2)/((c^2+d^2)^(1/2)+c)*ln(d*tan(f*x+e)+c+(c+d*tan(f* x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*(2*(c^2+d^2)...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 534 vs. \(2 (60) = 120\).
Time = 0.10 (sec) , antiderivative size = 534, normalized size of antiderivative = 7.03 \[ \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {{\left ({\left (c^{2} - 2 i \, c d - d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} + d^{2}\right )} f\right )} \sqrt {\frac {4 i \, a^{2}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} \log \left (\frac {{\left (2 \, a c + {\left ({\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, c^{2} + 2 \, c d - i \, d^{2}\right )} f\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 {4 i \, a^{2}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} + 2 \, {\left (a c - i \, a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) - {\left ({\left (c^{2} - 2 i \, c d - d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} + d^{2}\right )} f\right )} \sqrt {\frac {4 i \, a^{2}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} \log \left (\frac {{\left (2 \, a c + {\left ({\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} f\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 {4 i \, a^{2}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} + 2 \, {\left (a c - i \, a d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a}\right ) + 8 \, {\left (i \, a e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a\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}}}{4 \, {\left ({\left (c^{2} - 2 i \, c d - d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} + d^{2}\right )} f\right )}} \] Input:
integrate((a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")
Output:
1/4*(((c^2 - 2*I*c*d - d^2)*f*e^(2*I*f*x + 2*I*e) + (c^2 + d^2)*f)*sqrt(4* I*a^2/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*log((2*a*c + ((I*c^2 + 2 *c*d - I*d^2)*f*e^(2*I*f*x + 2*I*e) + (I*c^2 + 2*c*d - I*d^2)*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(4*I* a^2/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2)) + 2*(a*c - I*a*d)*e^(2*I*f *x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a) - ((c^2 - 2*I*c*d - d^2)*f*e^(2*I*f*x + 2*I*e) + (c^2 + d^2)*f)*sqrt(4*I*a^2/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d ^3)*f^2))*log((2*a*c + ((-I*c^2 - 2*c*d + I*d^2)*f*e^(2*I*f*x + 2*I*e) + ( -I*c^2 - 2*c*d + I*d^2)*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(4*I*a^2/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d ^3)*f^2)) + 2*(a*c - I*a*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a) + 8*(I*a*e^(2*I*f*x + 2*I*e) + I*a)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/((c^2 - 2*I*c*d - d^2)*f*e^(2*I*f*x + 2*I*e) + (c^2 + d^2)*f)
\[ \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=i a \left (\int \left (- \frac {i}{c \sqrt {c + d \tan {\left (e + f x \right )}} + d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}\right )\, dx + \int \frac {\tan {\left (e + f x \right )}}{c \sqrt {c + d \tan {\left (e + f x \right )}} + d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}\, dx\right ) \] Input:
integrate((a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))**(3/2),x)
Output:
I*a*(Integral(-I/(c*sqrt(c + d*tan(e + f*x)) + d*sqrt(c + d*tan(e + f*x))* tan(e + f*x)), x) + Integral(tan(e + f*x)/(c*sqrt(c + d*tan(e + f*x)) + d* sqrt(c + d*tan(e + f*x))*tan(e + f*x)), x))
Exception generated. \[ \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(((-(2*c*d^4)/((c^2-d^2)^2>0)', s ee `assume
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (60) = 120\).
Time = 0.51 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.49 \[ \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {2 i \, a {\left (\frac {\sqrt {2} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{{\left (c - i \, d\right )} \sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {1}{\sqrt {d \tan \left (f x + e\right ) + c} {\left (c - i \, d\right )}}\right )}}{f} \] Input:
integrate((a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")
Output:
2*I*a*(sqrt(2)*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt (d*tan(f*x + e) + c))/(sqrt(2)*c*sqrt(-c + sqrt(c^2 + d^2)) - I*sqrt(2)*sq rt(-c + sqrt(c^2 + d^2))*d - sqrt(2)*sqrt(c^2 + d^2)*sqrt(-c + sqrt(c^2 + d^2))))/((c - I*d)*sqrt(-c + sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) + 1/(sqrt(d*tan(f*x + e) + c)*(c - I*d)))/f
Time = 11.37 (sec) , antiderivative size = 4612, normalized size of antiderivative = 60.68 \[ \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:
int((a + a*tan(e + f*x)*1i)/(c + d*tan(e + f*x))^(3/2),x)
Output:
(log((a^3*c*d^2*8i)/(f^3*(c^2 + d^2)^2) - ((((16*c*d^2*(c + d*tan(e + f*x) )^(1/2)*((4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - 4*a^2*c^3*f^2 + 12*a^2* c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(1/2) - (32*a*d^2*(c^2*1i - d^2*1i))/(f*(c ^2 + d^2)))*((4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - 4*a^2*c^3*f^2 + 12* a^2*c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(1/2))/4 + (16*a^2*d^2*(c^2 - d^2)*(c + d*tan(e + f*x))^(1/2))/(f^2*(c^2 + d^2)^2))*((4*(-a^4*d^2*f^4*(3*c^2 - d ^2)^2)^(1/2) - 4*a^2*c^3*f^2 + 12*a^2*c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(1/2 ))/4)*(((96*a^4*c^2*d^4*f^4 - 16*a^4*d^6*f^4 - 144*a^4*c^4*d^2*f^4)^(1/2) - 4*a^2*c^3*f^2 + 12*a^2*c*d^2*f^2)/(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3 *c^4*d^2*f^4))^(1/2))/4 + (log((a^3*c*d^2*8i)/(f^3*(c^2 + d^2)^2) - ((((16 *c*d^2*(c + d*tan(e + f*x))^(1/2)*(-(4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2 ) + 4*a^2*c^3*f^2 - 12*a^2*c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(1/2) - (32*a*d ^2*(c^2*1i - d^2*1i))/(f*(c^2 + d^2)))*(-(4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2) ^(1/2) + 4*a^2*c^3*f^2 - 12*a^2*c*d^2*f^2)/(f^4*(c^2 + d^2)^3))^(1/2))/4 + (16*a^2*d^2*(c^2 - d^2)*(c + d*tan(e + f*x))^(1/2))/(f^2*(c^2 + d^2)^2))* (-(4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) + 4*a^2*c^3*f^2 - 12*a^2*c*d^2*f ^2)/(f^4*(c^2 + d^2)^3))^(1/2))/4)*(-((96*a^4*c^2*d^4*f^4 - 16*a^4*d^6*f^4 - 144*a^4*c^4*d^2*f^4)^(1/2) + 4*a^2*c^3*f^2 - 12*a^2*c*d^2*f^2)/(c^6*f^4 + d^6*f^4 + 3*c^2*d^4*f^4 + 3*c^4*d^2*f^4))^(1/2))/4 - log(((((16*c*d^2*( c + d*tan(e + f*x))^(1/2)*((4*(-a^4*d^2*f^4*(3*c^2 - d^2)^2)^(1/2) - 4*...
\[ \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {a \left (-2 \sqrt {d \tan \left (f x +e \right )+c}-\left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2} d^{2}+2 \tan \left (f x +e \right ) c d +c^{2}}d x \right ) \tan \left (f x +e \right ) d^{2} f -\left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2} d^{2}+2 \tan \left (f x +e \right ) c d +c^{2}}d x \right ) c d f +\left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2} d^{2}+2 \tan \left (f x +e \right ) c d +c^{2}}d x \right ) \tan \left (f x +e \right ) d^{2} f i +\left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{2} d^{2}+2 \tan \left (f x +e \right ) c d +c^{2}}d x \right ) c d f i \right )}{d f \left (d \tan \left (f x +e \right )+c \right )} \] Input:
int((a+I*a*tan(f*x+e))/(c+d*tan(f*x+e))^(3/2),x)
Output:
(a*( - 2*sqrt(tan(e + f*x)*d + c) - int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*tan(e + f*x )*d**2*f - int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)**2 *d**2 + 2*tan(e + f*x)*c*d + c**2),x)*c*d*f + int((sqrt(tan(e + f*x)*d + c )*tan(e + f*x))/(tan(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*tan( e + f*x)*d**2*f*i + int((sqrt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e + f *x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2),x)*c*d*f*i))/(d*f*(tan(e + f*x)*d + c))