Integrand size = 30, antiderivative size = 139 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {8 i a^3 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2} f}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{(c-i d) d f \sqrt {c+d \tan (e+f x)}}+\frac {4 a^3 c \sqrt {c+d \tan (e+f x)}}{d^2 (i c+d) f} \]
-8*I*a^3*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(3/2)/f+4*a ^3*c*(c+d*tan(f*x+e))^(1/2)/d^2/(I*c+d)/f+2*(c+I*d)*(a^3+I*a^3*tan(f*x+e)) /(c-I*d)/d/f/(c+d*tan(f*x+e))^(1/2)
Time = 2.32 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.82 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {2 a^3 \left (-\frac {4 i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{3/2}}+\frac {-2 i c^2+c d+i d^2-i (c-i d) d \tan (e+f x)}{(c-i d) d^2 \sqrt {c+d \tan (e+f x)}}\right )}{f} \]
(2*a^3*(((-4*I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(c - I*d) ^(3/2) + ((-2*I)*c^2 + c*d + I*d^2 - I*(c - I*d)*d*Tan[e + f*x])/((c - I*d )*d^2*Sqrt[c + d*Tan[e + f*x]])))/f
Time = 0.72 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4036, 25, 3042, 4075, 3042, 4020, 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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4036 |
| \(\displaystyle \frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}-\frac {2 \int -\frac {(i \tan (e+f x) a+a) \left (a^2 (c+2 i d)-i a^2 c \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}}dx}{d (d+i c)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2 \int \frac {(i \tan (e+f x) a+a) \left (a^2 (c+2 i d)-i a^2 c \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}}dx}{d (d+i c)}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \int \frac {(i \tan (e+f x) a+a) \left (a^2 (c+2 i d)-i a^2 c \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}}dx}{d (d+i c)}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4075 |
| \(\displaystyle \frac {2 \left (\frac {2 a^3 c \sqrt {c+d \tan (e+f x)}}{d f}+\int \frac {2 i a^3 d-2 a^3 d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx\right )}{d (d+i c)}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {2 a^3 c \sqrt {c+d \tan (e+f x)}}{d f}+\int \frac {2 i a^3 d-2 a^3 d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx\right )}{d (d+i c)}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4020 |
| \(\displaystyle \frac {2 \left (\frac {2 a^3 c \sqrt {c+d \tan (e+f x)}}{d f}-\frac {4 i a^6 d^2 \int \frac {1}{\sqrt {2} a^3 d \sqrt {2 c+2 d \tan (e+f x)} \left (2 a^3 d-2 i a^3 d \tan (e+f x)\right )}d\left (-2 a^3 d \tan (e+f x)\right )}{f}\right )}{d (d+i c)}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 \left (\frac {2 a^3 c \sqrt {c+d \tan (e+f x)}}{d f}-\frac {2 i \sqrt {2} a^3 d \int \frac {1}{\sqrt {2 c+2 d \tan (e+f x)} \left (2 a^3 d-2 i a^3 d \tan (e+f x)\right )}d\left (-2 a^3 d \tan (e+f x)\right )}{f}\right )}{d (d+i c)}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 \left (\frac {2 a^3 c \sqrt {c+d \tan (e+f x)}}{d f}+\frac {4 i \sqrt {2} a^6 d \int \frac {1}{2 a^3 (i c+d)-4 i a^9 d^2 \tan ^2(e+f x)}d\sqrt {2 c+2 d \tan (e+f x)}}{f}\right )}{d (d+i c)}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 \left (\frac {2 a^3 c \sqrt {c+d \tan (e+f x)}}{d f}-\frac {4 a^3 d \text {arctanh}\left (\frac {\sqrt {2} a^3 d \tan (e+f x)}{\sqrt {c-i d}}\right )}{f \sqrt {c-i d}}\right )}{d (d+i c)}+\frac {2 (c+i d) \left (a^3+i a^3 \tan (e+f x)\right )}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}\) |
(2*(c + I*d)*(a^3 + I*a^3*Tan[e + f*x]))/((c - I*d)*d*f*Sqrt[c + d*Tan[e + f*x]]) + (2*((-4*a^3*d*ArcTanh[(Sqrt[2]*a^3*d*Tan[e + f*x])/Sqrt[c - I*d] ])/(Sqrt[c - I*d]*f) + (2*a^3*c*Sqrt[c + d*Tan[e + f*x]])/(d*f)))/(d*(I*c + d))
3.12.25.3.1 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[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_)])^(n_), x_Symbol] :> Simp[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x] )^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] + Si mp[a/(d*(b*c + a*d)*(n + 1)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[ e + f*x])^(n + 1)*Simp[b*(b*c*(m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[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[m, 1] && LtQ[n, -1] && (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_)]), x_Symbol] :> Simp[B *d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f* x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && !LeQ[m, -1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2593 vs. \(2 (122 ) = 244\).
Time = 0.88 (sec) , antiderivative size = 2594, normalized size of antiderivative = 18.66
| method | result | size |
| derivativedivides | \(\text {Expression too large to display}\) | \(2594\) |
| default | \(\text {Expression too large to display}\) | \(2594\) |
| parts | \(\text {Expression too large to display}\) | \(7395\) |
-2*I/f*a^3/d^2*(c+d*tan(f*x+e))^(1/2)+4*I/f*a^3*d^2/(c^2+d^2)^(3/2)/((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-4*I/f*a ^3*d^2/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*a rctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2 )^(1/2)-2*c)^(1/2))*c+4*I/f*a^3/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)/(2*(c^ 2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f *x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c^3-4*I/f*a^3/(c^2+d^2)^(3/2) /((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^ 3-4*I/f*a^3/((c^2+d^2)^(1/2)+c)/(c^2+d^2)^(1/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1 /2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^ 2+d^2)^(1/2)-2*c)^(1/2))*c+4*I/f*a^3/((c^2+d^2)^(1/2)+c)/(c^2+d^2)^(1/2)/( 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-8/f*a^3*d/(c^2+d^2)^(3 /2)/((c^2+d^2)^(1/2)+c)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2) ^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)) *c^2+I/f*a^3*d^2/(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)+2/f*a^3*d/(c^2+d^2)^(3/2)/((c^2+d^2)^(1/2)+c)*ln...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 604 vs. \(2 (115) = 230\).
Time = 0.28 (sec) , antiderivative size = 604, normalized size of antiderivative = 4.35 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {\frac {64 i \, a^{6}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} {\left ({\left (c^{2} d^{2} - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d^{2} + d^{4}\right )} f\right )} \log \left (\frac {{\left (8 \, a^{3} c + \sqrt {\frac {64 i \, a^{6}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} {\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}} + 8 \, {\left (a^{3} c - i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) - \sqrt {\frac {64 i \, a^{6}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} {\left ({\left (c^{2} d^{2} - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d^{2} + d^{4}\right )} f\right )} \log \left (\frac {{\left (8 \, a^{3} c + \sqrt {\frac {64 i \, a^{6}}{{\left (-i \, c^{3} - 3 \, c^{2} d + 3 i \, c d^{2} + d^{3}\right )} f^{2}}} {\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}} + 8 \, {\left (a^{3} c - i \, a^{3} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a^{3}}\right ) + 16 \, {\left (-i \, a^{3} c^{2} + a^{3} c d + {\left (-i \, a^{3} c^{2} + i \, a^{3} 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}}}{4 \, {\left ({\left (c^{2} d^{2} - 2 i \, c d^{3} - d^{4}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (c^{2} d^{2} + d^{4}\right )} f\right )}} \]
1/4*(sqrt(64*I*a^6/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*((c^2*d^2 - 2*I*c*d^3 - d^4)*f*e^(2*I*f*x + 2*I*e) + (c^2*d^2 + d^4)*f)*log(1/4*(8*a^ 3*c + sqrt(64*I*a^6/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*((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)) + 8*(a^3 *c - I*a^3*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^3) - sqrt(64*I*a ^6/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*((c^2*d^2 - 2*I*c*d^3 - d^4 )*f*e^(2*I*f*x + 2*I*e) + (c^2*d^2 + d^4)*f)*log(1/4*(8*a^3*c + sqrt(64*I* a^6/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*((-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)) + 8*(a^3*c - I*a^3*d)* e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^3) + 16*(-I*a^3*c^2 + a^3*c*d + (-I*a^3*c^2 + I*a^3*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)))/((c^2*d^2 - 2*I*c*d^3 - d ^4)*f*e^(2*I*f*x + 2*I*e) + (c^2*d^2 + d^4)*f)
\[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=- i a^{3} \left (\int \frac {i}{c \sqrt {c + d \tan {\left (e + f x \right )}} + d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}\, dx + \int \left (- \frac {3 \tan {\left (e + f x \right )}}{c \sqrt {c + d \tan {\left (e + f x \right )}} + d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}\right )\, dx + \int \frac {\tan ^{3}{\left (e + f x \right )}}{c \sqrt {c + d \tan {\left (e + f x \right )}} + d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (e + f x \right )}}{c \sqrt {c + d \tan {\left (e + f x \right )}} + d \sqrt {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )}}\right )\, dx\right ) \]
-I*a**3*(Integral(I/(c*sqrt(c + d*tan(e + f*x)) + d*sqrt(c + d*tan(e + f*x ))*tan(e + f*x)), x) + Integral(-3*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) + Integral(tan(e + f*x)** 3/(c*sqrt(c + d*tan(e + f*x)) + d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)), x) + Integral(-3*I*tan(e + f*x)**2/(c*sqrt(c + d*tan(e + f*x)) + d*sqrt(c + d*tan(e + f*x))*tan(e + f*x)), x))
\[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (115) = 230\).
Time = 0.88 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.75 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {16 \, a^{3} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{{\left (-i \, c f - d f\right )} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {2 i \, \sqrt {d \tan \left (f x + e\right ) + c} a^{3}}{d^{2} f} + \frac {2 \, {\left (-i \, a^{3} c^{2} + 2 \, a^{3} c d + i \, a^{3} d^{2}\right )}}{{\left (c d^{2} f - i \, d^{3} f\right )} \sqrt {d \tan \left (f x + e\right ) + c}} \]
16*a^3*arctan(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))))/((-I*c*f - d *f)*sqrt(-2*c + 2*sqrt(c^2 + d^2))*(-I*d/(c - sqrt(c^2 + d^2)) + 1)) - 2*I *sqrt(d*tan(f*x + e) + c)*a^3/(d^2*f) + 2*(-I*a^3*c^2 + 2*a^3*c*d + I*a^3* d^2)/((c*d^2*f - I*d^3*f)*sqrt(d*tan(f*x + e) + c))
Time = 7.44 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.31 \[ \int \frac {(a+i a \tan (e+f x))^3}{(c+d \tan (e+f x))^{3/2}} \, dx=-\frac {a^3\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,2{}\mathrm {i}}{d^2\,f}+\frac {a^3\,\mathrm {atan}\left (\frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\left (2\,c^4\,f^2+4\,c^2\,d^2\,f^2+2\,d^4\,f^2\right )}{2\,f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{3/2}\,\left (f\,c^3+1{}\mathrm {i}\,f\,c^2\,d+f\,c\,d^2+1{}\mathrm {i}\,f\,d^3\right )}\right )\,8{}\mathrm {i}}{f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\left (a^3\,c^2+a^3\,c\,d\,2{}\mathrm {i}-a^3\,d^2\right )\,2{}\mathrm {i}}{d^2\,f\,\left (c-d\,1{}\mathrm {i}\right )\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \]
(a^3*atan(((c + d*tan(e + f*x))^(1/2)*(2*c^4*f^2 + 2*d^4*f^2 + 4*c^2*d^2*f ^2))/(2*f*(d*1i - c)^(3/2)*(c^3*f + d^3*f*1i + c*d^2*f + c^2*d*f*1i)))*8i) /(f*(d*1i - c)^(3/2)) - (a^3*(c + d*tan(e + f*x))^(1/2)*2i)/(d^2*f) - ((a^ 3*c^2 - a^3*d^2 + a^3*c*d*2i)*2i)/(d^2*f*(c - d*1i)*(c + d*tan(e + f*x))^( 1/2))