Integrand size = 33, antiderivative size = 95 \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=-\frac {i c^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{\sqrt {2} a f}+\frac {i c^2 \sqrt {c-i c \tan (e+f x)}}{a f (c+i c \tan (e+f x))} \]
-1/2*I*c^(3/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a/f*2 ^(1/2)+I*c^2*(c-I*c*tan(f*x+e))^(1/2)/a/f/(c+I*c*tan(f*x+e))
Time = 1.20 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.33 \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=\frac {2 c^2 (1-i \tan (e+f x))+\sqrt {2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) (-1-i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{2 a f (-i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \]
(2*c^2*(1 - I*Tan[e + f*x]) + Sqrt[2]*c^(3/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])]*(-1 - I*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]] )/(2*a*f*(-I + Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.39 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.83, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 4005, 3042, 3968, 51, 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.
\(\displaystyle \int \frac {(c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)}dx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle \frac {\int \cos ^2(e+f x) (c-i c \tan (e+f x))^{5/2}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(c-i c \tan (e+f x))^{5/2}}{\sec (e+f x)^2}dx}{a c}\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle \frac {i c^2 \int \frac {\sqrt {c-i c \tan (e+f x)}}{(i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))}{a f}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {i c^2 \left (\frac {\sqrt {c-i c \tan (e+f x)}}{c+i c \tan (e+f x)}-\frac {1}{2} \int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))\right )}{a f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {i c^2 \left (\frac {\sqrt {c-i c \tan (e+f x)}}{c+i c \tan (e+f x)}-\int \frac {1}{c^2 \tan ^2(e+f x)+2 c}d\sqrt {c-i c \tan (e+f x)}\right )}{a f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {i c^2 \left (\frac {i \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2}}\right )}{\sqrt {2} \sqrt {c}}+\frac {\sqrt {c-i c \tan (e+f x)}}{c+i c \tan (e+f x)}\right )}{a f}\) |
(I*c^2*((I*ArcTan[(Sqrt[c]*Tan[e + f*x])/Sqrt[2]])/(Sqrt[2]*Sqrt[c]) + Sqr t[c - I*c*Tan[e + f*x]]/(c + I*c*Tan[e + f*x])))/(a*f)
3.10.65.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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[(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[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] && !(IGtQ[n, 0] && (LtQ[ m, 0] || GtQ[m, n]))
Time = 0.64 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(\frac {2 i c^{2} \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{2 c +2 i c \tan \left (f x +e \right )}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}\right )}{f a}\) | \(77\) |
default | \(\frac {2 i c^{2} \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{2 c +2 i c \tan \left (f x +e \right )}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}\right )}{f a}\) | \(77\) |
2*I/f/a*c^2*(1/4*(c-I*c*tan(f*x+e))^(1/2)/(1/2*c+1/2*I*c*tan(f*x+e))-1/4*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. 256 vs. \(2 (74) = 148\).
Time = 0.24 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.69 \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=-\frac {{\left (\sqrt {2} a f \sqrt {-\frac {c^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, c^{2} + {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c^{3}}{a^{2} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) - \sqrt {2} a f \sqrt {-\frac {c^{3}}{a^{2} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {2 \, {\left (i \, c^{2} - {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {c^{3}}{a^{2} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) + 2 \, \sqrt {2} {\left (-i \, c e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a f} \]
-1/4*(sqrt(2)*a*f*sqrt(-c^3/(a^2*f^2))*e^(2*I*f*x + 2*I*e)*log(-2*(I*c^2 + (a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-c ^3/(a^2*f^2)))*e^(-I*f*x - I*e)/(a*f)) - sqrt(2)*a*f*sqrt(-c^3/(a^2*f^2))* e^(2*I*f*x + 2*I*e)*log(-2*(I*c^2 - (a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt(c /(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-c^3/(a^2*f^2)))*e^(-I*f*x - I*e)/(a*f)) + 2*sqrt(2)*(-I*c*e^(2*I*f*x + 2*I*e) - I*c)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a*f)
\[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=- \frac {i \left (\int \frac {c \sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan {\left (e + f x \right )} - i}\, dx + \int \left (- \frac {i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan {\left (e + f x \right )} - i}\right )\, dx\right )}{a} \]
-I*(Integral(c*sqrt(-I*c*tan(e + f*x) + c)/(tan(e + f*x) - I), x) + Integr al(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)/(tan(e + f*x) - I), x))/a
Time = 0.47 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.12 \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=\frac {i \, {\left (\frac {\sqrt {2} c^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a} - \frac {4 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} c^{3}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )} a - 2 \, a c}\right )}}{4 \, c f} \]
1/4*I*(sqrt(2)*c^(5/2)*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 - 4*sqrt(-I*c*tan(f*x + e) + c)*c^3/((-I*c*tan(f*x + e) + c)*a - 2*a*c))/(c*f)
\[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=\int { \frac {{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{i \, a \tan \left (f x + e\right ) + a} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.87 \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{a+i a \tan (e+f x)} \, dx=\frac {\sqrt {2}\,{\left (-c\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,1{}\mathrm {i}}{2\,a\,f}+\frac {c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{a\,f\,\left (c+c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )} \]