Integrand size = 35, antiderivative size = 136 \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i \sqrt {a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{15 c^2 f \sqrt {c-i c \tan (e+f x)}} \]
-2/15*I*(a+I*a*tan(f*x+e))^(1/2)/c^2/f/(c-I*c*tan(f*x+e))^(1/2)-1/5*I*(a+I *a*tan(f*x+e))^(1/2)/f/(c-I*c*tan(f*x+e))^(5/2)-2/15*I*(a+I*a*tan(f*x+e))^ (1/2)/c/f/(c-I*c*tan(f*x+e))^(3/2)
Time = 2.38 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {a (-i+\tan (e+f x)) \left (-7+6 i \tan (e+f x)+2 \tan ^2(e+f x)\right )}{15 c^2 f (i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
(a*(-I + Tan[e + f*x])*(-7 + (6*I)*Tan[e + f*x] + 2*Tan[e + f*x]^2))/(15*c ^2*f*(I + Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.30 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4006, 55, 55, 48}
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.
\(\displaystyle \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c-i c \tan (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c-i c \tan (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a c \int \frac {1}{\sqrt {i \tan (e+f x) a+a} (c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {a c \left (\frac {2 \int \frac {1}{\sqrt {i \tan (e+f x) a+a} (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{5 c}-\frac {i \sqrt {a+i a \tan (e+f x)}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {a c \left (\frac {2 \left (\frac {\int \frac {1}{\sqrt {i \tan (e+f x) a+a} (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{3 c}-\frac {i \sqrt {a+i a \tan (e+f x)}}{3 a c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {i \sqrt {a+i a \tan (e+f x)}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {a c \left (\frac {2 \left (-\frac {i \sqrt {a+i a \tan (e+f x)}}{3 a c^2 \sqrt {c-i c \tan (e+f x)}}-\frac {i \sqrt {a+i a \tan (e+f x)}}{3 a c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {i \sqrt {a+i a \tan (e+f x)}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
(a*c*(((-1/5*I)*Sqrt[a + I*a*Tan[e + f*x]])/(a*c*(c - I*c*Tan[e + f*x])^(5 /2)) + (2*(((-1/3*I)*Sqrt[a + I*a*Tan[e + f*x]])/(a*c*(c - I*c*Tan[e + f*x ])^(3/2)) - ((I/3)*Sqrt[a + I*a*Tan[e + f*x]])/(a*c^2*Sqrt[c - I*c*Tan[e + f*x]])))/(5*c)))/f
3.11.38.3.1 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 + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
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*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*( c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
Time = 0.90 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.57
method | result | size |
risch | \(-\frac {i \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (3 \,{\mathrm e}^{4 i \left (f x +e \right )}+10 \,{\mathrm e}^{2 i \left (f x +e \right )}+15\right )}{60 c^{2} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\) | \(77\) |
derivativedivides | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i \left (\tan ^{2}\left (f x +e \right )\right )+2 \left (\tan ^{3}\left (f x +e \right )\right )-7 i-13 \tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{4}}\) | \(83\) |
default | \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i \left (\tan ^{2}\left (f x +e \right )\right )+2 \left (\tan ^{3}\left (f x +e \right )\right )-7 i-13 \tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{4}}\) | \(83\) |
-1/60*I/c^2*(a*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))^(1/2)/(c/(exp(2*I*(f *x+e))+1))^(1/2)*(3*exp(4*I*(f*x+e))+10*exp(2*I*(f*x+e))+15)/f
Time = 0.23 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.64 \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-3 i \, e^{\left (7 i \, f x + 7 i \, e\right )} - 13 i \, e^{\left (5 i \, f x + 5 i \, e\right )} - 25 i \, e^{\left (3 i \, f x + 3 i \, e\right )} - 15 i \, e^{\left (i \, f x + i \, e\right )}\right )}}{60 \, c^{3} f} \]
1/60*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*( -3*I*e^(7*I*f*x + 7*I*e) - 13*I*e^(5*I*f*x + 5*I*e) - 25*I*e^(3*I*f*x + 3* I*e) - 15*I*e^(I*f*x + I*e))/(c^3*f)
\[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )}}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {\sqrt {i \, a \tan \left (f x + e\right ) + a}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
Time = 5.96 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {a+i a \tan (e+f x)}}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,10{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}-10\,\sin \left (2\,e+2\,f\,x\right )-3\,\sin \left (4\,e+4\,f\,x\right )+15{}\mathrm {i}\right )}{60\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]