Integrand size = 35, antiderivative size = 234 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {4 i \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}} \]
-8/15*I*(a+I*a*tan(f*x+e))^(1/2)/a^2/c^2/f/(c-I*c*tan(f*x+e))^(1/2)+4/3*I/ a/f/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(5/2)-4/5*I*(a+I*a*tan(f*x +e))^(1/2)/a^2/f/(c-I*c*tan(f*x+e))^(5/2)+1/3*I/f/(a+I*a*tan(f*x+e))^(3/2) /(c-I*c*tan(f*x+e))^(5/2)-8/15*I*(a+I*a*tan(f*x+e))^(1/2)/a^2/c/f/(c-I*c*t an(f*x+e))^(3/2)
Time = 3.54 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.49 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {3+12 i \tan (e+f x)+12 \tan ^2(e+f x)+8 i \tan ^3(e+f x)+8 \tan ^4(e+f x)}{15 a c^2 f (-i+\tan (e+f x)) (i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
(3 + (12*I)*Tan[e + f*x] + 12*Tan[e + f*x]^2 + (8*I)*Tan[e + f*x]^3 + 8*Ta n[e + f*x]^4)/(15*a*c^2*f*(-I + Tan[e + f*x])*(I + Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.35 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4006, 55, 55, 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.
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4006 |
| \(\displaystyle \frac {a c \int \frac {1}{(i \tan (e+f x) a+a)^{5/2} (c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle \frac {a c \left (\frac {4 \int \frac {1}{(i \tan (e+f x) a+a)^{3/2} (c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{3 a}+\frac {i}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle \frac {a c \left (\frac {4 \left (\frac {3 \int \frac {1}{\sqrt {i \tan (e+f x) a+a} (c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{a}+\frac {i}{a c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}\right )}{3 a}+\frac {i}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle \frac {a c \left (\frac {4 \left (\frac {3 \left (\frac {2 \int \frac {1}{\sqrt {i \tan (e+f x) a+a} (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{5 c}-\frac {i \sqrt {a+i a \tan (e+f x)}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{a}+\frac {i}{a c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}\right )}{3 a}+\frac {i}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle \frac {a c \left (\frac {4 \left (\frac {3 \left (\frac {2 \left (\frac {\int \frac {1}{\sqrt {i \tan (e+f x) a+a} (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{3 c}-\frac {i \sqrt {a+i a \tan (e+f x)}}{3 a c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {i \sqrt {a+i a \tan (e+f x)}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{a}+\frac {i}{a c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}\right )}{3 a}+\frac {i}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 48 |
| \(\displaystyle \frac {a c \left (\frac {4 \left (\frac {3 \left (\frac {2 \left (-\frac {i \sqrt {a+i a \tan (e+f x)}}{3 a c^2 \sqrt {c-i c \tan (e+f x)}}-\frac {i \sqrt {a+i a \tan (e+f x)}}{3 a c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {i \sqrt {a+i a \tan (e+f x)}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{a}+\frac {i}{a c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}\right )}{3 a}+\frac {i}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
(a*c*((I/3)/(a*c*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2) ) + (4*(I/(a*c*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(5/2)) + (3*(((-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)))/a))/(3*a)))/f
3.11.40.3.1 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 = 0.85 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.56
| method | result | size |
| derivativedivides | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i \left (\tan ^{5}\left (f x +e \right )\right )+8 \left (\tan ^{6}\left (f x +e \right )\right )+20 i \left (\tan ^{3}\left (f x +e \right )\right )+20 \left (\tan ^{4}\left (f x +e \right )\right )+12 i \tan \left (f x +e \right )+15 \left (\tan ^{2}\left (f x +e \right )\right )+3\right )}{15 f \,a^{2} c^{3} \left (\tan \left (f x +e \right )+i\right )^{4} \left (-\tan \left (f x +e \right )+i\right )^{3}}\) | \(130\) |
| default | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i \left (\tan ^{5}\left (f x +e \right )\right )+8 \left (\tan ^{6}\left (f x +e \right )\right )+20 i \left (\tan ^{3}\left (f x +e \right )\right )+20 \left (\tan ^{4}\left (f x +e \right )\right )+12 i \tan \left (f x +e \right )+15 \left (\tan ^{2}\left (f x +e \right )\right )+3\right )}{15 f \,a^{2} c^{3} \left (\tan \left (f x +e \right )+i\right )^{4} \left (-\tan \left (f x +e \right )+i\right )^{3}}\) | \(130\) |
-1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)/a^2/c^3*(8* I*tan(f*x+e)^5+8*tan(f*x+e)^6+20*I*tan(f*x+e)^3+20*tan(f*x+e)^4+12*I*tan(f *x+e)+15*tan(f*x+e)^2+3)/(tan(f*x+e)+I)^4/(-tan(f*x+e)+I)^3
Time = 0.26 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.57 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-3 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 23 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 110 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 48 i \, e^{\left (5 i \, f x + 5 i \, e\right )} - 30 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 48 i \, e^{\left (3 i \, f x + 3 i \, e\right )} + 65 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 5 i\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{240 \, a^{2} c^{3} f} \]
1/240*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^(10*I*f*x + 10*I*e) - 23*I*e^(8*I*f*x + 8*I*e) - 110*I*e^(6*I*f*x + 6*I*e) + 48*I*e^(5*I*f*x + 5*I*e) - 30*I*e^(4*I*f*x + 4*I*e) + 48*I*e^(3 *I*f*x + 3*I*e) + 65*I*e^(2*I*f*x + 2*I*e) + 5*I)*e^(-3*I*f*x - 3*I*e)/(a^ 2*c^3*f)
\[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]
Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
Time = 6.84 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.60 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,20{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+40\,\sin \left (2\,e+2\,f\,x\right )+4\,\sin \left (4\,e+4\,f\,x\right )-45{}\mathrm {i}\right )}{120\,a^2\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]