Integrand size = 30, antiderivative size = 470 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i e^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {i e^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {i e^{5/2} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{2 \sqrt {2} \sqrt {a} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac {i e^{5/2} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{2 \sqrt {2} \sqrt {a} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac {i \cos ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{a d (e \cos (c+d x))^{5/2}} \]
1/2*I*e^(5/2)*arctan(1-2^(1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a^(1/2)/(e *sec(d*x+c))^(1/2))/d/(e*cos(d*x+c))^(5/2)/(e*sec(d*x+c))^(5/2)*2^(1/2)/a^ (1/2)-1/2*I*e^(5/2)*arctan(1+2^(1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a^(1 /2)/(e*sec(d*x+c))^(1/2))/d/(e*cos(d*x+c))^(5/2)/(e*sec(d*x+c))^(5/2)*2^(1 /2)/a^(1/2)-1/4*I*e^(5/2)*ln(a-2^(1/2)*a^(1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^ (1/2)/(e*sec(d*x+c))^(1/2)+cos(d*x+c)*(a+I*a*tan(d*x+c)))/d/(e*cos(d*x+c)) ^(5/2)/(e*sec(d*x+c))^(5/2)*2^(1/2)/a^(1/2)+1/4*I*e^(5/2)*ln(a+2^(1/2)*a^( 1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(1/2)+cos(d*x+c)*(a+I *a*tan(d*x+c)))/d/(e*cos(d*x+c))^(5/2)/(e*sec(d*x+c))^(5/2)*2^(1/2)/a^(1/2 )-I*cos(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2)/a/d/(e*cos(d*x+c))^(5/2)
Time = 3.37 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i e^{i c-\frac {i d x}{2}} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (-2 e^{\frac {3 i d x}{2}}+\left (-e^{-2 i c}\right )^{3/4} \left (1+e^{2 i (c+d x)}\right ) \arctan \left (\frac {e^{\frac {i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )-\left (-e^{-2 i c}\right )^{3/4} \left (1+e^{2 i (c+d x)}\right ) \text {arctanh}\left (\frac {e^{\frac {i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )\right )}{d \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \sqrt {\cos (c+d x)} (e \cos (c+d x))^{5/2} \sec ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \]
(I*E^(I*c - (I/2)*d*x)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*(-2 *E^(((3*I)/2)*d*x) + (-E^((-2*I)*c))^(3/4)*(1 + E^((2*I)*(c + d*x)))*ArcTa n[E^((I/2)*d*x)/(-E^((-2*I)*c))^(1/4)] - (-E^((-2*I)*c))^(3/4)*(1 + E^((2* I)*(c + d*x)))*ArcTanh[E^((I/2)*d*x)/(-E^((-2*I)*c))^(1/4)]))/(d*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]*Sqrt[Cos[c + d*x]]*(e*Cos[c + d*x]) ^(5/2)*Sec[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])
Time = 0.83 (sec) , antiderivative size = 400, normalized size of antiderivative = 0.85, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 3998, 3042, 3982, 3042, 3976, 826, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
\(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3998 |
\(\displaystyle \frac {\int \frac {(e \sec (c+d x))^{5/2}}{\sqrt {i \tan (c+d x) a+a}}dx}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(e \sec (c+d x))^{5/2}}{\sqrt {i \tan (c+d x) a+a}}dx}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3982 |
\(\displaystyle \frac {\frac {e^2 \int \sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}dx}{2 a}-\frac {i e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{a d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {e^2 \int \sqrt {e \sec (c+d x)} \sqrt {i \tan (c+d x) a+a}dx}{2 a}-\frac {i e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{a d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3976 |
\(\displaystyle \frac {-\frac {2 i e^4 \int \frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e \left (a^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2\right )}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{d}-\frac {i e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{a d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 826 |
\(\displaystyle \frac {-\frac {2 i e^4 \left (\frac {\int \frac {a+\cos (c+d x) (i \tan (c+d x) a+a)}{a^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 e}-\frac {\int \frac {a-\cos (c+d x) (i \tan (c+d x) a+a)}{a^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 e}\right )}{d}-\frac {i e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{a d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {-\frac {2 i e^4 \left (\frac {\frac {\int \frac {1}{\frac {a}{e}-\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 e}+\frac {\int \frac {1}{\frac {a}{e}+\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 e}}{2 e}-\frac {\int \frac {a-\cos (c+d x) (i \tan (c+d x) a+a)}{a^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 e}\right )}{d}-\frac {i e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{a d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {-\frac {2 i e^4 \left (\frac {\frac {\int \frac {1}{-\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}-1}d\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\int \frac {1}{-\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}-1}d\left (\frac {\sqrt {2} \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {a} \sqrt {e \sec (c+d x)}}+1\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\int \frac {a-\cos (c+d x) (i \tan (c+d x) a+a)}{a^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 e}\right )}{d}-\frac {i e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{a d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {-\frac {2 i e^4 \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\int \frac {a-\cos (c+d x) (i \tan (c+d x) a+a)}{a^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 e}\right )}{d}-\frac {i e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{a d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {-\frac {2 i e^4 \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {-\frac {\int -\frac {\sqrt {2} \sqrt {a}-\frac {2 \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{\sqrt {e} \left (\frac {a}{e}-\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}\right )}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {a}+\frac {\sqrt {2} \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}\right )}{\sqrt {e} \left (\frac {a}{e}+\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}\right )}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}\right )}{d}-\frac {i e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{a d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {-\frac {2 i e^4 \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\frac {\int \frac {\sqrt {2} \sqrt {a}-\frac {2 \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{\sqrt {e} \left (\frac {a}{e}-\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}\right )}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} \sqrt {e}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {a}+\frac {\sqrt {2} \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}\right )}{\sqrt {e} \left (\frac {a}{e}+\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}\right )}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}\right )}{d}-\frac {i e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{a d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-\frac {2 i e^4 \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\frac {\int \frac {\sqrt {2} \sqrt {a}-\frac {2 \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{\frac {a}{e}-\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {2} \sqrt {a} e}+\frac {\int \frac {\sqrt {a}+\frac {\sqrt {2} \sqrt {e} \sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{\frac {a}{e}+\frac {\sqrt {2} \sqrt {i \tan (c+d x) a+a} \sqrt {a}}{\sqrt {e} \sqrt {e \sec (c+d x)}}+\frac {\cos (c+d x) (i \tan (c+d x) a+a)}{e}}d\frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {e \sec (c+d x)}}}{2 \sqrt {a} e}}{2 e}\right )}{d}-\frac {i e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{a d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {-\frac {2 i e^4 \left (\frac {\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}-\frac {\frac {\log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt {2} \sqrt {a} \sqrt {e}}}{2 e}\right )}{d}-\frac {i e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{a d}}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\) |
(((-2*I)*e^4*((-(ArcTan[1 - (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/( Sqrt[a]*Sqrt[e*Sec[c + d*x]])]/(Sqrt[2]*Sqrt[a]*Sqrt[e])) + ArcTan[1 + (Sq rt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]/ (Sqrt[2]*Sqrt[a]*Sqrt[e]))/(2*e) - (-1/2*Log[a - (Sqrt[2]*Sqrt[a]*Sqrt[e]* Sqrt[a + I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a + I*a*T an[c + d*x])]/(Sqrt[2]*Sqrt[a]*Sqrt[e]) + Log[a + (Sqrt[2]*Sqrt[a]*Sqrt[e] *Sqrt[a + I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a + I*a* Tan[c + d*x])]/(2*Sqrt[2]*Sqrt[a]*Sqrt[e]))/(2*e)))/d - (I*e^2*Sqrt[e*Sec[ c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(a*d))/((e*Cos[c + d*x])^(5/2)*(e*Se c[c + d*x])^(5/2))
3.7.86.3.1 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_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[Sqrt[(d_.)*sec[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.) *(x_)]], x_Symbol] :> Simp[-4*b*(d^2/f) Subst[Int[x^2/(a^2 + d^2*x^4), x] , x, Sqrt[a + b*Tan[e + f*x]]/Sqrt[d*Sec[e + f*x]]], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[d^2*(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Simp[d^2*((m - 2)/(a*(m + n - 1))) Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ [{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, 0] && GtQ[m, 1] && !IL tQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_.), x_Symbol] :> Simp[(d*Cos[e + f*x])^m*(d*Sec[e + f*x])^m Int[( a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m , n}, x] && !IntegerQ[m]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 764 vs. \(2 (370 ) = 740\).
Time = 12.39 (sec) , antiderivative size = 765, normalized size of antiderivative = 1.63
-1/4/d*(I*csc(d*x+c)^2*arctanh(1/2*(-cot(d*x+c)+csc(d*x+c)+1)*2^(1/2)/(csc (d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2))*2^(1/2)*(1-cos(d*x+c))^2+I*csc(d*x+c) ^2*arctanh(1/2*(-cot(d*x+c)+csc(d*x+c)-1)*2^(1/2)/(csc(d*x+c)^2*(1-cos(d*x +c))^2+1)^(1/2))*2^(1/2)*(1-cos(d*x+c))^2-csc(d*x+c)^2*arctanh(1/2*(-cot(d *x+c)+csc(d*x+c)+1)*2^(1/2)/(csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2))*2^(1/ 2)*(1-cos(d*x+c))^2+csc(d*x+c)^2*arctanh(1/2*(-cot(d*x+c)+csc(d*x+c)-1)*2^ (1/2)/(csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2))*2^(1/2)*(1-cos(d*x+c))^2-I* arctanh(1/2*(-cot(d*x+c)+csc(d*x+c)+1)*2^(1/2)/(csc(d*x+c)^2*(1-cos(d*x+c) )^2+1)^(1/2))*2^(1/2)-I*arctanh(1/2*(-cot(d*x+c)+csc(d*x+c)-1)*2^(1/2)/(cs c(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2))*2^(1/2)-4*I*(csc(d*x+c)^2*(1-cos(d*x +c))^2+1)^(1/2)*(csc(d*x+c)-cot(d*x+c))+2^(1/2)*arctanh(1/2*(-cot(d*x+c)+c sc(d*x+c)+1)*2^(1/2)/(csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2))-2^(1/2)*arct anh(1/2*(-cot(d*x+c)+csc(d*x+c)-1)*2^(1/2)/(csc(d*x+c)^2*(1-cos(d*x+c))^2+ 1)^(1/2))-4*(csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(1/2))*(csc(d*x+c)^2*(1-cos( d*x+c))^2-1)*(-csc(d*x+c)+cot(d*x+c)+I)/(-a*(2*I*(csc(d*x+c)-cot(d*x+c))-c sc(d*x+c)^2*(1-cos(d*x+c))^2+1)/(csc(d*x+c)^2*(1-cos(d*x+c))^2-1))^(1/2)/( csc(d*x+c)^2*(1-cos(d*x+c))^2+1)^(5/2)/(-e*(csc(d*x+c)^2*(1-cos(d*x+c))^2- 1)/(csc(d*x+c)^2*(1-cos(d*x+c))^2+1))^(5/2)
Time = 0.29 (sec) , antiderivative size = 502, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {-4 i \, \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {3}{2} i \, d x + \frac {3}{2} i \, c\right )} - {\left (a d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a d e^{3}\right )} \sqrt {\frac {i}{a d^{2} e^{5}}} \log \left (i \, a d e^{3} \sqrt {\frac {i}{a d^{2} e^{5}}} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}\right ) + {\left (a d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a d e^{3}\right )} \sqrt {\frac {i}{a d^{2} e^{5}}} \log \left (-i \, a d e^{3} \sqrt {\frac {i}{a d^{2} e^{5}}} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}\right ) - {\left (a d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a d e^{3}\right )} \sqrt {-\frac {i}{a d^{2} e^{5}}} \log \left (i \, a d e^{3} \sqrt {-\frac {i}{a d^{2} e^{5}}} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}\right ) + {\left (a d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a d e^{3}\right )} \sqrt {-\frac {i}{a d^{2} e^{5}}} \log \left (-i \, a d e^{3} \sqrt {-\frac {i}{a d^{2} e^{5}}} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}\right )}{2 \, {\left (a d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a d e^{3}\right )}} \]
1/2*(-4*I*sqrt(2)*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I *d*x + 2*I*c) + 1))*e^(3/2*I*d*x + 3/2*I*c) - (a*d*e^3*e^(2*I*d*x + 2*I*c) + a*d*e^3)*sqrt(I/(a*d^2*e^5))*log(I*a*d*e^3*sqrt(I/(a*d^2*e^5)) + sqrt(2 )*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I*c)) + (a*d*e^3*e^(2*I*d*x + 2*I*c) + a*d*e^3)*sq rt(I/(a*d^2*e^5))*log(-I*a*d*e^3*sqrt(I/(a*d^2*e^5)) + sqrt(2)*sqrt(1/2)*s qrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I *d*x - 1/2*I*c)) - (a*d*e^3*e^(2*I*d*x + 2*I*c) + a*d*e^3)*sqrt(-I/(a*d^2* e^5))*log(I*a*d*e^3*sqrt(-I/(a*d^2*e^5)) + sqrt(2)*sqrt(1/2)*sqrt(e*e^(2*I *d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I *c)) + (a*d*e^3*e^(2*I*d*x + 2*I*c) + a*d*e^3)*sqrt(-I/(a*d^2*e^5))*log(-I *a*d*e^3*sqrt(-I/(a*d^2*e^5)) + sqrt(2)*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I* c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I*c)))/(a*d* e^3*e^(2*I*d*x + 2*I*c) + a*d*e^3)
Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Timed out} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2147 vs. \(2 (358) = 716\).
Time = 0.49 (sec) , antiderivative size = 2147, normalized size of antiderivative = 4.57 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Too large to display} \]
-8*(2*(sqrt(2)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)) ) + I*sqrt(2)*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sqrt(2))*arctan2(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2* d*x + 3/2*c))) + 1, sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2* d*x + 3/2*c))) + 1) + 2*(sqrt(2)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos (3/2*d*x + 3/2*c))) + I*sqrt(2)*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos( 3/2*d*x + 3/2*c))) + sqrt(2))*arctan2(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1, -sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + 2*(sqrt(2)*cos(4/3*arctan2(sin(3/ 2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + I*sqrt(2)*sin(4/3*arctan2(sin(3/2 *d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sqrt(2))*arctan2(sqrt(2)*cos(1/3*a rctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 1, sqrt(2)*sin(1/3*a rctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + 2*(sqrt(2)*cos( 4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + I*sqrt(2)*sin(4 /3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sqrt(2))*arctan2 (sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 1, -sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1 ) + 2*(-I*sqrt(2)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2* c))) + sqrt(2)*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)) ) - I*sqrt(2))*arctan2(sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), co...
\[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{5/2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]