Integrand size = 30, antiderivative size = 335 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx=\frac {i \sqrt {2} \sqrt {a} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{d \sqrt {e}}-\frac {i \sqrt {2} \sqrt {a} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{d \sqrt {e}}-\frac {i \sqrt {a} \log \left (a \sqrt {e}-\sqrt {2} \sqrt {a} \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)}+\sqrt {e} \cos (c+d x) (a+i a \tan (c+d x))\right )}{\sqrt {2} d \sqrt {e}}+\frac {i \sqrt {a} \log \left (a \sqrt {e}+\sqrt {2} \sqrt {a} \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)}+\sqrt {e} \cos (c+d x) (a+i a \tan (c+d x))\right )}{\sqrt {2} d \sqrt {e}} \]
-1/2*I*ln(a*e^(1/2)-2^(1/2)*a^(1/2)*(e*cos(d*x+c))^(1/2)*(a+I*a*tan(d*x+c) )^(1/2)+cos(d*x+c)*e^(1/2)*(a+I*a*tan(d*x+c)))*a^(1/2)/d*2^(1/2)/e^(1/2)+1 /2*I*ln(a*e^(1/2)+2^(1/2)*a^(1/2)*(e*cos(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^ (1/2)+cos(d*x+c)*e^(1/2)*(a+I*a*tan(d*x+c)))*a^(1/2)/d*2^(1/2)/e^(1/2)+I*a rctan(1-2^(1/2)*(e*cos(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a^(1/2)/e^(1 /2))*2^(1/2)*a^(1/2)/d/e^(1/2)-I*arctan(1+2^(1/2)*(e*cos(d*x+c))^(1/2)*(a+ I*a*tan(d*x+c))^(1/2)/a^(1/2)/e^(1/2))*2^(1/2)*a^(1/2)/d/e^(1/2)
Time = 1.79 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.37 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx=\frac {i e^{-\frac {3}{2} i d x} \left (-e^{-2 i c}\right )^{3/4} \left (1+e^{2 i (c+d x)}\right ) \left (\arctan \left (\frac {e^{\frac {i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )-\text {arctanh}\left (\frac {e^{\frac {i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )\right ) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \cos (c+d x)}} \]
(I*(-E^((-2*I)*c))^(3/4)*(1 + E^((2*I)*(c + d*x)))*(ArcTan[E^((I/2)*d*x)/( -E^((-2*I)*c))^(1/4)] - ArcTanh[E^((I/2)*d*x)/(-E^((-2*I)*c))^(1/4)])*Sqrt [a + I*a*Tan[c + d*x]])/(d*E^(((3*I)/2)*d*x)*Sqrt[e*Cos[c + d*x]])
Time = 0.50 (sec) , antiderivative size = 330, normalized size of antiderivative = 0.99, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3996, 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.
\(\displaystyle \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \cos (c+d x)}}dx\) |
\(\Big \downarrow \) 3996 |
\(\displaystyle -\frac {4 i a \int \frac {e \cos (c+d x) (i \tan (c+d x) a+a)}{a^2 e^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2 e^2}d\left (\sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )}{d}\) |
\(\Big \downarrow \) 826 |
\(\displaystyle -\frac {4 i a \left (\frac {1}{2} \int \frac {a e+\cos (c+d x) (i \tan (c+d x) a+a) e}{a^2 e^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2 e^2}d\left (\sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )-\frac {1}{2} \int \frac {a e-e \cos (c+d x) (i \tan (c+d x) a+a)}{a^2 e^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2 e^2}d\left (\sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )\right )}{d}\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle -\frac {4 i a \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{a e+\cos (c+d x) (i \tan (c+d x) a+a) e-\sqrt {2} \sqrt {a} \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a} \sqrt {e}}d\left (\sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )+\frac {1}{2} \int \frac {1}{a e+\cos (c+d x) (i \tan (c+d x) a+a) e+\sqrt {2} \sqrt {a} \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a} \sqrt {e}}d\left (\sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )\right )-\frac {1}{2} \int \frac {a e-e \cos (c+d x) (i \tan (c+d x) a+a)}{a^2 e^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2 e^2}d\left (\sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )\right )}{d}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle -\frac {4 i a \left (\frac {1}{2} \left (\frac {\int \frac {1}{-e \cos (c+d x) (i \tan (c+d x) a+a)-1}d\left (1-\frac {\sqrt {2} \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\int \frac {1}{-e \cos (c+d x) (i \tan (c+d x) a+a)-1}d\left (\frac {\sqrt {2} \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}}{\sqrt {a} \sqrt {e}}+1\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}\right )-\frac {1}{2} \int \frac {a e-e \cos (c+d x) (i \tan (c+d x) a+a)}{a^2 e^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2 e^2}d\left (\sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )\right )}{d}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {4 i a \left (\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {a+i a \tan (c+d x)} \sqrt {e \cos (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {a+i a \tan (c+d x)} \sqrt {e \cos (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}\right )-\frac {1}{2} \int \frac {a e-e \cos (c+d x) (i \tan (c+d x) a+a)}{a^2 e^2+\cos ^2(c+d x) (i \tan (c+d x) a+a)^2 e^2}d\left (\sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )\right )}{d}\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle -\frac {4 i a \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2} \sqrt {a} \sqrt {e}-2 \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}}{a e+\cos (c+d x) (i \tan (c+d x) a+a) e-\sqrt {2} \sqrt {a} \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a} \sqrt {e}}d\left (\sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )}{2 \sqrt {2} \sqrt {a} \sqrt {e}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {a} \sqrt {e}+\sqrt {2} \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )}{a e+\cos (c+d x) (i \tan (c+d x) a+a) e+\sqrt {2} \sqrt {a} \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a} \sqrt {e}}d\left (\sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )}{2 \sqrt {2} \sqrt {a} \sqrt {e}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {a+i a \tan (c+d x)} \sqrt {e \cos (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {a+i a \tan (c+d x)} \sqrt {e \cos (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}\right )\right )}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {4 i a \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {a} \sqrt {e}-2 \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}}{a e+\cos (c+d x) (i \tan (c+d x) a+a) e-\sqrt {2} \sqrt {a} \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a} \sqrt {e}}d\left (\sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )}{2 \sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {a} \sqrt {e}+\sqrt {2} \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )}{a e+\cos (c+d x) (i \tan (c+d x) a+a) e+\sqrt {2} \sqrt {a} \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a} \sqrt {e}}d\left (\sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )}{2 \sqrt {2} \sqrt {a} \sqrt {e}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {a+i a \tan (c+d x)} \sqrt {e \cos (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {a+i a \tan (c+d x)} \sqrt {e \cos (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}\right )\right )}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4 i a \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {a} \sqrt {e}-2 \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}}{a e+\cos (c+d x) (i \tan (c+d x) a+a) e-\sqrt {2} \sqrt {a} \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a} \sqrt {e}}d\left (\sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )}{2 \sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\int \frac {\sqrt {a} \sqrt {e}+\sqrt {2} \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}}{a e+\cos (c+d x) (i \tan (c+d x) a+a) e+\sqrt {2} \sqrt {a} \sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a} \sqrt {e}}d\left (\sqrt {e \cos (c+d x)} \sqrt {i \tan (c+d x) a+a}\right )}{2 \sqrt {a} \sqrt {e}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {a+i a \tan (c+d x)} \sqrt {e \cos (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {a+i a \tan (c+d x)} \sqrt {e \cos (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}\right )\right )}{d}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle -\frac {4 i a \left (\frac {1}{2} \left (\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {a+i a \tan (c+d x)} \sqrt {e \cos (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {a+i a \tan (c+d x)} \sqrt {e \cos (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {2} \sqrt {a} \sqrt {e}}\right )+\frac {1}{2} \left (\frac {\log \left (e \cos (c+d x) (a+i a \tan (c+d x))-\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)} \sqrt {e \cos (c+d x)}+a e\right )}{2 \sqrt {2} \sqrt {a} \sqrt {e}}-\frac {\log \left (e \cos (c+d x) (a+i a \tan (c+d x))+\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)} \sqrt {e \cos (c+d x)}+a e\right )}{2 \sqrt {2} \sqrt {a} \sqrt {e}}\right )\right )}{d}\) |
((-4*I)*a*((-(ArcTan[1 - (Sqrt[2]*Sqrt[e*Cos[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e])]/(Sqrt[2]*Sqrt[a]*Sqrt[e])) + ArcTan[1 + (Sqrt[ 2]*Sqrt[e*Cos[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e])]/(Sq rt[2]*Sqrt[a]*Sqrt[e]))/2 + (Log[a*e - Sqrt[2]*Sqrt[a]*Sqrt[e]*Sqrt[e*Cos[ c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]] + e*Cos[c + d*x]*(a + I*a*Tan[c + d*x ])]/(2*Sqrt[2]*Sqrt[a]*Sqrt[e]) - Log[a*e + Sqrt[2]*Sqrt[a]*Sqrt[e]*Sqrt[e *Cos[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]] + e*Cos[c + d*x]*(a + I*a*Tan[c + d*x])]/(2*Sqrt[2]*Sqrt[a]*Sqrt[e]))/2))/d
3.7.77.3.1 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[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[cos[(e_.) + (f_.)*(x_)] *(d_.)], x_Symbol] :> Simp[-4*(b/f) Subst[Int[x^2/(a^2*d^2 + x^4), x], x, Sqrt[d*Cos[e + f*x]]*Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, d, e, f }, x] && EqQ[a^2 + b^2, 0]
Time = 9.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.46
method | result | size |
default | \(\frac {\left (-1-i\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (i \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+\operatorname {arctanh}\left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right ) \cos \left (d x +c \right )}{d \left (-i \cos \left (d x +c \right )+\sin \left (d x +c \right )-i\right ) \sqrt {e \cos \left (d x +c \right )}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\) | \(155\) |
(-1-I)/d*(a*(1+I*tan(d*x+c)))^(1/2)*(I*arctanh(1/2*(cos(d*x+c)+sin(d*x+c)+ 1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2))+arctanh(1/2*(-cos(d*x+c)+sin(d *x+c)-1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2)))*cos(d*x+c)/(-I*cos(d*x+ c)+sin(d*x+c)-I)/(e*cos(d*x+c))^(1/2)/(1/(cos(d*x+c)+1))^(1/2)
Time = 0.27 (sec) , antiderivative size = 313, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx=-\frac {1}{2} \, \sqrt {\frac {4 i \, a}{d^{2} e}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + \frac {1}{2} i \, d e \sqrt {\frac {4 i \, a}{d^{2} e}}\right ) + \frac {1}{2} \, \sqrt {\frac {4 i \, a}{d^{2} e}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - \frac {1}{2} i \, d e \sqrt {\frac {4 i \, a}{d^{2} e}}\right ) - \frac {1}{2} \, \sqrt {-\frac {4 i \, a}{d^{2} e}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + \frac {1}{2} i \, d e \sqrt {-\frac {4 i \, a}{d^{2} e}}\right ) + \frac {1}{2} \, \sqrt {-\frac {4 i \, a}{d^{2} e}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - \frac {1}{2} i \, d e \sqrt {-\frac {4 i \, a}{d^{2} e}}\right ) \]
-1/2*sqrt(4*I*a/(d^2*e))*log(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) + 1/2*I*d* e*sqrt(4*I*a/(d^2*e))) + 1/2*sqrt(4*I*a/(d^2*e))*log(sqrt(2)*sqrt(1/2)*sqr t(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) - 1/2*I*d*e*sqrt(4*I*a/(d^2*e))) - 1/2*sqrt(-4*I*a/(d^2*e))* log(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) + 1/2*I*d*e*sqrt(-4*I*a/(d^2*e))) + 1/2*sqrt(-4*I*a/(d^2*e))*log(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) - 1/2*I*d *e*sqrt(-4*I*a/(d^2*e)))
\[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx=\int \frac {\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}{\sqrt {e \cos {\left (c + d x \right )}}}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1400 vs. \(2 (247) = 494\).
Time = 0.50 (sec) , antiderivative size = 1400, normalized size of antiderivative = 4.18 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx=\text {Too large to display} \]
1/4*(-2*I*sqrt(2)*arctan2(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), co s(3/2*d*x + 3/2*c))) + 1, sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), co s(3/2*d*x + 3/2*c))) + 1) - 2*I*sqrt(2)*arctan2(sqrt(2)*cos(1/3*arctan2(si n(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1, -sqrt(2)*sin(1/3*arctan2(s in(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - 2*I*sqrt(2)*arctan2(sqr t(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 1, sqr t(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)*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)*arctan2(sqrt(2)*sin(1/3*arctan2(sin(3/2*d* x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))), sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos (3/2*d*x + 3/2*c))) + cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + 2*sqrt(2)*arctan2(-sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3 /2*c), cos(3/2*d*x + 3/2*c))) + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos( 3/2*d*x + 3/2*c))), -sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2 *d*x + 3/2*c))) + cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2* c))) + 1) + I*sqrt(2)*log(2*sqrt(2)*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*...
\[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx=\int { \frac {\sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {e \cos \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx=\int \frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {e\,\cos \left (c+d\,x\right )}} \,d x \]