\(\int \frac {1}{(e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx\) [685]

Optimal result

Integrand size = 30, antiderivative size = 495 \[ \int \frac {1}{(e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i \sqrt {2} \sqrt {a} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \cos (c+d x)} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e}}\right ) \sec (c+d x)}{d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i \sqrt {2} \sqrt {a} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \cos (c+d x)} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e}}\right ) \sec (c+d x)}{d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i \sqrt {a} \log \left (a \sqrt {e}-\sqrt {2} \sqrt {a} \sqrt {e \cos (c+d x)} \sqrt {a-i a \tan (c+d x)}+\sqrt {e} \cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{\sqrt {2} d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {i \sqrt {a} \log \left (a \sqrt {e}+\sqrt {2} \sqrt {a} \sqrt {e \cos (c+d x)} \sqrt {a-i a \tan (c+d x)}+\sqrt {e} \cos (c+d x) (a-i a \tan (c+d x))\right ) \sec (c+d x)}{\sqrt {2} d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[Out]

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)))*sec(d*x+c)*a^(1/2)/d/e^(3/2)*2^(1/2)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(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)))*sec
(d*x+c)*a^(1/2)/d/e^(3/2)*2^(1/2)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(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))*sec(d*x+c)*2^(1/2)*a^(1/2)/d/e^(3/2)/(a-I*a*tan(d*x+c)
)^(1/2)/(a+I*a*tan(d*x+c))^(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))*sec(d*x+c)*2^(1/2)*a^(1/2)/d/e^(3/2)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 495, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3595, 3594, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {1}{(e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i \sqrt {2} \sqrt {a} \sec (c+d x) \arctan \left (1-\frac {\sqrt {2} \sqrt {a-i a \tan (c+d x)} \sqrt {e \cos (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i \sqrt {2} \sqrt {a} \sec (c+d x) \arctan \left (1+\frac {\sqrt {2} \sqrt {a-i a \tan (c+d x)} \sqrt {e \cos (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i \sqrt {a} \sec (c+d x) \log \left (-\sqrt {2} \sqrt {a} \sqrt {a-i a \tan (c+d x)} \sqrt {e \cos (c+d x)}+\sqrt {e} \cos (c+d x) (a-i a \tan (c+d x))+a \sqrt {e}\right )}{\sqrt {2} d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {i \sqrt {a} \sec (c+d x) \log \left (\sqrt {2} \sqrt {a} \sqrt {a-i a \tan (c+d x)} \sqrt {e \cos (c+d x)}+\sqrt {e} \cos (c+d x) (a-i a \tan (c+d x))+a \sqrt {e}\right )}{\sqrt {2} d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[In]

Int[1/((e*Cos[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((-I)*Sqrt[2]*Sqrt[a]*ArcTan[1 - (Sqrt[2]*Sqrt[e*Cos[c + d*x]]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e])]*
Sec[c + d*x])/(d*e^(3/2)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (I*Sqrt[2]*Sqrt[a]*ArcTan[1
+ (Sqrt[2]*Sqrt[e*Cos[c + d*x]]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e])]*Sec[c + d*x])/(d*e^(3/2)*Sqrt[a
- I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (I*Sqrt[a]*Log[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])]*Sec[c + d*x])/(Sqrt[2]*d*e^(3/2
)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (I*Sqrt[a]*Log[a*Sqrt[e] + Sqrt[2]*Sqrt[a]*Sqrt[e*C
os[c + d*x]]*Sqrt[a - I*a*Tan[c + d*x]] + Sqrt[e]*Cos[c + d*x]*(a - I*a*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*
d*e^(3/2)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

Rule 210

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])

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 3594

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[cos[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[-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]

Rule 3595

Int[1/((cos[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)*Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[1/
(d*Cos[e + f*x]*Sqrt[a - b*Tan[e + f*x]]*Sqrt[a + b*Tan[e + f*x]]), Int[Sqrt[a - b*Tan[e + f*x]]/Sqrt[d*Cos[e
+ f*x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sec (c+d x) \int \frac {\sqrt {a-i a \tan (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx}{e \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {(4 i a \sec (c+d x)) \text {Subst}\left (\int \frac {x^2}{a^2 e^2+x^4} \, dx,x,\sqrt {e \cos (c+d x)} \sqrt {a-i a \tan (c+d x)}\right )}{d e \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = -\frac {(2 i a \sec (c+d x)) \text {Subst}\left (\int \frac {a e-x^2}{a^2 e^2+x^4} \, dx,x,\sqrt {e \cos (c+d x)} \sqrt {a-i a \tan (c+d x)}\right )}{d e \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {(2 i a \sec (c+d x)) \text {Subst}\left (\int \frac {a e+x^2}{a^2 e^2+x^4} \, dx,x,\sqrt {e \cos (c+d x)} \sqrt {a-i a \tan (c+d x)}\right )}{d e \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {\left (i \sqrt {a} \sec (c+d x)\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {a} \sqrt {e}+2 x}{-a e-\sqrt {2} \sqrt {a} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \cos (c+d x)} \sqrt {a-i a \tan (c+d x)}\right )}{\sqrt {2} d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (i \sqrt {a} \sec (c+d x)\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {a} \sqrt {e}-2 x}{-a e+\sqrt {2} \sqrt {a} \sqrt {e} x-x^2} \, dx,x,\sqrt {e \cos (c+d x)} \sqrt {a-i a \tan (c+d x)}\right )}{\sqrt {2} d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {(i a \sec (c+d x)) \text {Subst}\left (\int \frac {1}{a e-\sqrt {2} \sqrt {a} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \cos (c+d x)} \sqrt {a-i a \tan (c+d x)}\right )}{d e \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {(i a \sec (c+d x)) \text {Subst}\left (\int \frac {1}{a e+\sqrt {2} \sqrt {a} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \cos (c+d x)} \sqrt {a-i a \tan (c+d x)}\right )}{d e \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = \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 ) \sec (c+d x)}{\sqrt {2} d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\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 ) \sec (c+d x)}{\sqrt {2} d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (i \sqrt {2} \sqrt {a} \sec (c+d x)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e \cos (c+d x)} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (i \sqrt {2} \sqrt {a} \sec (c+d x)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e \cos (c+d x)} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e}}\right )}{d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ & = -\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 ) \sec (c+d x)}{d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\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 ) \sec (c+d x)}{d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\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 ) \sec (c+d x)}{\sqrt {2} d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\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 ) \sec (c+d x)}{\sqrt {2} d e^{3/2} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.59 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.42 \[ \int \frac {1}{(e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {i e^{\frac {1}{2} i (c+d x)} \left (2 \arctan \left (1-\sqrt {2} e^{\frac {1}{2} i (c+d x)}\right )-2 \arctan \left (1+\sqrt {2} e^{\frac {1}{2} i (c+d x)}\right )+\log \left (1-\sqrt {2} e^{\frac {1}{2} i (c+d x)}+e^{i (c+d x)}\right )-\log \left (1+\sqrt {2} e^{\frac {1}{2} i (c+d x)}+e^{i (c+d x)}\right )\right )}{\sqrt {2} d e \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {e e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}} \]

[In]

Integrate[1/((e*Cos[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

(I*E^((I/2)*(c + d*x))*(2*ArcTan[1 - Sqrt[2]*E^((I/2)*(c + d*x))] - 2*ArcTan[1 + Sqrt[2]*E^((I/2)*(c + d*x))]
+ Log[1 - Sqrt[2]*E^((I/2)*(c + d*x)) + E^(I*(c + d*x))] - Log[1 + Sqrt[2]*E^((I/2)*(c + d*x)) + E^(I*(c + d*x
))]))/(Sqrt[2]*d*e*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[(e*(1 + E^((2*I)*(c + d*x))))/
E^(I*(c + d*x))])

Maple [A] (verified)

Time = 12.92 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.33

method	result	size
default	\(\frac {\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-\operatorname {arctanh}\left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )\right ) \left (\cos \left (d x +c \right )+1+i \sin \left (d x +c \right )\right )}{d \left (\cos \left (d x +c \right )+1\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, e \sqrt {e \cos \left (d x +c \right )}}\)	\(161\)

[In]

int(1/(e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

(-1/2-1/2*I)/d*(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)+1+I*sin(d*x+c))/(cos(d*x+c)+1)
/(a*(1+I*tan(d*x+c)))^(1/2)/(1/(cos(d*x+c)+1))^(1/2)/e/(e*cos(d*x+c))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 341, normalized size of antiderivative = 0.69 \[ \int \frac {1}{(e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {1}{2} \, \sqrt {\frac {4 i}{a d^{2} e^{3}}} \log \left (\frac {1}{2} \, a d e^{2} \sqrt {\frac {4 i}{a d^{2} e^{3}}} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}\right ) + \frac {1}{2} \, \sqrt {\frac {4 i}{a d^{2} e^{3}}} \log \left (-\frac {1}{2} \, a d e^{2} \sqrt {\frac {4 i}{a d^{2} e^{3}}} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}\right ) + \frac {1}{2} \, \sqrt {-\frac {4 i}{a d^{2} e^{3}}} \log \left (\frac {1}{2} \, a d e^{2} \sqrt {-\frac {4 i}{a d^{2} e^{3}}} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}\right ) - \frac {1}{2} \, \sqrt {-\frac {4 i}{a d^{2} e^{3}}} \log \left (-\frac {1}{2} \, a d e^{2} \sqrt {-\frac {4 i}{a d^{2} e^{3}}} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}\right ) \]

[In]

integrate(1/(e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-1/2*sqrt(4*I/(a*d^2*e^3))*log(1/2*a*d*e^2*sqrt(4*I/(a*d^2*e^3)) + 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*sqrt(4*I/(a*d^2*e^3))*log(-1/2*a*d*e^
2*sqrt(4*I/(a*d^2*e^3)) + 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*sqrt(-4*I/(a*d^2*e^3))*log(1/2*a*d*e^2*sqrt(-4*I/(a*d^2*e^3)) + 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*sqrt(-4*
I/(a*d^2*e^3))*log(-1/2*a*d*e^2*sqrt(-4*I/(a*d^2*e^3)) + sqrt(2)*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqr
t(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(-1/2*I*d*x - 1/2*I*c))

Sympy [F]

\[ \int \frac {1}{(e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{\left (e \cos {\left (c + d x \right )}\right )^{\frac {3}{2}} \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

[In]

integrate(1/(e*cos(d*x+c))**(3/2)/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(1/((e*cos(c + d*x))**(3/2)*sqrt(I*a*(tan(c + d*x) - I))), x)

Maxima [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 714, normalized size of antiderivative = 1.44 \[ \int \frac {1}{(e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-1/4*(2*I*sqrt(2)*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 2*I*sqrt(2)*ar
ctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 2*I*sqrt(2)*arctan2(sqrt(2)*cos(1
/2*d*x + 1/2*c) - 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 2*I*sqrt(2)*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1,
-sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) - 2*sqrt(2)*arctan2(sqrt(2)*sin(1/2*d*x + 1/2*c) + sin(d*x + c), sqrt(2)*c
os(1/2*d*x + 1/2*c) + cos(d*x + c) + 1) + 2*sqrt(2)*arctan2(-sqrt(2)*sin(1/2*d*x + 1/2*c) + sin(d*x + c), -sqr
t(2)*cos(1/2*d*x + 1/2*c) + cos(d*x + c) + 1) + I*sqrt(2)*log(2*sqrt(2)*sin(d*x + c)*sin(1/2*d*x + 1/2*c) + 2*
(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1)*cos(d*x + c) + cos(d*x + c)^2 + 2*cos(1/2*d*x + 1/2*c)^2 + sin(d*x + c)^2 +
2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 1) - I*sqrt(2)*log(-2*sqrt(2)*sin(d*x + c)*sin(1/
2*d*x + 1/2*c) - 2*(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1)*cos(d*x + c) + cos(d*x + c)^2 + 2*cos(1/2*d*x + 1/2*c)^2
+ sin(d*x + c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 1) - sqrt(2)*log(2*cos(1/2*d*x
+ 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2)
+ sqrt(2)*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)
*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1
/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt(2)*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x +
1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))/(sqrt(a)*d*e^(3/2))

Giac [F]

\[ \int \frac {1}{(e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {1}{\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} \sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate(1/(e*cos(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(1/((e*cos(d*x + c))^(3/2)*sqrt(I*a*tan(d*x + c) + a)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]

[In]

int(1/((e*cos(c + d*x))^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2)),x)

[Out]

int(1/((e*cos(c + d*x))^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2)), x)