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

Optimal. Leaf size=524 \[ -\frac{i a^{3/2} \sec (c+d x) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{a-i a \tan (c+d x)} \sqrt{e \cos (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{\sqrt{2} d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i a^{3/2} \sec (c+d x) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{a-i a \tan (c+d x)} \sqrt{e \cos (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{\sqrt{2} d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i a^{3/2} \sec (c+d x) \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{a-i a \tan (c+d x)} \sqrt{e \cos (c+d x)}}{\sqrt{e}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt{2} d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{i a^{3/2} \sec (c+d x) \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{a-i a \tan (c+d x)} \sqrt{e \cos (c+d x)}}{\sqrt{e}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt{2} d e^{3/2} \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{i a}{d \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.590468, antiderivative size = 620, normalized size of antiderivative = 1.18, number of steps used = 13, number of rules used = 10, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3515, 3498, 3499, 3495, 297, 1162, 617, 204, 1165, 628} \[ -\frac{i a^{3/2} e^{3/2} \sec (c+d x) \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}+\frac{i a^{3/2} e^{3/2} \sec (c+d x) \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}+\frac{i a^{3/2} e^{3/2} \sec (c+d x) \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}-\frac{i a^{3/2} e^{3/2} \sec (c+d x) \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a-i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{2 \sqrt{2} d \sqrt{a-i a \tan (c+d x)} \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}+\frac{i a}{d \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*a)/(d*(e*Cos[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) - (I*a^(3/2)*e^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e]*S
qrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]*Sec[c + d*x])/(Sqrt[2]*d*(e*Cos[c + d*x])^(3/2)*(e*
Sec[c + d*x])^(3/2)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (I*a^(3/2)*e^(3/2)*ArcTan[1 + (Sq
rt[2]*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]*Sec[c + d*x])/(Sqrt[2]*d*(e*Cos[c +
d*x])^(3/2)*(e*Sec[c + d*x])^(3/2)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + ((I/2)*a^(3/2)*e^(
3/2)*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
*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*d*(e*Cos[c + d*x])^(3/2)*(e*Sec[c + d*x])^(3/2)*Sqrt[a - I*a*Tan[c + d*
x]]*Sqrt[a + I*a*Tan[c + d*x]]) - ((I/2)*a^(3/2)*e^(3/2)*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*Tan[c + d*x])]*Sec[c + d*x])/(Sqrt[2]*d*(e*Cos[c + d*x])^
(3/2)*(e*Sec[c + d*x])^(3/2)*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

Rule 3515

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[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]

Rule 3498

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] + Dist[(a*(m + 2*n - 2))/(m + n - 1), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] &&
GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 3499

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

Rule 3495

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

Rule 297

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 1162

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 617

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

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 628

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]

Rubi steps

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

Mathematica [A]  time = 3.89304, size = 274, normalized size = 0.52 \[ \frac{i e^{-\frac{1}{2} i (c+d x)} \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)} (\cos (c+d x)+i \sin (c+d x)) \left (-2 i \sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )+2 \sqrt{2} \cos \left (\frac{1}{2} (c+d x)\right )+\log \left (-\sqrt{2} e^{\frac{1}{2} i (c+d x)}+e^{i (c+d x)}+1\right ) \cos (c+d x)-\log \left (\sqrt{2} e^{\frac{1}{2} i (c+d x)}+e^{i (c+d x)}+1\right ) \cos (c+d x)+2 \cos (c+d x) \tan ^{-1}\left (1-\sqrt{2} e^{\frac{1}{2} i (c+d x)}\right )-2 \cos (c+d x) \tan ^{-1}\left (1+\sqrt{2} e^{\frac{1}{2} i (c+d x)}\right )\right )}{\sqrt{2} d \left (1+e^{2 i (c+d x)}\right ) (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*Cos[c + d*x]^2*(2*Sqrt[2]*Cos[(c + d*x)/2] + 2*ArcTan[1 - Sqrt[2]*E^((I/2)*(c + d*x))]*Cos[c + d*x] - 2*Arc
Tan[1 + Sqrt[2]*E^((I/2)*(c + d*x))]*Cos[c + d*x] + Cos[c + d*x]*Log[1 - Sqrt[2]*E^((I/2)*(c + d*x)) + E^(I*(c
+ d*x))] - Cos[c + d*x]*Log[1 + Sqrt[2]*E^((I/2)*(c + d*x)) + E^(I*(c + d*x))] - (2*I)*Sqrt[2]*Sin[(c + d*x)/
2])*(Cos[c + d*x] + I*Sin[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[2]*d*E^((I/2)*(c + d*x))*(1 + E^((2*I)*(
c + d*x)))*(e*Cos[c + d*x])^(3/2))

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Maple [A]  time = 0.412, size = 308, normalized size = 0.6 \begin{align*}{\frac{\cos \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) -1 \right ) ^{2}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) }\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( 2\,i\sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+i\cos \left ( dx+c \right ){\it Artanh} \left ({\frac{-\cos \left ( dx+c \right ) -1+\sin \left ( dx+c \right ) }{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) +i{\it Artanh} \left ({\frac{\cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) }{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \cos \left ( dx+c \right ) -2\,\cos \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-\cos \left ( dx+c \right ){\it Artanh} \left ({\frac{-\cos \left ( dx+c \right ) -1+\sin \left ( dx+c \right ) }{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) +{\it Artanh} \left ({\frac{\cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) }{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \cos \left ( dx+c \right ) -2\,\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \left ( \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1} \right ) ^{-{\frac{3}{2}}} \left ( e\cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)*(cos(d*x+c)-1)^2*(2*I*sin(d*x+c)*(1/(cos(d*x+c
)+1))^(1/2)+I*cos(d*x+c)*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)-1+sin(d*x+c)))+I*arctanh(1/2*(1/(co
s(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*cos(d*x+c)-2*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)-cos(d*x+c)*arct
anh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)-1+sin(d*x+c)))+arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+
1+sin(d*x+c)))*cos(d*x+c)-2*(1/(cos(d*x+c)+1))^(1/2))/sin(d*x+c)^3/(I*sin(d*x+c)+cos(d*x+c)-1)/(1/(cos(d*x+c)+
1))^(3/2)/(e*cos(d*x+c))^(3/2)

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Maxima [B]  time = 3.33966, size = 2476, normalized size = 4.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(16*(sqrt(2)*cos(2*d*x + 2*c) + I*sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*arctan2(sqrt(2)*cos(1/4*arctan2(sin(2*d
*x + 2*c), cos(2*d*x + 2*c))) + 1, sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 16*(sqr
t(2)*cos(2*d*x + 2*c) + I*sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*arctan2(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c)
, cos(2*d*x + 2*c))) + 1, -sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 16*(sqrt(2)*cos
(2*d*x + 2*c) + I*sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*arctan2(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*
d*x + 2*c))) - 1, sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 16*(sqrt(2)*cos(2*d*x +
2*c) + I*sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*arctan2(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
))) - 1, -sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (16*I*sqrt(2)*cos(2*d*x + 2*c) -
16*sqrt(2)*sin(2*d*x + 2*c) + 16*I*sqrt(2))*arctan2(sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))), sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*
x + 2*c))) + cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (-16*I*sqrt(2)*cos(2*d*x + 2*c) + 16*
sqrt(2)*sin(2*d*x + 2*c) - 16*I*sqrt(2))*arctan2(-sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))
+ sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))), -sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c))) + cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 8*(sqrt(2)*cos(2*d*x + 2*c) + I*sqrt(2)
*sin(2*d*x + 2*c) + sqrt(2))*log(2*sqrt(2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/4*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1)*
cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2
+ 2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c))) + 1) - 8*(sqrt(2)*cos(2*d*x + 2*c) + I*sqrt(2)*sin(2*d*x + 2*c) + sqrt(2))*log(-2*sqrt(2)*
sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 2*
(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c))) + cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c)))^2 - 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (8*I*sqrt(2)*cos
(2*d*x + 2*c) - 8*sqrt(2)*sin(2*d*x + 2*c) + 8*I*sqrt(2))*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*
c), cos(2*d*x + 2*c))) + 2*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + (-8*I*sqrt(2)*c
os(2*d*x + 2*c) + 8*sqrt(2)*sin(2*d*x + 2*c) - 8*I*sqrt(2))*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c))) - 2*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + (8*I*sqrt(2)*
cos(2*d*x + 2*c) - 8*sqrt(2)*sin(2*d*x + 2*c) + 8*I*sqrt(2))*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x
+ 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c))) + 2*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + (-8*I*sqrt(2
)*cos(2*d*x + 2*c) + 8*sqrt(2)*sin(2*d*x + 2*c) - 8*I*sqrt(2))*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d
*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c))) - 2*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) - 128*cos(1/
4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 128*I*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*s
qrt(a)*sqrt(e)/((-64*I*e^2*cos(2*d*x + 2*c) + 64*e^2*sin(2*d*x + 2*c) - 64*I*e^2)*d)

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Fricas [A]  time = 2.24811, size = 1303, normalized size = 2.49 \begin{align*} \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{1}{2} i \, d x + \frac{1}{2} i \, c\right )} +{\left (d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{2}\right )} \sqrt{\frac{i \, a}{d^{2} e^{3}}} \log \left (i \, d e^{2} \sqrt{\frac{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 ) -{\left (d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{2}\right )} \sqrt{\frac{i \, a}{d^{2} e^{3}}} \log \left (-i \, d e^{2} \sqrt{\frac{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 ) +{\left (d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{2}\right )} \sqrt{-\frac{i \, a}{d^{2} e^{3}}} \log \left (i \, d e^{2} \sqrt{-\frac{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 ) -{\left (d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{2}\right )} \sqrt{-\frac{i \, a}{d^{2} e^{3}}} \log \left (-i \, d e^{2} \sqrt{-\frac{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 )}{2 \,{\left (d e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + d e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{i \, a \tan \left (d x + c\right ) + a}}{\left (e \cos \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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