3.677 \(\int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{e \cos (c+d x)}} \, dx\)

Optimal. Leaf size=335 \[ \frac{i \sqrt{2} \sqrt{a} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{a+i a \tan (c+d x)} \sqrt{e \cos (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{d \sqrt{e}}-\frac{i \sqrt{2} \sqrt{a} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{a+i a \tan (c+d x)} \sqrt{e \cos (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{d \sqrt{e}}-\frac{i \sqrt{a} \log \left (-\sqrt{2} \sqrt{a} \sqrt{a+i a \tan (c+d x)} \sqrt{e \cos (c+d x)}+\sqrt{e} \cos (c+d x) (a+i a \tan (c+d x))+a \sqrt{e}\right )}{\sqrt{2} d \sqrt{e}}+\frac{i \sqrt{a} \log \left (\sqrt{2} \sqrt{a} \sqrt{a+i a \tan (c+d x)} \sqrt{e \cos (c+d x)}+\sqrt{e} \cos (c+d x) (a+i a \tan (c+d x))+a \sqrt{e}\right )}{\sqrt{2} d \sqrt{e}} \]

[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])])/(d
*Sqrt[e]) - (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]*S
qrt[e])])/(d*Sqrt[e]) - (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])])/(Sqrt[2]*d*Sqrt[e]) + (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])])/(Sqrt[
2]*d*Sqrt[e])

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Rubi [A]  time = 0.214058, antiderivative size = 335, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.233, Rules used = {3513, 297, 1162, 617, 204, 1165, 628} \[ \frac{i \sqrt{2} \sqrt{a} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{a+i a \tan (c+d x)} \sqrt{e \cos (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{d \sqrt{e}}-\frac{i \sqrt{2} \sqrt{a} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{a+i a \tan (c+d x)} \sqrt{e \cos (c+d x)}}{\sqrt{a} \sqrt{e}}\right )}{d \sqrt{e}}-\frac{i \sqrt{a} \log \left (-\sqrt{2} \sqrt{a} \sqrt{a+i a \tan (c+d x)} \sqrt{e \cos (c+d x)}+\sqrt{e} \cos (c+d x) (a+i a \tan (c+d x))+a \sqrt{e}\right )}{\sqrt{2} d \sqrt{e}}+\frac{i \sqrt{a} \log \left (\sqrt{2} \sqrt{a} \sqrt{a+i a \tan (c+d x)} \sqrt{e \cos (c+d x)}+\sqrt{e} \cos (c+d x) (a+i a \tan (c+d x))+a \sqrt{e}\right )}{\sqrt{2} d \sqrt{e}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[e*Cos[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])])/(d
*Sqrt[e]) - (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]*S
qrt[e])])/(d*Sqrt[e]) - (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])])/(Sqrt[2]*d*Sqrt[e]) + (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])])/(Sqrt[
2]*d*Sqrt[e])

Rule 3513

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

Mathematica [A]  time = 0.849434, size = 125, normalized size = 0.37 \[ \frac{i \left (-e^{-2 i c}\right )^{3/4} e^{-\frac{3}{2} i d x} \left (1+e^{2 i (c+d x)}\right ) \sqrt{a+i a \tan (c+d x)} \left (\tan ^{-1}\left (\frac{e^{\frac{i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )-\tanh ^{-1}\left (\frac{e^{\frac{i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )\right )}{d \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[e*Cos[c + d*x]],x]

[Out]

(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]])

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Maple [A]  time = 0.371, size = 226, normalized size = 0.7 \begin{align*} -{\frac{\cos \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) -1 \right ) }{d\sin \left ( dx+c \right ) \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 ( 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 ) +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 ) +{\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 ) \right ){\frac{1}{\sqrt{e\cos \left ( dx+c \right ) }}}{\frac{1}{\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}}}} \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))^(1/2),x)

[Out]

-1/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)*(I*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/(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)))-arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+si
n(d*x+c))))/sin(d*x+c)/(e*cos(d*x+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c)-1)/(1/(cos(d*x+c)+1))^(1/2)

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Maxima [B]  time = 3.13678, size = 1890, normalized size = 5.64 \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))^(1/2),x, algorithm="maxima")

[Out]

1/4*(-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)*si
n(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*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*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), co
s(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)*ar
ctan2(-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)*c
os(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1)*cos(2/3*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)))^2 + 2*cos(1/3*arctan2(sin(3/2*d
*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin
(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*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*x + 3/2*c))) - 1)*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + cos(2/3*arc
tan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/
2*c)))^2 + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/
2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*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)*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x
+ 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) +
2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - sqrt(2)*log(2*cos(1/3*arctan2(si
n(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^
2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2
*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) + sqrt(2)*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x +
3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(sin
(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*
c))) + 2) - sqrt(2)*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(s
in(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x +
3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2))*sqrt(a)/(d*sqrt(e))

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Fricas [A]  time = 2.26279, size = 917, normalized size = 2.74 \begin{align*} \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} \, 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} \, 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} \, 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} \, d e \sqrt{-\frac{4 i \, a}{d^{2} e}}\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))^(1/2),x, algorithm="fricas")

[Out]

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*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*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*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*d*e*sqrt
(-4*I*a/(d^2*e)))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )}}{\sqrt{e \cos{\left (c + d x \right )}}}\, dx \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))**(1/2),x)

[Out]

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

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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}}{\sqrt{e \cos \left (d x + c\right )}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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