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]

-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*ta
n(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*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)-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)

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Rubi [A]  time = 0.21, antiderivative size = 335, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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 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 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]

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*}

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Mathematica [A]  time = 0.89, 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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fricas [A]  time = 0.48, size = 313, normalized size = 0.93 \[ -\frac {1}{2} \, \sqrt {\frac {4 i \, a}{d^{2} e}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + \frac {1}{2} i \, d e \sqrt {\frac {4 i \, a}{d^{2} e}}\right ) + \frac {1}{2} \, \sqrt {\frac {4 i \, a}{d^{2} e}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - \frac {1}{2} i \, d e \sqrt {\frac {4 i \, a}{d^{2} e}}\right ) - \frac {1}{2} \, \sqrt {-\frac {4 i \, a}{d^{2} e}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} + \frac {1}{2} i \, d e \sqrt {-\frac {4 i \, a}{d^{2} e}}\right ) + \frac {1}{2} \, \sqrt {-\frac {4 i \, a}{d^{2} e}} \log \left (\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )} - \frac {1}{2} i \, d e \sqrt {-\frac {4 i \, a}{d^{2} e}}\right ) \]

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*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)*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) - 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(sq
rt(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)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {e \cos \left (d x + c\right )}}\,{d x} \]

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)

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maple [A]  time = 1.54, size = 226, normalized size = 0.67 \[ -\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \left (-1+\cos \left (d x +c \right )\right ) \left (i \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right )+i \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (-\cos \left (d x +c \right )-1+\sin \left (d x +c \right )\right )}{2}\right )-\arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right )+\arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (-\cos \left (d x +c \right )-1+\sin \left (d x +c \right )\right )}{2}\right )\right )}{d \sin \left (d x +c \right ) \sqrt {e \cos \left (d x +c \right )}\, \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}} \]

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)*(-1+cos(d*x+c))*(I*arctanh(1/2*(1/(1+cos(d*x+c)
))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))+I*arctanh(1/2*(1/(1+cos(d*x+c)))^(1/2)*(-cos(d*x+c)-1+sin(d*x+c)))-arctanh
(1/2*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))+arctanh(1/2*(1/(1+cos(d*x+c)))^(1/2)*(-cos(d*x+c)-1+s
in(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/(1+cos(d*x+c)))^(1/2)

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maxima [B]  time = 0.88, size = 1400, normalized size = 4.18 \[ \text {result too large to display} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {e\,\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}{\sqrt {e \cos {\left (c + d x \right )}}}\, dx \]

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

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