3.2.68 \(\int \frac {1}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx\) [168]

Optimal. Leaf size=262 \[ -\frac {\left (\frac {3}{4}-\frac {i}{4}\right ) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a \sqrt {d} f}+\frac {\left (\frac {3}{4}-\frac {i}{4}\right ) \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a \sqrt {d} f}-\frac {\left (\frac {3}{8}+\frac {i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a \sqrt {d} f}+\frac {\left (\frac {3}{8}+\frac {i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))} \]

[Out]

(-3/8+1/8*I)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))/a/f*2^(1/2)/d^(1/2)+(3/8-1/8*I)*arctan(1+2^(1/2)*(
d*tan(f*x+e))^(1/2)/d^(1/2))/a/f*2^(1/2)/d^(1/2)-(3/16+1/16*I)*ln(d^(1/2)-2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)
*tan(f*x+e))/a/f*2^(1/2)/d^(1/2)+(3/16+1/16*I)*ln(d^(1/2)+2^(1/2)*(d*tan(f*x+e))^(1/2)+d^(1/2)*tan(f*x+e))/a/f
*2^(1/2)/d^(1/2)+1/2*(d*tan(f*x+e))^(1/2)/d/f/(a+I*a*tan(f*x+e))

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Rubi [A]
time = 0.15, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3633, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {\left (\frac {3}{4}-\frac {i}{4}\right ) \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a \sqrt {d} f}+\frac {\left (\frac {3}{4}-\frac {i}{4}\right ) \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} a \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}-\frac {\left (\frac {3}{8}+\frac {i}{8}\right ) \log \left (\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a \sqrt {d} f}+\frac {\left (\frac {3}{8}+\frac {i}{8}\right ) \log \left (\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}+\sqrt {d}\right )}{\sqrt {2} a \sqrt {d} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d*Tan[e + f*x]]*(a + I*a*Tan[e + f*x])),x]

[Out]

((-3/4 + I/4)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a*Sqrt[d]*f) + ((3/4 - I/4)*ArcTan[
1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[2]*a*Sqrt[d]*f) - ((3/8 + I/8)*Log[Sqrt[d] + Sqrt[d]*Tan[e
+ f*x] - Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a*Sqrt[d]*f) + ((3/8 + I/8)*Log[Sqrt[d] + Sqrt[d]*Tan[e + f*x
] + Sqrt[2]*Sqrt[d*Tan[e + f*x]]])/(Sqrt[2]*a*Sqrt[d]*f) + Sqrt[d*Tan[e + f*x]]/(2*d*f*(a + I*a*Tan[e + f*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 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 1182

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

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3633

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

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \, dx &=\frac {\sqrt {d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}-\frac {\int \frac {-\frac {3 a d}{2}+\frac {1}{2} i a d \tan (e+f x)}{\sqrt {d \tan (e+f x)}} \, dx}{2 a^2 d}\\ &=\frac {\sqrt {d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}-\frac {\text {Subst}\left (\int \frac {-\frac {3 a d^2}{2}+\frac {1}{2} i a d x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a^2 d f}\\ &=\frac {\sqrt {d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}--\frac {\left (\frac {3}{4}+\frac {i}{4}\right ) \text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a f}--\frac {\left (\frac {3}{4}-\frac {i}{4}\right ) \text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a f}\\ &=\frac {\sqrt {d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}--\frac {\left (\frac {3}{8}-\frac {i}{8}\right ) \text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a f}--\frac {\left (\frac {3}{8}-\frac {i}{8}\right ) \text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{a f}-\frac {\left (\frac {3}{8}+\frac {i}{8}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a \sqrt {d} f}-\frac {\left (\frac {3}{8}+\frac {i}{8}\right ) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a \sqrt {d} f}\\ &=-\frac {\left (\frac {3}{8}+\frac {i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a \sqrt {d} f}+\frac {\left (\frac {3}{8}+\frac {i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}--\frac {\left (\frac {3}{4}-\frac {i}{4}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a \sqrt {d} f}-\frac {\left (\frac {3}{4}-\frac {i}{4}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a \sqrt {d} f}\\ &=-\frac {\left (\frac {3}{4}-\frac {i}{4}\right ) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a \sqrt {d} f}+\frac {\left (\frac {3}{4}-\frac {i}{4}\right ) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a \sqrt {d} f}-\frac {\left (\frac {3}{8}+\frac {i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)-\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a \sqrt {d} f}+\frac {\left (\frac {3}{8}+\frac {i}{8}\right ) \log \left (\sqrt {d}+\sqrt {d} \tan (e+f x)+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\sqrt {2} a \sqrt {d} f}+\frac {\sqrt {d \tan (e+f x)}}{2 d f (a+i a \tan (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 0.87, size = 147, normalized size = 0.56 \begin {gather*} \frac {\sec (e+f x) \sqrt {\sin (2 (e+f x))} \left (-2 i \sec (e+f x) \sqrt {\sin (2 (e+f x))}+(1+3 i) \text {ArcSin}(\cos (e+f x)-\sin (e+f x)) (1+i \tan (e+f x))+(3+i) \log \left (\cos (e+f x)+\sin (e+f x)+\sqrt {\sin (2 (e+f x))}\right ) (-i+\tan (e+f x))\right )}{8 a f \sqrt {d \tan (e+f x)} (-i+\tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[d*Tan[e + f*x]]*(a + I*a*Tan[e + f*x])),x]

[Out]

(Sec[e + f*x]*Sqrt[Sin[2*(e + f*x)]]*((-2*I)*Sec[e + f*x]*Sqrt[Sin[2*(e + f*x)]] + (1 + 3*I)*ArcSin[Cos[e + f*
x] - Sin[e + f*x]]*(1 + I*Tan[e + f*x]) + (3 + I)*Log[Cos[e + f*x] + Sin[e + f*x] + Sqrt[Sin[2*(e + f*x)]]]*(-
I + Tan[e + f*x])))/(8*a*f*Sqrt[d*Tan[e + f*x]]*(-I + Tan[e + f*x]))

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Maple [A]
time = 0.17, size = 102, normalized size = 0.39

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (-\frac {-\frac {\sqrt {d \tan \left (f x +e \right )}}{i d \tan \left (f x +e \right )+d}+\frac {2 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{4 d^{2}}+\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 d^{2} \sqrt {i d}}\right )}{f a}\)	\(102\)
default	\(\frac {2 d^{2} \left (-\frac {-\frac {\sqrt {d \tan \left (f x +e \right )}}{i d \tan \left (f x +e \right )+d}+\frac {2 i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{4 d^{2}}+\frac {i \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 d^{2} \sqrt {i d}}\right )}{f a}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f/a*d^2*(-1/4/d^2*(-(d*tan(f*x+e))^(1/2)/(I*d*tan(f*x+e)+d)+2*I/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I
*d)^(1/2)))+1/4*I/d^2/(I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(I*d)^(1/2)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 541 vs. \(2 (196) = 392\).
time = 0.40, size = 541, normalized size = 2.06 \begin {gather*} -\frac {{\left (a d f \sqrt {-\frac {i}{4 \, a^{2} d f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-2 \, {\left (2 \, {\left (a d f e^{\left (2 i \, f x + 2 i \, e\right )} + a d f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {i}{4 \, a^{2} d f^{2}}} + i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - a d f \sqrt {-\frac {i}{4 \, a^{2} d f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (2 \, {\left (2 \, {\left (a d f e^{\left (2 i \, f x + 2 i \, e\right )} + a d f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {i}{4 \, a^{2} d f^{2}}} - i \, d e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - a d f \sqrt {\frac {i}{a^{2} d f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left ({\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i}{a^{2} d f^{2}}} + i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a f}\right ) + a d f \sqrt {\frac {i}{a^{2} d f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left ({\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i}{a^{2} d f^{2}}} - i\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{a f}\right ) - \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{4 \, a d f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a*d*f*sqrt(-1/4*I/(a^2*d*f^2))*e^(2*I*f*x + 2*I*e)*log(-2*(2*(a*d*f*e^(2*I*f*x + 2*I*e) + a*d*f)*sqrt((-
I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-1/4*I/(a^2*d*f^2)) + I*d*e^(2*I*f*x + 2*I*e))*
e^(-2*I*f*x - 2*I*e)) - a*d*f*sqrt(-1/4*I/(a^2*d*f^2))*e^(2*I*f*x + 2*I*e)*log(2*(2*(a*d*f*e^(2*I*f*x + 2*I*e)
+ a*d*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-1/4*I/(a^2*d*f^2)) - I*d*e^(2
*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - a*d*f*sqrt(I/(a^2*d*f^2))*e^(2*I*f*x + 2*I*e)*log(((a*f*e^(2*I*f*x +
2*I*e) + a*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(I/(a^2*d*f^2)) + I)*e^(-2*
I*f*x - 2*I*e)/(a*f)) + a*d*f*sqrt(I/(a^2*d*f^2))*e^(2*I*f*x + 2*I*e)*log(-((a*f*e^(2*I*f*x + 2*I*e) + a*f)*sq
rt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(I/(a^2*d*f^2)) - I)*e^(-2*I*f*x - 2*I*e)/(
a*f)) - sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I*e) + 1))*e^(-2*I*f*
x - 2*I*e)/(a*d*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \int \frac {1}{\sqrt {d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )} - i \sqrt {d \tan {\left (e + f x \right )}}}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e)),x)

[Out]

-I*Integral(1/(sqrt(d*tan(e + f*x))*tan(e + f*x) - I*sqrt(d*tan(e + f*x))), x)/a

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Giac [A]
time = 0.60, size = 173, normalized size = 0.66 \begin {gather*} \frac {i \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{2 \, a \sqrt {d} f {\left (\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {i \, \sqrt {2} \arctan \left (\frac {8 \, \sqrt {d^{2}} \sqrt {d \tan \left (f x + e\right )}}{-4 i \, \sqrt {2} d^{\frac {3}{2}} + 4 \, \sqrt {2} \sqrt {d^{2}} \sqrt {d}}\right )}{a \sqrt {d} f {\left (-\frac {i \, d}{\sqrt {d^{2}}} + 1\right )}} - \frac {i \, \sqrt {d \tan \left (f x + e\right )}}{2 \, {\left (d \tan \left (f x + e\right ) - i \, d\right )} a f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*sqrt(2)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(4*I*sqrt(2)*d^(3/2) + 4*sqrt(2)*sqrt(d^2)*sqrt(d)))/(a*
sqrt(d)*f*(I*d/sqrt(d^2) + 1)) - I*sqrt(2)*arctan(8*sqrt(d^2)*sqrt(d*tan(f*x + e))/(-4*I*sqrt(2)*d^(3/2) + 4*s
qrt(2)*sqrt(d^2)*sqrt(d)))/(a*sqrt(d)*f*(-I*d/sqrt(d^2) + 1)) - 1/2*I*sqrt(d*tan(f*x + e))/((d*tan(f*x + e) -
I*d)*a*f)

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Mupad [B]
time = 5.92, size = 128, normalized size = 0.49 \begin {gather*} -\mathrm {atan}\left (2\,a\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {1{}\mathrm {i}}{4\,a^2\,d\,f^2}}\right )\,\sqrt {\frac {1{}\mathrm {i}}{4\,a^2\,d\,f^2}}\,2{}\mathrm {i}+\mathrm {atan}\left (4\,a\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {1{}\mathrm {i}}{16\,a^2\,d\,f^2}}\right )\,\sqrt {-\frac {1{}\mathrm {i}}{16\,a^2\,d\,f^2}}\,2{}\mathrm {i}+\frac {\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}{2\,a\,f\,\left (-d\,\mathrm {tan}\left (e+f\,x\right )+d\,1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(4*a*f*(d*tan(e + f*x))^(1/2)*(-1i/(16*a^2*d*f^2))^(1/2))*(-1i/(16*a^2*d*f^2))^(1/2)*2i - atan(2*a*f*(d*ta
n(e + f*x))^(1/2)*(1i/(4*a^2*d*f^2))^(1/2))*(1i/(4*a^2*d*f^2))^(1/2)*2i + ((d*tan(e + f*x))^(1/2)*1i)/(2*a*f*(
d*1i - d*tan(e + f*x)))

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