3.12.22 \(\int \frac {1}{(a+i a \tan (e+f x)) \sqrt {c+d \tan (e+f x)}} \, dx\) [1122]

Optimal. Leaf size=155 \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 a \sqrt {c-i d} f}+\frac {(i c-2 d) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 a (c+i d)^{3/2} f}-\frac {\sqrt {c+d \tan (e+f x)}}{2 (i c-d) f (a+i a \tan (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3633, 3620, 3618, 65, 214} \begin {gather*} -\frac {\sqrt {c+d \tan (e+f x)}}{2 f (-d+i c) (a+i a \tan (e+f x))}-\frac {i \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{2 a f \sqrt {c-i d}}+\frac {(-2 d+i c) \tanh ^{-1}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{2 a f (c+i d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/2*I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(a*Sqrt[c - I*d]*f) + ((I*c - 2*d)*ArcTanh[Sqrt[c +
d*Tan[e + f*x]]/Sqrt[c + I*d]])/(2*a*(c + I*d)^(3/2)*f) - Sqrt[c + d*Tan[e + f*x]]/(2*(I*c - d)*f*(a + I*a*Ta
n[e + f*x]))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 3618

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

Rule 3620

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

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

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Mathematica [A]
time = 1.55, size = 222, normalized size = 1.43 \begin {gather*} \frac {\sec (e+f x) (\cos (f x)+i \sin (f x)) \left (-\frac {2 \left (\sqrt {-c+i d} (-i c+2 d) \text {ArcTan}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {-c-i d}}\right )-i (-c-i d)^{3/2} \text {ArcTan}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {-c+i d}}\right )\right ) (\cos (e)+i \sin (e))}{(-c-i d)^{3/2} \sqrt {-c+i d}}+\frac {2 \cos (e+f x) (i \cos (f x)+\sin (f x)) \sqrt {c+d \tan (e+f x)}}{c+i d}\right )}{4 f (a+i a \tan (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])*((-2*(Sqrt[-c + I*d]*((-I)*c + 2*d)*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt
[-c - I*d]] - I*(-c - I*d)^(3/2)*ArcTan[Sqrt[c + d*Tan[e + f*x]]/Sqrt[-c + I*d]])*(Cos[e] + I*Sin[e]))/((-c -
I*d)^(3/2)*Sqrt[-c + I*d]) + (2*Cos[e + f*x]*(I*Cos[f*x] + Sin[f*x])*Sqrt[c + d*Tan[e + f*x]])/(c + I*d)))/(4*
f*(a + I*a*Tan[e + f*x]))

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Maple [A]
time = 0.40, size = 150, normalized size = 0.97

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (\frac {-\frac {d \sqrt {c +d \tan \left (f x +e \right )}}{\left (i d +c \right ) \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {\left (i c -2 d \right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\left (i d +c \right ) \sqrt {-i d -c}}}{4 d^{2}}+\frac {i \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 d^{2} \sqrt {i d -c}}\right )}{f a}\)	\(150\)
default	\(\frac {2 d^{2} \left (\frac {-\frac {d \sqrt {c +d \tan \left (f x +e \right )}}{\left (i d +c \right ) \left (-d \tan \left (f x +e \right )+i d \right )}-\frac {\left (i c -2 d \right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{\left (i d +c \right ) \sqrt {-i d -c}}}{4 d^{2}}+\frac {i \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{4 d^{2} \sqrt {i d -c}}\right )}{f a}\)	\(150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c+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*(-1/(c+I*d)*d*(c+d*tan(f*x+e))^(1/2)/(-d*tan(f*x+e)+I*d)-(I*c-2*d)/(c+I*d)/(-I*d-c)^(1/2)*a
rctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2)))+1/4*I/d^2/(I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(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/(c+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. 1018 vs. \(2 (123) = 246\).
time = 1.42, size = 1018, normalized size = 6.57 \begin {gather*} -\frac {{\left (2 \, {\left (i \, a c - a d\right )} f \sqrt {\frac {i}{4 \, {\left (-i \, a^{2} c - a^{2} d\right )} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-2 \, {\left (2 \, {\left ({\left (i \, a c + a d\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, a c + a d\right )} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i}{4 \, {\left (-i \, a^{2} c - a^{2} d\right )} f^{2}}} - {\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - c\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) + 2 \, {\left (-i \, a c + a d\right )} f \sqrt {\frac {i}{4 \, {\left (-i \, a^{2} c - a^{2} d\right )} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-2 \, {\left (2 \, {\left ({\left (-i \, a c - a d\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, a c - a d\right )} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {i}{4 \, {\left (-i \, a^{2} c - a^{2} d\right )} f^{2}}} - {\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - c\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) + {\left (i \, a c - a d\right )} f \sqrt {-\frac {-i \, c^{2} + 4 \, c d + 4 i \, d^{2}}{{\left (-i \, a^{2} c^{3} + 3 \, a^{2} c^{2} d + 3 i \, a^{2} c d^{2} - a^{2} d^{3}\right )} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (-i \, c^{2} + 3 \, c d + 2 i \, d^{2} + {\left ({\left (a c^{2} + 2 i \, a c d - a d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (a c^{2} + 2 i \, a c d - a d^{2}\right )} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {-i \, c^{2} + 4 \, c d + 4 i \, d^{2}}{{\left (-i \, a^{2} c^{3} + 3 \, a^{2} c^{2} d + 3 i \, a^{2} c d^{2} - a^{2} d^{3}\right )} f^{2}}} + {\left (-i \, c^{2} + 2 \, c d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, {\left (a c^{2} + 2 i \, a c d - a d^{2}\right )} f}\right ) + {\left (-i \, a c + a d\right )} f \sqrt {-\frac {-i \, c^{2} + 4 \, c d + 4 i \, d^{2}}{{\left (-i \, a^{2} c^{3} + 3 \, a^{2} c^{2} d + 3 i \, a^{2} c d^{2} - a^{2} d^{3}\right )} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (-i \, c^{2} + 3 \, c d + 2 i \, d^{2} - {\left ({\left (a c^{2} + 2 i \, a c d - a d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (a c^{2} + 2 i \, a c d - a d^{2}\right )} f\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {-i \, c^{2} + 4 \, c d + 4 i \, d^{2}}{{\left (-i \, a^{2} c^{3} + 3 \, a^{2} c^{2} d + 3 i \, a^{2} c d^{2} - a^{2} d^{3}\right )} f^{2}}} + {\left (-i \, c^{2} + 2 \, c d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, {\left (a c^{2} + 2 i \, a c d - a d^{2}\right )} f}\right ) + 2 \, \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + 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 )}}{8 \, {\left (i \, a c - a d\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*(I*a*c - a*d)*f*sqrt(1/4*I/((-I*a^2*c - a^2*d)*f^2))*e^(2*I*f*x + 2*I*e)*log(-2*(2*((I*a*c + a*d)*f*e^
(2*I*f*x + 2*I*e) + (I*a*c + a*d)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))
*sqrt(1/4*I/((-I*a^2*c - a^2*d)*f^2)) - (c - I*d)*e^(2*I*f*x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) + 2*(-I*a*c +
a*d)*f*sqrt(1/4*I/((-I*a^2*c - a^2*d)*f^2))*e^(2*I*f*x + 2*I*e)*log(-2*(2*((-I*a*c - a*d)*f*e^(2*I*f*x + 2*I*
e) + (-I*a*c - a*d)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/4*I/((
-I*a^2*c - a^2*d)*f^2)) - (c - I*d)*e^(2*I*f*x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) + (I*a*c - a*d)*f*sqrt(-(-I
*c^2 + 4*c*d + 4*I*d^2)/((-I*a^2*c^3 + 3*a^2*c^2*d + 3*I*a^2*c*d^2 - a^2*d^3)*f^2))*e^(2*I*f*x + 2*I*e)*log(-1
/2*(-I*c^2 + 3*c*d + 2*I*d^2 + ((a*c^2 + 2*I*a*c*d - a*d^2)*f*e^(2*I*f*x + 2*I*e) + (a*c^2 + 2*I*a*c*d - a*d^2
)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(-I*c^2 + 4*c*d + 4*I*d^2
)/((-I*a^2*c^3 + 3*a^2*c^2*d + 3*I*a^2*c*d^2 - a^2*d^3)*f^2)) + (-I*c^2 + 2*c*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*
f*x - 2*I*e)/((a*c^2 + 2*I*a*c*d - a*d^2)*f)) + (-I*a*c + a*d)*f*sqrt(-(-I*c^2 + 4*c*d + 4*I*d^2)/((-I*a^2*c^3
+ 3*a^2*c^2*d + 3*I*a^2*c*d^2 - a^2*d^3)*f^2))*e^(2*I*f*x + 2*I*e)*log(-1/2*(-I*c^2 + 3*c*d + 2*I*d^2 - ((a*c
^2 + 2*I*a*c*d - a*d^2)*f*e^(2*I*f*x + 2*I*e) + (a*c^2 + 2*I*a*c*d - a*d^2)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*
I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(-I*c^2 + 4*c*d + 4*I*d^2)/((-I*a^2*c^3 + 3*a^2*c^2*d + 3*I*a
^2*c*d^2 - a^2*d^3)*f^2)) + (-I*c^2 + 2*c*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/((a*c^2 + 2*I*a*c*d - a
*d^2)*f)) + 2*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + 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)/((I*a*c - 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 {c + d \tan {\left (e + f x \right )}} \tan {\left (e + f x \right )} - i \sqrt {c + d \tan {\left (e + f x \right )}}}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (123) = 246\).
time = 0.57, size = 376, normalized size = 2.43 \begin {gather*} \frac {\sqrt {d \tan \left (f x + e\right ) + c} d}{2 \, {\left (a c f + i \, a d f\right )} {\left (d \tan \left (f x + e\right ) - i \, d\right )}} + \frac {2 \, {\left (c + 2 i \, d\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} + i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{-2 \, {\left (-i \, a c f + a d f\right )} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {i \, \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{a \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 8.31, size = 2500, normalized size = 16.13 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(a*d^6*f*1i - ((-(c*d^6*48i + 48*d^7 + 96*c^2*d^5 - c^3*d^4*32i - a^2*c^2*f^2*((((48*c^2*d^7 - 48*d^9 + 32*
c^4*d^5)*1i)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2) + (144*c*d^8 + 112*c^3*d^6 + 32*c^5*d^4)/(a^2*c^4
*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2))^2 - 4*((4*d^8 + 3*c^2*d^6 + c^4*d^4)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a
^4*c^2*d^2*f^4) + ((4*c*d^7 + 2*c^3*d^5)*1i)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4))*(256*d^6 + 256*c
^2*d^4))^(1/2)*1i + a^2*d^2*f^2*((((48*c^2*d^7 - 48*d^9 + 32*c^4*d^5)*1i)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c
^2*d^2*f^2) + (144*c*d^8 + 112*c^3*d^6 + 32*c^5*d^4)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2))^2 - 4*((
4*d^8 + 3*c^2*d^6 + c^4*d^4)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4) + ((4*c*d^7 + 2*c^3*d^5)*1i)/(a^4
*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4))*(256*d^6 + 256*c^2*d^4))^(1/2)*1i + 2*a^2*c*d*f^2*((((48*c^2*d^7
- 48*d^9 + 32*c^4*d^5)*1i)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2) + (144*c*d^8 + 112*c^3*d^6 + 32*c^5
*d^4)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2))^2 - 4*((4*d^8 + 3*c^2*d^6 + c^4*d^4)/(a^4*c^4*f^4 + a^4
*d^4*f^4 + 2*a^4*c^2*d^2*f^4) + ((4*c*d^7 + 2*c^3*d^5)*1i)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4))*(2
56*d^6 + 256*c^2*d^4))^(1/2))/(512*(d^6 + c^2*d^4)*(a^2*d^2*f^2*1i - a^2*c^2*f^2*1i + 2*a^2*c*d*f^2)))^(1/2)*(
24*a^3*d^7*f^3 + a^3*c*d^6*f^3*16i + 32*a^3*c^2*d^5*f^3 + a^3*c^3*d^4*f^3*16i + 8*a^3*c^4*d^3*f^3 - 2*(c + d*t
an(e + f*x))^(1/2)*(a^2*c^2*d^3*f^2*64i - 32*a^2*c*d^4*f^2 + 32*a^2*c^3*d^2*f^2)*(a^2*d^2*f^2 - a^2*c^2*f^2 +
a^2*c*d*f^2*2i)*(-(c*d^6*48i + 48*d^7 + 96*c^2*d^5 - c^3*d^4*32i - a^2*c^2*f^2*((((48*c^2*d^7 - 48*d^9 + 32*c^
4*d^5)*1i)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2) + (144*c*d^8 + 112*c^3*d^6 + 32*c^5*d^4)/(a^2*c^4*f
^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2))^2 - 4*((4*d^8 + 3*c^2*d^6 + c^4*d^4)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4
*c^2*d^2*f^4) + ((4*c*d^7 + 2*c^3*d^5)*1i)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4))*(256*d^6 + 256*c^2
*d^4))^(1/2)*1i + a^2*d^2*f^2*((((48*c^2*d^7 - 48*d^9 + 32*c^4*d^5)*1i)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2
*d^2*f^2) + (144*c*d^8 + 112*c^3*d^6 + 32*c^5*d^4)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2))^2 - 4*((4*
d^8 + 3*c^2*d^6 + c^4*d^4)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4) + ((4*c*d^7 + 2*c^3*d^5)*1i)/(a^4*c
^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4))*(256*d^6 + 256*c^2*d^4))^(1/2)*1i + 2*a^2*c*d*f^2*((((48*c^2*d^7 -
48*d^9 + 32*c^4*d^5)*1i)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2) + (144*c*d^8 + 112*c^3*d^6 + 32*c^5*d
^4)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2))^2 - 4*((4*d^8 + 3*c^2*d^6 + c^4*d^4)/(a^4*c^4*f^4 + a^4*d
^4*f^4 + 2*a^4*c^2*d^2*f^4) + ((4*c*d^7 + 2*c^3*d^5)*1i)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4))*(256
*d^6 + 256*c^2*d^4))^(1/2))/(512*(d^6 + c^2*d^4)*(a^2*d^2*f^2*1i - a^2*c^2*f^2*1i + 2*a^2*c*d*f^2)))^(1/2)) -
2*(c + d*tan(e + f*x))^(1/2)*(a^2*d^2*f^2 - a^2*c^2*f^2 + a^2*c*d*f^2*2i)*(c*d^3*6i - 5*d^4 + 2*c^2*d^2))*(-(c
*d^6*48i + 48*d^7 + 96*c^2*d^5 - c^3*d^4*32i - a^2*c^2*f^2*((((48*c^2*d^7 - 48*d^9 + 32*c^4*d^5)*1i)/(a^2*c^4*
f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2) + (144*c*d^8 + 112*c^3*d^6 + 32*c^5*d^4)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2
*a^2*c^2*d^2*f^2))^2 - 4*((4*d^8 + 3*c^2*d^6 + c^4*d^4)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4) + ((4*
c*d^7 + 2*c^3*d^5)*1i)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4))*(256*d^6 + 256*c^2*d^4))^(1/2)*1i + a^
2*d^2*f^2*((((48*c^2*d^7 - 48*d^9 + 32*c^4*d^5)*1i)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2) + (144*c*d
^8 + 112*c^3*d^6 + 32*c^5*d^4)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2))^2 - 4*((4*d^8 + 3*c^2*d^6 + c^
4*d^4)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4) + ((4*c*d^7 + 2*c^3*d^5)*1i)/(a^4*c^4*f^4 + a^4*d^4*f^4
+ 2*a^4*c^2*d^2*f^4))*(256*d^6 + 256*c^2*d^4))^(1/2)*1i + 2*a^2*c*d*f^2*((((48*c^2*d^7 - 48*d^9 + 32*c^4*d^5)
*1i)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2) + (144*c*d^8 + 112*c^3*d^6 + 32*c^5*d^4)/(a^2*c^4*f^2 + a
^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2))^2 - 4*((4*d^8 + 3*c^2*d^6 + c^4*d^4)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d
^2*f^4) + ((4*c*d^7 + 2*c^3*d^5)*1i)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4))*(256*d^6 + 256*c^2*d^4))
^(1/2))/(512*(d^6 + c^2*d^4)*(a^2*d^2*f^2*1i - a^2*c^2*f^2*1i + 2*a^2*c*d*f^2)))^(1/2) - (3*a*c*d^5*f)/2 - (a*
c^3*d^3*f)/2)*(-(c*d^6*48i + 48*d^7 + 96*c^2*d^5 - c^3*d^4*32i - a^2*c^2*f^2*((((48*c^2*d^7 - 48*d^9 + 32*c^4*
d^5)*1i)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2) + (144*c*d^8 + 112*c^3*d^6 + 32*c^5*d^4)/(a^2*c^4*f^2
+ a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2))^2 - 4*((4*d^8 + 3*c^2*d^6 + c^4*d^4)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c
^2*d^2*f^4) + ((4*c*d^7 + 2*c^3*d^5)*1i)/(a^4*c^4*f^4 + a^4*d^4*f^4 + 2*a^4*c^2*d^2*f^4))*(256*d^6 + 256*c^2*d
^4))^(1/2)*1i + a^2*d^2*f^2*((((48*c^2*d^7 - 48*d^9 + 32*c^4*d^5)*1i)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d
^2*f^2) + (144*c*d^8 + 112*c^3*d^6 + 32*c^5*d^4)/(a^2*c^4*f^2 + a^2*d^4*f^2 + 2*a^2*c^2*d^2*f^2))^2 - 4*((4*d^
8 + 3*c^2*d^6 + c^4*d^4)/(a^4*c^4*f^4 + a^4*d^4...

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