3.1122 \(\int \frac{1}{(a+i a \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx\)

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

[Out]

((-I/2)*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*Tan[
e + f*x]))

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Rubi [A]  time = 0.303445, antiderivative size = 155, normalized size of antiderivative = 1., 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 = {3552, 3539, 3537, 63, 208} \[ -\frac{\sqrt{c+d \tan (e+f x)}}{2 f (-d+i c) (a+i a \tan (e+f x))}-\frac{i \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{2 a f \sqrt{c-i d}}+\frac{(-2 d+i c) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{2 a f (c+i d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/2)*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*Tan[
e + f*x]))

Rule 3552

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]

Rule 3539

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 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^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 63

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 208

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

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 \operatorname{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)) \operatorname{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{\operatorname{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) \operatorname{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*}

Mathematica [A]  time = 1.39183, size = 222, normalized size = 1.43 \[ \frac{\sec (e+f x) (\cos (f x)+i \sin (f x)) \left (\frac{2 \cos (e+f x) (\sin (f x)+i \cos (f x)) \sqrt{c+d \tan (e+f x)}}{c+i d}-\frac{2 (\cos (e)+i \sin (e)) \left (\sqrt{-c+i d} (2 d-i c) \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c-i d}}\right )-i (-c-i d)^{3/2} \tan ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{-c+i d}}\right )\right )}{(-c-i d)^{3/2} \sqrt{-c+i d}}\right )}{4 f (a+i a \tan (e+f x))} \]

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 = 0.076, size = 191, normalized size = 1.2 \begin{align*}{\frac{{\frac{i}{2}}}{af}\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{id-c}}}} \right ){\frac{1}{\sqrt{id-c}}}}+{\frac{d}{2\,af \left ( c+id \right ) \left ( -id+d\tan \left ( fx+e \right ) \right ) }\sqrt{c+d\tan \left ( fx+e \right ) }}-{\frac{{\frac{i}{2}}c}{af \left ( c+id \right ) }\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id-c}}}} \right ){\frac{1}{\sqrt{-id-c}}}}+{\frac{d}{af \left ( c+id \right ) }\arctan \left ({\sqrt{c+d\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{-id-c}}}} \right ){\frac{1}{\sqrt{-id-c}}}} \end{align*}

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)

[Out]

1/2*I/f/a/(I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(I*d-c)^(1/2))+1/2/f/a*d/(c+I*d)*(c+d*tan(f*x+e))^(1/2)/
(-I*d+d*tan(f*x+e))-1/2*I/f/a/(c+I*d)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e))^(1/2)/(-I*d-c)^(1/2))*c+1/f/a*d/(
c+I*d)/(-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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

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

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Fricas [B]  time = 2.43867, size = 2472, normalized size = 15.95 \begin{align*} \text{result too large to display} \end{align*}

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]

-1/4*((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))*s
qrt(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(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)/((4*I*a^2*c^3 - 12*a^2*c^2*d - 12*I*a^2*c*d^2 + 4*a^2*d^3)*f^2))*e^(2*I*f*x + 2*I*e)*log(-(
-I*c^2 + 3*c*d + 2*I*d^2 + ((2*a*c^2 + 4*I*a*c*d - 2*a*d^2)*f*e^(2*I*f*x + 2*I*e) + (2*a*c^2 + 4*I*a*c*d - 2*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)/((4*I*a^2*c^3 - 12*a^2*c^2*d - 12*I*a^2*c*d^2 + 4*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)/((2*a*c^2 + 4*I*a*c*d - 2*a*d^2)*f)) + (-I*a*c + a*d)*f*sqrt((-I*c^2 + 4*c*d + 4*I*d^2)/(
(4*I*a^2*c^3 - 12*a^2*c^2*d - 12*I*a^2*c*d^2 + 4*a^2*d^3)*f^2))*e^(2*I*f*x + 2*I*e)*log(-(-I*c^2 + 3*c*d + 2*I
*d^2 - ((2*a*c^2 + 4*I*a*c*d - 2*a*d^2)*f*e^(2*I*f*x + 2*I*e) + (2*a*c^2 + 4*I*a*c*d - 2*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)/((4*I*a^2*c^3 -
12*a^2*c^2*d - 12*I*a^2*c*d^2 + 4*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)
/((2*a*c^2 + 4*I*a*c*d - 2*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))*(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(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

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]

Exception raised: AttributeError

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Giac [B]  time = 1.43241, size = 528, normalized size = 3.41 \begin{align*} -\frac{1}{2} \, d^{2}{\left (\frac{8 \,{\left (c + 2 i \, d\right )} \arctan \left (\frac{4 \,{\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{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} + i \, \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d - \sqrt{c^{2} + d^{2}} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}}\right )}{{\left (-2 i \, a c d^{2} f + 2 \, a d^{3} f\right )} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}{\left (\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}} - \frac{\sqrt{d \tan \left (f x + e\right ) + c}}{{\left (a c d f + i \, a d^{2} f\right )}{\left (d \tan \left (f x + e\right ) - i \, d\right )}} - \frac{4 i \, \arctan \left (\frac{4 \,{\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{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} - i \, \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d - \sqrt{c^{2} + d^{2}} \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}}}\right )}{a \sqrt{-8 \, c + 8 \, \sqrt{c^{2} + d^{2}}} d^{2} f{\left (-\frac{i \, d}{c - \sqrt{c^{2} + d^{2}}} + 1\right )}}\right )} \end{align*}

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