3.1163 \(\int \frac{\sqrt{a+i a \tan (e+f x)}}{(c+d \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=129 \[ -\frac{2 d \sqrt{a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{i \sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f (c-i d)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.212067, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {3548, 3544, 208} \[ -\frac{2 d \sqrt{a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt{c+d \tan (e+f x)}}-\frac{i \sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f (c-i d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + I*a*Tan[e + f*x]]/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

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

Rule 3548

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

Rule 3544

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

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

Mathematica [B]  time = 4.23238, size = 337, normalized size = 2.61 \[ \frac{\sqrt{2} \sqrt{e^{i f x}} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \sqrt{a+i a \tan (e+f x)} \left (-\frac{2 d \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}}{(c-i d) (c+i d) \left (c \left (1+e^{2 i (e+f x)}\right )-i d \left (-1+e^{2 i (e+f x)}\right )\right )}-\frac{i e^{-i (e+f x)} \log \left (2 \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{(c-i d)^{3/2}}\right )}{f \sqrt{\sec (e+f x)} \sqrt{\cos (f x)+i \sin (f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + I*a*Tan[e + f*x]]/(c + d*Tan[e + f*x])^(3/2),x]

[Out]

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

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Maple [B]  time = 0.122, size = 1289, normalized size = 10. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.62371, size = 1565, normalized size = 12.13 \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(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

(sqrt(2)*(-4*I*d*e^(2*I*f*x + 2*I*e) - 4*I*d)*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(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) + ((I*c^3 + c^2*d + I*c*d^2 + d^3)*f*e^(2*I*f*x +
2*I*e) + (I*c^3 - c^2*d + I*c*d^2 - d^3)*f)*sqrt(2*I*a/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*log(((I*c^2
+ 2*c*d - I*d^2)*f*sqrt(2*I*a/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*e^(2*I*f*x + 2*I*e) + sqrt(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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I
*f*x + 2*I*e) + 1)*e^(I*f*x + I*e))*e^(-2*I*f*x - 2*I*e)) + ((-I*c^3 - c^2*d - I*c*d^2 - d^3)*f*e^(2*I*f*x + 2
*I*e) + (-I*c^3 + c^2*d - I*c*d^2 + d^3)*f)*sqrt(2*I*a/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*log(((-I*c^
2 - 2*c*d + I*d^2)*f*sqrt(2*I*a/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*e^(2*I*f*x + 2*I*e) + sqrt(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))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*
I*f*x + 2*I*e) + 1)*e^(I*f*x + I*e))*e^(-2*I*f*x - 2*I*e)))/((2*I*c^3 + 2*c^2*d + 2*I*c*d^2 + 2*d^3)*f*e^(2*I*
f*x + 2*I*e) + (2*I*c^3 - 2*c^2*d + 2*I*c*d^2 - 2*d^3)*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \left (i \tan{\left (e + f x \right )} + 1\right )}}{\left (c + d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(1/2)/(c+d*tan(f*x+e))**(3/2),x)

[Out]

Integral(sqrt(a*(I*tan(e + f*x) + 1))/(c + d*tan(e + f*x))**(3/2), x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out