3.1156 \(\int \frac{(a+i a \tan (e+f x))^{3/2}}{\sqrt{c+d \tan (e+f x)}} \, dx\)

Optimal. Leaf size=151 \[ -\frac{2 (-1)^{3/4} a^{3/2} \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{d} f}-\frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f \sqrt{c-i d}} \]

[Out]

(-2*(-1)^(3/4)*a^(3/2)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x
]])])/(Sqrt[d]*f) - ((2*I)*Sqrt[2]*a^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*S
qrt[a + I*a*Tan[e + f*x]])])/(Sqrt[c - I*d]*f)

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Rubi [A]  time = 0.434776, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.219, Rules used = {3555, 3544, 208, 3599, 63, 217, 206} \[ -\frac{2 (-1)^{3/4} a^{3/2} \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{d} f}-\frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f \sqrt{c-i d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-1)^(3/4)*a^(3/2)*ArcTanh[((-1)^(3/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x
]])])/(Sqrt[d]*f) - ((2*I)*Sqrt[2]*a^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*S
qrt[a + I*a*Tan[e + f*x]])])/(Sqrt[c - I*d]*f)

Rule 3555

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dis
t[2*a, Int[Sqrt[a + b*Tan[e + f*x]]/Sqrt[c + d*Tan[e + f*x]], x], x] + Dist[b/a, Int[((b + a*Tan[e + f*x])*Sqr
t[a + b*Tan[e + f*x]])/Sqrt[c + d*Tan[e + f*x]], 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 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]

Rule 3599

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*B)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [B]  time = 4.45608, size = 505, normalized size = 3.34 \[ \frac{\left (\frac{1}{2}-\frac{i}{2}\right ) \cos (e+f x) (\cos (f x)-i \sin (f x)) (a+i a \tan (e+f x))^{3/2} \sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1} (\cos (2 e+f x)-i \sin (2 e+f x)) \left (\sqrt{c-i d} \log \left (\frac{(2-2 i) e^{\frac{3 i e}{2}} \left (-(1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c \left (e^{i (e+f x)}+i\right )+d e^{i (e+f x)}-i d\right )}{\sqrt{d} \left (e^{i (e+f x)}+i\right )}\right )-\sqrt{c-i d} \log \left (\frac{(2+2 i) e^{\frac{3 i e}{2}} \left ((1-i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+c \left (e^{i (e+f x)}-i\right )-i d e^{i (e+f x)}+d\right )}{\sqrt{d} \left (e^{i (e+f x)}-i\right )}\right )+(2-2 i) \sqrt{d} \log \left (2 \left (i \sqrt{c-i d} \sin (e+f x)+\sqrt{c-i d} \cos (e+f x)+\sqrt{i \sin (2 (e+f x))+\cos (2 (e+f x))+1} \sqrt{c+d \tan (e+f x)}\right )\right )\right )}{\sqrt{d} f \sqrt{c-i d}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.072, size = 986, normalized size = 6.5 \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))^(3/2)/(c+d*tan(f*x+e))^(1/2),x)

[Out]

1/2/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*a^2*(I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*tan(f*
x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*c-I*2^(1/2)*(-a*(
I*d-c))^(1/2)*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d
)/(I*a*d)^(1/2))*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*ta
n(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(tan(f*x+e)+I))*(I*a*d)^(1/2)*c+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))*(I*a*d)^(1/
2)*d-2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)
*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*c^2-2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*t
an(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*d^2+2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*
I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*c+2^(1/
2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*tan(f*x+e)*d+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(
1/2)+a*d)/(I*a*d)^(1/2))*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))*(I*a*d)^(1/2)*c-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))*(I*a*d)
^(1/2)*d)/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(c^2+d^2)/(I*a*d)^(1/2)*2^(1/2)/(-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))^(3/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.43922, size = 1530, normalized size = 10.13 \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))^(3/2)/(c+d*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out