3.1168 \(\int \frac{(a+i a \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=179 \[ -\frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f (c-i d)^{5/2}}-\frac{2 d (a+i a \tan (e+f x))^{3/2}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac{2 i a \sqrt{a+i a \tan (e+f x)}}{f (c-i d)^2 \sqrt{c+d \tan (e+f x)}} \]

[Out]

((-2*I)*Sqrt[2]*a^(3/2)*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)^(5/2)*f) - (2*d*(a + I*a*Tan[e + f*x])^(3/2))/(3*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)
) + ((2*I)*a*Sqrt[a + I*a*Tan[e + f*x]])/((c - I*d)^2*f*Sqrt[c + d*Tan[e + f*x]])

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Rubi [A]  time = 0.341262, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3548, 3545, 3544, 208} \[ -\frac{2 i \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f (c-i d)^{5/2}}-\frac{2 d (a+i a \tan (e+f x))^{3/2}}{3 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^{3/2}}+\frac{2 i a \sqrt{a+i a \tan (e+f x)}}{f (c-i d)^2 \sqrt{c+d \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(3/2)/(c + d*Tan[e + f*x])^(5/2),x]

[Out]

((-2*I)*Sqrt[2]*a^(3/2)*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)^(5/2)*f) - (2*d*(a + I*a*Tan[e + f*x])^(3/2))/(3*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^(3/2)
) + ((2*I)*a*Sqrt[a + I*a*Tan[e + f*x]])/((c - I*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 3545

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*b*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m - 1)*(a*c - b*d)), x] + Dist[(2*a^2)/(
a*c - b*d), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), 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] && EqQ[m + n, 0] && GtQ[m, 1/2]

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{(a+i a \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac{2 d (a+i a \tan (e+f x))^{3/2}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{\int \frac{(a+i a \tan (e+f x))^{3/2}}{(c+d \tan (e+f x))^{3/2}} \, dx}{c-i d}\\ &=-\frac{2 d (a+i a \tan (e+f x))^{3/2}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 i a \sqrt{a+i a \tan (e+f x)}}{(c-i d)^2 f \sqrt{c+d \tan (e+f x)}}+\frac{(2 a) \int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx}{(c-i d)^2}\\ &=-\frac{2 d (a+i a \tan (e+f x))^{3/2}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 i a \sqrt{a+i a \tan (e+f x)}}{(c-i d)^2 f \sqrt{c+d \tan (e+f x)}}-\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 )}{(c-i d)^2 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 )}{(c-i d)^{5/2} f}-\frac{2 d (a+i a \tan (e+f x))^{3/2}}{3 \left (c^2+d^2\right ) f (c+d \tan (e+f x))^{3/2}}+\frac{2 i a \sqrt{a+i a \tan (e+f x)}}{(c-i d)^2 f \sqrt{c+d \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 5.78654, size = 278, normalized size = 1.55 \[ \frac{(a+i a \tan (e+f x))^{3/2} \left (\frac{2 (\tan (e+f x)+i) \left (3 c^2+2 d (c+2 i d) \tan (e+f x)+4 i c d+d^2\right )}{3 (c-i d)^2 (c+i d) \sqrt{\sec (e+f x)} (c+d \tan (e+f x))^{3/2}}-\frac{2 i \sqrt{2} \log \left (2 e^{-i e} \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{(c-i d)^{5/2} \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{3/2} \left (1+e^{2 i (e+f x)}\right )^{3/2}}\right )}{f \sec ^{\frac{3}{2}}(e+f x)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(3/2)/(c + d*Tan[e + f*x])^(5/2),x]

[Out]

((a + I*a*Tan[e + f*x])^(3/2)*(((-2*I)*Sqrt[2]*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)))]))/E^(I*e)])/((c - I*d)^(5/2)*(E^(I
*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(3/2)*(1 + E^((2*I)*(e + f*x)))^(3/2)) + (2*(I + Tan[e + f*x])*(3*c^2 +
(4*I)*c*d + d^2 + 2*(c + (2*I)*d)*d*Tan[e + f*x]))/(3*(c - I*d)^2*(c + I*d)*Sqrt[Sec[e + f*x]]*(c + d*Tan[e +
f*x])^(3/2))))/(f*Sec[e + f*x]^(3/2))

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Maple [B]  time = 0.062, size = 1791, 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))^(3/2)/(c+d*tan(f*x+e))^(5/2),x)

[Out]

1/6/f*2^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*(3*I*2^(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))*tan(f*x+e)^2*a*d^2*(-a*(I*d-c))^(1/2)-20*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)*(I*a*d)^(1/2)*c^3*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^2*(I*a*d)^(1/2)+3*I*2^(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))*a*c^2*(-a*(I*d-c))^(1/2)+8*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)*(I*a*d)^(1/2)*tan(f*x+e)*d^4-4*2^(1/2)*tan(f*x+e)*c^3*d*(a*(c+d*tan(f*x+e))*(1+I
*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*(-a*(I*d-c))^(1/2)+20*2^(1/2)*tan(f*x+e)*c*d^3*(a*(c+d*tan(f*x+e))*(1+I*tan(
f*x+e)))^(1/2)*(I*a*d)^(1/2)*(-a*(I*d-c))^(1/2)+4*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)*(I*a*d)^(1/2)*c*d^3-16*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)*(
I*a*d)^(1/2)*tan(f*x+e)*c^2*d^2-3*2^(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))*tan(f*x+e)^2*a*d^2*(-a*(I*d-c))^(1/2)-6*2^(1/2)*(-a*(I*d-c))^(1/2
)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*c^4+20*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+
e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*c^2*d^2+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)*(I*a*d)^(1/2)*d^4-6*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*d*(I*a*d)^(1/2)+6*I*2^(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))
*tan(f*x+e)*a*c*d*(-a*(I*d-c))^(1/2)-6*2^(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))*tan(f*x+e)*a*c*d*(-a*(I*d-c))^(1/2)+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)^2*a*d^2*(I*a*d)^(1/2)-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))*tan(f*x+e)^2*a*d^2*(I*a*d)^(1/2)-3*
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
))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c^2+6*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*(I*a*d)^(1/2)+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))*a*c^2*(I*a*d)^(1/2))*a/(c+d*tan(f*x+e))^(3/2)/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)
/(c^2+d^2)^2/(I*c-d)/(I*a*d)^(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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.56371, size = 2580, normalized size = 14.41 \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))^(5/2),x, algorithm="fricas")

[Out]

(sqrt(2)*(12*a*c^2 + 24*I*a*c*d - 12*a*d^2 + (12*a*c^2 + 8*I*a*c*d + 20*a*d^2)*e^(4*I*f*x + 4*I*e) + (24*a*c^2
+ 32*I*a*c*d + 8*a*d^2)*e^(2*I*f*x + 2*I*e))*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) + ((-3*I*c^5 - 9*c^4*d + 6*I*c^3*d^2 - 6*c^2*d^3 +
9*I*c*d^4 + 3*d^5)*f*e^(4*I*f*x + 4*I*e) + (-6*I*c^5 - 6*c^4*d - 12*I*c^3*d^2 - 12*c^2*d^3 - 6*I*c*d^4 - 6*d^
5)*f*e^(2*I*f*x + 2*I*e) + (-3*I*c^5 + 3*c^4*d - 6*I*c^3*d^2 + 6*c^2*d^3 - 3*I*c*d^4 + 3*d^5)*f)*sqrt(-8*I*a^3
/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log(1/2*((I*c^3 + 3*c^2*d - 3*I*c*d^2
- d^3)*f*sqrt(-8*I*a^3/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*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) + ((3*I*c^5 + 9*c^4*d - 6*I
*c^3*d^2 + 6*c^2*d^3 - 9*I*c*d^4 - 3*d^5)*f*e^(4*I*f*x + 4*I*e) + (6*I*c^5 + 6*c^4*d + 12*I*c^3*d^2 + 12*c^2*d
^3 + 6*I*c*d^4 + 6*d^5)*f*e^(2*I*f*x + 2*I*e) + (3*I*c^5 - 3*c^4*d + 6*I*c^3*d^2 - 6*c^2*d^3 + 3*I*c*d^4 - 3*d
^5)*f)*sqrt(-8*I*a^3/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log(1/2*((-I*c^3 -
3*c^2*d + 3*I*c*d^2 + d^3)*f*sqrt(-8*I*a^3/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*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))/((-6*
I*c^5 - 18*c^4*d + 12*I*c^3*d^2 - 12*c^2*d^3 + 18*I*c*d^4 + 6*d^5)*f*e^(4*I*f*x + 4*I*e) + (-12*I*c^5 - 12*c^4
*d - 24*I*c^3*d^2 - 24*c^2*d^3 - 12*I*c*d^4 - 12*d^5)*f*e^(2*I*f*x + 2*I*e) + (-6*I*c^5 + 6*c^4*d - 12*I*c^3*d
^2 + 12*c^2*d^3 - 6*I*c*d^4 + 6*d^5)*f)

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Sympy [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))**(5/2),x)

[Out]

Timed out

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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))^(5/2),x, algorithm="giac")

[Out]

Timed out