3.1167 \(\int \frac{(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=181 \[ \frac{4 i a^2 \sqrt{a+i a \tan (e+f x)}}{f (c-i d)^2 \sqrt{c+d \tan (e+f x)}}-\frac{4 i \sqrt{2} a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f (c-i d)^{5/2}}-\frac{2 a (a+i a \tan (e+f x))^{3/2}}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}} \]
[Out]
((-4*I)*Sqrt[2]*a^(5/2)*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*a*(a + I*a*Tan[e + f*x])^(3/2))/(3*(I*c + d)*f*(c + d*Tan[e + f*x])^(3/2))
+ ((4*I)*a^2*Sqrt[a + I*a*Tan[e + f*x]])/((c - I*d)^2*f*Sqrt[c + d*Tan[e + f*x]])
________________________________________________________________________________________
Rubi [A] time = 0.346542, antiderivative size = 181, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used =
{3545, 3544, 208} \[ \frac{4 i a^2 \sqrt{a+i a \tan (e+f x)}}{f (c-i d)^2 \sqrt{c+d \tan (e+f x)}}-\frac{4 i \sqrt{2} a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{f (c-i d)^{5/2}}-\frac{2 a (a+i a \tan (e+f x))^{3/2}}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(a + I*a*Tan[e + f*x])^(5/2)/(c + d*Tan[e + f*x])^(5/2),x]
[Out]
((-4*I)*Sqrt[2]*a^(5/2)*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*a*(a + I*a*Tan[e + f*x])^(3/2))/(3*(I*c + d)*f*(c + d*Tan[e + f*x])^(3/2))
+ ((4*I)*a^2*Sqrt[a + I*a*Tan[e + f*x]])/((c - I*d)^2*f*Sqrt[c + d*Tan[e + f*x]])
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))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx &=-\frac{2 a (a+i a \tan (e+f x))^{3/2}}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac{(2 a) \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 a (a+i a \tan (e+f x))^{3/2}}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac{4 i a^2 \sqrt{a+i a \tan (e+f x)}}{(c-i d)^2 f \sqrt{c+d \tan (e+f x)}}+\frac{\left (4 a^2\right ) \int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx}{(c-i d)^2}\\ &=-\frac{2 a (a+i a \tan (e+f x))^{3/2}}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac{4 i a^2 \sqrt{a+i a \tan (e+f x)}}{(c-i d)^2 f \sqrt{c+d \tan (e+f x)}}-\frac{\left (8 i a^4\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{4 i \sqrt{2} a^{5/2} \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 a (a+i a \tan (e+f x))^{3/2}}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac{4 i a^2 \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 = 6.3842, size = 283, normalized size = 1.56 \[ \frac{(a+i a \tan (e+f x))^{5/2} \left (-\frac{4 i \sqrt{2} e^{-3 i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \log \left (2 e^{-i e} \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{(c-i d)^{5/2}}-\frac{2 \sqrt{\sec (e+f x)} (\cos (2 (e+f x))-i \sin (2 (e+f x))) ((c-7 i d) \tan (e+f x)-7 i c-d)}{3 (c-i d)^2 (c+d \tan (e+f x))^{3/2}}\right )}{f \sec ^{\frac{5}{2}}(e+f x)} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + I*a*Tan[e + f*x])^(5/2)/(c + d*Tan[e + f*x])^(5/2),x]
[Out]
((a + I*a*Tan[e + f*x])^(5/2)*(((-4*I)*Sqrt[2]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*Sqrt[1 + E^((2*
I)*(e + f*x))]*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^((3*I)*(e + f*x))) - (2*Sqrt[Sec[e
+ f*x]]*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)])*((-7*I)*c - d + (c - (7*I)*d)*Tan[e + f*x]))/(3*(c - I*d)^2*(c
+ d*Tan[e + f*x])^(3/2))))/(f*Sec[e + f*x]^(5/2))
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Maple [B] time = 0.101, size = 1852, normalized size = 10.2 \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))^(5/2)/(c+d*tan(f*x+e))^(5/2),x)
[Out]
1/3/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)-18*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+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^3-3*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)+7*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*t
an(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)-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)
-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*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)-7*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+18*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^(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-I*2^(1/2)*tan(f*
x+e)*c^4*(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)+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))*ta
n(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^2/(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)
________________________________________________________________________________________
Maxima [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))^(5/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [B] time = 1.61908, size = 2223, normalized size = 12.28 \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))^(5/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
-(8*sqrt(2)*(-3*I*a^2*c + 3*a^2*d + 4*(-I*a^2*c - a^2*d)*e^(4*I*f*x + 4*I*e) + (-7*I*a^2*c - a^2*d)*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*c^4 - 12*I*c^3*d - 18*c^2*d^2 + 12*I*c*d^3 + 3*d^4)*f*e^(4*I*f*x + 4*I*e) + (6*
c^4 - 12*I*c^3*d - 12*I*c*d^3 - 6*d^4)*f*e^(2*I*f*x + 2*I*e) + 3*(c^4 + 2*c^2*d^2 + d^4)*f)*sqrt(-32*I*a^5/((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(1/4*((I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^
3)*sqrt(-32*I*a^5/((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))*f*e^(2*I*f*x + 2*I*e)
+ 4*sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^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^(I*f*x + I*e))*e^(-2*I*f*x - 2*I*e)/a^2) + ((3*c^4 - 12*I*c^3*d -
18*c^2*d^2 + 12*I*c*d^3 + 3*d^4)*f*e^(4*I*f*x + 4*I*e) + (6*c^4 - 12*I*c^3*d - 12*I*c*d^3 - 6*d^4)*f*e^(2*I*f
*x + 2*I*e) + 3*(c^4 + 2*c^2*d^2 + d^4)*f)*sqrt(-32*I*a^5/((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(1/4*((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*sqrt(-32*I*a^5/((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))*f*e^(2*I*f*x + 2*I*e) + 4*sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^2)*sqr
t(((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^2))/((6*c^4 - 24*I*c^3*d - 36*c^2*d^2 + 24*I*c*d^3 + 6*d^4)*f*e^(4*I*f*x +
4*I*e) + (12*c^4 - 24*I*c^3*d - 24*I*c*d^3 - 12*d^4)*f*e^(2*I*f*x + 2*I*e) + 6*(c^4 + 2*c^2*d^2 + d^4)*f)
________________________________________________________________________________________
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))**(5/2)/(c+d*tan(f*x+e))**(5/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="giac")
[Out]
Exception raised: TypeError