3.1148 \(\int \frac{(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=225 \[ \frac{(d+i c) \sqrt{c+d \tan (e+f x)}}{4 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{i (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{4 \sqrt{2} a^{5/2} f}-\frac{(c+d \tan (e+f x))^{5/2}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}+\frac{i (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}} \]
[Out]
((-I/4)*(c - I*d)^(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]])])/(Sqrt[2]*a^(5/2)*f) + ((I*c + d)*Sqrt[c + d*Tan[e + f*x]])/(4*a^2*f*Sqrt[a + I*a*Tan[e + f*x]]) + ((
I/6)*(c + d*Tan[e + f*x])^(3/2))/(a*f*(a + I*a*Tan[e + f*x])^(3/2)) - (c + d*Tan[e + f*x])^(5/2)/(5*(I*c - d)*
f*(a + I*a*Tan[e + f*x])^(5/2))
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Rubi [A] time = 0.458663, antiderivative size = 225, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used =
{3547, 3546, 3544, 208} \[ \frac{(d+i c) \sqrt{c+d \tan (e+f x)}}{4 a^2 f \sqrt{a+i a \tan (e+f x)}}-\frac{i (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{4 \sqrt{2} a^{5/2} f}-\frac{(c+d \tan (e+f x))^{5/2}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}}+\frac{i (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*Tan[e + f*x])^(3/2)/(a + I*a*Tan[e + f*x])^(5/2),x]
[Out]
((-I/4)*(c - I*d)^(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]])])/(Sqrt[2]*a^(5/2)*f) + ((I*c + d)*Sqrt[c + d*Tan[e + f*x]])/(4*a^2*f*Sqrt[a + I*a*Tan[e + f*x]]) + ((
I/6)*(c + d*Tan[e + f*x])^(3/2))/(a*f*(a + I*a*Tan[e + f*x])^(3/2)) - (c + d*Tan[e + f*x])^(5/2)/(5*(I*c - d)*
f*(a + I*a*Tan[e + f*x])^(5/2))
Rule 3547
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2*f*m*(b*c - a*d)), x] + Dist[1/(2*a), Int[(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Eq
Q[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n + 1, 0] && LtQ[m, -1]
Rule 3546
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(2*b*f*m), x] - Dist[(a*c - b*d)/(2*b^2), 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] && LeQ[m, -2^(-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{(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx &=-\frac{(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac{\int \frac{(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx}{2 a}\\ &=\frac{i (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}-\frac{(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac{(c-i d) \int \frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{a+i a \tan (e+f x)}} \, dx}{4 a^2}\\ &=\frac{(i c+d) \sqrt{c+d \tan (e+f x)}}{4 a^2 f \sqrt{a+i a \tan (e+f x)}}+\frac{i (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}-\frac{(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac{(c-i d)^2 \int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx}{8 a^3}\\ &=\frac{(i c+d) \sqrt{c+d \tan (e+f x)}}{4 a^2 f \sqrt{a+i a \tan (e+f x)}}+\frac{i (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}-\frac{(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}-\frac{\left (i (c-i d)^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 )}{4 a f}\\ &=-\frac{i (c-i d)^{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 )}{4 \sqrt{2} a^{5/2} f}+\frac{(i c+d) \sqrt{c+d \tan (e+f x)}}{4 a^2 f \sqrt{a+i a \tan (e+f x)}}+\frac{i (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}-\frac{(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}\\ \end{align*}
Mathematica [A] time = 5.5531, size = 302, normalized size = 1.34 \[ \frac{\sec ^{\frac{5}{2}}(e+f x) \left (\frac{2 i \sqrt{c+d \tan (e+f x)} \left (4 \left (5 i c^2+3 c d+5 i d^2\right ) \sin (2 (e+f x))+2 \left (13 c^2+7 d^2\right ) \cos (2 (e+f x))+11 c^2+10 i c d+d^2\right )}{15 (c+i d) \sqrt{\sec (e+f x)}}-i \sqrt{2} (c-i d)^{3/2} e^{2 i (e+f x)} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \log \left (2 \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )\right )}{8 f (a+i a \tan (e+f x))^{5/2}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(c + d*Tan[e + f*x])^(3/2)/(a + I*a*Tan[e + f*x])^(5/2),x]
[Out]
(Sec[e + f*x]^(5/2)*((-I)*Sqrt[2]*(c - I*d)^(3/2)*E^((2*I)*(e + f*x))*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))]*Sqr
t[c - (I*d*(-1 + E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))])] + (((2*I)/15)*(11*c^2 + (10*I)*c*d + d^2 +
2*(13*c^2 + 7*d^2)*Cos[2*(e + f*x)] + 4*((5*I)*c^2 + 3*c*d + (5*I)*d^2)*Sin[2*(e + f*x)])*Sqrt[c + d*Tan[e +
f*x]])/((c + I*d)*Sqrt[Sec[e + f*x]])))/(8*f*(a + I*a*Tan[e + f*x])^(5/2))
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Maple [B] time = 0.056, size = 1657, normalized size = 7.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x)
[Out]
-1/240/f/a^3*(-52*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^3*d^2-60*I*2^(1/2)*(-a*(I*d-c))^(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)*d^2+40*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c*d+136*I*(a*
(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*c*d-60*I*tan(f*x+e)^3*c^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f
*x+e)))^(1/2)-15*2^(1/2)*(-a*(I*d-c))^(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))*c^2-15*2^(1/2)*(-a*(I*d-c))^(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))*d^2-212*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^2*d^2-220*tan(f*x+e
)^2*c^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)+60*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^2-15*2^(1
/2)*(-a*(I*d-c))^(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)^4*c^2-15*2^(1/2)*(-a*(I*d-c))^(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)^4*d^2+90*2^(1/2)*(-a*(I*d-c))^(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*
c^2+90*2^(1/2)*(-a*(I*d-c))^(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*d^2-60*I*2^(1/2)*(-a*(I*d-c))^(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)*c^2-40*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)^3*c*
d+56*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*tan(f*x+e)*c*d+60*I*2^(1/2)*(-a*(I*d-c))^(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)^3*d^2+60*I*2^(1/2)*(-a*(I*d-c))^(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)^3*c^
2+308*I*tan(f*x+e)*c^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)+220*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))
^(1/2)*tan(f*x+e)*d^2+148*c^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan
(f*x+e))^(1/2)/(-tan(f*x+e)+I)^4/(I*c-d)/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(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((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
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Fricas [B] time = 1.92717, size = 1592, normalized size = 7.08 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
1/120*(15*sqrt(1/2)*(-I*a^3*c + a^3*d)*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^5*f^2))*e^(6*I*f*x + 6*I
*e)*log((2*sqrt(1/2)*a^3*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^5*f^2))*e^(2*I*f*x + 2*I*e) + sqrt(2)*
((I*c + d)*e^(2*I*f*x + 2*I*e) + I*c + 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))*e^(-I*f*x - I*e)/(I*c + d)) + 15*sqrt(1/2)*(I*a^3*c -
a^3*d)*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^5*f^2))*e^(6*I*f*x + 6*I*e)*log(-(2*sqrt(1/2)*a^3*f*sqr
t(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^5*f^2))*e^(2*I*f*x + 2*I*e) - sqrt(2)*((I*c + d)*e^(2*I*f*x + 2*I*e)
+ I*c + 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))*e^(-I*f*x - I*e)/(I*c + d)) - sqrt(2)*(3*c^2 + 6*I*c*d - 3*d^2 + (23*c^2 - 6*I*c*d
+ 17*d^2)*e^(6*I*f*x + 6*I*e) + 2*(17*c^2 + 2*I*c*d + 9*d^2)*e^(4*I*f*x + 4*I*e) + 2*(7*c^2 + 8*I*c*d - 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))*e^(-6*I*f*x - 6*I*e)/((I*a^3*c - a^3*d)*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((c+d*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e))**(5/2),x)
[Out]
Timed out
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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((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="giac")
[Out]
Exception raised: TypeError