3.1146 \(\int \frac{(c+d \tan (e+f x))^{3/2}}{\sqrt{a+i a \tan (e+f x)}} \, dx\)
Optimal. Leaf size=195 \[ \frac{2 (-1)^{3/4} d^{3/2} \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{(-d+i c) \sqrt{c+d \tan (e+f x)}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{i (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{\sqrt{2} \sqrt{a} f} \]
[Out]
(2*(-1)^(3/4)*d^(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[a]*f) - (I*(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]*Sqrt[a]*f) + ((I*c - d)*Sqrt[c + d*Tan[e + f*x]])/(f*Sqrt[a + I*a*Tan[e + f*x
]])
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Rubi [A] time = 0.69962, antiderivative size = 195, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 8, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used =
{3558, 3601, 3544, 208, 3599, 63, 217, 206} \[ \frac{2 (-1)^{3/4} d^{3/2} \tanh ^{-1}\left (\frac{(-1)^{3/4} \sqrt{d} \sqrt{a+i a \tan (e+f x)}}{\sqrt{a} \sqrt{c+d \tan (e+f x)}}\right )}{\sqrt{a} f}+\frac{(-d+i c) \sqrt{c+d \tan (e+f x)}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{i (c-i d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{a} \sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d} \sqrt{a+i a \tan (e+f x)}}\right )}{\sqrt{2} \sqrt{a} f} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*Tan[e + f*x])^(3/2)/Sqrt[a + I*a*Tan[e + f*x]],x]
[Out]
(2*(-1)^(3/4)*d^(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[a]*f) - (I*(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]*Sqrt[a]*f) + ((I*c - d)*Sqrt[c + d*Tan[e + f*x]])/(f*Sqrt[a + I*a*Tan[e + f*x
]])
Rule 3558
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(2*a*f*m), x] + Dist[1/(2*a^2*m), Int[(a
+ b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1))
- d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Tan[e + f*x], 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] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m,
2*n])
Rule 3601
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), 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] && NeQ[A*b + a*B, 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{(c+d \tan (e+f x))^{3/2}}{\sqrt{a+i a \tan (e+f x)}} \, dx &=\frac{(i c-d) \sqrt{c+d \tan (e+f x)}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\int \frac{\sqrt{a+i a \tan (e+f x)} \left (-\frac{1}{2} a \left (c^2-2 i c d+d^2\right )+i a d^2 \tan (e+f x)\right )}{\sqrt{c+d \tan (e+f x)}} \, dx}{a^2}\\ &=\frac{(i c-d) \sqrt{c+d \tan (e+f x)}}{f \sqrt{a+i a \tan (e+f x)}}+\frac{(c-i d)^2 \int \frac{\sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 a}+\frac{d^2 \int \frac{(a-i a \tan (e+f x)) \sqrt{a+i a \tan (e+f x)}}{\sqrt{c+d \tan (e+f x)}} \, dx}{a^2}\\ &=\frac{(i c-d) \sqrt{c+d \tan (e+f x)}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\left (i a (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 )}{f}+\frac{d^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{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 )}{\sqrt{2} \sqrt{a} f}+\frac{(i c-d) \sqrt{c+d \tan (e+f x)}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\left (2 i d^2\right ) \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 )}{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 )}{\sqrt{2} \sqrt{a} f}+\frac{(i c-d) \sqrt{c+d \tan (e+f x)}}{f \sqrt{a+i a \tan (e+f x)}}-\frac{\left (2 i d^2\right ) \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 )}{a f}\\ &=\frac{2 (-1)^{3/4} d^{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{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 )}{\sqrt{2} \sqrt{a} f}+\frac{(i c-d) \sqrt{c+d \tan (e+f x)}}{f \sqrt{a+i a \tan (e+f x)}}\\ \end{align*}
Mathematica [B] time = 6.86787, size = 518, normalized size = 2.66 \[ \frac{\sqrt{\sec (e+f x)} \left (\sqrt{2} \sqrt{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt{1+e^{2 i (e+f x)}} \left (-(1-i) d^{3/2} \left (\log \left (\frac{\left (\frac{1}{2}+\frac{i}{2}\right ) e^{\frac{i e}{2}} \left ((1-i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}-c \left (e^{i (e+f x)}+i\right )+i d e^{i (e+f x)}+d\right )}{d^{5/2} \left (e^{i (e+f x)}+i\right )}\right )-\log \left (-\frac{\left (\frac{1}{2}-\frac{i}{2}\right ) e^{\frac{i e}{2}} \left ((1+i) \sqrt{d} \sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c e^{i (e+f x)}+c+d e^{i (e+f x)}+i d\right )}{d^{5/2} \left (e^{i (e+f x)}-i\right )}\right )\right )-i (c-i d)^{3/2} \log \left (2 \left (\sqrt{c-i d} e^{i (e+f x)}+\sqrt{1+e^{2 i (e+f x)}} \sqrt{c-\frac{i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )\right )+\frac{2 i (c+i d) \sqrt{c+d \tan (e+f x)}}{\sqrt{\sec (e+f x)}}\right )}{2 f \sqrt{a+i a \tan (e+f x)}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(c + d*Tan[e + f*x])^(3/2)/Sqrt[a + I*a*Tan[e + f*x]],x]
[Out]
(Sqrt[Sec[e + f*x]]*(Sqrt[2]*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*Sqrt[1 + E^((2*I)*(e + f*x))]*((-
I)*(c - I*d)^(3/2)*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)))])] - (1 - I)*d^(3/2)*(Log[((1/2 + I/2)*E^((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)))]))/(d^(5/2)*(I + E^(I*(e + f*x))))] - Log[((-1/2 + I/2)*E^((I/2)*e
)*(c + I*d + I*c*E^(I*(e + f*x)) + d*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)))]))/(d^(5/2)*(-I + E^(I*(e + f*x))))])) + ((2*I)*(c
+ I*d)*Sqrt[c + d*Tan[e + f*x]])/Sqrt[Sec[e + f*x]]))/(2*f*Sqrt[a + I*a*Tan[e + f*x]])
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Maple [B] time = 0.089, size = 1169, normalized size = 6. \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))^(1/2),x)
[Out]
-1/4/f/a*(I*(I*a*d)^(1/2)*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*I*(I*a*d)^
(1/2)*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+(I*a*d)^(1/2)*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-4*tan(f*x+e)*c*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)
*(I*a*d)^(1/2)-I*(I*a*d)^(1/2)*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+8*I*ln(1/2*(2*I*a*ta
n(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*
d^2+2*(I*a*d)^(1/2)*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-4*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^2-4*I*(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-(I*a*d)^(1/2)*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+4*I*c*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+4*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*d^2-4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*d)*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+
e))^(1/2)/(I*a*d)^(1/2)/(-tan(f*x+e)+I)^2/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d \tan \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{\sqrt{i \, a \tan \left (f x + e\right ) + a}}\,{d x} \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))^(1/2),x, algorithm="maxima")
[Out]
integrate((d*tan(f*x + e) + c)^(3/2)/sqrt(I*a*tan(f*x + e) + a), x)
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Fricas [B] time = 2.64247, size = 2390, normalized size = 12.26 \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))^(1/2),x, algorithm="fricas")
[Out]
-1/4*(a*f*sqrt(-(2*c^3 - 6*I*c^2*d - 6*c*d^2 + 2*I*d^3)/(a*f^2))*e^(2*I*f*x + 2*I*e)*log((a*f*sqrt(-(2*c^3 - 6
*I*c^2*d - 6*c*d^2 + 2*I*d^3)/(a*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)) - a*f*sqrt(-(2*c^3 - 6*I*c^2*d - 6*c*d^2 + 2*I*d^3)/(a*f^2))*e^(2*
I*f*x + 2*I*e)*log(-(a*f*sqrt(-(2*c^3 - 6*I*c^2*d - 6*c*d^2 + 2*I*d^3)/(a*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)) - a*f*sqrt(-4*I*d^3/(a*f^
2))*e^(2*I*f*x + 2*I*e)*log((8*sqrt(2)*(d^3*e^(2*I*f*x + 2*I*e) + d^3)*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) + ((2*I*a*c*d + 6*a*d^2)*
f*e^(2*I*f*x + 2*I*e) + (2*I*a*c*d - 2*a*d^2)*f)*sqrt(-4*I*d^3/(a*f^2)))/(I*c^3 + c^2*d + I*c*d^2 + d^3 + (I*c
^3 + c^2*d + I*c*d^2 + d^3)*e^(2*I*f*x + 2*I*e))) + a*f*sqrt(-4*I*d^3/(a*f^2))*e^(2*I*f*x + 2*I*e)*log((8*sqrt
(2)*(d^3*e^(2*I*f*x + 2*I*e) + d^3)*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) + ((-2*I*a*c*d - 6*a*d^2)*f*e^(2*I*f*x + 2*I*e) + (-2*I*a*c*
d + 2*a*d^2)*f)*sqrt(-4*I*d^3/(a*f^2)))/(I*c^3 + c^2*d + I*c*d^2 + d^3 + (I*c^3 + c^2*d + I*c*d^2 + d^3)*e^(2*
I*f*x + 2*I*e))) - sqrt(2)*((2*I*c - 2*d)*e^(2*I*f*x + 2*I*e) + 2*I*c - 2*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^(-2*I*f*x - 2*I*
e)/(a*f)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c + d \tan{\left (e + f x \right )}\right )^{\frac{3}{2}}}{\sqrt{a \left (i \tan{\left (e + f x \right )} + 1\right )}}\, dx \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))**(1/2),x)
[Out]
Integral((c + d*tan(e + f*x))**(3/2)/sqrt(a*(I*tan(e + f*x) + 1)), x)
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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))^(1/2),x, algorithm="giac")
[Out]
Exception raised: TypeError