3.12.45 \(\int \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2} \, dx\) [1145]

Optimal. Leaf size=196 \[ -\frac {\sqrt [4]{-1} \sqrt {a} (3 c-i d) \sqrt {d} \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{f}-\frac {i \sqrt {2} \sqrt {a} (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}+\frac {d \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f} \]

[Out]

-I*(c-I*d)^(3/2)*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2)/(a+I*a*tan(f*x+e))^(1/2))*2^(1/2
)*a^(1/2)/f-(-1)^(1/4)*(3*c-I*d)*arctanh((-1)^(3/4)*d^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c+d*tan(f*x+e))^
(1/2))*a^(1/2)*d^(1/2)/f+d*(a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.47, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3641, 3682, 3625, 214, 3680, 65, 223, 212} \begin {gather*} \frac {d \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f}-\frac {i \sqrt {2} \sqrt {a} (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 )}{f}-\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {d} (3 c-i d) \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 )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 212

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

Rule 214

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

Rule 223

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 3625

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 3641

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

Rule 3680

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 3682

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]

Rubi steps

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

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(507\) vs. \(2(196)=392\).
time = 7.85, size = 507, normalized size = 2.59 \begin {gather*} \frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {a+i a \tan (e+f x)} \left (-\frac {\cos (e+f x) \left ((3 c-i d) \sqrt {d} \log \left (\frac {(2+2 i) e^{\frac {i e}{2}} \left (d+i d e^{i (e+f x)}-c \left (i+e^{i (e+f x)}\right )+(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)}}}\right )}{d^{3/2} (3 i c+d) \left (i+e^{i (e+f x)}\right )}\right )+i \sqrt {d} (3 i c+d) \log \left (\frac {(2+2 i) e^{\frac {i e}{2}} \left (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-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )}{(3 c-i d) d^{3/2} \left (-i+e^{i (e+f x)}\right )}\right )+(2+2 i) (c-i d)^{3/2} \log \left (2 \left (\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)}\right )\right )\right )}{\sqrt {1+\cos (2 (e+f x))+i \sin (2 (e+f x))}}+(1-i) d \sqrt {c+d \tan (e+f x)}\right )}{f} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1701 vs. \(2 (153 ) = 306\).
time = 0.60, size = 1702, normalized size = 8.68

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1702\)
default	\(\text {Expression too large to display}\)	\(1702\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/4/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*(-2*I*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+
3*a*d*tan(f*x+e)+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
^3-2*I*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+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))*a*c^2*d*tan(f*x+e)-2*I*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+
e)*c-I*a*d+3*a*d*tan(f*x+e)+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*d^3*tan(f*x+e)+2*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+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^3*tan(f*x+e)-2*(I*a*d)^(1/2)*ln((3*
a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+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*d+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c
))^(1/2)*c*d+3*I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+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*d+I*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)
+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^3+2*I*(a*(c+d*tan(f
*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*c*d*tan(f*x+e)-2*I*2^(1/2)*(-a*(I*d-c)
)^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+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*d^2*tan(f*x+e)-2*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+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*d^2+2*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e))
)^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*d^2-2*I*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*
tan(f*x+e)+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*d^2-2
*(I*a*d)^(1/2)*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+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*d^3-2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+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^3*tan(f*x+e)-2*(a*(c+d*ta
n(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*d^2*tan(f*x+e)+2*(I*a*d)^(1/2)*ln((
3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+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*d^2*tan(f*x+e)-3*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+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*d*tan(f*x+e))/(a*(c+d*tan(f
*x+e))*(1+I*tan(f*x+e)))^(1/2)/(I*a*d)^(1/2)/(I*c-d)/(-tan(f*x+e)+I)*2^(1/2)/(-a*(I*d-c))^(1/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 796 vs. \(2 (152) = 304\).
time = 0.89, size = 796, normalized size = 4.06 \begin {gather*} \frac {2 \, \sqrt {2} d \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} - \sqrt {2} f \sqrt {-\frac {a c^{3} - 3 i \, a c^{2} d - 3 \, a c d^{2} + i \, a d^{3}}{f^{2}}} \log \left (\frac {{\left (\sqrt {2} f \sqrt {-\frac {a c^{3} - 3 i \, a c^{2} d - 3 \, a c d^{2} + i \, a d^{3}}{f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} {\left ({\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c + d\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{i \, c + d}\right ) + \sqrt {2} f \sqrt {-\frac {a c^{3} - 3 i \, a c^{2} d - 3 \, a c d^{2} + i \, a d^{3}}{f^{2}}} \log \left (-\frac {{\left (\sqrt {2} f \sqrt {-\frac {a c^{3} - 3 i \, a c^{2} d - 3 \, a c d^{2} + i \, a d^{3}}{f^{2}}} e^{\left (i \, f x + i \, e\right )} - \sqrt {2} {\left ({\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c + d\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{i \, c + d}\right ) + f \sqrt {\frac {9 i \, a c^{2} d + 6 \, a c d^{2} - i \, a d^{3}}{f^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (3 i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, c + d\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + 2 \, f \sqrt {\frac {9 i \, a c^{2} d + 6 \, a c d^{2} - i \, a d^{3}}{f^{2}}} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-i \, f x - i \, e\right )}}{3 i \, c + d}\right ) - f \sqrt {\frac {9 i \, a c^{2} d + 6 \, a c d^{2} - i \, a d^{3}}{f^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left ({\left (3 i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, c + d\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - 2 \, f \sqrt {\frac {9 i \, a c^{2} d + 6 \, a c d^{2} - i \, a d^{3}}{f^{2}}} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-i \, f x - i \, e\right )}}{3 i \, c + d}\right )}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*sqrt(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) - sqrt(2)*f*sqrt(-(a*c^3 - 3*I*a*c^2*d - 3*a*c*d^2 + I*a*d^3)/f^2)*log((sqrt(2)
*f*sqrt(-(a*c^3 - 3*I*a*c^2*d - 3*a*c*d^2 + I*a*d^3)/f^2)*e^(I*f*x + 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)/(I*c + d)) + sqrt(2)*f*sqrt(-(a*c^3 - 3*I*a*c^2*d - 3*a*c*d^2 + I*a*d^3)/f^2)*
log(-(sqrt(2)*f*sqrt(-(a*c^3 - 3*I*a*c^2*d - 3*a*c*d^2 + I*a*d^3)/f^2)*e^(I*f*x + 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)/(I*c + d)) + f*sqrt((9*I*a*c^2*d + 6*a*c*d^2 - I*a*d^3)/f^2)*log(
(sqrt(2)*((3*I*c + d)*e^(2*I*f*x + 2*I*e) + 3*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)) + 2*f*sqrt((9*I*a*c^2*d + 6*a*c*d^2 - I*a*d^3)/f^2)*e^(I*
f*x + I*e))*e^(-I*f*x - I*e)/(3*I*c + d)) - f*sqrt((9*I*a*c^2*d + 6*a*c*d^2 - I*a*d^3)/f^2)*log((sqrt(2)*((3*I
*c + d)*e^(2*I*f*x + 2*I*e) + 3*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)) - 2*f*sqrt((9*I*a*c^2*d + 6*a*c*d^2 - I*a*d^3)/f^2)*e^(I*f*x + I*e))*e^
(-I*f*x - I*e)/(3*I*c + d)))/f

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(I*a*(tan(e + f*x) - I))*(c + d*tan(e + f*x))**(3/2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^(1/2)*(c + d*tan(e + f*x))^(3/2),x)

[Out]

int((a + a*tan(e + f*x)*1i)^(1/2)*(c + d*tan(e + f*x))^(3/2), x)

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