3.12.46 \(\int \frac {(c+d \tan (e+f x))^{3/2}}{\sqrt {a+i a \tan (e+f x)}} \, dx\) [1146]

Optimal. Leaf size=195 \[ \frac {2 (-1)^{3/4} d^{3/2} \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {a} f}-\frac {i (c-i d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} f}+\frac {(i c-d) \sqrt {c+d \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.46, antiderivative size = 195, 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 = {3639, 3682, 3625, 214, 3680, 65, 223, 212} \begin {gather*} \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} \end {gather*}

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 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 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(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 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 \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 ) \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 {d^2 \text {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 ) \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 )}{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 ) \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 )}{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*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(518\) vs. \(2(195)=390\).
time = 7.88, size = 518, normalized size = 2.66 \begin {gather*} \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 (-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 )-(1-i) d^{3/2} \left (\log \left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) 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^{5/2} \left (i+e^{i (e+f x)}\right )}\right )-\log \left (-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) 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 )}{d^{5/2} \left (-i+e^{i (e+f x)}\right )}\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)}} \end {gather*}

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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1168 vs. \(2 (152 ) = 304\).
time = 0.52, size = 1169, normalized size = 5.99

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1169\)
default	\(\text {Expression too large to display}\)	\(1169\)

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,method=_RETURNVERBOSE)

[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*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))*c*tan(f*x+e)^2-2*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*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))*d*tan(f*x+e)+(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*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))*d*tan(f*x+e)^2-4*c*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1
/2)*tan(f*x+e)-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*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))*c+8*I*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^2*tan(f*x
+e)+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*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))*c*tan(f*x+e)-4*ln(1/2*(2*I*a*d*ta
n(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^2*tan(f*x+e
)^2-4*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)*d*tan(f*x+e)-(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*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))*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*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^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(-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((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(1/2),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. 928 vs. \(2 (151) = 302\).
time = 0.92, size = 928, normalized size = 4.76 \begin {gather*} -\frac {{\left (\sqrt {2} a f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (\frac {\sqrt {2} a f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a 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}}}{i \, c + d}\right ) - \sqrt {2} a f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (-\frac {\sqrt {2} a f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a 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}}}{i \, c + d}\right ) + a f \sqrt {-\frac {4 i \, d^{3}}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (\frac {2 \, {\left (4 \, \sqrt {2} {\left (d^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + d^{3} e^{\left (i \, f x + i \, e\right )}\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}} - {\left ({\left (i \, a c d + 3 \, a d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (i \, a c d - a d^{2}\right )} f\right )} \sqrt {-\frac {4 i \, d^{3}}{a f^{2}}}\right )}}{i \, c^{3} + c^{2} d + i \, c d^{2} + d^{3} + {\left (i \, c^{3} + c^{2} d + i \, c d^{2} + d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}\right ) - a f \sqrt {-\frac {4 i \, d^{3}}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (\frac {2 \, {\left (4 \, \sqrt {2} {\left (d^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + d^{3} e^{\left (i \, f x + i \, e\right )}\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}} - {\left ({\left (-i \, a c d - 3 \, a d^{2}\right )} f e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, a c d + a d^{2}\right )} f\right )} \sqrt {-\frac {4 i \, d^{3}}{a f^{2}}}\right )}}{i \, c^{3} + c^{2} d + i \, c d^{2} + d^{3} + {\left (i \, c^{3} + c^{2} d + i \, c d^{2} + d^{3}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}\right ) + 2 \, \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 )}}{4 \, a f} \end {gather*}

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*(sqrt(2)*a*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a*f^2))*e^(I*f*x + I*e)*log((sqrt(2)*a*f*sqrt(-(c
^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a*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)))/
(I*c + d)) - sqrt(2)*a*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a*f^2))*e^(I*f*x + I*e)*log(-(sqrt(2)*a*f*
sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a*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)))/(I*c + d)) + a*f*sqrt(-4*I*d^3/(a*f^2))*e^(I*f*x + I*e)*log(2*(4*sqrt(2)*(d^3*e^(3*I*f*x + 3*I*e) + d^
3*e^(I*f*x + 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)) - ((I*a*c*d + 3*a*d^2)*f*e^(2*I*f*x + 2*I*e) + (I*a*c*d - 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^(I*f*x + I*e)*log(2*(4*sqrt(2)*(d^3*e^(3*I*f*x + 3*I*e) + d^3*e^(I*f*x + 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)) - ((-I*a*c*d - 3*a*d^2)*f*
e^(2*I*f*x + 2*I*e) + (-I*a*c*d + 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))) + 2*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)/(a*f)

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

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(I*a*(tan(e + f*x) - I)), 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((c+d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(1/2),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.Non regu
lar value [

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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