3.12.56 \(\int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx\) [1156]

Optimal. Leaf size=151 \[ -\frac {2 (-1)^{3/4} a^{3/2} \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d} f}-\frac {2 i \sqrt {2} a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {c-i d} f} \]

[Out]

-2*I*a^(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)/f/
(c-I*d)^(1/2)-2*(-1)^(3/4)*a^(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/d^(1/2)

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Rubi [A]
time = 0.28, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3636, 3625, 214, 3680, 65, 223, 212} \begin {gather*} -\frac {2 (-1)^{3/4} a^{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 {d} f}-\frac {2 i \sqrt {2} a^{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 \sqrt {c-i d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-1)^(3/4)*a^(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[d]*f) - ((2*I)*Sqrt[2]*a^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*S
qrt[a + I*a*Tan[e + f*x]])])/(Sqrt[c - I*d]*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 3636

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dis
t[2*a, Int[Sqrt[a + b*Tan[e + f*x]]/Sqrt[c + d*Tan[e + f*x]], x], x] + Dist[b/a, Int[(b + a*Tan[e + f*x])*(Sqr
t[a + b*Tan[e + f*x]]/Sqrt[c + d*Tan[e + f*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]

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]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx &=i \int \frac {\sqrt {a+i a \tan (e+f x)} (i a+a \tan (e+f x))}{\sqrt {c+d \tan (e+f x)}} \, dx+(2 a) \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx\\ &=-\frac {a^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 {\left (4 i a^3\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 {2 i \sqrt {2} a^{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 {c-i d} f}+\frac {(2 i a) \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 {2 i \sqrt {2} a^{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 {c-i d} f}+\frac {(2 i a) \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 {2 (-1)^{3/4} a^{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 {d} f}-\frac {2 i \sqrt {2} a^{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 {c-i d} 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. \(505\) vs. \(2(151)=302\).
time = 5.73, size = 505, normalized size = 3.34 \begin {gather*} \frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \cos (e+f x) \left (\sqrt {c-i d} \log \left (\frac {(2-2 i) e^{\frac {3 i e}{2}} \left (-i d+d e^{i (e+f x)}+i 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 )}{\sqrt {d} \left (i+e^{i (e+f x)}\right )}\right )-\sqrt {c-i d} \log \left (\frac {(2+2 i) e^{\frac {3 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 )}{\sqrt {d} \left (-i+e^{i (e+f x)}\right )}\right )+(2-2 i) \sqrt {d} \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 ) (\cos (f x)-i \sin (f x)) \sqrt {1+\cos (2 (e+f x))+i \sin (2 (e+f x))} (\cos (2 e+f x)-i \sin (2 e+f x)) (a+i a \tan (e+f x))^{3/2}}{\sqrt {c-i d} \sqrt {d} f} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1/2 - I/2)*Cos[e + f*x]*(Sqrt[c - I*d]*Log[((2 - 2*I)*E^(((3*I)/2)*e)*((-I)*d + d*E^(I*(e + f*x)) + I*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)))]))/(Sqrt[d]*(I + E^(I*(e + f*x))))] - Sqrt[c - I*d]*Log[((2 + 2*I)*E^(((3*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)))]))/(Sqrt[d]*(-I + E^(I*(e + f*x))))] + (2 - 2*I)*Sqrt[d
]*Log[2*(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]])])*(Cos[f*x] - I*Sin[f*x])*Sqrt[1 + Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]]*(Co
s[2*e + f*x] - I*Sin[2*e + f*x])*(a + I*a*Tan[e + f*x])^(3/2))/(Sqrt[c - I*d]*Sqrt[d]*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. 982 vs. \(2 (114 ) = 228\).
time = 0.65, size = 983, normalized size = 6.51 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*a^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))*c+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))*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))*c-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
))*d+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))*c^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*t
an(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*d^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))*c+(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))*d+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))*c+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))*d)/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(c^2+d^2)/(I*a*d)^(1/2)*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: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume((d^2-2*c*d-c^2)>0)', see `assu
me?` for mor

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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. 557 vs. \(2 (113) = 226\).
time = 1.04, size = 557, normalized size = 3.69 \begin {gather*} \frac {1}{2} \, \sqrt {-\frac {8 i \, a^{3}}{{\left (i \, c + d\right )} f^{2}}} \log \left (\frac {{\left ({\left (i \, c + d\right )} f \sqrt {-\frac {8 i \, a^{3}}{{\left (i \, c + d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + 2 \, \sqrt {2} {\left (a e^{\left (2 i \, f x + 2 i \, e\right )} + a\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 )}}{2 \, a}\right ) - \frac {1}{2} \, \sqrt {-\frac {8 i \, a^{3}}{{\left (i \, c + d\right )} f^{2}}} \log \left (\frac {{\left ({\left (-i \, c - d\right )} f \sqrt {-\frac {8 i \, a^{3}}{{\left (i \, c + d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + 2 \, \sqrt {2} {\left (a e^{\left (2 i \, f x + 2 i \, e\right )} + a\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 )}}{2 \, a}\right ) - \frac {1}{2} \, \sqrt {-\frac {4 i \, a^{3}}{d f^{2}}} \log \left (\frac {{\left (d f \sqrt {-\frac {4 i \, a^{3}}{d f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} {\left (a e^{\left (2 i \, f x + 2 i \, e\right )} + a\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 )}}{a}\right ) + \frac {1}{2} \, \sqrt {-\frac {4 i \, a^{3}}{d f^{2}}} \log \left (-\frac {{\left (d f \sqrt {-\frac {4 i \, a^{3}}{d f^{2}}} e^{\left (i \, f x + i \, e\right )} - \sqrt {2} {\left (a e^{\left (2 i \, f x + 2 i \, e\right )} + a\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 )}}{a}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(-8*I*a^3/((I*c + d)*f^2))*log(1/2*((I*c + d)*f*sqrt(-8*I*a^3/((I*c + d)*f^2))*e^(I*f*x + I*e) + 2*sqr
t(2)*(a*e^(2*I*f*x + 2*I*e) + a)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqr
t(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/a) - 1/2*sqrt(-8*I*a^3/((I*c + d)*f^2))*log(1/2*((-I*c - d)*f
*sqrt(-8*I*a^3/((I*c + d)*f^2))*e^(I*f*x + I*e) + 2*sqrt(2)*(a*e^(2*I*f*x + 2*I*e) + a)*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) - 1
/2*sqrt(-4*I*a^3/(d*f^2))*log((d*f*sqrt(-4*I*a^3/(d*f^2))*e^(I*f*x + I*e) + sqrt(2)*(a*e^(2*I*f*x + 2*I*e) + a
)*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) + 1/2*sqrt(-4*I*a^3/(d*f^2))*log(-(d*f*sqrt(-4*I*a^3/(d*f^2))*e^(I*f*x + I*e) - sqrt(2)*(
a*e^(2*I*f*x + 2*I*e) + a)*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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((I*a*(tan(e + f*x) - I))**(3/2)/sqrt(c + d*tan(e + f*x)), 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))^(3/2)/(c+d*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:index.cc index_m operator + Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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