3.1138 \(\int (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)} \, dx\)

Optimal. Leaf size=250 \[ -\frac {\sqrt [4]{-1} a^{3/2} (3 d+i c) \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d} f}-\frac {2 i \sqrt {2} a^{3/2} \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}-\frac {a^2 (c+d \tan (e+f x))^{3/2}}{d f \sqrt {a+i a \tan (e+f x)}}+\frac {a^2 (c+i d) \sqrt {c+d \tan (e+f x)}}{d f \sqrt {a+i a \tan (e+f x)}} \]

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

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Rubi [A]  time = 0.96, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {3556, 3595, 3601, 3544, 208, 3599, 63, 217, 206} \[ -\frac {a^2 (c+d \tan (e+f x))^{3/2}}{d f \sqrt {a+i a \tan (e+f x)}}+\frac {a^2 (c+i d) \sqrt {c+d \tan (e+f x)}}{d f \sqrt {a+i a \tan (e+f x)}}-\frac {\sqrt [4]{-1} a^{3/2} (3 d+i c) \tanh ^{-1}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d} f}-\frac {2 i \sqrt {2} a^{3/2} \sqrt {c-i d} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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])

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 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 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 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3595

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

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 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]

Rubi steps

\begin {align*} \int (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)} \, dx &=-\frac {a^2 (c+d \tan (e+f x))^{3/2}}{d f \sqrt {a+i a \tan (e+f x)}}+\frac {a \int \frac {\left (-\frac {1}{2} a (i c-5 d)-\frac {1}{2} a (c-3 i d) \tan (e+f x)\right ) \sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx}{d}\\ &=\frac {a^2 (c+i d) \sqrt {c+d \tan (e+f x)}}{d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{3/2}}{d 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^2 (3 c-i d) d-\frac {1}{2} a^2 d (i c+3 d) \tan (e+f x)\right )}{\sqrt {c+d \tan (e+f x)}} \, dx}{a d}\\ &=\frac {a^2 (c+i d) \sqrt {c+d \tan (e+f x)}}{d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{3/2}}{d f \sqrt {a+i a \tan (e+f x)}}+(2 a (c-i d)) \int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx-\frac {1}{2} (c-3 i 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\\ &=\frac {a^2 (c+i d) \sqrt {c+d \tan (e+f x)}}{d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{3/2}}{d f \sqrt {a+i a \tan (e+f x)}}-\frac {\left (a^2 (c-3 i d)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c+d x}} \, dx,x,\tan (e+f x)\right )}{2 f}-\frac {\left (4 a^3 (i c+d)\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 {2 i \sqrt {2} a^{3/2} \sqrt {c-i d} \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 {a^2 (c+i d) \sqrt {c+d \tan (e+f x)}}{d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{3/2}}{d f \sqrt {a+i a \tan (e+f x)}}+\frac {(a (i c+3 d)) \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 )}{f}\\ &=-\frac {2 i \sqrt {2} a^{3/2} \sqrt {c-i d} \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 {a^2 (c+i d) \sqrt {c+d \tan (e+f x)}}{d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{3/2}}{d f \sqrt {a+i a \tan (e+f x)}}+\frac {(a (i c+3 d)) \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 )}{f}\\ &=-\frac {\sqrt [4]{-1} a^{3/2} (i c+3 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 )}{\sqrt {d} f}-\frac {2 i \sqrt {2} a^{3/2} \sqrt {c-i d} \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 {a^2 (c+i d) \sqrt {c+d \tan (e+f x)}}{d f \sqrt {a+i a \tan (e+f x)}}-\frac {a^2 (c+d \tan (e+f x))^{3/2}}{d f \sqrt {a+i a \tan (e+f x)}}\\ \end {align*}

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Mathematica [B]  time = 6.09, size = 559, normalized size = 2.24 \[ \frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \cos (e+f x) (\cos (f x)-i \sin (f x)) (a+i a \tan (e+f x))^{3/2} \left ((1-i) \sin (e) \sqrt {c+d \tan (e+f x)}+(1+i) \cos (e) \sqrt {c+d \tan (e+f x)}+\frac {(\cos (e)-i \sin (e)) \cos (e+f x) \left ((-3 d-i c) \log \left (\frac {(2+2 i) e^{\frac {i e}{2}} \left (-(1+i) \sqrt {d} \sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c \left (e^{i (e+f x)}+i\right )+d e^{i (e+f x)}-i d\right )}{\sqrt {d} (3 d+i c) \left (e^{i (e+f x)}+i\right )}\right )+(3 d+i c) \log \left (\frac {(2+2 i) e^{\frac {i e}{2}} \left ((1+i) \sqrt {d} \sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}+i c e^{i (e+f x)}+c+d e^{i (e+f x)}+i d\right )}{\sqrt {d} (3 d+i c) \left (e^{i (e+f x)}-i\right )}\right )-(4+4 i) \sqrt {d} \sqrt {c-i d} \log \left (2 \left (i \sqrt {c-i d} \sin (e+f x)+\sqrt {c-i d} \cos (e+f x)+\sqrt {i \sin (2 (e+f x))+\cos (2 (e+f x))+1} \sqrt {c+d \tan (e+f x)}\right )\right )\right )}{\sqrt {d} \sqrt {i \sin (2 (e+f x))+\cos (2 (e+f x))+1}}\right )}{f} \]

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]*(Cos[f*x] - I*Sin[f*x])*(a + I*a*Tan[e + f*x])^(3/2)*((Cos[e + f*x]*(((-I)*c - 3*d)*
Log[((2 + 2*I)*E^((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*c + 3*
d)*(I + E^(I*(e + f*x))))] + (I*c + 3*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)))]))/(Sqrt[d]*(I*c + 3*d)*(-I + E^(I*(e + f*x))))] - (4 + 4*I)*Sqrt[c - I*d]*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[e] - I*Sin[e]))/(Sqrt[d]*Sqrt[1 + Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]]) + (1 + I)*
Cos[e]*Sqrt[c + d*Tan[e + f*x]] + (1 - I)*Sin[e]*Sqrt[c + d*Tan[e + f*x]]))/f

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fricas [B]  time = 0.48, size = 736, normalized size = 2.94 \[ \frac {2 i \, \sqrt {2} a \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 )} - f \sqrt {\frac {-i \, a^{3} c^{2} - 6 \, a^{3} c d + 9 i \, a^{3} d^{2}}{d f^{2}}} \log \left (\frac {{\left (2 i \, d f \sqrt {\frac {-i \, a^{3} c^{2} - 6 \, a^{3} c d + 9 i \, a^{3} d^{2}}{d f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} {\left (i \, a c + 3 \, a d + {\left (i \, a c + 3 \, a d\right )} e^{\left (2 i \, f x + 2 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}}\right )} e^{\left (-i \, f x - i \, e\right )}}{i \, a c + 3 \, a d}\right ) + f \sqrt {\frac {-i \, a^{3} c^{2} - 6 \, a^{3} c d + 9 i \, a^{3} d^{2}}{d f^{2}}} \log \left (\frac {{\left (-2 i \, d f \sqrt {\frac {-i \, a^{3} c^{2} - 6 \, a^{3} c d + 9 i \, a^{3} d^{2}}{d f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} {\left (i \, a c + 3 \, a d + {\left (i \, a c + 3 \, a d\right )} e^{\left (2 i \, f x + 2 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}}\right )} e^{\left (-i \, f x - i \, e\right )}}{i \, a c + 3 \, a d}\right ) + f \sqrt {-\frac {8 \, a^{3} c - 8 i \, a^{3} d}{f^{2}}} \log \left (\frac {{\left (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}} + i \, f \sqrt {-\frac {8 \, a^{3} c - 8 i \, a^{3} d}{f^{2}}} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a}\right ) - f \sqrt {-\frac {8 \, a^{3} c - 8 i \, a^{3} d}{f^{2}}} \log \left (\frac {{\left (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}} - i \, f \sqrt {-\frac {8 \, a^{3} c - 8 i \, a^{3} d}{f^{2}}} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a}\right )}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*I*sqrt(2)*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) - f*sqrt((-I*a^3*c^2 - 6*a^3*c*d + 9*I*a^3*d^2)/(d*f^2))*log((2*I*d*f*sqrt((-
I*a^3*c^2 - 6*a^3*c*d + 9*I*a^3*d^2)/(d*f^2))*e^(I*f*x + I*e) + sqrt(2)*(I*a*c + 3*a*d + (I*a*c + 3*a*d)*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)/(I*a*c + 3*a*d)) + f*sqrt((-I*a^3*c^2 - 6*a^3*c*d + 9*I*a^3*d^2)/(d*f^2))*log(
(-2*I*d*f*sqrt((-I*a^3*c^2 - 6*a^3*c*d + 9*I*a^3*d^2)/(d*f^2))*e^(I*f*x + I*e) + sqrt(2)*(I*a*c + 3*a*d + (I*a
*c + 3*a*d)*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))*sqr
t(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(I*a*c + 3*a*d)) + f*sqrt(-(8*a^3*c - 8*I*a^3*d)/f^2)*log(1/2
*(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)) + I*f*sqrt(-(8*a^3*c - 8*I*a^3*d)/f^2)*e^(I*f*x + I*e))*e^(-I*f*x - I*e)
/a) - f*sqrt(-(8*a^3*c - 8*I*a^3*d)/f^2)*log(1/2*(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)) - I*f*sqrt(-(8*a^3*c - 8
*I*a^3*d)/f^2)*e^(I*f*x + I*e))*e^(-I*f*x - I*e)/a))/f

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giac [A]  time = 0.61, size = 194, normalized size = 0.78 \[ \frac {\sqrt {2 \, a d^{2} + 2 \, \sqrt {{\left (d \tan \left (f x + e\right ) + c\right )}^{2} - 2 \, {\left (d \tan \left (f x + e\right ) + c\right )} c + c^{2} + d^{2}} a d} {\left (d \tan \left (f x + e\right ) + c\right )} a {\left (\frac {i \, {\left (d \tan \left (f x + e\right ) + c\right )} a d - i \, a c d}{a d^{2} + \sqrt {{\left (d \tan \left (f x + e\right ) + c\right )}^{2} a^{2} d^{2} - 2 \, {\left (d \tan \left (f x + e\right ) + c\right )} a^{2} c d^{2} + a^{2} c^{2} d^{2} + a^{2} d^{4}}} + 1\right )} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{2 \, {\left ({\left (-i \, d \tan \left (f x + e\right ) - i \, c\right )} d + i \, c d + d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.37, size = 866, normalized size = 3.46 \[ \frac {\sqrt {c +d \tan \left (f x +e \right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a \left (-\ln \left (\frac {2 i a \tan \left (f x +e \right ) d +i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i d a}+d a}{2 \sqrt {i d a}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, a c +3 i \ln \left (\frac {2 i a \tan \left (f x +e \right ) d +i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i d a}+d a}{2 \sqrt {i d a}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, a d +2 i \sqrt {i d a}\, \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}+4 \ln \left (\frac {2 i a \tan \left (f x +e \right ) d +i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i d a}+d a}{2 \sqrt {i d a}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, a d +4 i \ln \left (\frac {2 i a \tan \left (f x +e \right ) d +i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i d a}+d a}{2 \sqrt {i d a}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, a c +2 i \sqrt {i d a}\, \ln \left (\frac {3 c a +i a \tan \left (f x +e \right ) c -i d a +3 a \tan \left (f x +e \right ) d +2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) a c +2 i \sqrt {i d a}\, \ln \left (\frac {3 c a +i a \tan \left (f x +e \right ) c -i d a +3 a \tan \left (f x +e \right ) d +2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) a d -2 \sqrt {i d a}\, \ln \left (\frac {3 c a +i a \tan \left (f x +e \right ) c -i d a +3 a \tan \left (f x +e \right ) d +2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) a c +2 \ln \left (\frac {3 c a +i a \tan \left (f x +e \right ) c -i d a +3 a \tan \left (f x +e \right ) d +2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{\tan \left (f x +e \right )+i}\right ) d a \sqrt {i d a}\right ) \sqrt {2}}{4 f \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i d a}\, \sqrt {-a \left (i d -c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/f*(c+d*tan(f*x+e))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*a*(-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*d*a)^(1/2)+d*a)/(I*d*a)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c+3*I*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*d*a)^(1/2)+d*a)/(I*d*a)^(1/2))*2^(1/2)*(
-a*(I*d-c))^(1/2)*a*d+2*I*(I*d*a)^(1/2)*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)
+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*d*a)^(1/2)+d*a)/(I*d*a)^(
1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*d+4*I*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*d*a)^(1/2)+d*a)/(I*d*a)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*a*c+2*I*(I*d*a)^(1/2)*ln((3*c*a+I*a*tan
(f*x+e)*c-I*d*a+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))/(ta
n(f*x+e)+I))*a*c+2*I*(I*d*a)^(1/2)*ln((3*c*a+I*a*tan(f*x+e)*c-I*d*a+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))*a*d-2*(I*d*a)^(1/2)*ln((3*c*a+I*a*tan(f*x+e)*
c-I*d*a+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))*a*c+2*ln((3*c*a+I*a*tan(f*x+e)*c-I*d*a+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*a*(I*d*a)^(1/2))/(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)/(I*d*a
)^(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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(3/2),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(3*d-c>0)', see `assume?` for m
ore details)Is 3*d-c positive, negative or zero?

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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