\(\int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx\) [402]

Optimal result

Integrand size = 30, antiderivative size = 364 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx=\frac {3 i a^{3/2} \sqrt {e} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} d}-\frac {3 i a^{3/2} \sqrt {e} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} d}-\frac {3 i a^{3/2} \sqrt {e} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{2 \sqrt {2} d}+\frac {3 i a^{3/2} \sqrt {e} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{2 \sqrt {2} d}+\frac {i a \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

3/2*I*a^(3/2)*arctan(1-2^(1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a^(1/2)/(e*sec(d*x+c))^(1/2))*e^(1/2)/d*2^(1/2
)-3/2*I*a^(3/2)*arctan(1+2^(1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a^(1/2)/(e*sec(d*x+c))^(1/2))*e^(1/2)/d*2^(1
/2)-3/4*I*a^(3/2)*ln(a-2^(1/2)*a^(1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x+c))^(1/2)+cos(d*x+c)*(a+I*a
*tan(d*x+c)))*e^(1/2)/d*2^(1/2)+3/4*I*a^(3/2)*ln(a+2^(1/2)*a^(1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/(e*sec(d*x
+c))^(1/2)+cos(d*x+c)*(a+I*a*tan(d*x+c)))*e^(1/2)/d*2^(1/2)+I*a*(e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/
d

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3579, 3576, 303, 1176, 631, 210, 1179, 642} \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx=\frac {3 i a^{3/2} \sqrt {e} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} d}-\frac {3 i a^{3/2} \sqrt {e} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} d}-\frac {3 i a^{3/2} \sqrt {e} \log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt {2} d}+\frac {3 i a^{3/2} \sqrt {e} \log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt {2} d}+\frac {i a \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{d} \]

[In]

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

[Out]

((3*I)*a^(3/2)*Sqrt[e]*ArcTan[1 - (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]
)/(Sqrt[2]*d) - ((3*I)*a^(3/2)*Sqrt[e]*ArcTan[1 + (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e
*Sec[c + d*x]])])/(Sqrt[2]*d) - (((3*I)/2)*a^(3/2)*Sqrt[e]*Log[a - (Sqrt[2]*Sqrt[a]*Sqrt[e]*Sqrt[a + I*a*Tan[c
+ d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a + I*a*Tan[c + d*x])])/(Sqrt[2]*d) + (((3*I)/2)*a^(3/2)*Sqrt[e
]*Log[a + (Sqrt[2]*Sqrt[a]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/Sqrt[e*Sec[c + d*x]] + Cos[c + d*x]*(a + I*a*Ta
n[c + d*x])])/(Sqrt[2]*d) + (I*a*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d

Rule 210

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

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 3576

Int[Sqrt[(d_.)*sec[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-4*b*(d^
2/f), Subst[Int[x^2/(a^2 + d^2*x^4), x], x, Sqrt[a + b*Tan[e + f*x]]/Sqrt[d*Sec[e + f*x]]], x] /; FreeQ[{a, b,
d, e, f}, x] && EqQ[a^2 + b^2, 0]

Rule 3579

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

Rubi steps \begin{align*} \text {integral}& = \frac {i a \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}+\frac {1}{2} (3 a) \int \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx \\ & = \frac {i a \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\left (6 i a^2 e^2\right ) \text {Subst}\left (\int \frac {x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{d} \\ & = \frac {i a \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}+\frac {\left (3 i a^2 e\right ) \text {Subst}\left (\int \frac {a-e x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{d}-\frac {\left (3 i a^2 e\right ) \text {Subst}\left (\int \frac {a+e x^2}{a^2+e^2 x^4} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{d} \\ & = \frac {i a \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\left (3 i a^2\right ) \text {Subst}\left (\int \frac {1}{\frac {a}{e}-\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}+x^2} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{2 d}-\frac {\left (3 i a^2\right ) \text {Subst}\left (\int \frac {1}{\frac {a}{e}+\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}+x^2} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{2 d}-\frac {\left (3 i a^{3/2} \sqrt {e}\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {a}}{\sqrt {e}}+2 x}{-\frac {a}{e}-\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}-x^2} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{2 \sqrt {2} d}-\frac {\left (3 i a^{3/2} \sqrt {e}\right ) \text {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt {a}}{\sqrt {e}}-2 x}{-\frac {a}{e}+\frac {\sqrt {2} \sqrt {a} x}{\sqrt {e}}-x^2} \, dx,x,\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}\right )}{2 \sqrt {2} d} \\ & = -\frac {3 i a^{3/2} \sqrt {e} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{2 \sqrt {2} d}+\frac {3 i a^{3/2} \sqrt {e} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{2 \sqrt {2} d}+\frac {i a \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {\left (3 i a^{3/2} \sqrt {e}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} d}+\frac {\left (3 i a^{3/2} \sqrt {e}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} d} \\ & = \frac {3 i a^{3/2} \sqrt {e} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} d}-\frac {3 i a^{3/2} \sqrt {e} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{\sqrt {2} d}-\frac {3 i a^{3/2} \sqrt {e} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{2 \sqrt {2} d}+\frac {3 i a^{3/2} \sqrt {e} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{2 \sqrt {2} d}+\frac {i a \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.27 (sec) , antiderivative size = 338, normalized size of antiderivative = 0.93 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx=\frac {a e \left (i \sec (c+d x) \sqrt {1+\cos (2 c)+i \sin (2 c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}-3 \text {arctanh}\left (\frac {\sqrt {1-i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {-1-i \cos (c)-\sin (c)} \sqrt {1+i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}+3 \text {arctanh}\left (\frac {\sqrt {1+i \cos (c)-\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \sqrt {1-i \cos (c)+\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}\right ) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \sec (c+d x)} \sqrt {1+\cos (2 c)+i \sin (2 c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}} \]

[In]

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

[Out]

(a*e*(I*Sec[c + d*x]*Sqrt[1 + Cos[2*c] + I*Sin[2*c]]*Sqrt[I - Tan[(d*x)/2]] - 3*ArcTanh[(Sqrt[1 - I*Cos[c] + S
in[c]]*Sqrt[I - Tan[(d*x)/2]])/(Sqrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[I + Tan[(d*x)/2]])]*Sqrt[-1 - I*Cos[c] - Sin
[c]]*Sqrt[1 + I*Cos[c] - Sin[c]]*Sqrt[I + Tan[(d*x)/2]] + 3*ArcTanh[(Sqrt[1 + I*Cos[c] - Sin[c]]*Sqrt[I - Tan[
(d*x)/2]])/(Sqrt[-1 + I*Cos[c] + Sin[c]]*Sqrt[I + Tan[(d*x)/2]])]*Sqrt[1 - I*Cos[c] + Sin[c]]*Sqrt[-1 + I*Cos[
c] + Sin[c]]*Sqrt[I + Tan[(d*x)/2]])*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[e*Sec[c + d*x]]*Sqrt[1 + Cos[2*c] + I
*Sin[2*c]]*Sqrt[I - Tan[(d*x)/2]])

Maple [A] (verified)

Time = 10.48 (sec) , antiderivative size = 359, normalized size of antiderivative = 0.99

method	result	size
default	\(-\frac {\left (-\tan \left (d x +c \right )+i\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {e \sec \left (d x +c \right )}\, a \cos \left (d x +c \right ) \left (3 i \cos \left (d x +c \right ) \operatorname {arctanh}\left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )-3 i \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )-2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )-3 \,\operatorname {arctanh}\left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )-3 \,\operatorname {arctanh}\left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )-2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+2 \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\right )}{2 d \left (2 i \left (\cos ^{2}\left (d x +c \right )\right )+i \cos \left (d x +c \right )-2 \sin \left (d x +c \right ) \cos \left (d x +c \right )-i-\sin \left (d x +c \right )\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\)	\(359\)

[In]

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

[Out]

-1/2/d*(-tan(d*x+c)+I)*(a*(1+I*tan(d*x+c)))^(1/2)*(e*sec(d*x+c))^(1/2)*a*cos(d*x+c)*(3*I*cos(d*x+c)*arctanh(1/
2*(-cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2))-3*I*arctanh(1/2*(cos(d*x+c)+sin(d*x+c)+1
)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)-2*I*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)-3*arctanh(1/2*(-
cos(d*x+c)+sin(d*x+c)-1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)-3*arctanh(1/2*(cos(d*x+c)+sin(d*x
+c)+1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)-2*I*(1/(cos(d*x+c)+1))^(1/2)+2*sin(d*x+c)*(1/(cos(d
*x+c)+1))^(1/2))/(2*I*cos(d*x+c)^2+I*cos(d*x+c)-2*sin(d*x+c)*cos(d*x+c)-I-sin(d*x+c))/(1/(cos(d*x+c)+1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.15 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx=\frac {4 i \, a \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {3}{2} i \, d x + \frac {3}{2} i \, c\right )} + \sqrt {\frac {9 i \, a^{3} e}{d^{2}}} d \log \left (\frac {2 \, {\left (3 \, {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + \sqrt {\frac {9 i \, a^{3} e}{d^{2}}} d\right )}}{3 \, a}\right ) - \sqrt {\frac {9 i \, a^{3} e}{d^{2}}} d \log \left (\frac {2 \, {\left (3 \, {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - \sqrt {\frac {9 i \, a^{3} e}{d^{2}}} d\right )}}{3 \, a}\right ) - \sqrt {-\frac {9 i \, a^{3} e}{d^{2}}} d \log \left (\frac {2 \, {\left (3 \, {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + \sqrt {-\frac {9 i \, a^{3} e}{d^{2}}} d\right )}}{3 \, a}\right ) + \sqrt {-\frac {9 i \, a^{3} e}{d^{2}}} d \log \left (\frac {2 \, {\left (3 \, {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - \sqrt {-\frac {9 i \, a^{3} e}{d^{2}}} d\right )}}{3 \, a}\right )}{2 \, d} \]

[In]

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

[Out]

1/2*(4*I*a*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(3/2*I*d*x + 3/2*I*c) + sqrt(
9*I*a^3*e/d^2)*d*log(2/3*(3*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x +
2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + sqrt(9*I*a^3*e/d^2)*d)/a) - sqrt(9*I*a^3*e/d^2)*d*log(2/3*(3*(a*e^(2*I
*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)
- sqrt(9*I*a^3*e/d^2)*d)/a) - sqrt(-9*I*a^3*e/d^2)*d*log(2/3*(3*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*
x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + sqrt(-9*I*a^3*e/d^2)*d)/a) + sqrt
(-9*I*a^3*e/d^2)*d*log(2/3*(3*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x
+ 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) - sqrt(-9*I*a^3*e/d^2)*d)/a))/d

Sympy [F]

\[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx=\int \sqrt {e \sec {\left (c + d x \right )}} \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1871 vs. \(2 (268) = 536\).

Time = 0.49 (sec) , antiderivative size = 1871, normalized size of antiderivative = 5.14 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-8*(6*(sqrt(2)*a*cos(2*d*x + 2*c) + I*sqrt(2)*a*sin(2*d*x + 2*c) + sqrt(2)*a)*arctan2(sqrt(2)*cos(1/4*arctan2(
sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1, sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) +
6*(sqrt(2)*a*cos(2*d*x + 2*c) + I*sqrt(2)*a*sin(2*d*x + 2*c) + sqrt(2)*a)*arctan2(sqrt(2)*cos(1/4*arctan2(sin(
2*d*x + 2*c), cos(2*d*x + 2*c))) + 1, -sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 6*(
sqrt(2)*a*cos(2*d*x + 2*c) + I*sqrt(2)*a*sin(2*d*x + 2*c) + sqrt(2)*a)*arctan2(sqrt(2)*cos(1/4*arctan2(sin(2*d
*x + 2*c), cos(2*d*x + 2*c))) - 1, sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 6*(sqrt
(2)*a*cos(2*d*x + 2*c) + I*sqrt(2)*a*sin(2*d*x + 2*c) + sqrt(2)*a)*arctan2(sqrt(2)*cos(1/4*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c))) - 1, -sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 6*(-I*sqrt
(2)*a*cos(2*d*x + 2*c) + sqrt(2)*a*sin(2*d*x + 2*c) - I*sqrt(2)*a)*arctan2(sqrt(2)*sin(1/4*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))), sqrt(2)*cos(1/4*arctan2(sin(
2*d*x + 2*c), cos(2*d*x + 2*c))) + cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 6*(I*sqrt(2)*a*
cos(2*d*x + 2*c) - sqrt(2)*a*sin(2*d*x + 2*c) + I*sqrt(2)*a)*arctan2(-sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c)
, cos(2*d*x + 2*c))) + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))), -sqrt(2)*cos(1/4*arctan2(sin(2*d*
x + 2*c), cos(2*d*x + 2*c))) + cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) - 16*a*cos(3/4*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*(sqrt(2)*a*cos(2*d*x + 2*c) + I*sqrt(2)*a*sin(2*d*x + 2*c) + sqrt(2
)*a)*log(2*sqrt(2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(
2*d*x + 2*c))) + 2*(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1)*cos(1/2*arctan2(sin(2*d*
x + 2*c), cos(2*d*x + 2*c))) + cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*cos(1/4*arctan2(sin(
2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1)
+ 3*(sqrt(2)*a*cos(2*d*x + 2*c) + I*sqrt(2)*a*sin(2*d*x + 2*c) + sqrt(2)*a)*log(-2*sqrt(2)*sin(1/2*arctan2(si
n(2*d*x + 2*c), cos(2*d*x + 2*c)))*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 2*(sqrt(2)*cos(1/4*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 1)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + cos(1/
2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 +
sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))
)^2 - 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 3*(I*sqrt(2)*a*cos(2*d*x + 2*c) -
sqrt(2)*a*sin(2*d*x + 2*c) + I*sqrt(2)*a)*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin
(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x +
2*c))) + 2*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + 3*(-I*sqrt(2)*a*cos(2*d*x + 2*c
) + sqrt(2)*a*sin(2*d*x + 2*c) - I*sqrt(2)*a)*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2
*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*
x + 2*c))) - 2*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + 3*(I*sqrt(2)*a*cos(2*d*x +
2*c) - sqrt(2)*a*sin(2*d*x + 2*c) + I*sqrt(2)*a)*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2
+ 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2
*d*x + 2*c))) + 2*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + 3*(-I*sqrt(2)*a*cos(2*d*
x + 2*c) + sqrt(2)*a*sin(2*d*x + 2*c) - I*sqrt(2)*a)*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))
)^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 - 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), c
os(2*d*x + 2*c))) - 2*sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) - 16*I*a*sin(3/4*arcta
n2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sqrt(a)*sqrt(e)/(d*(-64*I*cos(2*d*x + 2*c) + 64*sin(2*d*x + 2*c) - 64
*I))

Giac [F]

\[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx=\int { \sqrt {e \sec \left (d x + c\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \,d x } \]

[In]

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

[Out]

integrate(sqrt(e*sec(d*x + c))*(I*a*tan(d*x + c) + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx=\int \sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

[In]

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

[Out]

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