\(\int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx\) [409]

Optimal result

Integrand size = 30, antiderivative size = 411 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx=\frac {21 i a^{5/2} \sqrt {e} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{4 \sqrt {2} d}-\frac {21 i a^{5/2} \sqrt {e} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{4 \sqrt {2} d}-\frac {21 i a^{5/2} \sqrt {e} \log \left (a-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{8 \sqrt {2} d}+\frac {21 i a^{5/2} \sqrt {e} \log \left (a+\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))\right )}{8 \sqrt {2} d}+\frac {7 i a^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d} \]

[Out]

21/8*I*a^(5/2)*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)-21/8*I*a^(5/2)*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)-21/16*I*a^(5/2)*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)+21/16*I*a^(5/2)*ln(a+2^(1/2)*a^(1/2)*e^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/(e*s
ec(d*x+c))^(1/2)+cos(d*x+c)*(a+I*a*tan(d*x+c)))*e^(1/2)/d*2^(1/2)+7/4*I*a^2*(e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*
x+c))^(1/2)/d+1/2*I*a*(e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^(3/2)/d

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.00, number of steps used = 12, 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} \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx=\frac {21 i a^{5/2} \sqrt {e} \arctan \left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{4 \sqrt {2} d}-\frac {21 i a^{5/2} \sqrt {e} \arctan \left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{4 \sqrt {2} d}-\frac {21 i a^{5/2} \sqrt {e} \log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{8 \sqrt {2} d}+\frac {21 i a^{5/2} \sqrt {e} \log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{8 \sqrt {2} d}+\frac {7 i a^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}{4 d}+\frac {i a (a+i a \tan (c+d x))^{3/2} \sqrt {e \sec (c+d x)}}{2 d} \]

[In]

Int[Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((21*I)/4)*a^(5/2)*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) - (((21*I)/4)*a^(5/2)*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) - (((21*I)/8)*a^(5/2)*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) + (((21*I)/8)*a^
(5/2)*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) + (((7*I)/4)*a^2*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d + ((
I/2)*a*Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2))/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)} (a+i a \tan (c+d x))^{3/2}}{2 d}+\frac {1}{4} (7 a) \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2} \, dx \\ & = \frac {7 i a^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}+\frac {1}{8} \left (21 a^2\right ) \int \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx \\ & = \frac {7 i a^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}-\frac {\left (21 i a^3 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 )}{2 d} \\ & = \frac {7 i a^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}+\frac {\left (21 i a^3 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 )}{4 d}-\frac {\left (21 i a^3 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 )}{4 d} \\ & = \frac {7 i a^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}-\frac {\left (21 i a^3\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 )}{8 d}-\frac {\left (21 i a^3\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 )}{8 d}-\frac {\left (21 i a^{5/2} \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 )}{8 \sqrt {2} d}-\frac {\left (21 i a^{5/2} \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 )}{8 \sqrt {2} d} \\ & = -\frac {21 i a^{5/2} \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 )}{8 \sqrt {2} d}+\frac {21 i a^{5/2} \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 )}{8 \sqrt {2} d}+\frac {7 i a^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d}-\frac {\left (21 i a^{5/2} \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 )}{4 \sqrt {2} d}+\frac {\left (21 i a^{5/2} \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 )}{4 \sqrt {2} d} \\ & = \frac {21 i a^{5/2} \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 )}{4 \sqrt {2} d}-\frac {21 i a^{5/2} \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 )}{4 \sqrt {2} d}-\frac {21 i a^{5/2} \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 )}{8 \sqrt {2} d}+\frac {21 i a^{5/2} \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 )}{8 \sqrt {2} d}+\frac {7 i a^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}{4 d}+\frac {i a \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{3/2}}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.09 (sec) , antiderivative size = 387, normalized size of antiderivative = 0.94 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {a^2 \sqrt {e \sec (c+d x)} (\cos (2 d x)+i \sin (2 d x)) \sqrt {a+i a \tan (c+d x)} \left (21 \text {arctanh}\left (\frac {\sqrt {1-i \cos (c)+\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1-i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \cos (c+d x) \sqrt {-1-i \cos (c)-\sin (c)} \sqrt {1+i \cos (c)-\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}-21 \text {arctanh}\left (\frac {\sqrt {1+i \cos (c)-\sin (c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )}}{\sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}}\right ) \cos (c+d x) \sqrt {1-i \cos (c)+\sin (c)} \sqrt {-1+i \cos (c)+\sin (c)} \sqrt {i+\tan \left (\frac {d x}{2}\right )}+\sqrt {1+\cos (2 c)+i \sin (2 c)} \sqrt {i-\tan \left (\frac {d x}{2}\right )} (-9 i+2 \tan (c+d x))\right )}{4 d \sqrt {1+\cos (2 c)+i \sin (2 c)} (\cos (d x)+i \sin (d x))^2 \sqrt {i-\tan \left (\frac {d x}{2}\right )}} \]

[In]

Integrate[Sqrt[e*Sec[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

-1/4*(a^2*Sqrt[e*Sec[c + d*x]]*(Cos[2*d*x] + I*Sin[2*d*x])*Sqrt[a + I*a*Tan[c + d*x]]*(21*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]])]*Cos[c + d*x]*S
qrt[-1 - I*Cos[c] - Sin[c]]*Sqrt[1 + I*Cos[c] - Sin[c]]*Sqrt[I + Tan[(d*x)/2]] - 21*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]])]*Cos[c + d*x]*Sqrt[1
- I*Cos[c] + Sin[c]]*Sqrt[-1 + I*Cos[c] + Sin[c]]*Sqrt[I + Tan[(d*x)/2]] + Sqrt[1 + Cos[2*c] + I*Sin[2*c]]*Sqr
t[I - Tan[(d*x)/2]]*(-9*I + 2*Tan[c + d*x])))/(d*Sqrt[1 + Cos[2*c] + I*Sin[2*c]]*(Cos[d*x] + I*Sin[d*x])^2*Sqr
t[I - Tan[(d*x)/2]])

Maple [A] (verified)

Time = 10.24 (sec) , antiderivative size = 449, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \left (\tan \left (d x +c \right )-i\right )^{2} \sqrt {e \sec \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \cos \left (d x +c \right ) \left (11 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-11 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos ^{2}\left (d x +c \right )\right )-21 i \left (\cos ^{2}\left (d x +c \right )\right ) \operatorname {arctanh}\left (\frac {\cos \left (d x +c \right )+\sin \left (d x +c \right )+1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+11 \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+2 i \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+11 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos ^{2}\left (d x +c \right )\right )-9 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )-21 \left (\cos ^{2}\left (d x +c \right )\right ) \operatorname {arctanh}\left (\frac {-\cos \left (d x +c \right )+\sin \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}\right )+2 \sin \left (d x +c \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+9 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right )+2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-2 \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\right )}{d \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (4 \left (\cos ^{3}\left (d x +c \right )\right )+2 \left (\cos ^{2}\left (d x +c \right )\right )+4 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-3 \cos \left (d x +c \right )+2 i \cos \left (d x +c \right ) \sin \left (d x +c \right )-1-i \sin \left (d x +c \right )\right )}\)	\(449\)

[In]

int((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

(1/8-1/8*I)/d*(tan(d*x+c)-I)^2*(e*sec(d*x+c))^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*a^2*cos(d*x+c)*(11*I*(1/(cos(d*
x+c)+1))^(1/2)*sin(d*x+c)*cos(d*x+c)-11*I*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2-21*I*arctanh(1/2*(cos(d*x+c)+s
in(d*x+c)+1)/(cos(d*x+c)+1)/(1/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2+11*sin(d*x+c)*cos(d*x+c)*(1/(cos(d*x+c)+1))
^(1/2)+2*I*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+11*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2-9*I*(1/(cos(d*x+c)+1))
^(1/2)*cos(d*x+c)-21*cos(d*x+c)^2*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))+2*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)+9*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+2*I*(1/(cos(d*x+c)+1))^(1/2)-2
*(1/(cos(d*x+c)+1))^(1/2))/(1/(cos(d*x+c)+1))^(1/2)/(4*cos(d*x+c)^3+2*cos(d*x+c)^2+4*I*cos(d*x+c)^2*sin(d*x+c)
-3*cos(d*x+c)+2*I*cos(d*x+c)*sin(d*x+c)-1-I*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 525, normalized size of antiderivative = 1.28 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx=\frac {{\left (11 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 7 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\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 {441 i \, a^{5} e}{16 \, d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (21 \, {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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 )} + 4 \, \sqrt {\frac {441 i \, a^{5} e}{16 \, d^{2}}} d\right )}}{21 \, a^{2}}\right ) - \sqrt {\frac {441 i \, a^{5} e}{16 \, d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (21 \, {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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 )} - 4 \, \sqrt {\frac {441 i \, a^{5} e}{16 \, d^{2}}} d\right )}}{21 \, a^{2}}\right ) - \sqrt {-\frac {441 i \, a^{5} e}{16 \, d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (21 \, {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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 )} + 4 \, \sqrt {-\frac {441 i \, a^{5} e}{16 \, d^{2}}} d\right )}}{21 \, a^{2}}\right ) + \sqrt {-\frac {441 i \, a^{5} e}{16 \, d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {2 \, {\left (21 \, {\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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 )} - 4 \, \sqrt {-\frac {441 i \, a^{5} e}{16 \, d^{2}}} d\right )}}{21 \, a^{2}}\right )}{2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/2*((11*I*a^2*e^(3*I*d*x + 3*I*c) + 7*I*a^2*e^(I*d*x + I*c))*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(441/16*I*a^5*e/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(2/21*(2
1*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*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) + 4*sqrt(441/16*I*a^5*e/d^2)*d)/a^2) - sqrt(441/16*I*a^5*e/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*lo
g(2/21*(21*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*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) - 4*sqrt(441/16*I*a^5*e/d^2)*d)/a^2) - sqrt(-441/16*I*a^5*e/d^2)*(d*e^(2*I*d*x + 2*I*
c) + d)*log(2/21*(21*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*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) + 4*sqrt(-441/16*I*a^5*e/d^2)*d)/a^2) + sqrt(-441/16*I*a^5*e/d^2)*(d*e^(2*I
*d*x + 2*I*c) + d)*log(2/21*(21*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*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) - 4*sqrt(-441/16*I*a^5*e/d^2)*d)/a^2))/(d*e^(2*I*d*x + 2*I*c) +
d)

Sympy [F(-1)]

Timed out. \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate((e*sec(d*x+c))**(1/2)*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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. 2431 vs. \(2 (299) = 598\).

Time = 0.87 (sec) , antiderivative size = 2431, normalized size of antiderivative = 5.91 \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

32*(176*a^2*cos(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 112*a^2*cos(3/4*arctan2(sin(2*d*x + 2*c), c
os(2*d*x + 2*c))) + 176*I*a^2*sin(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 112*I*a^2*sin(3/4*arctan2
(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 42*(sqrt(2)*a^2*cos(4*d*x + 4*c) + 2*sqrt(2)*a^2*cos(2*d*x + 2*c) + I*
sqrt(2)*a^2*sin(4*d*x + 4*c) + 2*I*sqrt(2)*a^2*sin(2*d*x + 2*c) + sqrt(2)*a^2)*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) -
42*(sqrt(2)*a^2*cos(4*d*x + 4*c) + 2*sqrt(2)*a^2*cos(2*d*x + 2*c) + I*sqrt(2)*a^2*sin(4*d*x + 4*c) + 2*I*sqrt
(2)*a^2*sin(2*d*x + 2*c) + sqrt(2)*a^2)*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) - 42*(sqrt(2)*a^2*cos(4*d*x + 4*c) + 2*
sqrt(2)*a^2*cos(2*d*x + 2*c) + I*sqrt(2)*a^2*sin(4*d*x + 4*c) + 2*I*sqrt(2)*a^2*sin(2*d*x + 2*c) + sqrt(2)*a^2
)*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) - 42*(sqrt(2)*a^2*cos(4*d*x + 4*c) + 2*sqrt(2)*a^2*cos(2*d*x + 2*c) + I*sqrt(2
)*a^2*sin(4*d*x + 4*c) + 2*I*sqrt(2)*a^2*sin(2*d*x + 2*c) + sqrt(2)*a^2)*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) - 42*(
-I*sqrt(2)*a^2*cos(4*d*x + 4*c) - 2*I*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2*sin(4*d*x + 4*c) + 2*sqrt(2)*
a^2*sin(2*d*x + 2*c) - I*sqrt(2)*a^2)*arctan2(sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + s
in(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) - 42*(I*sqrt(2)*a^2*cos(4*d*x + 4*c) + 2*I*sqr
t(2)*a^2*cos(2*d*x + 2*c) - sqrt(2)*a^2*sin(4*d*x + 4*c) - 2*sqrt(2)*a^2*sin(2*d*x + 2*c) + I*sqrt(2)*a^2)*arc
tan2(-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) + 21*(sqrt(2)*a^2*cos(4*d*x + 4*c) + 2*sqrt(2)*a^2*cos(2*d*x + 2*c) + I*sqrt(2)*a^2*s
in(4*d*x + 4*c) + 2*I*sqrt(2)*a^2*sin(2*d*x + 2*c) + sqrt(2)*a^2)*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(si
n(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*arct
an2(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) - 21*(sqrt(2)*a^2*cos(4*d*x + 4*c) + 2*sqrt(2)*a
^2*cos(2*d*x + 2*c) + I*sqrt(2)*a^2*sin(4*d*x + 4*c) + 2*I*sqrt(2)*a^2*sin(2*d*x + 2*c) + sqrt(2)*a^2)*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), c
os(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) - 21*(I*sq
rt(2)*a^2*cos(4*d*x + 4*c) + 2*I*sqrt(2)*a^2*cos(2*d*x + 2*c) - sqrt(2)*a^2*sin(4*d*x + 4*c) - 2*sqrt(2)*a^2*s
in(2*d*x + 2*c) + I*sqrt(2)*a^2)*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arct
an2(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) - 21*(-I*sqrt(2)*a^2*cos(4*d*x + 4*c) - 2*
I*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2*sin(4*d*x + 4*c) + 2*sqrt(2)*a^2*sin(2*d*x + 2*c) - I*sqrt(2)*a^2
)*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) - 21*(I*sqrt(2)*a^2*cos(4*d*x + 4*c) + 2*I*sqrt(2)*a^2*cos(2*d*x + 2*c) -
sqrt(2)*a^2*sin(4*d*x + 4*c) - 2*sqrt(2)*a^2*sin(2*d*x + 2*c) + I*sqrt(2)*a^2)*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) - 21*(-I*sqrt(2)*a^2*cos(4*d*x + 4*c) - 2*I*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2*sin(4*d*x + 4*c) +
2*sqrt(2)*a^2*sin(2*d*x + 2*c) - I*sqrt(2)*a^2)*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))*sqrt(a)*sqrt(e)/(d*(-1024
*I*cos(4*d*x + 4*c) - 2048*I*cos(2*d*x + 2*c) + 1024*sin(4*d*x + 4*c) + 2048*sin(2*d*x + 2*c) - 1024*I))

Giac [F]

\[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx=\int { \sqrt {e \sec \left (d x + c\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \,d x } \]

[In]

integrate((e*sec(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(e*sec(d*x + c))*(I*a*tan(d*x + c) + a)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx=\int \sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]

[In]

int((e/cos(c + d*x))^(1/2)*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

int((e/cos(c + d*x))^(1/2)*(a + a*tan(c + d*x)*1i)^(5/2), x)