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

Optimal. Leaf size=520 \[ \frac {i \sqrt {2} a^{5/2} \sec (c+d x) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d \sqrt {e} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {i \sqrt {2} a^{5/2} \sec (c+d x) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d \sqrt {e} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {i a^{5/2} \sec (c+d x) \log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{\sqrt {2} d \sqrt {e} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i a^{5/2} \sec (c+d x) \log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{\sqrt {2} d \sqrt {e} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \sec (c+d x)}} \]

[Out]

-1/2*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*ta
n(d*x+c)))*sec(d*x+c)/d*2^(1/2)/e^(1/2)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)+1/2*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)))*sec(d*x+c)
/d*2^(1/2)/e^(1/2)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)+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))*sec(d*x+c)*2^(1/2)/d/e^(1/2)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*
tan(d*x+c))^(1/2)-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))*se
c(d*x+c)*2^(1/2)/d/e^(1/2)/(a-I*a*tan(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-4*I*a*(a+I*a*tan(d*x+c))^(1/2)/d/
(e*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.43, antiderivative size = 520, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3496, 3499, 3495, 297, 1162, 617, 204, 1165, 628} \[ \frac {i \sqrt {2} a^{5/2} \sec (c+d x) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d \sqrt {e} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {i \sqrt {2} a^{5/2} \sec (c+d x) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {a} \sqrt {e \sec (c+d x)}}\right )}{d \sqrt {e} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {i a^{5/2} \sec (c+d x) \log \left (-\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{\sqrt {2} d \sqrt {e} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i a^{5/2} \sec (c+d x) \log \left (\frac {\sqrt {2} \sqrt {a} \sqrt {e} \sqrt {a-i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}}+\cos (c+d x) (a-i a \tan (c+d x))+a\right )}{\sqrt {2} d \sqrt {e} \sqrt {a-i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {4 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {e \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*Sqrt[2]*a^(5/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]*Sec
[c + d*x])/(d*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (I*Sqrt[2]*a^(5/2)*ArcTan[1 + (
Sqrt[2]*Sqrt[e]*Sqrt[a - I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])]*Sec[c + d*x])/(d*Sqrt[e]*Sqrt[a -
I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (I*a^(5/2)*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])]*Sec[c + d*x])/(Sqrt[2]*d*Sqrt[e]*Sqrt[a
- I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (I*a^(5/2)*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])]*Sec[c + d*x])/(Sqrt[2]*d*Sqrt[e]*Sqrt[
a - I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - ((4*I)*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[e*Sec[c + d*x
]])

Rule 204

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

Rule 297

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 617

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 628

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 1162

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 1165

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 3495

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 3496

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] - Dist[(b^2*(m + 2*n - 2))/(d^2*m), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3499

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(3/2)/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(d*Sec
[e + f*x])/(Sqrt[a - b*Tan[e + f*x]]*Sqrt[a + b*Tan[e + f*x]]), Int[Sqrt[d*Sec[e + f*x]]*Sqrt[a - b*Tan[e + f*
x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [B]  time = 6.17, size = 11314, normalized size = 21.76 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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fricas [A]  time = 0.57, size = 460, normalized size = 0.88 \[ -\frac {d e \sqrt {\frac {4 i \, a^{3}}{d^{2} e}} \log \left (\frac {2 \, {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + i \, d e \sqrt {\frac {4 i \, a^{3}}{d^{2} e}}}{a}\right ) - d e \sqrt {\frac {4 i \, a^{3}}{d^{2} e}} \log \left (\frac {2 \, {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - i \, d e \sqrt {\frac {4 i \, a^{3}}{d^{2} e}}}{a}\right ) + d e \sqrt {-\frac {4 i \, a^{3}}{d^{2} e}} \log \left (\frac {2 \, {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + i \, d e \sqrt {-\frac {4 i \, a^{3}}{d^{2} e}}}{a}\right ) - d e \sqrt {-\frac {4 i \, a^{3}}{d^{2} e}} \log \left (\frac {2 \, {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - i \, d e \sqrt {-\frac {4 i \, a^{3}}{d^{2} e}}}{a}\right ) - 2 \, {\left (-4 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{2 \, d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 1.24, size = 286, normalized size = 0.55 \[ -\frac {\left (i \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-i \arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right )}{2}\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right )}{2}\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\arctanh \left (\frac {\sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right )}{2}\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+4 i \cos \left (d x +c \right )-4 i-4 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \sqrt {\frac {e}{\cos \left (d x +c \right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*(I*arctanh(1/2*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-I*
arctanh(1/2*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+arctanh(1/
2*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+arctanh(1/2*(1/(1+co
s(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+4*I*cos(d*x+c)-4*I-4*sin(d*x+c
))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c)-1)/(e/cos(d*x+c))^(1/2)*a

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maxima [B]  time = 0.90, size = 1462, normalized size = 2.81 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*I*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*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + 2*I*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) + 2*I*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) + 2*I*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) - 2*sqrt(2)*a*arctan2(sqrt(2)*s
in(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))), sq
rt(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) + 2*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) + I*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*arcta
n2(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) - I*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*arc
tan2(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 + si
n(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) - 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) + 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)*si
n(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) - sqrt(2)*a*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), co
s(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(
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(si
n(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) - 16*I*a*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 16*a*s
in(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sqrt(a)/(d*sqrt(e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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