3.679 \(\int \frac{\sqrt{a+i a \tan (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=512 \[ \frac{3 i \sqrt{a} e^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{4 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{3 i \sqrt{a} e^{5/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{4 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{3 i \sqrt{a} e^{5/2} \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{8 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac{3 i \sqrt{a} e^{5/2} \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{8 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{3 i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}+\frac{i a}{2 d \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}} \]
[Out]
(((3*I)/4)*Sqrt[a]*e^(5/2)*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*(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/2)) - (((3*I)/4)*Sqrt[a]*e^(5/2)*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*(e*Cos[c + d*x])^(5/2)*(e*S
ec[c + d*x])^(5/2)) - (((3*I)/8)*Sqrt[a]*e^(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])])/(Sqrt[2]*d*(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x]
)^(5/2)) + (((3*I)/8)*Sqrt[a]*e^(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])])/(Sqrt[2]*d*(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/2)) +
((I/2)*a)/(d*(e*Cos[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) - (((3*I)/4)*Cos[c + d*x]^2*Sqrt[a + I*a*Tan[c
+ d*x]])/(d*(e*Cos[c + d*x])^(5/2))
________________________________________________________________________________________
Rubi [A] time = 0.564371, antiderivative size = 512, normalized size of antiderivative = 1.,
number of steps used = 13, number of rules used = 10, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used
= {3515, 3498, 3501, 3495, 297, 1162, 617, 204, 1165, 628} \[ \frac{3 i \sqrt{a} e^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{4 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{3 i \sqrt{a} e^{5/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{4 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{3 i \sqrt{a} e^{5/2} \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{8 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac{3 i \sqrt{a} e^{5/2} \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{8 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{3 i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}+\frac{i a}{2 d \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + I*a*Tan[c + d*x]]/(e*Cos[c + d*x])^(5/2),x]
[Out]
(((3*I)/4)*Sqrt[a]*e^(5/2)*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*(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/2)) - (((3*I)/4)*Sqrt[a]*e^(5/2)*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*(e*Cos[c + d*x])^(5/2)*(e*S
ec[c + d*x])^(5/2)) - (((3*I)/8)*Sqrt[a]*e^(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])])/(Sqrt[2]*d*(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x]
)^(5/2)) + (((3*I)/8)*Sqrt[a]*e^(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])])/(Sqrt[2]*d*(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/2)) +
((I/2)*a)/(d*(e*Cos[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) - (((3*I)/4)*Cos[c + d*x]^2*Sqrt[a + I*a*Tan[c
+ d*x]])/(d*(e*Cos[c + d*x])^(5/2))
Rule 3515
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] && !IntegerQ[m]
Rule 3498
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]
Rule 3501
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d^2*
(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(d^2*(m - 2))/(a*(m + n -
1)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2
+ b^2, 0] && LtQ[n, 0] && GtQ[m, 1] && !ILtQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]
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 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 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 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 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 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 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]
Rubi steps
\begin{align*} \int \frac{\sqrt{a+i a \tan (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx &=\frac{\int (e \sec (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)} \, dx}{(e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac{i a}{2 d (e \cos (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}}+\frac{(3 a) \int \frac{(e \sec (c+d x))^{5/2}}{\sqrt{a+i a \tan (c+d x)}} \, dx}{4 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac{i a}{2 d (e \cos (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}}-\frac{3 i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}+\frac{\left (3 e^2\right ) \int \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)} \, dx}{8 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac{i a}{2 d (e \cos (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}}-\frac{3 i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}-\frac{\left (3 i a e^4\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 )}{2 d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac{i a}{2 d (e \cos (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}}-\frac{3 i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}+\frac{\left (3 i a e^3\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 )}{4 d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{\left (3 i a e^3\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 )}{4 d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac{i a}{2 d (e \cos (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}}-\frac{3 i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}-\frac{\left (3 i a e^2\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 )}{8 d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{\left (3 i a e^2\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 )}{8 d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{\left (3 i \sqrt{a} e^{5/2}\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 )}{8 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{\left (3 i \sqrt{a} e^{5/2}\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 )}{8 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=-\frac{3 i \sqrt{a} e^{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 )}{8 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac{3 i \sqrt{a} e^{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 )}{8 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac{i a}{2 d (e \cos (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}}-\frac{3 i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}-\frac{\left (3 i \sqrt{a} e^{5/2}\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 )}{4 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac{\left (3 i \sqrt{a} e^{5/2}\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 )}{4 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\frac{3 i \sqrt{a} e^{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 )}{4 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{3 i \sqrt{a} e^{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 )}{4 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{3 i \sqrt{a} e^{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 )}{8 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac{3 i \sqrt{a} e^{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 )}{8 \sqrt{2} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac{i a}{2 d (e \cos (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}}-\frac{3 i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d (e \cos (c+d x))^{5/2}}\\ \end{align*}
Mathematica [A] time = 2.3867, size = 227, normalized size = 0.44 \[ \frac{\sqrt{\cos (c+d x)} \sqrt{a+i a \tan (c+d x)} \left (-3 i \cos ^{\frac{3}{2}}(c+d x)+2 \sqrt{\cos (c+d x)} (\sin (c+d x)+i \cos (c+d x))+\frac{3 i \left (-e^{-2 i c}\right )^{3/4} e^{-\frac{1}{2} i (2 c+5 d x)} \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (1+e^{2 i (c+d x)}\right )^2 \left (\tan ^{-1}\left (\frac{e^{\frac{i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )-\tanh ^{-1}\left (\frac{e^{\frac{i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )\right )}{4 \sqrt{2}}\right )}{4 d (e \cos (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + I*a*Tan[c + d*x]]/(e*Cos[c + d*x])^(5/2),x]
[Out]
(Sqrt[Cos[c + d*x]]*((((3*I)/4)*(-E^((-2*I)*c))^(3/4)*(1 + E^((2*I)*(c + d*x)))^2*Sqrt[(1 + E^((2*I)*(c + d*x)
))/E^(I*(c + d*x))]*(ArcTan[E^((I/2)*d*x)/(-E^((-2*I)*c))^(1/4)] - ArcTanh[E^((I/2)*d*x)/(-E^((-2*I)*c))^(1/4)
]))/(Sqrt[2]*E^((I/2)*(2*c + 5*d*x))) - (3*I)*Cos[c + d*x]^(3/2) + 2*Sqrt[Cos[c + d*x]]*(I*Cos[c + d*x] + Sin[
c + d*x]))*Sqrt[a + I*a*Tan[c + d*x]])/(4*d*(e*Cos[c + d*x])^(5/2))
________________________________________________________________________________________
Maple [A] time = 0.432, size = 366, normalized size = 0.7 \begin{align*} -{\frac{\cos \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) -1 \right ) ^{3}}{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5} \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) }\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( 6\,i\sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\cos \left ( dx+c \right ) +3\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}{\it Artanh} \left ({\frac{\cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) }{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) +3\,i{\it Artanh} \left ({\frac{-\cos \left ( dx+c \right ) -1+\sin \left ( dx+c \right ) }{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+4\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) +6\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-3\,{\it Artanh} \left ( 1/2\,\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+3\,{\it Artanh} \left ( 1/2\,\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( -\cos \left ( dx+c \right ) -1+\sin \left ( dx+c \right ) \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2\,\cos \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-4\,\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \left ( \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1} \right ) ^{-{\frac{5}{2}}} \left ( e\cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+I*a*tan(d*x+c))^(1/2)/(e*cos(d*x+c))^(5/2),x)
[Out]
-1/8/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)*(cos(d*x+c)-1)^3*(6*I*sin(d*x+c)*(1/(cos(d*x+
c)+1))^(1/2)*cos(d*x+c)+3*I*cos(d*x+c)^2*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))+3*I*a
rctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)-1+sin(d*x+c)))*cos(d*x+c)^2+4*I*(1/(cos(d*x+c)+1))^(1/2)*sin(
d*x+c)+6*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)-3*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)
))*cos(d*x+c)^2+3*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)-1+sin(d*x+c)))*cos(d*x+c)^2+2*cos(d*x+c)*(
1/(cos(d*x+c)+1))^(1/2)-4*(1/(cos(d*x+c)+1))^(1/2))/sin(d*x+c)^5/(I*sin(d*x+c)+cos(d*x+c)-1)/(1/(cos(d*x+c)+1)
)^(5/2)/(e*cos(d*x+c))^(5/2)
________________________________________________________________________________________
Maxima [B] time = 3.60334, size = 3058, normalized size = 5.97 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))^(1/2)/(e*cos(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
-((192*sqrt(2)*cos(4*d*x + 4*c) + 384*sqrt(2)*cos(2*d*x + 2*c) + 192*I*sqrt(2)*sin(4*d*x + 4*c) + 384*I*sqrt(2
)*sin(2*d*x + 2*c) + 192*sqrt(2))*arctan2(sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1, sq
rt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (192*sqrt(2)*cos(4*d*x + 4*c) + 384*sqrt(2)*
cos(2*d*x + 2*c) + 192*I*sqrt(2)*sin(4*d*x + 4*c) + 384*I*sqrt(2)*sin(2*d*x + 2*c) + 192*sqrt(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) + (192*sqrt(2)*cos(4*d*x + 4*c) + 384*sqrt(2)*cos(2*d*x + 2*c) + 192*I*sqrt(2)*sin(4*d*x +
4*c) + 384*I*sqrt(2)*sin(2*d*x + 2*c) + 192*sqrt(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) + (192*sqrt(2)*cos(4*d*x +
4*c) + 384*sqrt(2)*cos(2*d*x + 2*c) + 192*I*sqrt(2)*sin(4*d*x + 4*c) + 384*I*sqrt(2)*sin(2*d*x + 2*c) + 192*sq
rt(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) - (192*I*sqrt(2)*cos(4*d*x + 4*c) + 384*I*sqrt(2)*cos(2*d*x + 2*c) - 192
*sqrt(2)*sin(4*d*x + 4*c) - 384*sqrt(2)*sin(2*d*x + 2*c) + 192*I*sqrt(2))*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*arcta
n2(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) - (-192*I*
sqrt(2)*cos(4*d*x + 4*c) - 384*I*sqrt(2)*cos(2*d*x + 2*c) + 192*sqrt(2)*sin(4*d*x + 4*c) + 384*sqrt(2)*sin(2*d
*x + 2*c) - 192*I*sqrt(2))*arctan2(-sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + sin(1/2*arc
tan2(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) - (96*sqrt(2)*cos(4*d*x + 4*c) + 192*sqrt(2)*cos(2*d*x
+ 2*c) + 96*I*sqrt(2)*sin(4*d*x + 4*c) + 192*I*sqrt(2)*sin(2*d*x + 2*c) + 96*sqrt(2))*log(2*sqrt(2)*sin(1/2*ar
ctan2(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)*c
os(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*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) + (96*sqrt(2)*cos(4*d*x + 4*
c) + 192*sqrt(2)*cos(2*d*x + 2*c) + 96*I*sqrt(2)*sin(4*d*x + 4*c) + 192*I*sqrt(2)*sin(2*d*x + 2*c) + 96*sqrt(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(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)
- (-96*I*sqrt(2)*cos(4*d*x + 4*c) - 192*I*sqrt(2)*cos(2*d*x + 2*c) + 96*sqrt(2)*sin(4*d*x + 4*c) + 192*sqrt(2)
*sin(2*d*x + 2*c) - 96*I*sqrt(2))*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*sin(1/4*arc
tan2(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) - (96*I*sqrt(2)*cos(4*d*x + 4*c) + 192*I*
sqrt(2)*cos(2*d*x + 2*c) - 96*sqrt(2)*sin(4*d*x + 4*c) - 192*sqrt(2)*sin(2*d*x + 2*c) + 96*I*sqrt(2))*log(2*co
s(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) - (-96*I*sqrt(2)*cos(4*d*x + 4*c) - 192*I*sqrt(2)*cos(2*d*x + 2*c) + 96*sqrt(2)*sin
(4*d*x + 4*c) + 192*sqrt(2)*sin(2*d*x + 2*c) - 96*I*sqrt(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) - (96*I*sqrt(2
)*cos(4*d*x + 4*c) + 192*I*sqrt(2)*cos(2*d*x + 2*c) - 96*sqrt(2)*sin(4*d*x + 4*c) - 192*sqrt(2)*sin(2*d*x + 2*
c) + 96*I*sqrt(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) + 1536*cos(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
))) - 512*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1536*I*sin(7/4*arctan2(sin(2*d*x + 2*c), cos(
2*d*x + 2*c))) - 512*I*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*sqrt(a)*sqrt(e)/((-1024*I*e^3*cos
(4*d*x + 4*c) - 2048*I*e^3*cos(2*d*x + 2*c) + 1024*e^3*sin(4*d*x + 4*c) + 2048*e^3*sin(2*d*x + 2*c) - 1024*I*e
^3)*d)
________________________________________________________________________________________
Fricas [A] time = 2.57679, size = 1628, normalized size = 3.18 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))^(1/2)/(e*cos(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
1/2*(sqrt(2)*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-3*I*e^(3*I*d*x + 3*
I*c) + I*e^(I*d*x + I*c))*e^(1/2*I*d*x + 1/2*I*c) + (d*e^3*e^(4*I*d*x + 4*I*c) + 2*d*e^3*e^(2*I*d*x + 2*I*c) +
d*e^3)*sqrt(9/16*I*a/(d^2*e^5))*log(4/3*d*e^3*sqrt(9/16*I*a/(d^2*e^5)) + sqrt(2)*sqrt(1/2)*sqrt(e*e^(2*I*d*x
+ 2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)) - (d*e^3*e^(4*I*d*x + 4*I*c) + 2*d*e^
3*e^(2*I*d*x + 2*I*c) + d*e^3)*sqrt(9/16*I*a/(d^2*e^5))*log(-4/3*d*e^3*sqrt(9/16*I*a/(d^2*e^5)) + sqrt(2)*sqrt
(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)) - (d*e^3*e^(4
*I*d*x + 4*I*c) + 2*d*e^3*e^(2*I*d*x + 2*I*c) + d*e^3)*sqrt(-9/16*I*a/(d^2*e^5))*log(4/3*d*e^3*sqrt(-9/16*I*a/
(d^2*e^5)) + sqrt(2)*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x
+ 1/2*I*c)) + (d*e^3*e^(4*I*d*x + 4*I*c) + 2*d*e^3*e^(2*I*d*x + 2*I*c) + d*e^3)*sqrt(-9/16*I*a/(d^2*e^5))*log(
-4/3*d*e^3*sqrt(-9/16*I*a/(d^2*e^5)) + sqrt(2)*sqrt(1/2)*sqrt(e*e^(2*I*d*x + 2*I*c) + e)*sqrt(a/(e^(2*I*d*x +
2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)))/(d*e^3*e^(4*I*d*x + 4*I*c) + 2*d*e^3*e^(2*I*d*x + 2*I*c) + d*e^3)
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))**(1/2)/(e*cos(d*x+c))**(5/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{i \, a \tan \left (d x + c\right ) + a}}{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+I*a*tan(d*x+c))^(1/2)/(e*cos(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate(sqrt(I*a*tan(d*x + c) + a)/(e*cos(d*x + c))^(5/2), x)