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

Optimal. Leaf size=411 \[ \frac{21 i a^{5/2} \sqrt{e} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{4 \sqrt{2} d}-\frac{21 i a^{5/2} \sqrt{e} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{4 \sqrt{2} d}+\frac{7 i a^2 \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}{4 d}-\frac{21 i a^{5/2} \sqrt{e} \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{8 \sqrt{2} d}+\frac{21 i a^{5/2} \sqrt{e} \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{8 \sqrt{2} d}+\frac{i a (a+i a \tan (c+d x))^{3/2} \sqrt{e \sec (c+d x)}}{2 d} \]

[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

________________________________________________________________________________________

Rubi [A]  time = 0.453239, antiderivative size = 411, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {3498, 3495, 297, 1162, 617, 204, 1165, 628} \[ \frac{21 i a^{5/2} \sqrt{e} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{4 \sqrt{2} d}-\frac{21 i a^{5/2} \sqrt{e} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{4 \sqrt{2} d}+\frac{7 i a^2 \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}{4 d}-\frac{21 i a^{5/2} \sqrt{e} \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{8 \sqrt{2} d}+\frac{21 i a^{5/2} \sqrt{e} \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{8 \sqrt{2} d}+\frac{i a (a+i a \tan (c+d x))^{3/2} \sqrt{e \sec (c+d x)}}{2 d} \]

Antiderivative was successfully verified.

[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 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 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 \sqrt{e \sec (c+d x)} (a+i a \tan (c+d x))^{5/2} \, dx &=\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 ) \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}\\ &=\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 ) \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}-\frac{\left (21 i a^3 e\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}\\ &=\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 ) \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}-\frac{\left (21 i a^3\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}-\frac{\left (21 i a^{5/2} \sqrt{e}\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}-\frac{\left (21 i a^{5/2} \sqrt{e}\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}\\ &=-\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 ) \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}+\frac{\left (21 i a^{5/2} \sqrt{e}\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}\\ &=\frac{21 i a^{5/2} \sqrt{e} \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}-\frac{21 i a^{5/2} \sqrt{e} \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}-\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]  time = 2.56917, size = 397, normalized size = 0.97 \[ \frac{a^2 (\cos (2 d x)+i \sin (2 d x)) \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)} \left (\sqrt{-\sin (c)+i \cos (c)-1} \left (21 \sqrt{\sin (c)+i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i} \cos (c+d x) \tan ^{-1}\left (\frac{\sqrt{\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{\sin (c)+i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )+\sqrt{\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i} (-2 \tan (c+d x)+9 i)\right )-21 \sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i} \cos (c+d x) \tan ^{-1}\left (\frac{\sqrt{-\sin (c)+i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )\right )}{4 d \sqrt{-\sin (c)+i \cos (c)-1} \sqrt{\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i} (\cos (d x)+i \sin (d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.359, size = 371, normalized size = 0.9 \begin{align*}{\frac{{a}^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) }{8\,d \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) } \left ( 21\,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}-21\,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}+22\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +4\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) +21\,{\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}+21\,{\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}+22\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+18\,\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 ) \sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}\sqrt{{\frac{e}{\cos \left ( dx+c \right ) }}}{\frac{1}{\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/d*a^2*(cos(d*x+c)-1)*(21*I*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*cos(d*x+c)^2-21
*I*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+22*I*(1/(cos(d*x+c)+1))^(1/2)*
cos(d*x+c)*sin(d*x+c)+4*I*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+21*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+21*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+22*cos(d*x+c)^2*(1/(cos(d*x+c)+1))^(1/2)+18*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)-4*(1/(cos(d*x+c)+1))^(1/2))*
(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(e/cos(d*x+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c)-1)/(1/(cos(d*x+c)
+1))^(1/2)/cos(d*x+c)/sin(d*x+c)

________________________________________________________________________________________

Maxima [B]  time = 2.75543, size = 3291, normalized size = 8.01 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5632*a^2*cos(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 3584*a^2*cos(3/4*arctan2(sin(2*d*x + 2*c), co
s(2*d*x + 2*c))) + 5632*I*a^2*sin(7/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 3584*I*a^2*sin(3/4*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - (1344*sqrt(2)*a^2*cos(4*d*x + 4*c) + 2688*sqrt(2)*a^2*cos(2*d*x + 2*c
) + 1344*I*sqrt(2)*a^2*sin(4*d*x + 4*c) + 2688*I*sqrt(2)*a^2*sin(2*d*x + 2*c) + 1344*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) - (1344*sqrt(2)*a^2*cos(4*d*x + 4*c) + 2688*sqrt(2)*a^2*cos(2*d*x + 2*c) + 1344*I*sqrt(2)*a^
2*sin(4*d*x + 4*c) + 2688*I*sqrt(2)*a^2*sin(2*d*x + 2*c) + 1344*sqrt(2)*a^2)*arctan2(sqrt(2)*cos(1/4*arctan2(s
in(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) -
(1344*sqrt(2)*a^2*cos(4*d*x + 4*c) + 2688*sqrt(2)*a^2*cos(2*d*x + 2*c) + 1344*I*sqrt(2)*a^2*sin(4*d*x + 4*c) +
2688*I*sqrt(2)*a^2*sin(2*d*x + 2*c) + 1344*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) - (1344*sqrt(2)*a^2*cos
(4*d*x + 4*c) + 2688*sqrt(2)*a^2*cos(2*d*x + 2*c) + 1344*I*sqrt(2)*a^2*sin(4*d*x + 4*c) + 2688*I*sqrt(2)*a^2*s
in(2*d*x + 2*c) + 1344*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) + (1344*I*sqrt(2)*a^2*cos(4*d*x + 4*c) + 26
88*I*sqrt(2)*a^2*cos(2*d*x + 2*c) - 1344*sqrt(2)*a^2*sin(4*d*x + 4*c) - 2688*sqrt(2)*a^2*sin(2*d*x + 2*c) + 13
44*I*sqrt(2)*a^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*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + cos(1/2*arctan
2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 1) + (-1344*I*sqrt(2)*a^2*cos(4*d*x + 4*c) - 2688*I*sqrt(2)*a^2*cos(2
*d*x + 2*c) + 1344*sqrt(2)*a^2*sin(4*d*x + 4*c) + 2688*sqrt(2)*a^2*sin(2*d*x + 2*c) - 1344*I*sqrt(2)*a^2)*arct
an2(-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) + (672*sqrt(2)*a^2*cos(4*d*x + 4*c) + 1344*sqrt(2)*a^2*cos(2*d*x + 2*c) + 672*I*sqrt(2
)*a^2*sin(4*d*x + 4*c) + 1344*I*sqrt(2)*a^2*sin(2*d*x + 2*c) + 672*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(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) - (672*sqrt(2)*a^2*cos(4*d*x + 4*c
) + 1344*sqrt(2)*a^2*cos(2*d*x + 2*c) + 672*I*sqrt(2)*a^2*sin(4*d*x + 4*c) + 1344*I*sqrt(2)*a^2*sin(2*d*x + 2*
c) + 672*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(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 2*co
s(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) + (-672*I*sqrt(2)*a^2*cos(4*d*x + 4*c) - 1344*I*sqrt(2)*a^2*cos(2*d*x + 2*c) + 672*sqrt(2)*
a^2*sin(4*d*x + 4*c) + 1344*sqrt(2)*a^2*sin(2*d*x + 2*c) - 672*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*ar
ctan2(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) + (672*I*sqrt(2)*a^2*cos(4*d*x + 4*c) + 1344*I*sqrt(2)*a^2*cos(2*d*x + 2*c) - 672*sqrt(2)*a^2*sin(4*d*x + 4
*c) - 1344*sqrt(2)*a^2*sin(2*d*x + 2*c) + 672*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) + (-672*I*sqrt
(2)*a^2*cos(4*d*x + 4*c) - 1344*I*sqrt(2)*a^2*cos(2*d*x + 2*c) + 672*sqrt(2)*a^2*sin(4*d*x + 4*c) + 1344*sqrt(
2)*a^2*sin(2*d*x + 2*c) - 672*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) + (672*I*sqrt(2)*a^2*cos(4*d*x
+ 4*c) + 1344*I*sqrt(2)*a^2*cos(2*d*x + 2*c) - 672*sqrt(2)*a^2*sin(4*d*x + 4*c) - 1344*sqrt(2)*a^2*sin(2*d*x
+ 2*c) + 672*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(s
in(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*sqr
t(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))

________________________________________________________________________________________

Fricas [A]  time = 2.3617, size = 1701, normalized size = 4.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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^(2*I*d*x + 2*I*c) + 7*I*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^(3/2*I*d*x + 3/2*I*c) - sqrt(441/16*I*a^5*e/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(1/21*(42*(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^(3/2*I*d*x + 3/2*I*c)
+ 8*I*sqrt(441/16*I*a^5*e/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) + sqrt(441/16*I*a^5*e/d^2)*(d*
e^(2*I*d*x + 2*I*c) + d)*log(1/21*(42*(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^(3/2*I*d*x + 3/2*I*c) - 8*I*sqrt(441/16*I*a^5*e/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-
2*I*d*x - 2*I*c)/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^(3/2*I*d*x + 3/2*I*c) + 4*I
*sqrt(-441/16*I*a^5*e/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/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^(3/2*I*d*x + 3/2*I*c) - 4*I*sqrt(-441/16*I*a^5*e/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*
I*d*x - 2*I*c)/a^2))/(d*e^(2*I*d*x + 2*I*c) + d)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)