3.416 \(\int \frac{(e \sec (c+d x))^{5/2}}{\sqrt{a+i a \tan (c+d x)}} \, dx\)
Optimal. Leaf size=369 \[ \frac{i e^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} \sqrt{a} d}-\frac{i e^{5/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} \sqrt{a} d}-\frac{i e^2 \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}{a d}-\frac{i e^{5/2} \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} \sqrt{a} d}+\frac{i e^{5/2} \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} \sqrt{a} d} \]
[Out]
(I*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]*S
qrt[a]*d) - (I*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]*Sqrt[a]*d) - ((I/2)*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]*Sqrt[a]*d) + ((I/2)*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]*S
qrt[a]*d) - (I*e^2*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(a*d)
________________________________________________________________________________________
Rubi [A] time = 0.307687, antiderivative size = 369, normalized size of antiderivative = 1.,
number of steps used = 11, number of rules used = 8, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used
= {3501, 3495, 297, 1162, 617, 204, 1165, 628} \[ \frac{i e^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} \sqrt{a} d}-\frac{i e^{5/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} \sqrt{a} d}-\frac{i e^2 \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}{a d}-\frac{i e^{5/2} \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} \sqrt{a} d}+\frac{i e^{5/2} \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} \sqrt{a} d} \]
Antiderivative was successfully verified.
[In]
Int[(e*Sec[c + d*x])^(5/2)/Sqrt[a + I*a*Tan[c + d*x]],x]
[Out]
(I*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]*S
qrt[a]*d) - (I*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]*Sqrt[a]*d) - ((I/2)*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]*Sqrt[a]*d) + ((I/2)*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]*S
qrt[a]*d) - (I*e^2*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(a*d)
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{(e \sec (c+d x))^{5/2}}{\sqrt{a+i a \tan (c+d x)}} \, dx &=-\frac{i e^2 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{e^2 \int \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)} \, dx}{2 a}\\ &=-\frac{i e^2 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}-\frac{\left (2 i 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 )}{d}\\ &=-\frac{i e^2 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}+\frac{\left (i 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 )}{d}-\frac{\left (i 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 )}{d}\\ &=-\frac{i e^2 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}-\frac{\left (i 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 )}{2 d}-\frac{\left (i 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 )}{2 d}-\frac{\left (i 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 )}{2 \sqrt{2} \sqrt{a} d}-\frac{\left (i 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 )}{2 \sqrt{2} \sqrt{a} d}\\ &=-\frac{i 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 )}{2 \sqrt{2} \sqrt{a} d}+\frac{i 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 )}{2 \sqrt{2} \sqrt{a} d}-\frac{i e^2 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}-\frac{\left (i 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 )}{\sqrt{2} \sqrt{a} d}+\frac{\left (i 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 )}{\sqrt{2} \sqrt{a} d}\\ &=\frac{i 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 )}{\sqrt{2} \sqrt{a} d}-\frac{i 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 )}{\sqrt{2} \sqrt{a} d}-\frac{i 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 )}{2 \sqrt{2} \sqrt{a} d}+\frac{i 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 )}{2 \sqrt{2} \sqrt{a} d}-\frac{i e^2 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a d}\\ \end{align*}
Mathematica [A] time = 1.85106, size = 366, normalized size = 0.99 \[ \frac{e^2 (\tan (c+d x)-i) \sqrt{e \sec (c+d x)} \left (\sqrt{-\sin (c)+i \cos (c)-1} \left (i \sqrt{\sin (c)+i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i} \cos (c+d x) \tan ^{-1}\left (\frac{\sqrt{\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{\sin (c)+i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )+\sqrt{\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}\right )-i \sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i} \cos (c+d x) \tan ^{-1}\left (\frac{\sqrt{-\sin (c)+i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )\right )}{d \sqrt{-\sin (c)+i \cos (c)-1} \sqrt{\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i} \sqrt{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(e*Sec[c + d*x])^(5/2)/Sqrt[a + I*a*Tan[c + d*x]],x]
[Out]
(e^2*Sqrt[e*Sec[c + d*x]]*((-I)*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[-1 - I*Cos[c] + Sin[c]]*S
qrt[I + Tan[(d*x)/2]] + Sqrt[-1 + I*Cos[c] - Sin[c]]*(Sqrt[-1 - I*Cos[c] + Sin[c]]*Sqrt[I - Tan[(d*x)/2]] + I*
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]]))*(-I + Tan[c + d*x]))/(d*Sqrt[-1 + I*C
os[c] - Sin[c]]*Sqrt[-1 - I*Cos[c] + Sin[c]]*Sqrt[I - Tan[(d*x)/2]]*Sqrt[a + I*a*Tan[c + d*x]])
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Maple [A] time = 0.35, size = 316, normalized size = 0.9 \begin{align*} -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ( \cos \left ( dx+c \right ) -1 \right ) ^{3}}{2\,ad \left ( \sin \left ( dx+c \right ) \right ) ^{5} \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) } \left ({\frac{e}{\cos \left ( dx+c \right ) }} \right ) ^{{\frac{5}{2}}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( 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 ) \cos \left ( dx+c \right ) -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 ) \cos \left ( dx+c \right ) +2\,i\sin \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-{\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 ) \cos \left ( dx+c \right ) -{\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 ) \cos \left ( dx+c \right ) +2\,\cos \left ( dx+c \right ) \sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+2\,\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}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x)
[Out]
-1/2/d/a*(e/cos(d*x+c))^(5/2)*cos(d*x+c)^2*(cos(d*x+c)-1)^3*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(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)-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*I*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)-arctanh(1/2*(1/(cos(d*x+c)+1
))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*cos(d*x+c)-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*cos(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)+2*(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)
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Maxima [B] time = 2.26565, size = 3069, normalized size = 8.32 \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))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
-(128*e^2*cos(3/2*d*x + 3/2*c) + 128*I*e^2*sin(3/2*d*x + 3/2*c) + (16*sqrt(2)*e^2*cos(4/3*arctan2(sin(3/2*d*x
+ 3/2*c), cos(3/2*d*x + 3/2*c))) + 16*I*sqrt(2)*e^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)
)) + 16*sqrt(2)*e^2)*arctan2(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1, sqrt(2)
*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + (16*sqrt(2)*e^2*cos(4/3*arctan2(sin(3/2*d
*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 16*I*sqrt(2)*e^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2
*c))) + 16*sqrt(2)*e^2)*arctan2(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1, -sqr
t(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + (16*sqrt(2)*e^2*cos(4/3*arctan2(sin(3
/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 16*I*sqrt(2)*e^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x +
3/2*c))) + 16*sqrt(2)*e^2)*arctan2(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 1,
sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + (16*sqrt(2)*e^2*cos(4/3*arctan2(si
n(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 16*I*sqrt(2)*e^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*
x + 3/2*c))) + 16*sqrt(2)*e^2)*arctan2(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) -
1, -sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - (16*I*sqrt(2)*e^2*cos(4/3*arct
an2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 16*sqrt(2)*e^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/
2*d*x + 3/2*c))) + 16*I*sqrt(2)*e^2)*arctan2(sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c
))) + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))), sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*
c), cos(3/2*d*x + 3/2*c))) + cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - (-16*I*sqrt(2
)*e^2*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 16*sqrt(2)*e^2*sin(4/3*arctan2(sin(3/2*d*
x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 16*I*sqrt(2)*e^2)*arctan2(-sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c),
cos(3/2*d*x + 3/2*c))) + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))), -sqrt(2)*cos(1/3*arctan
2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))
+ 1) - (8*sqrt(2)*e^2*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 8*I*sqrt(2)*e^2*sin(4/3*a
rctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 8*sqrt(2)*e^2)*log(2*sqrt(2)*sin(2/3*arctan2(sin(3/2*d*x
+ 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*(sqrt(2)*co
s(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1)*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*
d*x + 3/2*c))) + cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*cos(1/3*arctan2(sin(3/2*d*
x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(
1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), c
os(3/2*d*x + 3/2*c))) + 1) + (8*sqrt(2)*e^2*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 8*I
*sqrt(2)*e^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 8*sqrt(2)*e^2)*log(-2*sqrt(2)*sin(
2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3
/2*c))) - 2*(sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 1)*cos(2/3*arctan2(sin(3/2
*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*cos
(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*
x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(
sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - (-8*I*sqrt(2)*e^2*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), co
s(3/2*d*x + 3/2*c))) + 8*sqrt(2)*e^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 8*I*sqrt(2
)*e^2)*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x +
3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*
sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - (8*I*sqrt(2)*e^2*cos(4/3*arctan2(s
in(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 8*sqrt(2)*e^2*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x
+ 3/2*c))) + 8*I*sqrt(2)*e^2)*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3
*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(
3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - (-8*I*sqrt(
2)*e^2*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 8*sqrt(2)*e^2*sin(4/3*arctan2(sin(3/2*d*
x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 8*I*sqrt(2)*e^2)*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x
+ 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(si
n(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2
*c))) + 2) - (8*I*sqrt(2)*e^2*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 8*sqrt(2)*e^2*sin
(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 8*I*sqrt(2)*e^2)*log(2*cos(1/3*arctan2(sin(3/2*d*x
+ 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqr
t(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/
2*c), cos(3/2*d*x + 3/2*c))) + 2))*sqrt(a)*sqrt(e)/((-64*I*a*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x
+ 3/2*c))) + 64*a*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 64*I*a)*d)
________________________________________________________________________________________
Fricas [A] time = 2.29068, size = 1420, normalized size = 3.85 \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))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
1/2*(-4*I*e^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) + sq
rt(I*e^5/(a*d^2))*a*d*log(-2*(I*sqrt(I*e^5/(a*d^2))*a*d*e^(2*I*d*x + 2*I*c) - (e^2*e^(2*I*d*x + 2*I*c) + e^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))*e^(-2*I*d*x - 2*I
*c)/e^2) - sqrt(I*e^5/(a*d^2))*a*d*log(-2*(-I*sqrt(I*e^5/(a*d^2))*a*d*e^(2*I*d*x + 2*I*c) - (e^2*e^(2*I*d*x +
2*I*c) + e^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))*e^(
-2*I*d*x - 2*I*c)/e^2) + sqrt(-I*e^5/(a*d^2))*a*d*log(-2*(I*sqrt(-I*e^5/(a*d^2))*a*d*e^(2*I*d*x + 2*I*c) - (e^
2*e^(2*I*d*x + 2*I*c) + e^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))*e^(-2*I*d*x - 2*I*c)/e^2) - sqrt(-I*e^5/(a*d^2))*a*d*log(-2*(-I*sqrt(-I*e^5/(a*d^2))*a*d*e^(2*I*d*
x + 2*I*c) - (e^2*e^(2*I*d*x + 2*I*c) + e^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))*e^(-2*I*d*x - 2*I*c)/e^2))/(a*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))**(5/2)/(a+I*a*tan(d*x+c))**(1/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sec \left (d x + c\right )\right )^{\frac{5}{2}}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*sec(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((e*sec(d*x + c))^(5/2)/sqrt(I*a*tan(d*x + c) + a), x)