3.430 \(\int \frac{(e \sec (c+d x))^{9/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=411 \[ -\frac{5 i e^{9/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} a^{5/2} d}+\frac{5 i e^{9/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} a^{5/2} d}+\frac{5 i e^4 \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}{a^3 d}+\frac{5 i e^{9/2} \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} a^{5/2} d}-\frac{5 i e^{9/2} \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} a^{5/2} d}+\frac{4 i e^2 (e \sec (c+d x))^{5/2}}{a d (a+i a \tan (c+d x))^{3/2}} \]
[Out]
((-5*I)*e^(9/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])])/(Sqrt
[2]*a^(5/2)*d) + ((5*I)*e^(9/2)*ArcTan[1 + (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c
+ d*x]])])/(Sqrt[2]*a^(5/2)*d) + (((5*I)/2)*e^(9/2)*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]*a^(5/2)*d) - (((5*I)/2)*e^(9/2)*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]*a^(5/2)*d) + ((4*I)*e^2*(e*Sec[c + d*x])^(5/2))/(a*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((5*I)*e^
4*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(a^3*d)
________________________________________________________________________________________
Rubi [A] time = 0.433811, antiderivative size = 411, normalized size of antiderivative = 1.,
number of steps used = 12, number of rules used = 9, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used =
{3500, 3501, 3495, 297, 1162, 617, 204, 1165, 628} \[ -\frac{5 i e^{9/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} a^{5/2} d}+\frac{5 i e^{9/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} a^{5/2} d}+\frac{5 i e^4 \sqrt{a+i a \tan (c+d x)} \sqrt{e \sec (c+d x)}}{a^3 d}+\frac{5 i e^{9/2} \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} a^{5/2} d}-\frac{5 i e^{9/2} \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} a^{5/2} d}+\frac{4 i e^2 (e \sec (c+d x))^{5/2}}{a d (a+i a \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(e*Sec[c + d*x])^(9/2)/(a + I*a*Tan[c + d*x])^(5/2),x]
[Out]
((-5*I)*e^(9/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c + d*x]])])/(Sqrt
[2]*a^(5/2)*d) + ((5*I)*e^(9/2)*ArcTan[1 + (Sqrt[2]*Sqrt[e]*Sqrt[a + I*a*Tan[c + d*x]])/(Sqrt[a]*Sqrt[e*Sec[c
+ d*x]])])/(Sqrt[2]*a^(5/2)*d) + (((5*I)/2)*e^(9/2)*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]*a^(5/2)*d) - (((5*I)/2)*e^(9/2)*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]*a^(5/2)*d) + ((4*I)*e^2*(e*Sec[c + d*x])^(5/2))/(a*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((5*I)*e^
4*Sqrt[e*Sec[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(a^3*d)
Rule 3500
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*d^2
*(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + 2*n)), x] - Dist[(d^2*(m - 2))/(b^2*(m + 2*n
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]
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))^{9/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{4 i e^2 (e \sec (c+d x))^{5/2}}{a d (a+i a \tan (c+d x))^{3/2}}-\frac{\left (5 e^2\right ) \int \frac{(e \sec (c+d x))^{5/2}}{\sqrt{a+i a \tan (c+d x)}} \, dx}{a^2}\\ &=\frac{4 i e^2 (e \sec (c+d x))^{5/2}}{a d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i e^4 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a^3 d}-\frac{\left (5 e^4\right ) \int \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)} \, dx}{2 a^3}\\ &=\frac{4 i e^2 (e \sec (c+d x))^{5/2}}{a d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i e^4 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a^3 d}+\frac{\left (10 i e^6\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 )}{a^2 d}\\ &=\frac{4 i e^2 (e \sec (c+d x))^{5/2}}{a d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i e^4 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a^3 d}-\frac{\left (5 i e^5\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 )}{a^2 d}+\frac{\left (5 i e^5\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 )}{a^2 d}\\ &=\frac{4 i e^2 (e \sec (c+d x))^{5/2}}{a d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i e^4 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a^3 d}+\frac{\left (5 i e^4\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 a^2 d}+\frac{\left (5 i e^4\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 a^2 d}+\frac{\left (5 i e^{9/2}\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} a^{5/2} d}+\frac{\left (5 i e^{9/2}\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} a^{5/2} d}\\ &=\frac{5 i e^{9/2} \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} a^{5/2} d}-\frac{5 i e^{9/2} \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} a^{5/2} d}+\frac{4 i e^2 (e \sec (c+d x))^{5/2}}{a d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i e^4 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a^3 d}+\frac{\left (5 i e^{9/2}\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} a^{5/2} d}-\frac{\left (5 i e^{9/2}\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} a^{5/2} d}\\ &=-\frac{5 i e^{9/2} \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} a^{5/2} d}+\frac{5 i e^{9/2} \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} a^{5/2} d}+\frac{5 i e^{9/2} \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} a^{5/2} d}-\frac{5 i e^{9/2} \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} a^{5/2} d}+\frac{4 i e^2 (e \sec (c+d x))^{5/2}}{a d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i e^4 \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)}}{a^3 d}\\ \end{align*}
Mathematica [A] time = 5.43652, size = 369, normalized size = 0.9 \[ \frac{e^2 (\cos (d x)+i \sin (d x))^3 (e \sec (c+d x))^{5/2} \left (-\frac{5 \sqrt{\sin (c)+i \cos (c)-1} (\cos (3 c)+i \sin (3 c)) \sqrt{\tan \left (\frac{d x}{2}\right )+i} \tan ^{-1}\left (\frac{\sqrt{\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{\sin (c)+i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )}{\sqrt{\sin (c)-i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}+\frac{5 \sqrt{-\sin (c)-i \cos (c)-1} (\cos (3 c)+i \sin (3 c)) \sqrt{\tan \left (\frac{d x}{2}\right )+i} \tan ^{-1}\left (\frac{\sqrt{-\sin (c)+i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}{\sqrt{-\sin (c)-i \cos (c)-1} \sqrt{\tan \left (\frac{d x}{2}\right )+i}}\right )}{\sqrt{-\sin (c)+i \cos (c)-1} \sqrt{-\tan \left (\frac{d x}{2}\right )+i}}-(\tan (c+d x)-9 i) (\cos (2 c-d x)+i \sin (2 c-d x))\right )}{d (a+i a \tan (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(e*Sec[c + d*x])^(9/2)/(a + I*a*Tan[c + d*x])^(5/2),x]
[Out]
(e^2*(e*Sec[c + d*x])^(5/2)*(Cos[d*x] + I*Sin[d*x])^3*((5*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]])]*Sqrt[-1 - I*Cos[c] - Sin[c]]*(Cos[3*c] + I*Sin
[3*c])*Sqrt[I + Tan[(d*x)/2]])/(Sqrt[-1 + I*Cos[c] - Sin[c]]*Sqrt[I - Tan[(d*x)/2]]) - (5*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]])]*Sqrt[-1 + I*Co
s[c] + Sin[c]]*(Cos[3*c] + I*Sin[3*c])*Sqrt[I + Tan[(d*x)/2]])/(Sqrt[-1 - I*Cos[c] + Sin[c]]*Sqrt[I - Tan[(d*x
)/2]]) - (Cos[2*c - d*x] + I*Sin[2*c - d*x])*(-9*I + Tan[c + d*x])))/(d*(a + I*a*Tan[c + d*x])^(5/2))
________________________________________________________________________________________
Maple [B] time = 0.346, size = 1439, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*sec(d*x+c))^(9/2)/(a+I*a*tan(d*x+c))^(5/2),x)
[Out]
-1/4/d/a^3*(cos(d*x+c)-1)^4*(10*cos(d*x+c)^2*sin(d*x+c)*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin
(d*x+c)))-10*cos(d*x+c)^2*sin(d*x+c)*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))+5*cos(d*x
+c)*sin(d*x+c)*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))-5*cos(d*x+c)*sin(d*x+c)*arctanh
(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))+20*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1
-sin(d*x+c)))*cos(d*x+c)^4+20*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*cos(d*x+c)^4-80*
(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^4-10*cos(d*x+c)^3*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d
*x+c)))-10*cos(d*x+c)^3*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))+84*cos(d*x+c)^2*(1/(co
s(d*x+c)+1))^(1/2)+5*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*cos(d*x+c)+5*arctanh(1/2*
(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*cos(d*x+c)-20*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*
x+c)+1-sin(d*x+c)))*cos(d*x+c)^3*sin(d*x+c)+20*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)^3*sin(d*x+c)+20*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)^3*s
in(d*x+c)-80*I*(1/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3*sin(d*x+c)-10*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*sin(d*x+c)-10*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*sin(d*x+c)-5*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)*s
in(d*x+c)-5*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)*sin(d*x+c)+44*I*(1/(c
os(d*x+c)+1))^(1/2)*cos(d*x+c)*sin(d*x+c)-4*(1/(cos(d*x+c)+1))^(1/2)+20*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(
cos(d*x+c)+1+sin(d*x+c)))*cos(d*x+c)^3*sin(d*x+c)+20*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)^4-20*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)^4-10*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)^3+10*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)^3-15*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+15*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+5*I*arct
anh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*cos(d*x+c)-5*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)-15*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*cos
(d*x+c)^2-15*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)*(a*(I*sin(d*x+c)+co
s(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^4*(e/cos(d*x+c))^(9/2)/(4*I*sin(d*x+c)*cos(d*x+c)^2+4*cos(d*x+c)^3-I*si
n(d*x+c)-3*cos(d*x+c))/(1/(cos(d*x+c)+1))^(9/2)/sin(d*x+c)^9
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Maxima [B] time = 2.25316, size = 3312, normalized size = 8.06 \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))^(9/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
-(64*e^4*cos(1/2*d*x + 1/2*c)^2 + 64*e^4*sin(1/2*d*x + 1/2*c)^2 + 16*e^4 + (10*I*sqrt(2)*e^4*cos(3/2*d*x + 3/2
*c) + 10*I*sqrt(2)*e^4*cos(1/2*d*x + 1/2*c) + 10*sqrt(2)*e^4*sin(3/2*d*x + 3/2*c) - 10*sqrt(2)*e^4*sin(1/2*d*x
+ 1/2*c))*arctan2(sqrt(2)*sin(1/2*d*x + 1/2*c) + sin(d*x + c), sqrt(2)*cos(1/2*d*x + 1/2*c) + cos(d*x + c) +
1) + (-10*I*sqrt(2)*e^4*cos(3/2*d*x + 3/2*c) - 10*I*sqrt(2)*e^4*cos(1/2*d*x + 1/2*c) - 10*sqrt(2)*e^4*sin(3/2*
d*x + 3/2*c) + 10*sqrt(2)*e^4*sin(1/2*d*x + 1/2*c))*arctan2(-sqrt(2)*sin(1/2*d*x + 1/2*c) + sin(d*x + c), -sqr
t(2)*cos(1/2*d*x + 1/2*c) + cos(d*x + c) + 1) - (10*sqrt(2)*e^4*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, sqrt
(2)*sin(1/2*d*x + 1/2*c) + 1) + 10*sqrt(2)*e^4*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, -sqrt(2)*sin(1/2*d*x
+ 1/2*c) + 1) + 10*sqrt(2)*e^4*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 1
0*sqrt(2)*e^4*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) - 5*I*sqrt(2)*e^4*l
og(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*
x + 1/2*c) + 2) + 5*I*sqrt(2)*e^4*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*
d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 5*I*sqrt(2)*e^4*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*
d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 5*I*sqrt(2)*e^4*log(2*
cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1
/2*c) + 2) - 64*e^4*cos(1/2*d*x + 1/2*c) + 64*I*e^4*sin(1/2*d*x + 1/2*c))*cos(3/2*d*x + 3/2*c) - (10*sqrt(2)*e
^4*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 10*sqrt(2)*e^4*arctan2(sqrt(2
)*cos(1/2*d*x + 1/2*c) + 1, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 10*sqrt(2)*e^4*arctan2(sqrt(2)*cos(1/2*d*x +
1/2*c) - 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 10*sqrt(2)*e^4*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1, -sqrt
(2)*sin(1/2*d*x + 1/2*c) + 1) - 5*I*sqrt(2)*e^4*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sq
rt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 5*I*sqrt(2)*e^4*log(2*cos(1/2*d*x + 1/2*c)^
2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 5*I*sqrt
(2)*e^4*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*s
in(1/2*d*x + 1/2*c) + 2) + 5*I*sqrt(2)*e^4*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)
*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(1/2*d*x + 1/2*c) - (5*sqrt(2)*e^4*cos(3/2*d*x
+ 3/2*c) + 5*sqrt(2)*e^4*cos(1/2*d*x + 1/2*c) - 5*I*sqrt(2)*e^4*sin(3/2*d*x + 3/2*c) + 5*I*sqrt(2)*e^4*sin(1/
2*d*x + 1/2*c))*log(2*sqrt(2)*sin(d*x + c)*sin(1/2*d*x + 1/2*c) + 2*(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1)*cos(d*x
+ c) + cos(d*x + c)^2 + 2*cos(1/2*d*x + 1/2*c)^2 + sin(d*x + c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(
1/2*d*x + 1/2*c) + 1) + (5*sqrt(2)*e^4*cos(3/2*d*x + 3/2*c) + 5*sqrt(2)*e^4*cos(1/2*d*x + 1/2*c) - 5*I*sqrt(2)
*e^4*sin(3/2*d*x + 3/2*c) + 5*I*sqrt(2)*e^4*sin(1/2*d*x + 1/2*c))*log(-2*sqrt(2)*sin(d*x + c)*sin(1/2*d*x + 1/
2*c) - 2*(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1)*cos(d*x + c) + cos(d*x + c)^2 + 2*cos(1/2*d*x + 1/2*c)^2 + sin(d*x
+ c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 1) + (10*I*sqrt(2)*e^4*arctan2(sqrt(2)*c
os(1/2*d*x + 1/2*c) + 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 10*I*sqrt(2)*e^4*arctan2(sqrt(2)*cos(1/2*d*x + 1/
2*c) + 1, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) + 10*I*sqrt(2)*e^4*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1, sqrt
(2)*sin(1/2*d*x + 1/2*c) + 1) + 10*I*sqrt(2)*e^4*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1, -sqrt(2)*sin(1/2*d*
x + 1/2*c) + 1) + 5*sqrt(2)*e^4*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*
x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 5*sqrt(2)*e^4*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x
+ 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 5*sqrt(2)*e^4*log(2*cos(1/
2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c)
+ 2) - 5*sqrt(2)*e^4*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c)
- 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 64*I*e^4*cos(1/2*d*x + 1/2*c) - 64*e^4*sin(1/2*d*x + 1/2*c))*sin(3/2*d
*x + 3/2*c) + (-10*I*sqrt(2)*e^4*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) -
10*I*sqrt(2)*e^4*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) + 1, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) - 10*I*sqrt(2)*
e^4*arctan2(sqrt(2)*cos(1/2*d*x + 1/2*c) - 1, sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) - 10*I*sqrt(2)*e^4*arctan2(sqr
t(2)*cos(1/2*d*x + 1/2*c) - 1, -sqrt(2)*sin(1/2*d*x + 1/2*c) + 1) - 5*sqrt(2)*e^4*log(2*cos(1/2*d*x + 1/2*c)^2
+ 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 5*sqrt(2)
*e^4*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(
1/2*d*x + 1/2*c) + 2) - 5*sqrt(2)*e^4*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(
1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 5*sqrt(2)*e^4*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/
2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*sin(1/2*d*x + 1/2*c))
*sqrt(a)*sqrt(e)/((8*I*a^3*cos(3/2*d*x + 3/2*c) + 8*I*a^3*cos(1/2*d*x + 1/2*c) + 8*a^3*sin(3/2*d*x + 3/2*c) -
8*a^3*sin(1/2*d*x + 1/2*c))*d)
________________________________________________________________________________________
Fricas [B] time = 2.51361, size = 1704, normalized size = 4.15 \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))^(9/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
-1/2*(sqrt(25*I*e^9/(a^5*d^2))*a^3*d*e^(2*I*d*x + 2*I*c)*log(-2/5*(I*sqrt(25*I*e^9/(a^5*d^2))*a^3*d*e^(2*I*d*x
+ 2*I*c) - 5*(e^4*e^(2*I*d*x + 2*I*c) + e^4)*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^4) - sqrt(25*I*e^9/(a^5*d^2))*a^3*d*e^(2*I*d*x + 2*I*c)*lo
g(-2/5*(-I*sqrt(25*I*e^9/(a^5*d^2))*a^3*d*e^(2*I*d*x + 2*I*c) - 5*(e^4*e^(2*I*d*x + 2*I*c) + e^4)*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^4) + s
qrt(-25*I*e^9/(a^5*d^2))*a^3*d*e^(2*I*d*x + 2*I*c)*log(-2/5*(I*sqrt(-25*I*e^9/(a^5*d^2))*a^3*d*e^(2*I*d*x + 2*
I*c) - 5*(e^4*e^(2*I*d*x + 2*I*c) + e^4)*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^4) - sqrt(-25*I*e^9/(a^5*d^2))*a^3*d*e^(2*I*d*x + 2*I*c)*log(-2
/5*(-I*sqrt(-25*I*e^9/(a^5*d^2))*a^3*d*e^(2*I*d*x + 2*I*c) - 5*(e^4*e^(2*I*d*x + 2*I*c) + e^4)*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^4) - 2*(1
0*I*e^4*e^(2*I*d*x + 2*I*c) + 8*I*e^4)*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)/(a^3*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))**(9/2)/(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 \frac{\left (e \sec \left (d x + c\right )\right )^{\frac{9}{2}}}{{\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))^(9/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((e*sec(d*x + c))^(9/2)/(I*a*tan(d*x + c) + a)^(5/2), x)