3.686 \(\int \frac{1}{(e \cos (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}} \, dx\)
Optimal. Leaf size=470 \[ \frac{i e^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} \sqrt{a} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{i e^{5/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} \sqrt{a} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{i e^{5/2} \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} \sqrt{a} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac{i e^{5/2} \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} \sqrt{a} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{a d (e \cos (c+d x))^{5/2}} \]
[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*(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/2)) - (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*(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/
2)) - ((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*(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/2)) + ((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*(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/2)) - (I*Cos[c + d*x]^2*Sqrt[a
+ I*a*Tan[c + d*x]])/(a*d*(e*Cos[c + d*x])^(5/2))
________________________________________________________________________________________
Rubi [A] time = 0.440715, antiderivative size = 470, 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 =
{3515, 3501, 3495, 297, 1162, 617, 204, 1165, 628} \[ \frac{i e^{5/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} \sqrt{a} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{i e^{5/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{a} \sqrt{e \sec (c+d x)}}\right )}{\sqrt{2} \sqrt{a} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{i e^{5/2} \log \left (-\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} \sqrt{a} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\frac{i e^{5/2} \log \left (\frac{\sqrt{2} \sqrt{a} \sqrt{e} \sqrt{a+i a \tan (c+d x)}}{\sqrt{e \sec (c+d x)}}+\cos (c+d x) (a+i a \tan (c+d x))+a\right )}{2 \sqrt{2} \sqrt{a} d (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{a d (e \cos (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[1/((e*Cos[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*(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/2)) - (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*(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/
2)) - ((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*(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/2)) + ((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*(e*Cos[c + d*x])^(5/2)*(e*Sec[c + d*x])^(5/2)) - (I*Cos[c + d*x]^2*Sqrt[a
+ I*a*Tan[c + d*x]])/(a*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 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{1}{(e \cos (c+d x))^{5/2} \sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{\int \frac{(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 \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{a d (e \cos (c+d x))^{5/2}}+\frac{e^2 \int \sqrt{e \sec (c+d x)} \sqrt{a+i a \tan (c+d x)} \, dx}{2 a (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=-\frac{i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{a d (e \cos (c+d x))^{5/2}}-\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=-\frac{i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{a d (e \cos (c+d x))^{5/2}}+\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=-\frac{i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{a d (e \cos (c+d x))^{5/2}}-\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=-\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{a d (e \cos (c+d x))^{5/2}}-\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}\\ &=\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}+\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 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}}-\frac{i \cos ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{a d (e \cos (c+d x))^{5/2}}\\ \end{align*}
Mathematica [A] time = 1.6913, size = 250, normalized size = 0.53 \[ \frac{i e^{i c-\frac{i d x}{2}} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (\left (-e^{-2 i c}\right )^{3/4} \left (1+e^{2 i (c+d x)}\right ) \tan ^{-1}\left (\frac{e^{\frac{i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )-\left (-e^{-2 i c}\right )^{3/4} \left (1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac{e^{\frac{i d x}{2}}}{\sqrt [4]{-e^{-2 i c}}}\right )-2 e^{\frac{3 i d x}{2}}\right )}{d \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \sqrt{\cos (c+d x)} \sec ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[1/((e*Cos[c + d*x])^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]
[Out]
(I*E^(I*c - (I/2)*d*x)*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*(-2*E^(((3*I)/2)*d*x) + (-E^((-2*I)*c))
^(3/4)*(1 + E^((2*I)*(c + d*x)))*ArcTan[E^((I/2)*d*x)/(-E^((-2*I)*c))^(1/4)] - (-E^((-2*I)*c))^(3/4)*(1 + E^((
2*I)*(c + d*x)))*ArcTanh[E^((I/2)*d*x)/(-E^((-2*I)*c))^(1/4)]))/(d*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*
x))]*Sqrt[Cos[c + d*x]]*(e*Cos[c + d*x])^(5/2)*Sec[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]])
________________________________________________________________________________________
Maple [A] time = 0.356, size = 313, normalized size = 0.7 \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 ) }\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( i\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 ) +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}}+\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 ) -{\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}}} \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(1/(e*cos(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x)
[Out]
-1/2/d/a*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*cos(d*x+c)*arctanh(1/
2*(1/(cos(d*x+c)+1))^(1/2)*(-cos(d*x+c)-1+sin(d*x+c)))+I*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+si
n(d*x+c)))*cos(d*x+c)+2*I*sin(d*x+c)*(1/(cos(d*x+c)+1))^(1/2)+cos(d*x+c)*arctanh(1/2*(1/(cos(d*x+c)+1))^(1/2)*
(-cos(d*x+c)-1+sin(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)/(e*cos(d*x+c))^(5/2)
________________________________________________________________________________________
Maxima [B] time = 3.32576, size = 2909, normalized size = 6.19 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*cos(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
-(16*(sqrt(2)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + I*sqrt(2)*sin(4/3*arctan2(sin(3/2
*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sqrt(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)*c
os(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + I*sqrt(2)*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c),
cos(3/2*d*x + 3/2*c))) + sqrt(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)*cos(4/3*arctan2
(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + I*sqrt(2)*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x +
3/2*c))) + sqrt(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)*cos(4/3*arctan2(sin(3/2*d*x +
3/2*c), cos(3/2*d*x + 3/2*c))) + I*sqrt(2)*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sqrt
(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*arctan
2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - (16*I*sqrt(2)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(
3/2*d*x + 3/2*c))) - 16*sqrt(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))*a
rctan2(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*a
rctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - (-16*I*sqrt(2)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c)
, cos(3/2*d*x + 3/2*c))) + 16*sqrt(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))*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))) + c
os(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - 8*(sqrt(2)*cos(4/3*arctan2(sin(3/2*d*x + 3/
2*c), cos(3/2*d*x + 3/2*c))) + I*sqrt(2)*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sqrt(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)*co
s(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*sqr
t(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) + 8*(sqrt(2)*cos(4/3*arctan2(sin(3/2*d*
x + 3/2*c), cos(3/2*d*x + 3/2*c))) + I*sqrt(2)*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) +
sqrt(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)*cos(4/3*arctan2(
sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 8*sqrt(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))*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*arctan
2(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)*cos(4
/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 8*sqrt(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))*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*s
qrt(2)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 8*sqrt(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))*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)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 8*sqrt(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))*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) + 128*cos(3/2*d*x + 3/2*c) + 128*I*sin(3/2*d*x + 3/2*c))*sqrt(a)*sqrt(e)/((-64*I*a*e^3*cos(4/3*a
rctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 64*a*e^3*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d
*x + 3/2*c))) - 64*I*a*e^3)*d)
________________________________________________________________________________________
Fricas [A] time = 2.25852, size = 1331, normalized size = 2.83 \begin{align*} \frac{-4 i \, \sqrt{2} \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac{3}{2} i \, d x + \frac{3}{2} i \, c\right )} +{\left (a d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a d e^{3}\right )} \sqrt{\frac{i}{a d^{2} e^{5}}} \log \left (a d e^{3} \sqrt{\frac{i}{a d^{2} e^{5}}} + \sqrt{2} \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )}\right ) -{\left (a d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a d e^{3}\right )} \sqrt{\frac{i}{a d^{2} e^{5}}} \log \left (-a d e^{3} \sqrt{\frac{i}{a d^{2} e^{5}}} + \sqrt{2} \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )}\right ) -{\left (a d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a d e^{3}\right )} \sqrt{-\frac{i}{a d^{2} e^{5}}} \log \left (a d e^{3} \sqrt{-\frac{i}{a d^{2} e^{5}}} + \sqrt{2} \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )}\right ) +{\left (a d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a d e^{3}\right )} \sqrt{-\frac{i}{a d^{2} e^{5}}} \log \left (-a d e^{3} \sqrt{-\frac{i}{a d^{2} e^{5}}} + \sqrt{2} \sqrt{\frac{1}{2}} \sqrt{e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac{1}{2} i \, d x + \frac{1}{2} i \, c\right )}\right )}{2 \,{\left (a d e^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a d e^{3}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*cos(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
1/2*(-4*I*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^(3/2*I*d*x + 3
/2*I*c) + (a*d*e^3*e^(2*I*d*x + 2*I*c) + a*d*e^3)*sqrt(I/(a*d^2*e^5))*log(a*d*e^3*sqrt(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)) - (a*d*
e^3*e^(2*I*d*x + 2*I*c) + a*d*e^3)*sqrt(I/(a*d^2*e^5))*log(-a*d*e^3*sqrt(I/(a*d^2*e^5)) + sqrt(2)*sqrt(1/2)*sq
rt(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)) - (a*d*e^3*e^(2*I*d*x
+ 2*I*c) + a*d*e^3)*sqrt(-I/(a*d^2*e^5))*log(a*d*e^3*sqrt(-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)) + (a*d*e^3*e^(2*I*d*x + 2*I*c) + a
*d*e^3)*sqrt(-I/(a*d^2*e^5))*log(-a*d*e^3*sqrt(-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)))/(a*d*e^3*e^(2*I*d*x + 2*I*c) + a*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(1/(e*cos(d*x+c))**(5/2)/(a+I*a*tan(d*x+c))**(1/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (e \cos \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(1/(e*cos(d*x+c))^(5/2)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate(1/((e*cos(d*x + c))^(5/2)*sqrt(I*a*tan(d*x + c) + a)), x)