3.744 \(\int \frac{1}{\cot ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx\)
Optimal. Leaf size=234 \[ \frac{5 i \sqrt{\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}-\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}-\frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}-\frac{\sqrt{\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2} \]
[Out]
((-9/16 - (5*I)/16)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^2*d) + ((9/16 + (5*I)/16)*ArcTan[1 + Sq
rt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^2*d) + (((5*I)/8)*Sqrt[Cot[c + d*x]])/(a^2*d*(I + Cot[c + d*x])) - Sqrt[
Cot[c + d*x]]/(4*d*(I*a + a*Cot[c + d*x])^2) - ((9/32 - (5*I)/32)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c +
d*x]])/(Sqrt[2]*a^2*d) + ((9/32 - (5*I)/32)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a^2*
d)
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Rubi [A] time = 0.329752, antiderivative size = 234, normalized size of antiderivative = 1.,
number of steps used = 13, number of rules used = 10, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used
= {3673, 3559, 3596, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{5 i \sqrt{\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}-\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}-\frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}-\frac{\sqrt{\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2} \]
Antiderivative was successfully verified.
[In]
Int[1/(Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^2),x]
[Out]
((-9/16 - (5*I)/16)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^2*d) + ((9/16 + (5*I)/16)*ArcTan[1 + Sq
rt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^2*d) + (((5*I)/8)*Sqrt[Cot[c + d*x]])/(a^2*d*(I + Cot[c + d*x])) - Sqrt[
Cot[c + d*x]]/(4*d*(I*a + a*Cot[c + d*x])^2) - ((9/32 - (5*I)/32)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c +
d*x]])/(Sqrt[2]*a^2*d) + ((9/32 - (5*I)/32)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a^2*
d)
Rule 3673
Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] && !IntegerQ[m] && IntegersQ[n, p]
Rule 3559
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d))
, Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan
[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2
+ d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Rule 3596
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n,
0]
Rule 3534
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]
Rule 1168
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]
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}{\cot ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx &=\int \frac{1}{\sqrt{\cot (c+d x)} (i a+a \cot (c+d x))^2} \, dx\\ &=-\frac{\sqrt{\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac{\int \frac{-\frac{7 i a}{2}+\frac{3}{2} a \cot (c+d x)}{\sqrt{\cot (c+d x)} (i a+a \cot (c+d x))} \, dx}{4 a^2}\\ &=\frac{5 i \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}-\frac{\sqrt{\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac{\int \frac{-\frac{9 a^2}{2}-\frac{5}{2} i a^2 \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx}{8 a^4}\\ &=\frac{5 i \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}-\frac{\sqrt{\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac{\operatorname{Subst}\left (\int \frac{\frac{9 a^2}{2}+\frac{5}{2} i a^2 x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{4 a^4 d}\\ &=\frac{5 i \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}-\frac{\sqrt{\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac{\left (\frac{9}{16}-\frac{5 i}{16}\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a^2 d}+\frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a^2 d}\\ &=\frac{5 i \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}-\frac{\sqrt{\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac{\left (\frac{9}{32}+\frac{5 i}{32}\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a^2 d}+\frac{\left (\frac{9}{32}+\frac{5 i}{32}\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a^2 d}+-\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{\sqrt{2} a^2 d}+-\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{\sqrt{2} a^2 d}\\ &=\frac{5 i \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}-\frac{\sqrt{\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}-\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt{2} a^2 d}+-\frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a^2 d}\\ &=-\frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{9}{16}+\frac{5 i}{16}\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a^2 d}+\frac{5 i \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}-\frac{\sqrt{\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}-\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{9}{32}-\frac{5 i}{32}\right ) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt{2} a^2 d}\\ \end{align*}
Mathematica [A] time = 1.18814, size = 232, normalized size = 0.99 \[ \frac{\cot ^{\frac{3}{2}}(c+d x) \csc (c+d x) \sec ^2(c+d x) \left (5 i \sin (c+d x)+5 i \sin (3 (c+d x))-7 \cos (c+d x)+7 \cos (3 (c+d x))+(9+5 i) \sqrt{\sin (2 (c+d x))} \sin ^{-1}(\cos (c+d x)-\sin (c+d x)) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+(5+9 i) \sin ^{\frac{3}{2}}(2 (c+d x)) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )+(9-5 i) \sqrt{\sin (2 (c+d x))} \cos (2 (c+d x)) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )\right )}{32 a^2 d (\cot (c+d x)+i)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^2),x]
[Out]
(Cot[c + d*x]^(3/2)*Csc[c + d*x]*Sec[c + d*x]^2*(-7*Cos[c + d*x] + 7*Cos[3*(c + d*x)] + (5*I)*Sin[c + d*x] + (
9 - 5*I)*Cos[2*(c + d*x)]*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Sin[2*(c + d*x)]] + (
9 + 5*I)*ArcSin[Cos[c + d*x] - Sin[c + d*x]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*Sqrt[Sin[2*(c + d*x)]] +
(5 + 9*I)*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sin[2*(c + d*x)]^(3/2) + (5*I)*Sin[3*(c +
d*x)]))/(32*a^2*d*(I + Cot[c + d*x])^2)
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Maple [C] time = 0.282, size = 2545, normalized size = 10.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^2,x)
[Out]
1/16/a^2/d*2^(1/2)*(cos(d*x+c)-1)*(5*I*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*si
n(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*
x+c))/sin(d*x+c))^(1/2)+14*I*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*c
os(d*x+c)^2*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d
*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)-5*I*cos(d*x+c)^2*2^(1/2)-10*I*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(
d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)^2*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c)
)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)-4*I*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^
(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-
1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)+14*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/si
n(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)^3*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c
))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)-14*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/s
in(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)^2*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)
-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)-4*(-(cos(d*x+c)-1-sin(d*x+c))/s
in(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(
cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)^3-4*((cos(d*x+c)-1)/sin(d*x+c))^(
1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*E
llipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*x+c)-10*EllipticF((-(cos(
d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)^3*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c
)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)-14*I*EllipticPi((-(cos(d*x+c)-
1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+
c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)-7*I*EllipticPi((-(cos(d*x+c)-
1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+
c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)+14*I*EllipticPi((-(cos(d*x+c)
-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)^3*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d
*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)-4*I*EllipticPi((-(cos(d*x+
c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)^2*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^
(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)+5*I*cos(d*x+c
)^3*2^(1/2)+2*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1
-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))
*sin(d*x+c)-14*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-
sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*
cos(d*x+c)+7*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-si
n(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*si
n(d*x+c)+4*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*
x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(
d*x+c)+2*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+
c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*
x+c)+10*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)*((cos(d*x+c)-1)/sin(d*
x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)+4*I*Ell
ipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)^3*((cos(d*x+c)-1)/sin(
d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)-7*sin
(d*x+c)*2^(1/2)*cos(d*x+c)^2+7*cos(d*x+c)*sin(d*x+c)*2^(1/2))*cos(d*x+c)^2*(cos(d*x+c)+1)^2/(2*I*cos(d*x+c)*si
n(d*x+c)+2*cos(d*x+c)^2-1)/(cos(d*x+c)/sin(d*x+c))^(5/2)/sin(d*x+c)^6
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
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Fricas [B] time = 1.71443, size = 1424, normalized size = 6.09 \begin{align*} \frac{{\left (4 \, a^{2} d \sqrt{\frac{i}{16 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (2 \,{\left (4 \,{\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt{\frac{i}{16 \, a^{4} d^{2}}} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - 4 \, a^{2} d \sqrt{\frac{i}{16 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-2 \,{\left (4 \,{\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt{\frac{i}{16 \, a^{4} d^{2}}} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 4 \, a^{2} d \sqrt{-\frac{49 i}{64 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac{{\left (8 \,{\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt{-\frac{49 i}{64 \, a^{4} d^{2}}} + 7 i\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) - 4 \, a^{2} d \sqrt{-\frac{49 i}{64 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac{{\left (8 \,{\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} d\right )} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt{-\frac{49 i}{64 \, a^{4} d^{2}}} - 7 i\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}{\left (6 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 7 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
[Out]
1/16*(4*a^2*d*sqrt(1/16*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(2*(4*(a^2*d*e^(2*I*d*x + 2*I*c) - a^2*d)*sqrt((I*
e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(1/16*I/(a^4*d^2)) + I*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*
x - 2*I*c)) - 4*a^2*d*sqrt(1/16*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(-2*(4*(a^2*d*e^(2*I*d*x + 2*I*c) - a^2*d)
*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(1/16*I/(a^4*d^2)) - I*e^(2*I*d*x + 2*I*c))*e
^(-2*I*d*x - 2*I*c)) + 4*a^2*d*sqrt(-49/64*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(1/8*(8*(a^2*d*e^(2*I*d*x + 2*I
*c) - a^2*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-49/64*I/(a^4*d^2)) + 7*I)*e^(-2
*I*d*x - 2*I*c)/(a^2*d)) - 4*a^2*d*sqrt(-49/64*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(-1/8*(8*(a^2*d*e^(2*I*d*x
+ 2*I*c) - a^2*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-49/64*I/(a^4*d^2)) - 7*I)*
e^(-2*I*d*x - 2*I*c)/(a^2*d)) + sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(6*e^(4*I*d*x + 4*
I*c) - 7*e^(2*I*d*x + 2*I*c) + 1))*e^(-4*I*d*x - 4*I*c)/(a^2*d)
________________________________________________________________________________________
Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)**(5/2)/(a+I*a*tan(d*x+c))**2,x)
[Out]
Exception raised: AttributeError
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \cot \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
[Out]
integrate(1/((I*a*tan(d*x + c) + a)^2*cot(d*x + c)^(5/2)), x)