3.2.6 \(\int \cot ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [106]

Optimal. Leaf size=190 \[ \frac {45 i a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{8 d}-\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {19 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

[Out]

45/8*I*a^(5/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/d-4*I*a^(5/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^
(1/2)/a^(1/2))*2^(1/2)/d+19/8*a^2*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-13/12*I*a^2*cot(d*x+c)^2*(a+I*a*tan(d*
x+c))^(1/2)/d-1/3*a^2*cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.39, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3634, 3679, 3681, 3561, 212, 3680, 65, 214} \begin {gather*} \frac {45 i a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{8 d}-\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {19 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((45*I)/8)*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d - ((4*I)*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a + I
*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (19*a^2*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/(8*d) - (((13*I)/12)
*a^2*Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/d - (a^2*Cot[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/(3*d)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3561

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 3634

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x]
+ Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(
m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && Lt
Q[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3679

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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*
(n + 1)*(c^2 + d^2))), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3680

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b*(B/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]

Rule 3681

Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(A*b + a*B)/(b*c + a*d), Int[(a + b*Tan[e + f*x])^m, x], x] - Dist[(B*c
- A*d)/(b*c + a*d), Int[(a + b*Tan[e + f*x])^m*((a - b*Tan[e + f*x])/(c + d*Tan[e + f*x])), x], x] /; FreeQ[{
a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]

Rubi steps

\begin {align*} \int \cot ^4(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {1}{3} \int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {13 i a^2}{2}+\frac {11}{2} a^2 \tan (c+d x)\right ) \, dx\\ &=-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\int \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {57 a^3}{4}+\frac {39}{4} i a^3 \tan (c+d x)\right ) \, dx}{6 a}\\ &=\frac {19 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {135 i a^4}{8}-\frac {57}{8} a^4 \tan (c+d x)\right ) \, dx}{6 a^2}\\ &=\frac {19 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {1}{16} (45 i a) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx+\left (4 a^2\right ) \int \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {19 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\left (45 i a^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{16 d}-\frac {\left (8 i a^3\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}\\ &=-\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {19 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {\left (45 a^2\right ) \text {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{8 d}\\ &=\frac {45 i a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{8 d}-\frac {4 i \sqrt {2} a^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {19 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {13 i a^2 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {a^2 \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\\ \end {align*}

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Mathematica [A]
time = 2.65, size = 200, normalized size = 1.05 \begin {gather*} \frac {a^2 e^{-i (c+2 d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} (\cos (d x)+i \sin (d x)) \left (-384 i \sinh ^{-1}\left (e^{i (c+d x)}\right )+270 i \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {1+e^{2 i (c+d x)}}}\right )+\sqrt {1+e^{2 i (c+d x)}} \csc ^3(c+d x) (49-65 \cos (2 (c+d x))-26 i \sin (2 (c+d x)))\right )}{48 \sqrt {2} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(a^2*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(Cos[d*x] + I*Sin[d
*x])*((-384*I)*ArcSinh[E^(I*(c + d*x))] + (270*I)*Sqrt[2]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[1 + E^((2*I)*
(c + d*x))]] + Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c + d*x]^3*(49 - 65*Cos[2*(c + d*x)] - (26*I)*Sin[2*(c + d*x)
])))/(48*Sqrt[2]*d*E^(I*(c + 2*d*x)))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 925 vs. \(2 (153 ) = 306\).
time = 0.86, size = 926, normalized size = 4.87

method	result	size
default	\(\text {Expression too large to display}\)	\(926\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/48/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(-114*I*cos(d*x+c)*sin(d*x+c)-270*I*cos(d*x+c)^2*(-2*cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))+19
2*cos(d*x+c)^4*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2
^(1/2)+135*cos(d*x+c)^4*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+13
5*I*cos(d*x+c)^4*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(
d*x+c)+1)/sin(d*x+c))+192*I*cos(d*x+c)^4*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)-52*I*cos(d*x+c)^2*sin(d*x+c)-384*2^(1/2)*cos(d*x+c)^2*
(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+182*cos(d*x+c)^4
+182*I*cos(d*x+c)^3*sin(d*x+c)-270*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(
1+cos(d*x+c)))^(1/2))+135*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(
d*x+c)-cos(d*x+c)+1)/sin(d*x+c))-130*cos(d*x+c)^3-384*I*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arct
anh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+192*I*(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+192*2
^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-166*cos(d
*x+c)^2+135*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+114*cos(d*x+c)
)/(-1+cos(d*x+c))/(I*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c)/(1+cos(d*x+c))*a^2

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Maxima [A]
time = 0.50, size = 216, normalized size = 1.14 \begin {gather*} \frac {i \, a^{3} {\left (\frac {96 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{\sqrt {a}} - \frac {135 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{\sqrt {a}} + \frac {2 \, {\left (57 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 88 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 39 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} - 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a + 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - a^{3}}\right )}}{48 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/48*I*a^3*(96*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x
+ c) + a)))/sqrt(a) - 135*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/
sqrt(a) + 2*(57*(I*a*tan(d*x + c) + a)^(5/2) - 88*(I*a*tan(d*x + c) + a)^(3/2)*a + 39*sqrt(I*a*tan(d*x + c) +
a)*a^2)/((I*a*tan(d*x + c) + a)^3 - 3*(I*a*tan(d*x + c) + a)^2*a + 3*(I*a*tan(d*x + c) + a)*a^2 - a^3))/d

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 659 vs. \(2 (145) = 290\).
time = 0.44, size = 659, normalized size = 3.47 \begin {gather*} \frac {192 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{5}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 192 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {a^{5}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) + 135 \, \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {16 \, {\left (3 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} - 2 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (i \, d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) - 135 \, \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {16 \, {\left (3 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} - 2 \, \sqrt {2} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (-i \, d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) + 4 \, \sqrt {2} {\left (91 i \, a^{2} e^{\left (7 i \, d x + 7 i \, c\right )} - 7 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 59 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 39 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{96 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/96*(192*sqrt(2)*sqrt(-a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) -
d)*log(4*(a^3*e^(I*d*x + I*c) + sqrt(-a^5/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1
)))*e^(-I*d*x - I*c)/a^2) - 192*sqrt(2)*sqrt(-a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*
e^(2*I*d*x + 2*I*c) - d)*log(4*(a^3*e^(I*d*x + I*c) + sqrt(-a^5/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(
e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) + 135*sqrt(-a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x
+ 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)*log(16*(3*a^3*e^(2*I*d*x + 2*I*c) + a^3 - 2*sqrt(2)*sqrt(-a^5/d^2)*(I*
d*e^(3*I*d*x + 3*I*c) + I*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/a) - 135*
sqrt(-a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)*log(16*(3*a^3*e
^(2*I*d*x + 2*I*c) + a^3 - 2*sqrt(2)*sqrt(-a^5/d^2)*(-I*d*e^(3*I*d*x + 3*I*c) - I*d*e^(I*d*x + I*c))*sqrt(a/(e
^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/a) + 4*sqrt(2)*(91*I*a^2*e^(7*I*d*x + 7*I*c) - 7*I*a^2*e^(5*I*d
*x + 5*I*c) - 59*I*a^2*e^(3*I*d*x + 3*I*c) + 39*I*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e
^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3061 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^4, x)

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Mupad [B]
time = 0.22, size = 171, normalized size = 0.90 \begin {gather*} -\frac {19\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{8\,d\,{\mathrm {tan}\left (c+d\,x\right )}^3}-\frac {\mathrm {atan}\left (\frac {\sqrt {-a^5}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^3}\right )\,\sqrt {-a^5}\,45{}\mathrm {i}}{8\,d}-\frac {13\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{8\,d\,{\mathrm {tan}\left (c+d\,x\right )}^3}+\frac {11\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d\,{\mathrm {tan}\left (c+d\,x\right )}^3}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {-a^5}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,a^3}\right )\,\sqrt {-a^5}\,4{}\mathrm {i}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

(11*a*(a + a*tan(c + d*x)*1i)^(3/2))/(3*d*tan(c + d*x)^3) - (atan(((-a^5)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))
/a^3)*(-a^5)^(1/2)*45i)/(8*d) - (13*a^2*(a + a*tan(c + d*x)*1i)^(1/2))/(8*d*tan(c + d*x)^3) - (19*(a + a*tan(c
+ d*x)*1i)^(5/2))/(8*d*tan(c + d*x)^3) + (2^(1/2)*atan((2^(1/2)*(-a^5)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(
2*a^3))*(-a^5)^(1/2)*4i)/d

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