3.8.50 \(\int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx\) [750]

Optimal. Leaf size=275 \[ -\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{3 a d (i a+a \cot (c+d x))^2}+\frac {5 \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a^3 d} \]

[Out]

(5/32-7/32*I)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)+(5/32-7/32*I)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2
))/a^3/d*2^(1/2)+(5/64+7/64*I)*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)-(5/64+7/64*I)*ln(1+cot(
d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)-1/6*cot(d*x+c)^(1/2)/d/(I*a+a*cot(d*x+c))^3+1/3*I*cot(d*x+c)^(1
/2)/a/d/(I*a+a*cot(d*x+c))^2+5/8*cot(d*x+c)^(1/2)/d/(I*a^3+a^3*cot(d*x+c))

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Rubi [A]
time = 0.32, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3754, 3640, 3677, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) \text {ArcTan}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {5 \sqrt {\cot (c+d x)}}{8 d \left (a^3 \cot (c+d x)+i a^3\right )}+\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^3 d}+\frac {i \sqrt {\cot (c+d x)}}{3 a d (a \cot (c+d x)+i a)^2}-\frac {\sqrt {\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^3),x]

[Out]

((-5/16 + (7*I)/16)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^3*d) + ((5/16 - (7*I)/16)*ArcTan[1 + Sq
rt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^3*d) - Sqrt[Cot[c + d*x]]/(6*d*(I*a + a*Cot[c + d*x])^3) + ((I/3)*Sqrt[C
ot[c + d*x]])/(a*d*(I*a + a*Cot[c + d*x])^2) + (5*Sqrt[Cot[c + d*x]])/(8*d*(I*a^3 + a^3*Cot[c + d*x])) + ((5/3
2 + (7*I)/32)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a^3*d) - ((5/32 + (7*I)/32)*Log[1 +
Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a^3*d)

Rule 210

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

Rule 631

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 642

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]

Rule 1176

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 1179

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 1182

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 3615

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 3640

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 3677

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 3754

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]

Rubi steps

\begin {align*} \int \frac {1}{\cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx &=\int \frac {1}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^3} \, dx\\ &=-\frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {\int \frac {-\frac {11 i a}{2}+\frac {5}{2} a \cot (c+d x)}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^2} \, dx}{6 a^2}\\ &=-\frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{3 a d (i a+a \cot (c+d x))^2}+\frac {\int \frac {-18 a^2-12 i a^2 \cot (c+d x)}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))} \, dx}{24 a^4}\\ &=-\frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{3 a d (i a+a \cot (c+d x))^2}+\frac {5 \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {\int \frac {21 i a^3-15 a^3 \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{48 a^6}\\ &=-\frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{3 a d (i a+a \cot (c+d x))^2}+\frac {5 \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {-21 i a^3+15 a^3 x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{24 a^6 d}\\ &=-\frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{3 a d (i a+a \cot (c+d x))^2}+\frac {5 \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+-\frac {\left (\frac {5}{16}+\frac {7 i}{16}\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^3 d}+\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^3 d}\\ &=-\frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{3 a d (i a+a \cot (c+d x))^2}+\frac {5 \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {\left (\frac {5}{32}-\frac {7 i}{32}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^3 d}+\frac {\left (\frac {5}{32}-\frac {7 i}{32}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^3 d}+\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) \text {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^3 d}+\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) \text {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^3 d}\\ &=-\frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{3 a d (i a+a \cot (c+d x))^2}+\frac {5 \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a^3 d}+-\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^3 d}\\ &=-\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^3 d}+\frac {\left (\frac {5}{16}-\frac {7 i}{16}\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^3 d}-\frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{3 a d (i a+a \cot (c+d x))^2}+\frac {5 \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a^3 d}-\frac {\left (\frac {5}{32}+\frac {7 i}{32}\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a^3 d}\\ \end {align*}

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Mathematica [A]
time = 2.08, size = 235, normalized size = 0.85 \begin {gather*} \frac {\cot ^{\frac {5}{2}}(c+d x) \csc (c+d x) \sec ^3(c+d x) \left (19 i-19 i \cos (4 (c+d x))-(15+21 i) \cos (3 (c+d x)) \log \left (\cos (c+d x)+\sin (c+d x)+\sqrt {\sin (2 (c+d x))}\right ) \sqrt {\sin (2 (c+d x))}-12 \sin (2 (c+d x))+(21-15 i) \log \left (\cos (c+d x)+\sin (c+d x)+\sqrt {\sin (2 (c+d x))}\right ) \sqrt {\sin (2 (c+d x))} \sin (3 (c+d x))+(21+15 i) \text {ArcSin}(\cos (c+d x)-\sin (c+d x)) \sqrt {\sin (2 (c+d x))} (-i \cos (3 (c+d x))+\sin (3 (c+d x)))+21 \sin (4 (c+d x))\right )}{96 a^3 d (i+\cot (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^3),x]

[Out]

(Cot[c + d*x]^(5/2)*Csc[c + d*x]*Sec[c + d*x]^3*(19*I - (19*I)*Cos[4*(c + d*x)] - (15 + 21*I)*Cos[3*(c + d*x)]
*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Sin[2*(c + d*x)]] - 12*Sin[2*(c + d*x)] + (21
- 15*I)*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Sin[2*(c + d*x)]]*Sin[3*(c + d*x)] + (2
1 + 15*I)*ArcSin[Cos[c + d*x] - Sin[c + d*x]]*Sqrt[Sin[2*(c + d*x)]]*((-I)*Cos[3*(c + d*x)] + Sin[3*(c + d*x)]
) + 21*Sin[4*(c + d*x)]))/(96*a^3*d*(I + Cot[c + d*x])^3)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 13.63, size = 9482, normalized size = 34.48

method	result	size
default	\(\text {Expression too large to display}\)	\(9482\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cot(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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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. 522 vs. \(2 (206) = 412\).
time = 1.04, size = 522, normalized size = 1.90 \begin {gather*} -\frac {{\left (12 \, a^{3} d \sqrt {-\frac {i}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-2 \, {\left (8 \, {\left (i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3} 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}{64 \, a^{6} 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 ) - 12 \, a^{3} d \sqrt {-\frac {i}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-2 \, {\left (8 \, {\left (-i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} 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}{64 \, a^{6} 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 ) - 12 \, a^{3} d \sqrt {\frac {9 i}{16 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left (4 \, {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} 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 {9 i}{16 \, a^{6} d^{2}}} + 3\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{3} d}\right ) + 12 \, a^{3} d \sqrt {\frac {9 i}{16 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {{\left (4 \, {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} 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 {9 i}{16 \, a^{6} d^{2}}} - 3\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{3} 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 (-20 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 26 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 7 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{48 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/48*(12*a^3*d*sqrt(-1/64*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-2*(8*(I*a^3*d*e^(2*I*d*x + 2*I*c) - I*a^3*d)*
sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/64*I/(a^6*d^2)) - I*e^(2*I*d*x + 2*I*c))*e
^(-2*I*d*x - 2*I*c)) - 12*a^3*d*sqrt(-1/64*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-2*(8*(-I*a^3*d*e^(2*I*d*x + 2
*I*c) + I*a^3*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/64*I/(a^6*d^2)) - I*e^(2*
I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)) - 12*a^3*d*sqrt(9/16*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(1/4*(4*(a^3*d*
e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(9/16*I/(a^6*d^2)
) + 3)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) + 12*a^3*d*sqrt(9/16*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-1/4*(4*(a^3*d*
e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(9/16*I/(a^6*d^2)
) - 3)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) - sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(-20*I*e^(6
*I*d*x + 6*I*c) + 26*I*e^(4*I*d*x + 4*I*c) - 7*I*e^(2*I*d*x + 2*I*c) + I))*e^(-6*I*d*x - 6*I*c)/(a^3*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(1/cot(d*x+c)**(7/2)/(a+I*a*tan(d*x+c))**3,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3063 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(1/cot(d*x+c)^(7/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(1/((I*a*tan(d*x + c) + a)^3*cot(d*x + c)^(7/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cot(c + d*x)^(7/2)*(a + a*tan(c + d*x)*1i)^3),x)

[Out]

int(1/(cot(c + d*x)^(7/2)*(a + a*tan(c + d*x)*1i)^3), x)

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