3.8.30 \(\int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 \, dx\) [730]

Optimal. Leaf size=106 \[ \frac {8 \sqrt [4]{-1} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {8 a^3 \sqrt {\cot (c+d x)}}{d}-\frac {8 i a^3 \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 \cot ^{\frac {3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{5 d} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3754, 3637, 3673, 3609, 3614, 214} \begin {gather*} -\frac {8 i a^3 \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 \cot ^{\frac {3}{2}}(c+d x) \left (a^3 \cot (c+d x)+i a^3\right )}{5 d}+\frac {8 a^3 \sqrt {\cot (c+d x)}}{d}+\frac {8 \sqrt [4]{-1} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(-1)^(1/4)*a^3*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d + (8*a^3*Sqrt[Cot[c + d*x]])/d - (((8*I)/5)*a^3*Co
t[c + d*x]^(3/2))/d - (2*Cot[c + d*x]^(3/2)*(I*a^3 + a^3*Cot[c + d*x]))/(5*d)

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 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3614

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2*(c^2/f), S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 3637

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*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] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3673

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

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 \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^3 \, dx &=\int \sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^3 \, dx\\ &=-\frac {2 \cot ^{\frac {3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{5 d}-\frac {1}{5} (2 i a) \int \sqrt {\cot (c+d x)} (-4 i a-6 a \cot (c+d x)) (i a+a \cot (c+d x)) \, dx\\ &=-\frac {8 i a^3 \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 \cot ^{\frac {3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{5 d}-\frac {1}{5} (2 i a) \int \sqrt {\cot (c+d x)} \left (10 a^2-10 i a^2 \cot (c+d x)\right ) \, dx\\ &=\frac {8 a^3 \sqrt {\cot (c+d x)}}{d}-\frac {8 i a^3 \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 \cot ^{\frac {3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{5 d}-\frac {1}{5} (2 i a) \int \frac {10 i a^2+10 a^2 \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=\frac {8 a^3 \sqrt {\cot (c+d x)}}{d}-\frac {8 i a^3 \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 \cot ^{\frac {3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{5 d}+\frac {\left (80 i a^5\right ) \text {Subst}\left (\int \frac {1}{-10 i a^2+10 a^2 x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {8 \sqrt [4]{-1} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {8 a^3 \sqrt {\cot (c+d x)}}{d}-\frac {8 i a^3 \cot ^{\frac {3}{2}}(c+d x)}{5 d}-\frac {2 \cot ^{\frac {3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{5 d}\\ \end {align*}

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Mathematica [A]
time = 2.21, size = 147, normalized size = 1.39 \begin {gather*} -\frac {a^3 e^{-3 i c} \sqrt {\cot (c+d x)} (\cos (3 (c+d x))+i \sin (3 (c+d x))) \left (\csc ^2(c+d x) (-19+21 \cos (2 (c+d x))+5 i \sin (2 (c+d x)))+40 \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right ) \sqrt {i \tan (c+d x)}\right )}{5 d (\cos (d x)+i \sin (d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(a^3*Sqrt[Cot[c + d*x]]*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)])*(Csc[c + d*x]^2*(-19 + 21*Cos[2*(c + d*x)
] + (5*I)*Sin[2*(c + d*x)]) + 40*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Sqrt[I*Ta
n[c + d*x]]))/(d*E^((3*I)*c)*(Cos[d*x] + I*Sin[d*x])^3)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 14.35, size = 1482, normalized size = 13.98

method	result	size
default	\(\text {Expression too large to display}\)	\(1482\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*a^3/d*(-20*I*cos(d*x+c)^3*(-(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)*((-1+cos(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))+20*I*cos(d*x+c)^3*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/si
n(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2
*2^(1/2))-20*cos(d*x+c)^3*(-(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)*((-1+cos(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))-20*I*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)*((-1+cos(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))+20*I*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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*
2^(1/2))-20*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)*((-1+cos(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))+20*I*cos(d*x+c)*(-(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)*((-1+cos(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))-20*I*cos(d*x+c)*(-(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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/
2))+5*I*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)+20*cos(d*x+c)*(-(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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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))+20*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)*((-1+cos(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))-20*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))/s
in(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/
2*2^(1/2))+21*cos(d*x+c)^3*2^(1/2)+20*(-(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)*((-1+cos(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))-20*cos(d*x+c)*2^(1/2))*sin(d*x+c)*(cos(d*x+c)/sin(d*x+c))^(7/2)/cos(d*x+c)^4*2^(1/2)

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Maxima [A]
time = 0.49, size = 158, normalized size = 1.49 \begin {gather*} \frac {5 \, {\left (\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \left (i + 1\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \left (i + 1\right ) \, \sqrt {2} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{3} + \frac {40 \, a^{3}}{\sqrt {\tan \left (d x + c\right )}} - \frac {10 i \, a^{3}}{\tan \left (d x + c\right )^{\frac {3}{2}}} - \frac {2 \, a^{3}}{\tan \left (d x + c\right )^{\frac {5}{2}}}}{5 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(5*((2*I - 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + (2*I - 2)*sqrt(2)*arctan(-1/2
*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) - (I + 1)*sqrt(2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) +
1) + (I + 1)*sqrt(2)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^3 + 40*a^3/sqrt(tan(d*x + c)) -
10*I*a^3/tan(d*x + c)^(3/2) - 2*a^3/tan(d*x + c)^(5/2))/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. 340 vs. \(2 (86) = 172\).
time = 0.65, size = 340, normalized size = 3.21 \begin {gather*} -\frac {5 \, \sqrt {\frac {64 i \, a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {\frac {64 i \, a^{6}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - 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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{3}}\right ) - 5 \, \sqrt {\frac {64 i \, a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {64 i \, a^{6}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - 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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{3}}\right ) - 16 \, {\left (13 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 19 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 8 \, a^{3}\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}}}{20 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, 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)^(7/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/20*(5*sqrt(64*I*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(8*I*a^3*e^(2*I*d*x
+ 2*I*c) + sqrt(64*I*a^6/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c
) - 1)))*e^(-2*I*d*x - 2*I*c)/a^3) - 5*sqrt(64*I*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d
)*log(1/4*(8*I*a^3*e^(2*I*d*x + 2*I*c) - sqrt(64*I*a^6/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt((I*e^(2*I*d*x + 2
*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d*x - 2*I*c)/a^3) - 16*(13*a^3*e^(4*I*d*x + 4*I*c) - 19*a^3*e^(
2*I*d*x + 2*I*c) + 8*a^3)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(d*e^(4*I*d*x + 4*I*c)
- 2*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)**(7/2)*(a+I*a*tan(d*x+c))**3,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 5987 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)^(7/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((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 = -0.01 \begin {gather*} \int {\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(cot(c + d*x)^(7/2)*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

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

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