3.8.48 \(\int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx\) [748]

Optimal. Leaf size=141 \[ -\frac {(-1)^{3/4} \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{8 a^3 d}+\frac {\sqrt {\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac {i \sqrt {\cot (c+d x)}}{6 a d (i a+a \cot (c+d x))^2}+\frac {\sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )} \]

[Out]

-1/8*(-1)^(3/4)*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/a^3/d+1/6*cot(d*x+c)^(1/2)/d/(I*a+a*cot(d*x+c))^3+1/6*I*c
ot(d*x+c)^(1/2)/a/d/(I*a+a*cot(d*x+c))^2+1/8*cot(d*x+c)^(1/2)/d/(I*a^3+a^3*cot(d*x+c))

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Rubi [A]
time = 0.20, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3754, 3639, 3677, 12, 3630, 3614, 214} \begin {gather*} \frac {\sqrt {\cot (c+d x)}}{8 d \left (a^3 \cot (c+d x)+i a^3\right )}-\frac {(-1)^{3/4} \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{8 a^3 d}+\frac {i \sqrt {\cot (c+d x)}}{6 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]^(3/2)*(a + I*a*Tan[c + d*x])^3),x]

[Out]

-1/8*((-1)^(3/4)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/(a^3*d) + Sqrt[Cot[c + d*x]]/(6*d*(I*a + a*Cot[c + d*
x])^3) + ((I/6)*Sqrt[Cot[c + d*x]])/(a*d*(I*a + a*Cot[c + d*x])^2) + Sqrt[Cot[c + d*x]]/(8*d*(I*a^3 + a^3*Cot[
c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 3630

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
a*c + b*d))*((c + d*Tan[e + f*x])^n/(2*(b*c - a*d)*f*(a + b*Tan[e + f*x]))), x] + Dist[1/(2*a*(b*c - a*d)), In
t[(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c*d*(n - 1) + b*c^2 + b*d^2*n - d*(b*c - a*d)*(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] && LtQ[0
, n, 1]

Rule 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Dist[1/(2*a^2*m), Int[(
a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)
) - d*(b*d*(m - n + 1) - a*c*(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] && LtQ[m, 0] && GtQ[n, 1] && (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 {3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx &=\int \frac {\cot ^{\frac {3}{2}}(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 {i a}{2}+\frac {7}{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)}}{6 a d (i a+a \cot (c+d x))^2}+\frac {\int -\frac {6 i a^2 \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)}}{6 a d (i a+a \cot (c+d x))^2}-\frac {i \int \frac {\sqrt {\cot (c+d x)}}{i a+a \cot (c+d x)} \, dx}{4 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)}}{6 a d (i a+a \cot (c+d x))^2}+\frac {\sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {i \int \frac {-\frac {a}{2}+\frac {1}{2} i a \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{8 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)}}{6 a d (i a+a \cot (c+d x))^2}+\frac {\sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {i \text {Subst}\left (\int \frac {1}{\frac {a}{2}+\frac {1}{2} i a x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{16 a^2 d}\\ &=-\frac {(-1)^{3/4} \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{8 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)}}{6 a d (i a+a \cot (c+d x))^2}+\frac {\sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 1.67, size = 154, normalized size = 1.09 \begin {gather*} \frac {i \sqrt {\cot (c+d x)} \csc ^3(c+d x) \left (5 \cos (c+d x)-5 \cos (3 (c+d x))+3 i \sin (c+d x)-3 i \sin (3 (c+d x))+6 \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x))) \sqrt {i \tan (c+d x)}\right )}{48 a^3 d (i+\cot (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/48)*Sqrt[Cot[c + d*x]]*Csc[c + d*x]^3*(5*Cos[c + d*x] - 5*Cos[3*(c + d*x)] + (3*I)*Sin[c + d*x] - (3*I)*Si
n[3*(c + d*x)] + 6*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*(Cos[3*(c + d*x)] + I*S
in[3*(c + d*x)])*Sqrt[I*Tan[c + d*x]]))/(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 = 15.08, size = 5851, normalized size = 41.50

method	result	size
default	\(\text {Expression too large to display}\)	\(5851\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cot(d*x+c)^(3/2)/(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)^(3/2)/(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. 307 vs. \(2 (113) = 226\).
time = 1.68, size = 307, normalized size = 2.18 \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 ) + \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 (-4 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 4 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 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)^(3/2)/(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)*s
qrt((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)) + sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(-4*I*e^(6*
I*d*x + 6*I*c) + 4*I*e^(4*I*d*x + 4*I*c) + 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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {i \int \frac {1}{\tan ^{3}{\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} \cot ^{\frac {3}{2}}{\left (c + d x \right )} + i \cot ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*Integral(1/(tan(c + d*x)**3*cot(c + d*x)**(3/2) - 3*I*tan(c + d*x)**2*cot(c + d*x)**(3/2) - 3*tan(c + d*x)*c
ot(c + d*x)**(3/2) + I*cot(c + d*x)**(3/2)), x)/a**3

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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)^(3/2)/(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)^(3/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,{\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)^(3/2)*(a + a*tan(c + d*x)*1i)^3),x)

[Out]

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

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