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

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

[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)

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Rubi [A]  time = 0.589992, antiderivative size = 190, normalized size of antiderivative = 1., 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 = {3553, 3598, 3600, 3480, 206, 3599, 63, 208} \[ \frac{45 i a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{8 d}-\frac{4 i \sqrt{2} a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{a^2 \cot ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}-\frac{13 i a^2 \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)}}{12 d}+\frac{19 a^2 \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{8 d} \]

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 3553

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(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] /; Fr
eeQ[{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] && LtQ[
n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3598

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 3600

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]

Rule 3480

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 206

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

Rule 3599

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 63

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 208

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

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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 2.34033, size = 200, normalized size = 1.05 \[ \frac{a^2 e^{-i (c+2 d x)} \sqrt{1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} (\cos (d x)+i \sin (d x)) \left (-384 i \sinh ^{-1}\left (e^{i (c+d x)}\right )+270 i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{1+e^{2 i (c+d x)}}}\right )+\sqrt{1+e^{2 i (c+d x)}} \csc ^3(c+d x) (-26 i \sin (2 (c+d x))-65 \cos (2 (c+d x))+49)\right )}{48 \sqrt{2} d} \]

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]  time = 0.388, size = 926, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

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)

[Out]

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

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

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]

Exception raised: ValueError

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Fricas [B]  time = 2.13505, size = 1858, normalized size = 9.78 \begin{align*} \text{result too large to display} \end{align*}

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]

1/48*(2*sqrt(2)*(91*I*a^2*e^(6*I*d*x + 6*I*c) - 7*I*a^2*e^(4*I*d*x + 4*I*c) - 59*I*a^2*e^(2*I*d*x + 2*I*c) + 3
9*I*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + 96*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(1/4*(4*I*sqrt(2)*sqrt(-a^5/d^2)*d*e^(2*I*d*x + 2*
I*c) + 4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-2*I*d*
x - 2*I*c)/a^2) - 96*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(1/4*(-4*I*sqrt(2)*sqrt(-a^5/d^2)*d*e^(2*I*d*x + 2*I*c) + 4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c)
+ a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-2*I*d*x - 2*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(1/45*(45*sqrt(2)*(a^2*e^(2*I*d*
x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + 90*I*sqrt(-a^5/d^2)*d*e^(2*I*d*x + 2*I*c
))*e^(-2*I*d*x - 2*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(1/45*(45*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(
I*d*x + I*c) - 90*I*sqrt(-a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cot \left (d x + c\right )^{4}\,{d x} \end{align*}

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)