3.654 \(\int \frac{1}{\tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=298 \[ \frac{4 b^2 \left (15 a^2 b^2+4 a^4+8 b^4\right ) \sqrt{\tan (c+d x)}}{3 a^4 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}+\frac{2 b^2 \left (7 a^2+8 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{4 b}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{\tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}-\frac{2}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]

[Out]

ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/((I*a - b)^(5/2)*d) + ArcTanh[(Sqrt[I*a +
b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/((I*a + b)^(5/2)*d) - 2/(3*a*d*Tan[c + d*x]^(3/2)*(a + b*Tan[
c + d*x])^(3/2)) + (4*b)/(a^2*d*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2)) + (2*b^2*(7*a^2 + 8*b^2)*Sqrt[T
an[c + d*x]])/(3*a^3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + (4*b^2*(4*a^4 + 15*a^2*b^2 + 8*b^4)*Sqrt[Tan[
c + d*x]])/(3*a^4*(a^2 + b^2)^2*d*Sqrt[a + b*Tan[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.18063, antiderivative size = 298, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3569, 3649, 3650, 3616, 3615, 93, 203, 206} \[ \frac{4 b^2 \left (15 a^2 b^2+4 a^4+8 b^4\right ) \sqrt{\tan (c+d x)}}{3 a^4 d \left (a^2+b^2\right )^2 \sqrt{a+b \tan (c+d x)}}+\frac{2 b^2 \left (7 a^2+8 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac{4 b}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{\tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}-\frac{2}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])^(5/2)),x]

[Out]

ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/((I*a - b)^(5/2)*d) + ArcTanh[(Sqrt[I*a +
b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/((I*a + b)^(5/2)*d) - 2/(3*a*d*Tan[c + d*x]^(3/2)*(a + b*Tan[
c + d*x])^(3/2)) + (4*b)/(a^2*d*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2)) + (2*b^2*(7*a^2 + 8*b^2)*Sqrt[T
an[c + d*x]])/(3*a^3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + (4*b^2*(4*a^4 + 15*a^2*b^2 + 8*b^4)*Sqrt[Tan[
c + d*x]])/(3*a^4*(a^2 + b^2)^2*d*Sqrt[a + b*Tan[c + d*x]])

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3649

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3650

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*t
an[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 + a^2*C)*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x]
)^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[
e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2)) - a*C*(b*c*(m + 1)
+ a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - b*C)*Tan[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 2)*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^
2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3616

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

Rule 3615

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 203

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

Rubi steps

\begin{align*} \int \frac{1}{\tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx &=-\frac{2}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}-\frac{2 \int \frac{3 b+\frac{3}{2} a \tan (c+d x)+3 b \tan ^2(c+d x)}{\tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx}{3 a}\\ &=-\frac{2}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{4 b}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{4 \int \frac{-\frac{3}{4} \left (a^2-8 b^2\right )+6 b^2 \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}} \, dx}{3 a^2}\\ &=-\frac{2}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{4 b}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{2 b^2 \left (7 a^2+8 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{8 \int \frac{-\frac{3}{8} \left (3 a^4-14 a^2 b^2-16 b^4\right )+\frac{9}{8} a^3 b \tan (c+d x)+\frac{3}{4} b^2 \left (7 a^2+8 b^2\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{9 a^3 \left (a^2+b^2\right )}\\ &=-\frac{2}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{4 b}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{2 b^2 \left (7 a^2+8 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right ) \sqrt{\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}+\frac{16 \int \frac{-\frac{9}{16} a^4 \left (a^2-b^2\right )+\frac{9}{8} a^5 b \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{9 a^4 \left (a^2+b^2\right )^2}\\ &=-\frac{2}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{4 b}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{2 b^2 \left (7 a^2+8 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right ) \sqrt{\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{\int \frac{1+i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}-\frac{\int \frac{1-i \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}\\ &=-\frac{2}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{4 b}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{2 b^2 \left (7 a^2+8 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right ) \sqrt{\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{(1-i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b)^2 d}-\frac{\operatorname{Subst}\left (\int \frac{1}{(1+i x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b)^2 d}\\ &=-\frac{2}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{4 b}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{2 b^2 \left (7 a^2+8 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right ) \sqrt{\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{1-(i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(a-i b)^2 d}-\frac{\operatorname{Subst}\left (\int \frac{1}{1-(-i a+b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(a+i b)^2 d}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac{\tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac{2}{3 a d \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{4 b}{a^2 d \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac{2 b^2 \left (7 a^2+8 b^2\right ) \sqrt{\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac{4 b^2 \left (4 a^4+15 a^2 b^2+8 b^4\right ) \sqrt{\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt{a+b \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 6.16777, size = 270, normalized size = 0.91 \[ \frac{\frac{6 a b^2 \left (15 a^2 b^2+5 a^4+8 b^4\right ) \tan ^2(c+d x)+4 b^3 \left (15 a^2 b^2+4 a^4+8 b^4\right ) \tan ^3(c+d x)+12 a^2 b \left (a^2+b^2\right )^2 \tan (c+d x)-2 a^3 \left (a^2+b^2\right )^2}{a^4 \left (a^2+b^2\right )^2 \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac{3 (-1)^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(-a-i b)^{5/2}}+\frac{3 (-1)^{3/4} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{(a-i b)^{5/2}}}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])^(5/2)),x]

[Out]

((3*(-1)^(3/4)*ArcTanh[((-1)^(1/4)*Sqrt[-a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(-a - I*b)^(5
/2) + (3*(-1)^(3/4)*ArcTanh[((-1)^(1/4)*Sqrt[a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(a - I*b)
^(5/2) + (-2*a^3*(a^2 + b^2)^2 + 12*a^2*b*(a^2 + b^2)^2*Tan[c + d*x] + 6*a*b^2*(5*a^4 + 15*a^2*b^2 + 8*b^4)*Ta
n[c + d*x]^2 + 4*b^3*(4*a^4 + 15*a^2*b^2 + 8*b^4)*Tan[c + d*x]^3)/(a^4*(a^2 + b^2)^2*Tan[c + d*x]^(3/2)*(a + b
*Tan[c + d*x])^(3/2)))/(3*d)

________________________________________________________________________________________

Maple [B]  time = 0.907, size = 1489930, normalized size = 4999.8 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \tan \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((b*tan(d*x + c) + a)^(5/2)*tan(d*x + c)^(5/2)), x)

________________________________________________________________________________________

Fricas [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(1/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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(1/tan(d*x+c)**(5/2)/(a+b*tan(d*x+c))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/tan(d*x+c)^(5/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError