3.261 \(\int \cot ^2(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=175 \[ \frac {a^3 (a B+4 A b) \log (\sin (c+d x))}{d}+\frac {b^2 \left (a^2 A+3 a b B+A b^2\right ) \tan (c+d x)}{d}-\frac {b^2 \left (6 a^2 B+4 a A b-b^2 B\right ) \log (\cos (c+d x))}{d}-x \left (a^4 A-4 a^3 b B-6 a^2 A b^2+4 a b^3 B+A b^4\right )+\frac {b (2 a A+b B) (a+b \tan (c+d x))^2}{2 d}-\frac {a A \cot (c+d x) (a+b \tan (c+d x))^3}{d} \]

[Out]

-(A*a^4-6*A*a^2*b^2+A*b^4-4*B*a^3*b+4*B*a*b^3)*x-b^2*(4*A*a*b+6*B*a^2-B*b^2)*ln(cos(d*x+c))/d+a^3*(4*A*b+B*a)*
ln(sin(d*x+c))/d+b^2*(A*a^2+A*b^2+3*B*a*b)*tan(d*x+c)/d+1/2*b*(2*A*a+B*b)*(a+b*tan(d*x+c))^2/d-a*A*cot(d*x+c)*
(a+b*tan(d*x+c))^3/d

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Rubi [A]  time = 0.48, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3605, 3647, 3637, 3624, 3475} \[ \frac {b^2 \left (a^2 A+3 a b B+A b^2\right ) \tan (c+d x)}{d}-\frac {b^2 \left (6 a^2 B+4 a A b-b^2 B\right ) \log (\cos (c+d x))}{d}-x \left (-6 a^2 A b^2+a^4 A-4 a^3 b B+4 a b^3 B+A b^4\right )+\frac {a^3 (a B+4 A b) \log (\sin (c+d x))}{d}+\frac {b (2 a A+b B) (a+b \tan (c+d x))^2}{2 d}-\frac {a A \cot (c+d x) (a+b \tan (c+d x))^3}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2*(a + b*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

-((a^4*A - 6*a^2*A*b^2 + A*b^4 - 4*a^3*b*B + 4*a*b^3*B)*x) - (b^2*(4*a*A*b + 6*a^2*B - b^2*B)*Log[Cos[c + d*x]
])/d + (a^3*(4*A*b + a*B)*Log[Sin[c + d*x]])/d + (b^2*(a^2*A + A*b^2 + 3*a*b*B)*Tan[c + d*x])/d + (b*(2*a*A +
b*B)*(a + b*Tan[c + d*x])^2)/(2*d) - (a*A*Cot[c + d*x]*(a + b*Tan[c + d*x])^3)/d

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3605

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

Rule 3624

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/tan[(e_.) + (f_.)*(x_)], x_Symbol
] :> Simp[B*x, x] + (Dist[A, Int[1/Tan[e + f*x], x], x] + Dist[C, Int[Tan[e + f*x], x], x]) /; FreeQ[{e, f, A,
B, C}, x] && NeQ[A, C]

Rule 3637

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

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^m*(c + d
*Tan[e + f*x])^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int \cot ^2(c+d x) (a+b \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx &=-\frac {a A \cot (c+d x) (a+b \tan (c+d x))^3}{d}+\int \cot (c+d x) (a+b \tan (c+d x))^2 \left (a (4 A b+a B)-\left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+b (2 a A+b B) \tan ^2(c+d x)\right ) \, dx\\ &=\frac {b (2 a A+b B) (a+b \tan (c+d x))^2}{2 d}-\frac {a A \cot (c+d x) (a+b \tan (c+d x))^3}{d}+\frac {1}{2} \int \cot (c+d x) (a+b \tan (c+d x)) \left (2 a^2 (4 A b+a B)-2 \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right ) \tan (c+d x)+2 b \left (a^2 A+A b^2+3 a b B\right ) \tan ^2(c+d x)\right ) \, dx\\ &=\frac {b^2 \left (a^2 A+A b^2+3 a b B\right ) \tan (c+d x)}{d}+\frac {b (2 a A+b B) (a+b \tan (c+d x))^2}{2 d}-\frac {a A \cot (c+d x) (a+b \tan (c+d x))^3}{d}-\frac {1}{2} \int \cot (c+d x) \left (-2 a^3 (4 A b+a B)+2 \left (a^4 A-6 a^2 A b^2+A b^4-4 a^3 b B+4 a b^3 B\right ) \tan (c+d x)-2 b^2 \left (4 a A b+6 a^2 B-b^2 B\right ) \tan ^2(c+d x)\right ) \, dx\\ &=-\left (a^4 A-6 a^2 A b^2+A b^4-4 a^3 b B+4 a b^3 B\right ) x+\frac {b^2 \left (a^2 A+A b^2+3 a b B\right ) \tan (c+d x)}{d}+\frac {b (2 a A+b B) (a+b \tan (c+d x))^2}{2 d}-\frac {a A \cot (c+d x) (a+b \tan (c+d x))^3}{d}+\left (a^3 (4 A b+a B)\right ) \int \cot (c+d x) \, dx+\left (b^2 \left (4 a A b+6 a^2 B-b^2 B\right )\right ) \int \tan (c+d x) \, dx\\ &=-\left (a^4 A-6 a^2 A b^2+A b^4-4 a^3 b B+4 a b^3 B\right ) x-\frac {b^2 \left (4 a A b+6 a^2 B-b^2 B\right ) \log (\cos (c+d x))}{d}+\frac {a^3 (4 A b+a B) \log (\sin (c+d x))}{d}+\frac {b^2 \left (a^2 A+A b^2+3 a b B\right ) \tan (c+d x)}{d}+\frac {b (2 a A+b B) (a+b \tan (c+d x))^2}{2 d}-\frac {a A \cot (c+d x) (a+b \tan (c+d x))^3}{d}\\ \end {align*}

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Mathematica [C]  time = 1.02, size = 134, normalized size = 0.77 \[ \frac {-2 a^4 A \cot (c+d x)+2 a^3 (a B+4 A b) \log (\tan (c+d x))+2 b^3 (4 a B+A b) \tan (c+d x)+i (a+i b)^4 (A+i B) \log (-\tan (c+d x)+i)-(a-i b)^4 (B+i A) \log (\tan (c+d x)+i)+b^4 B \tan ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2*(a + b*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

(-2*a^4*A*Cot[c + d*x] + I*(a + I*b)^4*(A + I*B)*Log[I - Tan[c + d*x]] + 2*a^3*(4*A*b + a*B)*Log[Tan[c + d*x]]
- (a - I*b)^4*(I*A + B)*Log[I + Tan[c + d*x]] + 2*b^3*(A*b + 4*a*B)*Tan[c + d*x] + b^4*B*Tan[c + d*x]^2)/(2*d
)

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fricas [A]  time = 0.63, size = 193, normalized size = 1.10 \[ \frac {B b^{4} \tan \left (d x + c\right )^{3} - 2 \, A a^{4} + {\left (B a^{4} + 4 \, A a^{3} b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) - {\left (6 \, B a^{2} b^{2} + 4 \, A a b^{3} - B b^{4}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) + 2 \, {\left (4 \, B a b^{3} + A b^{4}\right )} \tan \left (d x + c\right )^{2} + {\left (B b^{4} - 2 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} d x\right )} \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(B*b^4*tan(d*x + c)^3 - 2*A*a^4 + (B*a^4 + 4*A*a^3*b)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c
) - (6*B*a^2*b^2 + 4*A*a*b^3 - B*b^4)*log(1/(tan(d*x + c)^2 + 1))*tan(d*x + c) + 2*(4*B*a*b^3 + A*b^4)*tan(d*x
+ c)^2 + (B*b^4 - 2*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2 + 4*B*a*b^3 + A*b^4)*d*x)*tan(d*x + c))/(d*tan(d*x + c))

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giac [A]  time = 4.07, size = 195, normalized size = 1.11 \[ \frac {B b^{4} \tan \left (d x + c\right )^{2} + 8 \, B a b^{3} \tan \left (d x + c\right ) + 2 \, A b^{4} \tan \left (d x + c\right ) - 2 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} {\left (d x + c\right )} - {\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac {2 \, {\left (B a^{4} \tan \left (d x + c\right ) + 4 \, A a^{3} b \tan \left (d x + c\right ) + A a^{4}\right )}}{\tan \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

1/2*(B*b^4*tan(d*x + c)^2 + 8*B*a*b^3*tan(d*x + c) + 2*A*b^4*tan(d*x + c) - 2*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2
+ 4*B*a*b^3 + A*b^4)*(d*x + c) - (B*a^4 + 4*A*a^3*b - 6*B*a^2*b^2 - 4*A*a*b^3 + B*b^4)*log(tan(d*x + c)^2 + 1
) + 2*(B*a^4 + 4*A*a^3*b)*log(abs(tan(d*x + c))) - 2*(B*a^4*tan(d*x + c) + 4*A*a^3*b*tan(d*x + c) + A*a^4)/tan
(d*x + c))/d

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maple [A]  time = 0.40, size = 242, normalized size = 1.38 \[ -A \,a^{4} x -\frac {A \cot \left (d x +c \right ) a^{4}}{d}-\frac {A \,a^{4} c}{d}+\frac {a^{4} B \ln \left (\sin \left (d x +c \right )\right )}{d}+\frac {4 A \,a^{3} b \ln \left (\sin \left (d x +c \right )\right )}{d}+4 B x \,a^{3} b +\frac {4 B \,a^{3} b c}{d}+6 A x \,a^{2} b^{2}+\frac {6 A \,a^{2} b^{2} c}{d}-\frac {6 a^{2} b^{2} B \ln \left (\cos \left (d x +c \right )\right )}{d}-\frac {4 a A \,b^{3} \ln \left (\cos \left (d x +c \right )\right )}{d}-4 B x a \,b^{3}+\frac {4 B \tan \left (d x +c \right ) a \,b^{3}}{d}-\frac {4 B a \,b^{3} c}{d}-A x \,b^{4}+\frac {A \tan \left (d x +c \right ) b^{4}}{d}-\frac {A \,b^{4} c}{d}+\frac {B \,b^{4} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {b^{4} B \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x)

[Out]

-A*a^4*x-1/d*A*cot(d*x+c)*a^4-1/d*A*a^4*c+1/d*a^4*B*ln(sin(d*x+c))+4/d*A*a^3*b*ln(sin(d*x+c))+4*B*x*a^3*b+4/d*
B*a^3*b*c+6*A*x*a^2*b^2+6/d*A*a^2*b^2*c-6/d*a^2*b^2*B*ln(cos(d*x+c))-4/d*a*A*b^3*ln(cos(d*x+c))-4*B*x*a*b^3+4/
d*B*tan(d*x+c)*a*b^3-4/d*B*a*b^3*c-A*x*b^4+1/d*A*tan(d*x+c)*b^4-1/d*A*b^4*c+1/2/d*B*b^4*tan(d*x+c)^2+b^4*B*ln(
cos(d*x+c))/d

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maxima [A]  time = 0.69, size = 164, normalized size = 0.94 \[ \frac {B b^{4} \tan \left (d x + c\right )^{2} - \frac {2 \, A a^{4}}{\tan \left (d x + c\right )} - 2 \, {\left (A a^{4} - 4 \, B a^{3} b - 6 \, A a^{2} b^{2} + 4 \, B a b^{3} + A b^{4}\right )} {\left (d x + c\right )} - {\left (B a^{4} + 4 \, A a^{3} b - 6 \, B a^{2} b^{2} - 4 \, A a b^{3} + B b^{4}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \log \left (\tan \left (d x + c\right )\right ) + 2 \, {\left (4 \, B a b^{3} + A b^{4}\right )} \tan \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+b*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/2*(B*b^4*tan(d*x + c)^2 - 2*A*a^4/tan(d*x + c) - 2*(A*a^4 - 4*B*a^3*b - 6*A*a^2*b^2 + 4*B*a*b^3 + A*b^4)*(d*
x + c) - (B*a^4 + 4*A*a^3*b - 6*B*a^2*b^2 - 4*A*a*b^3 + B*b^4)*log(tan(d*x + c)^2 + 1) + 2*(B*a^4 + 4*A*a^3*b)
*log(tan(d*x + c)) + 2*(4*B*a*b^3 + A*b^4)*tan(d*x + c))/d

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mupad [B]  time = 6.44, size = 142, normalized size = 0.81 \[ \frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,b^4+4\,B\,a\,b^3\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (B\,a^4+4\,A\,b\,a^3\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^4}{2\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^4}{2\,d}-\frac {A\,a^4\,\mathrm {cot}\left (c+d\,x\right )}{d}+\frac {B\,b^4\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^2*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^4,x)

[Out]

(tan(c + d*x)*(A*b^4 + 4*B*a*b^3))/d + (log(tan(c + d*x))*(B*a^4 + 4*A*a^3*b))/d + (log(tan(c + d*x) - 1i)*(A*
1i - B)*(a*1i - b)^4)/(2*d) - (log(tan(c + d*x) + 1i)*(A*1i + B)*(a*1i + b)^4)/(2*d) - (A*a^4*cot(c + d*x))/d
+ (B*b^4*tan(c + d*x)^2)/(2*d)

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sympy [A]  time = 2.61, size = 289, normalized size = 1.65 \[ \begin {cases} \tilde {\infty } A a^{4} x & \text {for}\: c = 0 \wedge d = 0 \\x \left (A + B \tan {\relax (c )}\right ) \left (a + b \tan {\relax (c )}\right )^{4} \cot ^{2}{\relax (c )} & \text {for}\: d = 0 \\\tilde {\infty } A a^{4} x & \text {for}\: c = - d x \\- A a^{4} x - \frac {A a^{4}}{d \tan {\left (c + d x \right )}} - \frac {2 A a^{3} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {4 A a^{3} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 6 A a^{2} b^{2} x + \frac {2 A a b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - A b^{4} x + \frac {A b^{4} \tan {\left (c + d x \right )}}{d} - \frac {B a^{4} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B a^{4} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 4 B a^{3} b x + \frac {3 B a^{2} b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - 4 B a b^{3} x + \frac {4 B a b^{3} \tan {\left (c + d x \right )}}{d} - \frac {B b^{4} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b^{4} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2*(a+b*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)

[Out]

Piecewise((zoo*A*a**4*x, Eq(c, 0) & Eq(d, 0)), (x*(A + B*tan(c))*(a + b*tan(c))**4*cot(c)**2, Eq(d, 0)), (zoo*
A*a**4*x, Eq(c, -d*x)), (-A*a**4*x - A*a**4/(d*tan(c + d*x)) - 2*A*a**3*b*log(tan(c + d*x)**2 + 1)/d + 4*A*a**
3*b*log(tan(c + d*x))/d + 6*A*a**2*b**2*x + 2*A*a*b**3*log(tan(c + d*x)**2 + 1)/d - A*b**4*x + A*b**4*tan(c +
d*x)/d - B*a**4*log(tan(c + d*x)**2 + 1)/(2*d) + B*a**4*log(tan(c + d*x))/d + 4*B*a**3*b*x + 3*B*a**2*b**2*log
(tan(c + d*x)**2 + 1)/d - 4*B*a*b**3*x + 4*B*a*b**3*tan(c + d*x)/d - B*b**4*log(tan(c + d*x)**2 + 1)/(2*d) + B
*b**4*tan(c + d*x)**2/(2*d), True))

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