∫ (c−c sec(e+fx))4 ____________________________

3.72 \(\int \frac {(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=203 \[ \frac {2 c^4 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{a^{3/2} f}+\frac {12 \sqrt {2} c^4 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a}}\right )}{a^{3/2} f}+\frac {8 c^4 \tan ^3(e+f x)}{3 f (a \sec (e+f x)+a)^{3/2}}-\frac {14 c^4 \tan (e+f x)}{a f \sqrt {a \sec (e+f x)+a}}-\frac {a c^4 \sin (e+f x) \tan ^4(e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{f (a \sec (e+f x)+a)^{5/2}} \]

[Out]

2*c^4*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/a^(3/2)/f+12*c^4*arctan(1/2*a^(1/2)*tan(f*x+e)*2^(1/2)

/(a+a*sec(f*x+e))^(1/2))*2^(1/2)/a^(3/2)/f-14*c^4*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^(1/2)+8/3*c^4*tan(f*x+e)^3/f

/(a+a*sec(f*x+e))^(3/2)-a*c^4*sec(1/2*e+1/2*f*x)^2*sin(f*x+e)*tan(f*x+e)^4/f/(a+a*sec(f*x+e))^(5/2)

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Rubi [A] time = 0.29, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3904, 3887, 470, 582, 522, 203} \[ \frac {2 c^4 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{a^{3/2} f}+\frac {12 \sqrt {2} c^4 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a}}\right )}{a^{3/2} f}+\frac {8 c^4 \tan ^3(e+f x)}{3 f (a \sec (e+f x)+a)^{3/2}}-\frac {14 c^4 \tan (e+f x)}{a f \sqrt {a \sec (e+f x)+a}}-\frac {a c^4 \sin (e+f x) \tan ^4(e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{f (a \sec (e+f x)+a)^{5/2}} \]

Antiderivative was successfully veriﬁed.

[In]

Int[(c - c*Sec[e + f*x])^4/(a + a*Sec[e + f*x])^(3/2),x]

[Out]

(2*c^4*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(a^(3/2)*f) + (12*Sqrt[2]*c^4*ArcTan[(Sqrt[a]*

Tan[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(a^(3/2)*f) - (14*c^4*Tan[e + f*x])/(a*f*Sqrt[a + a*Sec[e +

f*x]]) + (8*c^4*Tan[e + f*x]^3)/(3*f*(a + a*Sec[e + f*x])^(3/2)) - (a*c^4*Sec[(e + f*x)/2]^2*Sin[e + f*x]*Tan

[e + f*x]^4)/(f*(a + a*Sec[e + f*x])^(5/2))

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 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2

*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2

*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +

(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &

& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f

)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a

, b, c, d, e, f, n}, x]

Rule 582

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),

x_Symbol] :> Simp[(f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q

+ 1) + 1)), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*

f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]

/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]

Rule 3887

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[(-2*a^(m/2 +

n + 1/2))/d, Subst[Int[(x^m*(2 + a*x^2)^(m/2 + n - 1/2))/(1 + a*x^2), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +

d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]

Rule 3904

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di

st[(-(a*c))^m, Int[Cot[e + f*x]^(2*m)*(c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]

&& EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] && !(IntegerQ[n] && GtQ[m - n, 0])

Rubi steps

\begin {align*} \int \frac {(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^{3/2}} \, dx &=\left (a^4 c^4\right ) \int \frac {\tan ^8(e+f x)}{(a+a \sec (e+f x))^{11/2}} \, dx\\ &=-\frac {\left (2 a^3 c^4\right ) \operatorname {Subst}\left (\int \frac {x^8}{\left (1+a x^2\right ) \left (2+a x^2\right )^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}\\ &=-\frac {a c^4 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^4(e+f x)}{f (a+a \sec (e+f x))^{5/2}}-\frac {\left (a c^4\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (10+8 a x^2\right )}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{f}\\ &=\frac {8 c^4 \tan ^3(e+f x)}{3 f (a+a \sec (e+f x))^{3/2}}-\frac {a c^4 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^4(e+f x)}{f (a+a \sec (e+f x))^{5/2}}+\frac {c^4 \operatorname {Subst}\left (\int \frac {x^2 \left (48 a+42 a^2 x^2\right )}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{3 a f}\\ &=-\frac {14 c^4 \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)}}+\frac {8 c^4 \tan ^3(e+f x)}{3 f (a+a \sec (e+f x))^{3/2}}-\frac {a c^4 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^4(e+f x)}{f (a+a \sec (e+f x))^{5/2}}-\frac {c^4 \operatorname {Subst}\left (\int \frac {84 a^2+78 a^3 x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{3 a^3 f}\\ &=-\frac {14 c^4 \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)}}+\frac {8 c^4 \tan ^3(e+f x)}{3 f (a+a \sec (e+f x))^{3/2}}-\frac {a c^4 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^4(e+f x)}{f (a+a \sec (e+f x))^{5/2}}-\frac {\left (2 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a f}-\frac {\left (24 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{2+a x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a f}\\ &=\frac {2 c^4 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^{3/2} f}+\frac {12 \sqrt {2} c^4 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{a^{3/2} f}-\frac {14 c^4 \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)}}+\frac {8 c^4 \tan ^3(e+f x)}{3 f (a+a \sec (e+f x))^{3/2}}-\frac {a c^4 \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x) \tan ^4(e+f x)}{f (a+a \sec (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [A] time = 1.58, size = 196, normalized size = 0.97 \[ \frac {c^4 \csc \left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {1}{2} (e+f x)\right ) \sec ^2(e+f x) \left (20 \cos (e+f x)-26 \cos (2 (e+f x))+28 \cos (3 (e+f x))+6 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {3}{2} (e+f x)\right )\right )^2 \sqrt {\sec (e+f x)-1} \tan ^{-1}\left (\sqrt {\sec (e+f x)-1}\right )+36 \sqrt {2} \left (\cos \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {3}{2} (e+f x)\right )\right )^2 \sqrt {\sec (e+f x)-1} \tan ^{-1}\left (\frac {\sqrt {\sec (e+f x)-1}}{\sqrt {2}}\right )-22\right )}{12 a f \sqrt {a (\sec (e+f x)+1)}} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[(c - c*Sec[e + f*x])^4/(a + a*Sec[e + f*x])^(3/2),x]

[Out]

(c^4*Csc[(e + f*x)/2]*Sec[(e + f*x)/2]*(-22 + 20*Cos[e + f*x] - 26*Cos[2*(e + f*x)] + 28*Cos[3*(e + f*x)] + 6*

ArcTan[Sqrt[-1 + Sec[e + f*x]]]*(Cos[(e + f*x)/2] + Cos[(3*(e + f*x))/2])^2*Sqrt[-1 + Sec[e + f*x]] + 36*Sqrt[

2]*ArcTan[Sqrt[-1 + Sec[e + f*x]]/Sqrt[2]]*(Cos[(e + f*x)/2] + Cos[(3*(e + f*x))/2])^2*Sqrt[-1 + Sec[e + f*x]]

)*Sec[e + f*x]^2)/(12*a*f*Sqrt[a*(1 + Sec[e + f*x])])

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fricas [A] time = 1.87, size = 634, normalized size = 3.12 \[ \left [\frac {18 \, \sqrt {2} {\left (a c^{4} \cos \left (f x + e\right )^{3} + 2 \, a c^{4} \cos \left (f x + e\right )^{2} + a c^{4} \cos \left (f x + e\right )\right )} \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {-\frac {1}{a}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, \cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) - 3 \, {\left (c^{4} \cos \left (f x + e\right )^{3} + 2 \, c^{4} \cos \left (f x + e\right )^{2} + c^{4} \cos \left (f x + e\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) - 2 \, {\left (28 \, c^{4} \cos \left (f x + e\right )^{2} + 15 \, c^{4} \cos \left (f x + e\right ) - c^{4}\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{3} + 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f \cos \left (f x + e\right )\right )}}, -\frac {2 \, {\left (3 \, {\left (c^{4} \cos \left (f x + e\right )^{3} + 2 \, c^{4} \cos \left (f x + e\right )^{2} + c^{4} \cos \left (f x + e\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) + {\left (28 \, c^{4} \cos \left (f x + e\right )^{2} + 15 \, c^{4} \cos \left (f x + e\right ) - c^{4}\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + \frac {18 \, \sqrt {2} {\left (a c^{4} \cos \left (f x + e\right )^{3} + 2 \, a c^{4} \cos \left (f x + e\right )^{2} + a c^{4} \cos \left (f x + e\right )\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right )}{\sqrt {a}}\right )}}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{3} + 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f \cos \left (f x + e\right )\right )}}\right ] \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^4/(a+a*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[1/3*(18*sqrt(2)*(a*c^4*cos(f*x + e)^3 + 2*a*c^4*cos(f*x + e)^2 + a*c^4*cos(f*x + e))*sqrt(-1/a)*log(-(2*sqrt(

2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt(-1/a)*cos(f*x + e)*sin(f*x + e) - 3*cos(f*x + e)^2 - 2*cos(f*x

+ e) + 1)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) - 3*(c^4*cos(f*x + e)^3 + 2*c^4*cos(f*x + e)^2 + c^4*cos(f*x

+ e))*sqrt(-a)*log((2*a*cos(f*x + e)^2 + 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(

f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) - 2*(28*c^4*cos(f*x + e)^2 + 15*c^4*cos(f*x + e) - c^4)*sqr

t((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(a^2*f*cos(f*x + e)^3 + 2*a^2*f*cos(f*x + e)^2 + a^2*f*cos(

f*x + e)), -2/3*(3*(c^4*cos(f*x + e)^3 + 2*c^4*cos(f*x + e)^2 + c^4*cos(f*x + e))*sqrt(a)*arctan(sqrt((a*cos(f

*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) + (28*c^4*cos(f*x + e)^2 + 15*c^4*cos(f*x + e)

- c^4)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) + 18*sqrt(2)*(a*c^4*cos(f*x + e)^3 + 2*a*c^4*cos(

f*x + e)^2 + a*c^4*cos(f*x + e))*arctan(sqrt(2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*

sin(f*x + e)))/sqrt(a))/(a^2*f*cos(f*x + e)^3 + 2*a^2*f*cos(f*x + e)^2 + a^2*f*cos(f*x + e))]

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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^4/(a+a*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab

le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p

i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab

le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p

i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab

le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p

i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab

le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to ch

eck sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable

to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)

Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warning, integration of abs or sign assumes constant

sign by intervals (correct if the argument is real):Check [abs(cos(f*t_nostep+exp(1)))]Unable to check sign:

(2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check

sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to

check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unab

le to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/

2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_n

ostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*

pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2

)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_no

step/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*p

i/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign

: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warning, assuming -2*a+a is positive. Hint: run assume to make assumptio

ns on a variableWarning, assuming -2*a+a is positive. Hint: run assume to make assumptions on a variableWarnin

g, assuming -2*a+a is positive. Hint: run assume to make assumptions on a variableWarning, assuming -2*a+a is

positive. Hint: run assume to make assumptions on a variableUnable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_n

ostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*

pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2

)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_no

step/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*p

i/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign

: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to chec

k sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable t

o check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Un

able to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_noste

p/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t

_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-

4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep

/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_

nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4

*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check si

gn: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to ch

eck sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable

to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)

Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nos

tep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi

/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>

(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nost

ep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/

t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign:

(4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check

sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to

check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unab

le to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/

2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_n