\(\int \frac {(c+d \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx\) [603]

Optimal result

Integrand size = 29, antiderivative size = 194 \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {2 d^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{a^{3/2} f}-\frac {\sqrt {c-d} (c+5 d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} f}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a+a \sin (e+f x))^{3/2}} \] Output:

-2*d^(3/2)*arctan(a^(1/2)*d^(1/2)*cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/(c+d*s 
in(f*x+e))^(1/2))/a^(3/2)/f-1/4*(c-d)^(1/2)*(c+5*d)*arctanh(1/2*a^(1/2)*(c 
-d)^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2) 
)*2^(1/2)/a^(3/2)/f-1/2*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/f/(a+a*sin 
(f*x+e))^(3/2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 15.61 (sec) , antiderivative size = 1625, normalized size of antiderivative = 8.38 \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

Integrate[(c + d*Sin[e + f*x])^(3/2)/(a + a*Sin[e + f*x])^(3/2),x]
 

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*((-c + d)/(2*(Cos[(e + f*x)/2] + 
Sin[(e + f*x)/2])) + (c*Sin[(e + f*x)/2] - d*Sin[(e + f*x)/2])/(Cos[(e + f 
*x)/2] + Sin[(e + f*x)/2])^2)*Sqrt[c + d*Sin[e + f*x]])/(f*(a*(1 + Sin[e + 
 f*x]))^(3/2)) + ((Sqrt[2]*(c^2 + 4*c*d - 5*d^2)*Log[1 + Tan[(e + f*x)/2]] 
 - Sqrt[2]*(c^2 + 4*c*d - 5*d^2)*Log[c - d + 2*Sqrt[c - d]*Sqrt[(1 + Cos[e 
 + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (-c + d)*Tan[(e + f*x)/2]] - (4* 
I)*Sqrt[c - d]*d^(3/2)*(Log[(c - I*(d + (1 + I)*Sqrt[2]*Sqrt[d]*Sqrt[(1 + 
Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]]) + ((-I)*c + d)*Tan[(e + f*x) 
/2])/(2*d^(5/2)*(I + Tan[(e + f*x)/2]))] - Log[(c + I*d + (1 + I)*Sqrt[2]* 
Sqrt[d]*Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[c + d*Sin[e + f*x]] + (I*c + d) 
*Tan[(e + f*x)/2])/(2*d^(5/2)*(-I + Tan[(e + f*x)/2]))]))*(Cos[(e + f*x)/2 
] + Sin[(e + f*x)/2])^3*(c^2/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt 
[c + d*Sin[e + f*x]]) + (c*d)/((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[ 
c + d*Sin[e + f*x]]) - d^2/(4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c 
 + d*Sin[e + f*x]]) + (d^2*Sin[e + f*x])/((Cos[(e + f*x)/2] + Sin[(e + f*x 
)/2])*Sqrt[c + d*Sin[e + f*x]])))/(f*(a*(1 + Sin[e + f*x]))^(3/2)*(((c^2 + 
 4*c*d - 5*d^2)*Sec[(e + f*x)/2]^2)/(Sqrt[2]*(1 + Tan[(e + f*x)/2])) - (Sq 
rt[2]*(c^2 + 4*c*d - 5*d^2)*(((-c + d)*Sec[(e + f*x)/2]^2)/2 + (Sqrt[c - d 
]*d*Cos[e + f*x]*Sqrt[(1 + Cos[e + f*x])^(-1)])/Sqrt[c + d*Sin[e + f*x]] + 
 Sqrt[c - d]*((1 + Cos[e + f*x])^(-1))^(3/2)*Sin[e + f*x]*Sqrt[c + d*Si...
 

Rubi [A] (verified)

Time = 1.00 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {3042, 3244, 27, 3042, 3461, 3042, 3254, 218, 3261, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(c+d \sin (e+f x))^{3/2}}{(a \sin (e+f x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c+d \sin (e+f x))^{3/2}}{(a \sin (e+f x)+a)^{3/2}}dx\)

\(\Big \downarrow \) 3244

\(\displaystyle -\frac {\int -\frac {4 a \sin (e+f x) d^2+a \left (c^2+4 d c-d^2\right )}{2 \sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{2 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {4 a \sin (e+f x) d^2+a \left (c^2+4 d c-d^2\right )}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {4 a \sin (e+f x) d^2+a \left (c^2+4 d c-d^2\right )}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3461

\(\displaystyle \frac {4 d^2 \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx+a (c-d) (c+5 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 d^2 \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c+d \sin (e+f x)}}dx+a (c-d) (c+5 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3254

\(\displaystyle \frac {a (c-d) (c+5 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {8 a d^2 \int \frac {1}{\frac {a^2 d \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}+a}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {a (c-d) (c+5 d) \int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}dx-\frac {8 \sqrt {a} d^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3261

\(\displaystyle \frac {-\frac {2 a^2 (c-d) (c+5 d) \int \frac {1}{2 a^2-\frac {a^3 (c-d) \cos ^2(e+f x)}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a} \sqrt {c+d \sin (e+f x)}}}{f}-\frac {8 \sqrt {a} d^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {8 \sqrt {a} d^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}-\frac {\sqrt {2} \sqrt {a} \sqrt {c-d} (c+5 d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c-d} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{f}}{4 a^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{2 f (a \sin (e+f x)+a)^{3/2}}\)

Input:

Int[(c + d*Sin[e + f*x])^(3/2)/(a + a*Sin[e + f*x])^(3/2),x]
 

Output:

((-8*Sqrt[a]*d^(3/2)*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin 
[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/f - (Sqrt[2]*Sqrt[a]*Sqrt[c - d]*(c 
 + 5*d)*ArcTanh[(Sqrt[a]*Sqrt[c - d]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin 
[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])])/f)/(4*a^2) - ((c - d)*Cos[e + f*x]* 
Sqrt[c + d*Sin[e + f*x]])/(2*f*(a + a*Sin[e + f*x])^(3/2))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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

rule 221

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3244

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e 
+ f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Simp[1/(a*b* 
(2*m + 1))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)* 
Simp[b*(c^2*(m + 1) + d^2*(n - 1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) 
 + b*c*(m + n))*Sin[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, -1] 
&& GtQ[n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
 

rule 3254

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*(b/f)   Subst[Int[1/(b + d*x^2), x], 
x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*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]
 

rule 3261

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e 
_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f)   Subst[Int[1/(2*b^2 - (a*c 
 - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S 
in[e + f*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]
 

rule 3461

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + 
(f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Sim 
p[(A*b - a*B)/b   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]) 
, x], x] + Simp[B/b   Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]] 
, x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

Maple [F(-1)]

Timed out.

hanged

Input:

int((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(3/2),x)
 

Output:

int((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(3/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 516 vs. \(2 (159) = 318\).

Time = 0.50 (sec) , antiderivative size = 2977, normalized size of antiderivative = 15.35 \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="fric 
as")
 

Output:

[1/4*(sqrt(1/2)*((a*c + 5*a*d)*cos(f*x + e)^2 - 2*a*c - 10*a*d - (a*c + 5* 
a*d)*cos(f*x + e) - (2*a*c + 10*a*d + (a*c + 5*a*d)*cos(f*x + e))*sin(f*x 
+ e))*sqrt((c - d)/a)*log((4*sqrt(1/2)*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin 
(f*x + e) + c)*sqrt((c - d)/a)*(cos(f*x + e) - sin(f*x + e) + 1) - (c - 3* 
d)*cos(f*x + e)^2 - (3*c - d)*cos(f*x + e) + ((c - 3*d)*cos(f*x + e) - 2*c 
 - 2*d)*sin(f*x + e) - 2*c - 2*d)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin 
(f*x + e) - cos(f*x + e) - 2)) + (a*d*cos(f*x + e)^2 - a*d*cos(f*x + e) - 
2*a*d - (a*d*cos(f*x + e) + 2*a*d)*sin(f*x + e))*sqrt(-d/a)*log((128*d^4*c 
os(f*x + e)^5 + 128*(2*c*d^3 - d^4)*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c^2 
*d^2 + 4*c*d^3 + d^4 - 32*(5*c^2*d^2 - 14*c*d^3 + 13*d^4)*cos(f*x + e)^3 - 
 32*(c^3*d - 2*c^2*d^2 + 9*c*d^3 - 4*d^4)*cos(f*x + e)^2 - 8*(16*d^3*cos(f 
*x + e)^4 + 24*(c*d^2 - d^3)*cos(f*x + e)^3 - c^3 + 17*c^2*d - 59*c*d^2 + 
51*d^3 - 2*(5*c^2*d - 26*c*d^2 + 33*d^3)*cos(f*x + e)^2 - (c^3 - 7*c^2*d + 
 31*c*d^2 - 25*d^3)*cos(f*x + e) + (16*d^3*cos(f*x + e)^3 + c^3 - 17*c^2*d 
 + 59*c*d^2 - 51*d^3 - 8*(3*c*d^2 - 5*d^3)*cos(f*x + e)^2 - 2*(5*c^2*d - 1 
4*c*d^2 + 13*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqr 
t(d*sin(f*x + e) + c)*sqrt(-d/a) + (c^4 - 28*c^3*d + 230*c^2*d^2 - 476*c*d 
^3 + 289*d^4)*cos(f*x + e) + (128*d^4*cos(f*x + e)^4 + c^4 + 4*c^3*d + 6*c 
^2*d^2 + 4*c*d^3 + d^4 - 256*(c*d^3 - d^4)*cos(f*x + e)^3 - 32*(5*c^2*d^2 
- 6*c*d^3 + 5*d^4)*cos(f*x + e)^2 + 32*(c^3*d - 7*c^2*d^2 + 15*c*d^3 - ...
 

Sympy [F]

\[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\int \frac {\left (c + d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((c+d*sin(f*x+e))**(3/2)/(a+a*sin(f*x+e))**(3/2),x)
 

Output:

Integral((c + d*sin(e + f*x))**(3/2)/(a*(sin(e + f*x) + 1))**(3/2), x)
 

Maxima [F]

\[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxi 
ma")
 

Output:

integrate((d*sin(f*x + e) + c)^(3/2)/(a*sin(f*x + e) + a)^(3/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac 
")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

int((c + d*sin(e + f*x))^(3/2)/(a + a*sin(e + f*x))^(3/2),x)
 

Output:

int((c + d*sin(e + f*x))^(3/2)/(a + a*sin(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1}d x \right ) d +\left (\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1}d x \right ) c \right )}{a^{2}} \] Input:

int((c+d*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(3/2),x)
 

Output:

(sqrt(a)*(int((sqrt(sin(e + f*x)*d + c)*sqrt(sin(e + f*x) + 1)*sin(e + f*x 
))/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1),x)*d + int((sqrt(sin(e + f*x)*d 
+ c)*sqrt(sin(e + f*x) + 1))/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1),x)*c)) 
/a**2