\(\int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [791]

Optimal result

Integrand size = 45, antiderivative size = 109 \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 \sqrt {a} B \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

2*B*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x+e))^(1/2))*a^(1/2)/f/c^(1/2)-(I*A+B)*(a+I*a
*tan(f*x+e))^(1/2)/f/(c-I*c*tan(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3669, 79, 65, 223, 209} \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 \sqrt {a} B \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {(B+i A) \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}} \]

[In]

Int[(Sqrt[a + I*a*Tan[e + f*x]]*(A + B*Tan[e + f*x]))/Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

(2*Sqrt[a]*B*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(Sqrt[c]*f) -
((I*A + B)*Sqrt[a + I*a*Tan[e + f*x]])/(f*Sqrt[c - I*c*Tan[e + f*x]])

Rule 65

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 209

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3669

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*(c/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {A+B x}{\sqrt {a+i a x} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}}+\frac {(i a B) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}}+\frac {(2 B) \text {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{f} \\ & = -\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}}+\frac {(2 B) \text {Subst}\left (\int \frac {1}{1+\frac {c x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c-i c \tan (e+f x)}}\right )}{f} \\ & = \frac {2 \sqrt {a} B \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.72 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.01 \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 \sqrt {a} B \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}+\frac {(-i A-B) \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}} \]

[In]

Integrate[(Sqrt[a + I*a*Tan[e + f*x]]*(A + B*Tan[e + f*x]))/Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

(2*Sqrt[a]*B*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/(Sqrt[c]*f) +
(((-I)*A - B)*Sqrt[a + I*a*Tan[e + f*x]])/(f*Sqrt[c - I*c*Tan[e + f*x]])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (88 ) = 176\).

Time = 0.39 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.94

method	result	size
derivativedivides	\(-\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (-2 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}+i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{f c \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{2} \sqrt {a c}}\)	\(321\)
default	\(-\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (-2 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}+i A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{f c \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{2} \sqrt {a c}}\)	\(321\)
parts	\(-\frac {i A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (i \tan \left (f x +e \right )-1\right )}{f c \left (i+\tan \left (f x +e \right )\right )^{2}}+\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right ) a c +\ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right )^{2} a c -a c \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right )-i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}-\tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{f c \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{2} \sqrt {a c}}\)	\(329\)

[In]

int((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)/c*(-2*I*B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+t
an(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)-B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)
)/(a*c)^(1/2))*a*c*tan(f*x+e)^2+I*A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)+I*B*(a*c)^(1/2)*(a*c*(
1+tan(f*x+e)^2))^(1/2)+B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2))*a*c+B*(a*c)
^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)-A*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c*(1+tan(f*x+e)^
2))^(1/2)/(I+tan(f*x+e))^2/(a*c)^(1/2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (83) = 166\).

Time = 0.27 (sec) , antiderivative size = 332, normalized size of antiderivative = 3.05 \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {c f \sqrt {-\frac {B^{2} a}{c f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (B e^{\left (3 i \, f x + 3 i \, e\right )} + B e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )} \sqrt {-\frac {B^{2} a}{c f^{2}}}\right )}}{B e^{\left (2 i \, f x + 2 i \, e\right )} + B}\right ) - c f \sqrt {-\frac {B^{2} a}{c f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (B e^{\left (3 i \, f x + 3 i \, e\right )} + B e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )} \sqrt {-\frac {B^{2} a}{c f^{2}}}\right )}}{B e^{\left (2 i \, f x + 2 i \, e\right )} + B}\right ) + 2 \, {\left ({\left (i \, A + B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (i \, A + B\right )} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{2 \, c f} \]

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/2*(c*f*sqrt(-B^2*a/(c*f^2))*log(4*(2*(B*e^(3*I*f*x + 3*I*e) + B*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e
) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) + (c*f*e^(2*I*f*x + 2*I*e) - c*f)*sqrt(-B^2*a/(c*f^2)))/(B*e^(2*I*f*
x + 2*I*e) + B)) - c*f*sqrt(-B^2*a/(c*f^2))*log(4*(2*(B*e^(3*I*f*x + 3*I*e) + B*e^(I*f*x + I*e))*sqrt(a/(e^(2*
I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) - (c*f*e^(2*I*f*x + 2*I*e) - c*f)*sqrt(-B^2*a/(c*f^2)))
/(B*e^(2*I*f*x + 2*I*e) + B)) + 2*((I*A + B)*e^(3*I*f*x + 3*I*e) + (I*A + B)*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f
*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(c*f)

Sympy [F]

\[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\int \frac {\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \left (A + B \tan {\left (e + f x \right )}\right )}{\sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}\, dx \]

[In]

integrate((a+I*a*tan(f*x+e))**(1/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(1/2),x)

[Out]

Integral(sqrt(I*a*(tan(e + f*x) - I))*(A + B*tan(e + f*x))/sqrt(-I*c*(tan(e + f*x) + I)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

\[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} \sqrt {i \, a \tan \left (f x + e\right ) + a}}{\sqrt {-i \, c \tan \left (f x + e\right ) + c}} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*sqrt(I*a*tan(f*x + e) + a)/sqrt(-I*c*tan(f*x + e) + c), x)

Mupad [B] (verification not implemented)

Time = 11.89 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.44 \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {4\,B\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{\sqrt {a}\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}\right )}{\sqrt {c}\,f}+\frac {A\,\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{c\,f\,\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {4\,B\,a\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{c\,f\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )\,\left (\frac {a}{c}-\frac {{\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}{{\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}^2}+\frac {2\,\sqrt {a}\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{\sqrt {c}\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}\right )} \]

[In]

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(1/2))/(c - c*tan(e + f*x)*1i)^(1/2),x)

[Out]

(4*B*a^(1/2)*atan((c^(1/2)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/(a^(1/2)*((c - c*tan(e + f*x)*1i)^(1/2)
- c^(1/2)))))/(c^(1/2)*f) + (A*(a + a*tan(e + f*x)*1i)^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(c*f*(tan(e + f*x)
+ 1i)) - (4*B*a*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/(c*f*((c - c*tan(e + f*x)*1i)^(1/2) - c^(1/2))*(a/
c - ((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2))^2/((c - c*tan(e + f*x)*1i)^(1/2) - c^(1/2))^2 + (2*a^(1/2)*((a +
a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/(c^(1/2)*((c - c*tan(e + f*x)*1i)^(1/2) - c^(1/2)))))