3.522 \(\int \frac{\sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=120 \[ \frac{(A+3 B-7 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(A-B+C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac{2 C \tan (c+d x)}{a d \sqrt{a \sec (c+d x)+a}} \]

[Out]

((A + 3*B - 7*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) + ((
A - B + C)*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + (2*C*Tan[c + d*x])/(a*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.246218, antiderivative size = 135, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {4078, 4001, 3795, 203} \[ \frac{(A+3 B-7 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(A-B+5 C) \tan (c+d x)}{2 a d \sqrt{a \sec (c+d x)+a}}-\frac{(A-B+C) \tan (c+d x) \sec (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

((A + 3*B - 7*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((
A - B + C)*Sec[c + d*x]*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ((A - B + 5*C)*Tan[c + d*x])/(2*a*d*S
qrt[a + a*Sec[c + d*x]])

Rule 4078

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*Csc[e + f*x]*(a + b*Cs
c[e + f*x])^m)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Sim
p[a*B - b*C - 2*A*b*(m + 1) - (b*B*(m + 2) - a*(A*(m + 2) - C*(m - 1)))*Csc[e + f*x], x], x], x] /; FreeQ[{a,
b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

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])

Rubi steps

\begin{align*} \int \frac{\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\sec (c+d x) \left (a (A+B-C)+\frac{1}{2} a (A-B+5 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(A-B+5 C) \tan (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}+\frac{(A+3 B-7 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(A-B+5 C) \tan (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}-\frac{(A+3 B-7 C) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{2 a d}\\ &=\frac{(A+3 B-7 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A-B+C) \sec (c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(A-B+5 C) \tan (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 6.48058, size = 748, normalized size = 6.23 \[ \frac{4 \sqrt{\frac{1}{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}} \sqrt{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )} \cos ^3\left (\frac{1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{(7 A-3 B-C) \sin \left (\frac{1}{2} (c+d x)\right ) \left (\frac{2 \sin ^2\left (\frac{1}{2} (c+d x)\right ) \cos ^2\left (\frac{1}{2} (c+d x)\right ) \text{Hypergeometric2F1}\left (2,\frac{5}{2},\frac{7}{2},-\frac{\sin ^2\left (\frac{1}{2} (c+d x)\right )}{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}\right )}{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}+5 \sqrt{-\frac{\sin ^2\left (\frac{1}{2} (c+d x)\right )}{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}} \left (1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )\right )^2 \left (3-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )\right ) \csc ^4\left (\frac{1}{2} (c+d x)\right ) \left (\sqrt{-\frac{\sin ^2\left (\frac{1}{2} (c+d x)\right )}{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}}-\tanh ^{-1}\left (\sqrt{-\frac{\sin ^2\left (\frac{1}{2} (c+d x)\right )}{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}}\right )\right )\right )}{10 \left (1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )\right )^{3/2}}+\frac{(A-B+C) \sqrt{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}}{1-\sin \left (\frac{1}{2} (c+d x)\right )}-\frac{(A-B+C) \sqrt{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}}{\sin \left (\frac{1}{2} (c+d x)\right )+1}+\frac{(A-B+C) \left (2 \sin \left (\frac{1}{2} (c+d x)\right )+1\right )}{4 \left (1-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \sqrt{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}}-\frac{(A-B+C) \left (1-2 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{4 \left (\sin \left (\frac{1}{2} (c+d x)\right )+1\right ) \sqrt{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}}+\frac{3}{2} (A-B+C) \tan ^{-1}\left (\frac{1-2 \sin \left (\frac{1}{2} (c+d x)\right )}{\sqrt{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}}\right )-\frac{3}{2} (A-B+C) \tan ^{-1}\left (\frac{2 \sin \left (\frac{1}{2} (c+d x)\right )+1}{\sqrt{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}}\right )+\frac{4 A \sin \left (\frac{1}{2} (c+d x)\right )}{\sqrt{1-2 \sin ^2\left (\frac{1}{2} (c+d x)\right )}}\right )}{d \sqrt{\sec (c+d x)} (a (\sec (c+d x)+1))^{3/2} (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(3/2),x]

[Out]

(4*Cos[(c + d*x)/2]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - 2*Sin[(c + d*x)/2]^2)^(-1)]*Sqrt[1 - 2
*Sin[(c + d*x)/2]^2]*((3*(A - B + C)*ArcTan[(1 - 2*Sin[(c + d*x)/2])/Sqrt[1 - 2*Sin[(c + d*x)/2]^2]])/2 - (3*(
A - B + C)*ArcTan[(1 + 2*Sin[(c + d*x)/2])/Sqrt[1 - 2*Sin[(c + d*x)/2]^2]])/2 + (4*A*Sin[(c + d*x)/2])/Sqrt[1
- 2*Sin[(c + d*x)/2]^2] - ((A - B + C)*(1 - 2*Sin[(c + d*x)/2]))/(4*(1 + Sin[(c + d*x)/2])*Sqrt[1 - 2*Sin[(c +
d*x)/2]^2]) + ((A - B + C)*(1 + 2*Sin[(c + d*x)/2]))/(4*(1 - Sin[(c + d*x)/2])*Sqrt[1 - 2*Sin[(c + d*x)/2]^2]
) + ((A - B + C)*Sqrt[1 - 2*Sin[(c + d*x)/2]^2])/(1 - Sin[(c + d*x)/2]) - ((A - B + C)*Sqrt[1 - 2*Sin[(c + d*x
)/2]^2])/(1 + Sin[(c + d*x)/2]) + ((7*A - 3*B - C)*Sin[(c + d*x)/2]*((2*Cos[(c + d*x)/2]^2*Hypergeometric2F1[2
, 5/2, 7/2, -(Sin[(c + d*x)/2]^2/(1 - 2*Sin[(c + d*x)/2]^2))]*Sin[(c + d*x)/2]^2)/(1 - 2*Sin[(c + d*x)/2]^2) +
5*Csc[(c + d*x)/2]^4*Sqrt[-(Sin[(c + d*x)/2]^2/(1 - 2*Sin[(c + d*x)/2]^2))]*(1 - 2*Sin[(c + d*x)/2]^2)^2*(3 -
2*Sin[(c + d*x)/2]^2)*(-ArcTanh[Sqrt[-(Sin[(c + d*x)/2]^2/(1 - 2*Sin[(c + d*x)/2]^2))]] + Sqrt[-(Sin[(c + d*x
)/2]^2/(1 - 2*Sin[(c + d*x)/2]^2))])))/(10*(1 - 2*Sin[(c + d*x)/2]^2)^(3/2))))/(d*(A + 2*C + 2*B*Cos[c + d*x]
+ A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [B]  time = 0.306, size = 583, normalized size = 4.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x)

[Out]

1/4/d/a^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(-A*cos(d*x+c)*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(
cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-3*B*cos(d*x+c)*
sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos
(d*x+c)+1))^(1/2)+7*C*cos(d*x+c)*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1
)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-A*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos
(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-3*B*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1)
)^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+7*C*ln(-(-(-2*cos
(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*
x+c)+2*A*cos(d*x+c)^2-2*B*cos(d*x+c)^2+10*C*cos(d*x+c)^2-2*A*cos(d*x+c)+2*B*cos(d*x+c)-2*C*cos(d*x+c)-8*C)/sin
(d*x+c)^3

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Maxima [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(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.611697, size = 1046, normalized size = 8.72 \begin{align*} \left [\frac{\sqrt{2}{\left ({\left (A + 3 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (A + 3 \, B - 7 \, C\right )} \cos \left (d x + c\right ) + A + 3 \, B - 7 \, C\right )} \sqrt{-a} \log \left (-\frac{2 \, \sqrt{2} \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \,{\left ({\left (A - B + 5 \, C\right )} \cos \left (d x + c\right ) + 4 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{8 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, -\frac{\sqrt{2}{\left ({\left (A + 3 \, B - 7 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (A + 3 \, B - 7 \, C\right )} \cos \left (d x + c\right ) + A + 3 \, B - 7 \, C\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) - 2 \,{\left ({\left (A - B + 5 \, C\right )} \cos \left (d x + c\right ) + 4 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{4 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/8*(sqrt(2)*((A + 3*B - 7*C)*cos(d*x + c)^2 + 2*(A + 3*B - 7*C)*cos(d*x + c) + A + 3*B - 7*C)*sqrt(-a)*log(-
(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) - 3*a*cos(d*x + c)^2 - 2
*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((A - B + 5*C)*cos(d*x + c) + 4*C)*sqrt((a*cos
(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), -1/4*(sqrt(2
)*((A + 3*B - 7*C)*cos(d*x + c)^2 + 2*(A + 3*B - 7*C)*cos(d*x + c) + A + 3*B - 7*C)*sqrt(a)*arctan(sqrt(2)*sqr
t((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 2*((A - B + 5*C)*cos(d*x + c) + 4*
C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)/(a*(sec(c + d*x) + 1))**(3/2), x)

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Giac [A]  time = 8.98315, size = 271, normalized size = 2.26 \begin{align*} \frac{\frac{{\left (\frac{\sqrt{2}{\left (A a^{2} - B a^{2} + C a^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{\sqrt{2}{\left (A a^{2} - B a^{2} + 9 \, C a^{2}\right )}}{a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}} + \frac{\sqrt{2}{\left (A + 3 \, B - 7 \, C\right )} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

1/4*((sqrt(2)*(A*a^2 - B*a^2 + C*a^2)*tan(1/2*d*x + 1/2*c)^2/(a^3*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - sqrt(2)*(
A*a^2 - B*a^2 + 9*C*a^2)/(a^3*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)/sqrt(-a*tan(1/2*d*x + 1/2
*c)^2 + a) + sqrt(2)*(A + 3*B - 7*C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 +
a)))/(sqrt(-a)*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d