3.342 \(\int \frac{x^{17/2}}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=242 \[ \frac{3 \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{5/4} c^{7/4}}-\frac{3 \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{5/4} c^{7/4}}-\frac{3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{5/4} c^{7/4}}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{5/4} c^{7/4}}+\frac{3 x^{3/2}}{16 b c \left (b+c x^2\right )}-\frac{x^{3/2}}{4 c \left (b+c x^2\right )^2} \]

[Out]

-x^(3/2)/(4*c*(b + c*x^2)^2) + (3*x^(3/2))/(16*b*c*(b + c*x^2)) - (3*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1
/4)])/(32*Sqrt[2]*b^(5/4)*c^(7/4)) + (3*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(5/4)*c^(
7/4)) + (3*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(5/4)*c^(7/4)) - (3*Log[S
qrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(5/4)*c^(7/4))

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Rubi [A]  time = 0.188428, antiderivative size = 242, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.526, Rules used = {1584, 288, 290, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{3 \log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{5/4} c^{7/4}}-\frac{3 \log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{64 \sqrt{2} b^{5/4} c^{7/4}}-\frac{3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{5/4} c^{7/4}}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt{2} b^{5/4} c^{7/4}}+\frac{3 x^{3/2}}{16 b c \left (b+c x^2\right )}-\frac{x^{3/2}}{4 c \left (b+c x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(17/2)/(b*x^2 + c*x^4)^3,x]

[Out]

-x^(3/2)/(4*c*(b + c*x^2)^2) + (3*x^(3/2))/(16*b*c*(b + c*x^2)) - (3*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1
/4)])/(32*Sqrt[2]*b^(5/4)*c^(7/4)) + (3*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(5/4)*c^(
7/4)) + (3*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(5/4)*c^(7/4)) - (3*Log[S
qrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(5/4)*c^(7/4))

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{x^{17/2}}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac{x^{5/2}}{\left (b+c x^2\right )^3} \, dx\\ &=-\frac{x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac{3 \int \frac{\sqrt{x}}{\left (b+c x^2\right )^2} \, dx}{8 c}\\ &=-\frac{x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac{3 x^{3/2}}{16 b c \left (b+c x^2\right )}+\frac{3 \int \frac{\sqrt{x}}{b+c x^2} \, dx}{32 b c}\\ &=-\frac{x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac{3 x^{3/2}}{16 b c \left (b+c x^2\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{16 b c}\\ &=-\frac{x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac{3 x^{3/2}}{16 b c \left (b+c x^2\right )}-\frac{3 \operatorname{Subst}\left (\int \frac{\sqrt{b}-\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{32 b c^{3/2}}+\frac{3 \operatorname{Subst}\left (\int \frac{\sqrt{b}+\sqrt{c} x^2}{b+c x^4} \, dx,x,\sqrt{x}\right )}{32 b c^{3/2}}\\ &=-\frac{x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac{3 x^{3/2}}{16 b c \left (b+c x^2\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{b}}{\sqrt{c}}-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt{x}\right )}{64 b c^2}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt{b}}{\sqrt{c}}+\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt{x}\right )}{64 b c^2}+\frac{3 \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{b}}{\sqrt [4]{c}}+2 x}{-\frac{\sqrt{b}}{\sqrt{c}}-\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt{x}\right )}{64 \sqrt{2} b^{5/4} c^{7/4}}+\frac{3 \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [4]{b}}{\sqrt [4]{c}}-2 x}{-\frac{\sqrt{b}}{\sqrt{c}}+\frac{\sqrt{2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt{x}\right )}{64 \sqrt{2} b^{5/4} c^{7/4}}\\ &=-\frac{x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac{3 x^{3/2}}{16 b c \left (b+c x^2\right )}+\frac{3 \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{5/4} c^{7/4}}-\frac{3 \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{5/4} c^{7/4}}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{5/4} c^{7/4}}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{5/4} c^{7/4}}\\ &=-\frac{x^{3/2}}{4 c \left (b+c x^2\right )^2}+\frac{3 x^{3/2}}{16 b c \left (b+c x^2\right )}-\frac{3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{5/4} c^{7/4}}+\frac{3 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{32 \sqrt{2} b^{5/4} c^{7/4}}+\frac{3 \log \left (\sqrt{b}-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{5/4} c^{7/4}}-\frac{3 \log \left (\sqrt{b}+\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{c} x\right )}{64 \sqrt{2} b^{5/4} c^{7/4}}\\ \end{align*}

Mathematica [C]  time = 0.0154096, size = 45, normalized size = 0.19 \[ \frac{2 x^{3/2} \left (\frac{\, _2F_1\left (\frac{3}{4},3;\frac{7}{4};-\frac{c x^2}{b}\right )}{b^2}-\frac{1}{\left (b+c x^2\right )^2}\right )}{5 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(17/2)/(b*x^2 + c*x^4)^3,x]

[Out]

(2*x^(3/2)*(-(b + c*x^2)^(-2) + Hypergeometric2F1[3/4, 3, 7/4, -((c*x^2)/b)]/b^2))/(5*c)

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Maple [A]  time = 0.061, size = 169, normalized size = 0.7 \begin{align*} 2\,{\frac{1}{ \left ( c{x}^{2}+b \right ) ^{2}} \left ({\frac{3\,{x}^{7/2}}{32\,b}}-1/32\,{\frac{{x}^{3/2}}{c}} \right ) }+{\frac{3\,\sqrt{2}}{128\,b{c}^{2}}\ln \left ({ \left ( x-\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) \left ( x+\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{3\,\sqrt{2}}{64\,b{c}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{3\,\sqrt{2}}{64\,b{c}^{2}}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(17/2)/(c*x^4+b*x^2)^3,x)

[Out]

2*(3/32/b*x^(7/2)-1/32*x^(3/2)/c)/(c*x^2+b)^2+3/128/b/c^2/(b/c)^(1/4)*2^(1/2)*ln((x-(b/c)^(1/4)*x^(1/2)*2^(1/2
)+(b/c)^(1/2))/(x+(b/c)^(1/4)*x^(1/2)*2^(1/2)+(b/c)^(1/2)))+3/64/b/c^2/(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c
)^(1/4)*x^(1/2)+1)+3/64/b/c^2/(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.3718, size = 613, normalized size = 2.53 \begin{align*} -\frac{12 \,{\left (b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} + b^{3} c\right )} \left (-\frac{1}{b^{5} c^{7}}\right )^{\frac{1}{4}} \arctan \left (\sqrt{-b^{3} c^{3} \sqrt{-\frac{1}{b^{5} c^{7}}} + x} b c^{2} \left (-\frac{1}{b^{5} c^{7}}\right )^{\frac{1}{4}} - b c^{2} \sqrt{x} \left (-\frac{1}{b^{5} c^{7}}\right )^{\frac{1}{4}}\right ) - 3 \,{\left (b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} + b^{3} c\right )} \left (-\frac{1}{b^{5} c^{7}}\right )^{\frac{1}{4}} \log \left (b^{4} c^{5} \left (-\frac{1}{b^{5} c^{7}}\right )^{\frac{3}{4}} + \sqrt{x}\right ) + 3 \,{\left (b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} + b^{3} c\right )} \left (-\frac{1}{b^{5} c^{7}}\right )^{\frac{1}{4}} \log \left (-b^{4} c^{5} \left (-\frac{1}{b^{5} c^{7}}\right )^{\frac{3}{4}} + \sqrt{x}\right ) - 4 \,{\left (3 \, c x^{3} - b x\right )} \sqrt{x}}{64 \,{\left (b c^{3} x^{4} + 2 \, b^{2} c^{2} x^{2} + b^{3} c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

-1/64*(12*(b*c^3*x^4 + 2*b^2*c^2*x^2 + b^3*c)*(-1/(b^5*c^7))^(1/4)*arctan(sqrt(-b^3*c^3*sqrt(-1/(b^5*c^7)) + x
)*b*c^2*(-1/(b^5*c^7))^(1/4) - b*c^2*sqrt(x)*(-1/(b^5*c^7))^(1/4)) - 3*(b*c^3*x^4 + 2*b^2*c^2*x^2 + b^3*c)*(-1
/(b^5*c^7))^(1/4)*log(b^4*c^5*(-1/(b^5*c^7))^(3/4) + sqrt(x)) + 3*(b*c^3*x^4 + 2*b^2*c^2*x^2 + b^3*c)*(-1/(b^5
*c^7))^(1/4)*log(-b^4*c^5*(-1/(b^5*c^7))^(3/4) + sqrt(x)) - 4*(3*c*x^3 - b*x)*sqrt(x))/(b*c^3*x^4 + 2*b^2*c^2*
x^2 + b^3*c)

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Sympy [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(x**(17/2)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.16774, size = 286, normalized size = 1.18 \begin{align*} \frac{3 \, c x^{\frac{7}{2}} - b x^{\frac{3}{2}}}{16 \,{\left (c x^{2} + b\right )}^{2} b c} + \frac{3 \, \sqrt{2} \left (b c^{3}\right )^{\frac{3}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{b}{c}\right )^{\frac{1}{4}} + 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{b}{c}\right )^{\frac{1}{4}}}\right )}{64 \, b^{2} c^{4}} + \frac{3 \, \sqrt{2} \left (b c^{3}\right )^{\frac{3}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (\frac{b}{c}\right )^{\frac{1}{4}} - 2 \, \sqrt{x}\right )}}{2 \, \left (\frac{b}{c}\right )^{\frac{1}{4}}}\right )}{64 \, b^{2} c^{4}} - \frac{3 \, \sqrt{2} \left (b c^{3}\right )^{\frac{3}{4}} \log \left (\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{128 \, b^{2} c^{4}} + \frac{3 \, \sqrt{2} \left (b c^{3}\right )^{\frac{3}{4}} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{128 \, b^{2} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(17/2)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

1/16*(3*c*x^(7/2) - b*x^(3/2))/((c*x^2 + b)^2*b*c) + 3/64*sqrt(2)*(b*c^3)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(b
/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/(b^2*c^4) + 3/64*sqrt(2)*(b*c^3)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^
(1/4) - 2*sqrt(x))/(b/c)^(1/4))/(b^2*c^4) - 3/128*sqrt(2)*(b*c^3)^(3/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x +
sqrt(b/c))/(b^2*c^4) + 3/128*sqrt(2)*(b*c^3)^(3/4)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^2*c^4)