3.469 \(\int \frac{x^2 (c+d x)^{5/2}}{(a+b x)^2} \, dx\)

Optimal. Leaf size=204 \[ -\frac{a^2 (c+d x)^{7/2}}{b^2 (a+b x) (b c-a d)}-\frac{a (c+d x)^{5/2} (4 b c-9 a d)}{5 b^3 (b c-a d)}-\frac{a (c+d x)^{3/2} (4 b c-9 a d)}{3 b^4}-\frac{a \sqrt{c+d x} (4 b c-9 a d) (b c-a d)}{b^5}+\frac{a (4 b c-9 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{11/2}}+\frac{2 (c+d x)^{7/2}}{7 b^2 d} \]

[Out]

-((a*(4*b*c - 9*a*d)*(b*c - a*d)*Sqrt[c + d*x])/b^5) - (a*(4*b*c - 9*a*d)*(c + d*x)^(3/2))/(3*b^4) - (a*(4*b*c
- 9*a*d)*(c + d*x)^(5/2))/(5*b^3*(b*c - a*d)) + (2*(c + d*x)^(7/2))/(7*b^2*d) - (a^2*(c + d*x)^(7/2))/(b^2*(b
*c - a*d)*(a + b*x)) + (a*(4*b*c - 9*a*d)*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/
b^(11/2)

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Rubi [A]  time = 0.224179, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {89, 80, 50, 63, 208} \[ -\frac{a^2 (c+d x)^{7/2}}{b^2 (a+b x) (b c-a d)}-\frac{a (c+d x)^{5/2} (4 b c-9 a d)}{5 b^3 (b c-a d)}-\frac{a (c+d x)^{3/2} (4 b c-9 a d)}{3 b^4}-\frac{a \sqrt{c+d x} (4 b c-9 a d) (b c-a d)}{b^5}+\frac{a (4 b c-9 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{11/2}}+\frac{2 (c+d x)^{7/2}}{7 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(c + d*x)^(5/2))/(a + b*x)^2,x]

[Out]

-((a*(4*b*c - 9*a*d)*(b*c - a*d)*Sqrt[c + d*x])/b^5) - (a*(4*b*c - 9*a*d)*(c + d*x)^(3/2))/(3*b^4) - (a*(4*b*c
- 9*a*d)*(c + d*x)^(5/2))/(5*b^3*(b*c - a*d)) + (2*(c + d*x)^(7/2))/(7*b^2*d) - (a^2*(c + d*x)^(7/2))/(b^2*(b
*c - a*d)*(a + b*x)) + (a*(4*b*c - 9*a*d)*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/
b^(11/2)

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 208

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

Rubi steps

\begin{align*} \int \frac{x^2 (c+d x)^{5/2}}{(a+b x)^2} \, dx &=-\frac{a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}+\frac{\int \frac{(c+d x)^{5/2} \left (-\frac{1}{2} a (2 b c-7 a d)+b (b c-a d) x\right )}{a+b x} \, dx}{b^2 (b c-a d)}\\ &=\frac{2 (c+d x)^{7/2}}{7 b^2 d}-\frac{a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}-\frac{(a (4 b c-9 a d)) \int \frac{(c+d x)^{5/2}}{a+b x} \, dx}{2 b^2 (b c-a d)}\\ &=-\frac{a (4 b c-9 a d) (c+d x)^{5/2}}{5 b^3 (b c-a d)}+\frac{2 (c+d x)^{7/2}}{7 b^2 d}-\frac{a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}-\frac{(a (4 b c-9 a d)) \int \frac{(c+d x)^{3/2}}{a+b x} \, dx}{2 b^3}\\ &=-\frac{a (4 b c-9 a d) (c+d x)^{3/2}}{3 b^4}-\frac{a (4 b c-9 a d) (c+d x)^{5/2}}{5 b^3 (b c-a d)}+\frac{2 (c+d x)^{7/2}}{7 b^2 d}-\frac{a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}-\frac{(a (4 b c-9 a d) (b c-a d)) \int \frac{\sqrt{c+d x}}{a+b x} \, dx}{2 b^4}\\ &=-\frac{a (4 b c-9 a d) (b c-a d) \sqrt{c+d x}}{b^5}-\frac{a (4 b c-9 a d) (c+d x)^{3/2}}{3 b^4}-\frac{a (4 b c-9 a d) (c+d x)^{5/2}}{5 b^3 (b c-a d)}+\frac{2 (c+d x)^{7/2}}{7 b^2 d}-\frac{a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}-\frac{\left (a (4 b c-9 a d) (b c-a d)^2\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{2 b^5}\\ &=-\frac{a (4 b c-9 a d) (b c-a d) \sqrt{c+d x}}{b^5}-\frac{a (4 b c-9 a d) (c+d x)^{3/2}}{3 b^4}-\frac{a (4 b c-9 a d) (c+d x)^{5/2}}{5 b^3 (b c-a d)}+\frac{2 (c+d x)^{7/2}}{7 b^2 d}-\frac{a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}-\frac{\left (a (4 b c-9 a d) (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{b^5 d}\\ &=-\frac{a (4 b c-9 a d) (b c-a d) \sqrt{c+d x}}{b^5}-\frac{a (4 b c-9 a d) (c+d x)^{3/2}}{3 b^4}-\frac{a (4 b c-9 a d) (c+d x)^{5/2}}{5 b^3 (b c-a d)}+\frac{2 (c+d x)^{7/2}}{7 b^2 d}-\frac{a^2 (c+d x)^{7/2}}{b^2 (b c-a d) (a+b x)}+\frac{a (4 b c-9 a d) (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.289241, size = 181, normalized size = 0.89 \[ \frac{\frac{a (9 a d-4 b c) \left (\sqrt{b} \sqrt{c+d x} \left (15 a^2 d^2-5 a b d (7 c+d x)+b^2 \left (23 c^2+11 c d x+3 d^2 x^2\right )\right )-15 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )\right )}{15 b^{7/2}}-\frac{a^2 (c+d x)^{7/2}}{a+b x}+\frac{2 (c+d x)^{7/2} (b c-a d)}{7 d}}{b^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(c + d*x)^(5/2))/(a + b*x)^2,x]

[Out]

((2*(b*c - a*d)*(c + d*x)^(7/2))/(7*d) - (a^2*(c + d*x)^(7/2))/(a + b*x) + (a*(-4*b*c + 9*a*d)*(Sqrt[b]*Sqrt[c
+ d*x]*(15*a^2*d^2 - 5*a*b*d*(7*c + d*x) + b^2*(23*c^2 + 11*c*d*x + 3*d^2*x^2)) - 15*(b*c - a*d)^(5/2)*ArcTan
h[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]]))/(15*b^(7/2)))/(b^2*(b*c - a*d))

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Maple [B]  time = 0.016, size = 377, normalized size = 1.9 \begin{align*}{\frac{2}{7\,{b}^{2}d} \left ( dx+c \right ) ^{{\frac{7}{2}}}}-{\frac{4\,a}{5\,{b}^{3}} \left ( dx+c \right ) ^{{\frac{5}{2}}}}+2\,{\frac{d \left ( dx+c \right ) ^{3/2}{a}^{2}}{{b}^{4}}}-{\frac{4\,ac}{3\,{b}^{3}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}-8\,{\frac{{d}^{2}{a}^{3}\sqrt{dx+c}}{{b}^{5}}}+12\,{\frac{{a}^{2}dc\sqrt{dx+c}}{{b}^{4}}}-4\,{\frac{a{c}^{2}\sqrt{dx+c}}{{b}^{3}}}-{\frac{{d}^{3}{a}^{4}}{{b}^{5} \left ( bdx+ad \right ) }\sqrt{dx+c}}+2\,{\frac{{d}^{2}{a}^{3}\sqrt{dx+c}c}{{b}^{4} \left ( bdx+ad \right ) }}-{\frac{{a}^{2}d{c}^{2}}{{b}^{3} \left ( bdx+ad \right ) }\sqrt{dx+c}}+9\,{\frac{{d}^{3}{a}^{4}}{{b}^{5}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-22\,{\frac{{d}^{2}{a}^{3}c}{{b}^{4}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+17\,{\frac{{a}^{2}d{c}^{2}}{{b}^{3}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-4\,{\frac{a{c}^{3}}{{b}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(d*x+c)^(5/2)/(b*x+a)^2,x)

[Out]

2/7*(d*x+c)^(7/2)/b^2/d-4/5/b^3*a*(d*x+c)^(5/2)+2*d/b^4*(d*x+c)^(3/2)*a^2-4/3/b^3*(d*x+c)^(3/2)*a*c-8*d^2/b^5*
a^3*(d*x+c)^(1/2)+12*d/b^4*a^2*c*(d*x+c)^(1/2)-4/b^3*a*c^2*(d*x+c)^(1/2)-d^3*a^4/b^5*(d*x+c)^(1/2)/(b*d*x+a*d)
+2*d^2*a^3/b^4*(d*x+c)^(1/2)/(b*d*x+a*d)*c-d*a^2/b^3*(d*x+c)^(1/2)/(b*d*x+a*d)*c^2+9*d^3*a^4/b^5/((a*d-b*c)*b)
^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))-22*d^2*a^3/b^4/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/(
(a*d-b*c)*b)^(1/2))*c+17*d*a^2/b^3/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*c^2-4*a/b^2
/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*c^3

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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^2*(d*x+c)^(5/2)/(b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.75509, size = 1331, normalized size = 6.52 \begin{align*} \left [\frac{105 \,{\left (4 \, a^{2} b^{2} c^{2} d - 13 \, a^{3} b c d^{2} + 9 \, a^{4} d^{3} +{\left (4 \, a b^{3} c^{2} d - 13 \, a^{2} b^{2} c d^{2} + 9 \, a^{3} b d^{3}\right )} x\right )} \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b d x + 2 \, b c - a d + 2 \, \sqrt{d x + c} b \sqrt{\frac{b c - a d}{b}}}{b x + a}\right ) + 2 \,{\left (30 \, b^{4} d^{3} x^{4} + 30 \, a b^{3} c^{3} - 749 \, a^{2} b^{2} c^{2} d + 1680 \, a^{3} b c d^{2} - 945 \, a^{4} d^{3} + 18 \,{\left (5 \, b^{4} c d^{2} - 3 \, a b^{3} d^{3}\right )} x^{3} + 2 \,{\left (45 \, b^{4} c^{2} d - 109 \, a b^{3} c d^{2} + 63 \, a^{2} b^{2} d^{3}\right )} x^{2} + 2 \,{\left (15 \, b^{4} c^{3} - 277 \, a b^{3} c^{2} d + 581 \, a^{2} b^{2} c d^{2} - 315 \, a^{3} b d^{3}\right )} x\right )} \sqrt{d x + c}}{210 \,{\left (b^{6} d x + a b^{5} d\right )}}, \frac{105 \,{\left (4 \, a^{2} b^{2} c^{2} d - 13 \, a^{3} b c d^{2} + 9 \, a^{4} d^{3} +{\left (4 \, a b^{3} c^{2} d - 13 \, a^{2} b^{2} c d^{2} + 9 \, a^{3} b d^{3}\right )} x\right )} \sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{\sqrt{d x + c} b \sqrt{-\frac{b c - a d}{b}}}{b c - a d}\right ) +{\left (30 \, b^{4} d^{3} x^{4} + 30 \, a b^{3} c^{3} - 749 \, a^{2} b^{2} c^{2} d + 1680 \, a^{3} b c d^{2} - 945 \, a^{4} d^{3} + 18 \,{\left (5 \, b^{4} c d^{2} - 3 \, a b^{3} d^{3}\right )} x^{3} + 2 \,{\left (45 \, b^{4} c^{2} d - 109 \, a b^{3} c d^{2} + 63 \, a^{2} b^{2} d^{3}\right )} x^{2} + 2 \,{\left (15 \, b^{4} c^{3} - 277 \, a b^{3} c^{2} d + 581 \, a^{2} b^{2} c d^{2} - 315 \, a^{3} b d^{3}\right )} x\right )} \sqrt{d x + c}}{105 \,{\left (b^{6} d x + a b^{5} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(d*x+c)^(5/2)/(b*x+a)^2,x, algorithm="fricas")

[Out]

[1/210*(105*(4*a^2*b^2*c^2*d - 13*a^3*b*c*d^2 + 9*a^4*d^3 + (4*a*b^3*c^2*d - 13*a^2*b^2*c*d^2 + 9*a^3*b*d^3)*x
)*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d + 2*sqrt(d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) + 2*(30*b^4
*d^3*x^4 + 30*a*b^3*c^3 - 749*a^2*b^2*c^2*d + 1680*a^3*b*c*d^2 - 945*a^4*d^3 + 18*(5*b^4*c*d^2 - 3*a*b^3*d^3)*
x^3 + 2*(45*b^4*c^2*d - 109*a*b^3*c*d^2 + 63*a^2*b^2*d^3)*x^2 + 2*(15*b^4*c^3 - 277*a*b^3*c^2*d + 581*a^2*b^2*
c*d^2 - 315*a^3*b*d^3)*x)*sqrt(d*x + c))/(b^6*d*x + a*b^5*d), 1/105*(105*(4*a^2*b^2*c^2*d - 13*a^3*b*c*d^2 + 9
*a^4*d^3 + (4*a*b^3*c^2*d - 13*a^2*b^2*c*d^2 + 9*a^3*b*d^3)*x)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sq
rt(-(b*c - a*d)/b)/(b*c - a*d)) + (30*b^4*d^3*x^4 + 30*a*b^3*c^3 - 749*a^2*b^2*c^2*d + 1680*a^3*b*c*d^2 - 945*
a^4*d^3 + 18*(5*b^4*c*d^2 - 3*a*b^3*d^3)*x^3 + 2*(45*b^4*c^2*d - 109*a*b^3*c*d^2 + 63*a^2*b^2*d^3)*x^2 + 2*(15
*b^4*c^3 - 277*a*b^3*c^2*d + 581*a^2*b^2*c*d^2 - 315*a^3*b*d^3)*x)*sqrt(d*x + c))/(b^6*d*x + a*b^5*d)]

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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**2*(d*x+c)**(5/2)/(b*x+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.19435, size = 385, normalized size = 1.89 \begin{align*} -\frac{{\left (4 \, a b^{3} c^{3} - 17 \, a^{2} b^{2} c^{2} d + 22 \, a^{3} b c d^{2} - 9 \, a^{4} d^{3}\right )} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} b^{5}} - \frac{\sqrt{d x + c} a^{2} b^{2} c^{2} d - 2 \, \sqrt{d x + c} a^{3} b c d^{2} + \sqrt{d x + c} a^{4} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} b^{5}} + \frac{2 \,{\left (15 \,{\left (d x + c\right )}^{\frac{7}{2}} b^{12} d^{6} - 42 \,{\left (d x + c\right )}^{\frac{5}{2}} a b^{11} d^{7} - 70 \,{\left (d x + c\right )}^{\frac{3}{2}} a b^{11} c d^{7} - 210 \, \sqrt{d x + c} a b^{11} c^{2} d^{7} + 105 \,{\left (d x + c\right )}^{\frac{3}{2}} a^{2} b^{10} d^{8} + 630 \, \sqrt{d x + c} a^{2} b^{10} c d^{8} - 420 \, \sqrt{d x + c} a^{3} b^{9} d^{9}\right )}}{105 \, b^{14} d^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(d*x+c)^(5/2)/(b*x+a)^2,x, algorithm="giac")

[Out]

-(4*a*b^3*c^3 - 17*a^2*b^2*c^2*d + 22*a^3*b*c*d^2 - 9*a^4*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(s
qrt(-b^2*c + a*b*d)*b^5) - (sqrt(d*x + c)*a^2*b^2*c^2*d - 2*sqrt(d*x + c)*a^3*b*c*d^2 + sqrt(d*x + c)*a^4*d^3)
/(((d*x + c)*b - b*c + a*d)*b^5) + 2/105*(15*(d*x + c)^(7/2)*b^12*d^6 - 42*(d*x + c)^(5/2)*a*b^11*d^7 - 70*(d*
x + c)^(3/2)*a*b^11*c*d^7 - 210*sqrt(d*x + c)*a*b^11*c^2*d^7 + 105*(d*x + c)^(3/2)*a^2*b^10*d^8 + 630*sqrt(d*x
+ c)*a^2*b^10*c*d^8 - 420*sqrt(d*x + c)*a^3*b^9*d^9)/(b^14*d^7)