3.408 \(\int \frac{(a+b x)^{3/2} (A+B x)}{x^6} \, dx\)

Optimal. Leaf size=172 \[ \frac{b^2 \sqrt{a+b x} (A b-2 a B)}{64 a^2 x^2}-\frac{3 b^3 \sqrt{a+b x} (A b-2 a B)}{128 a^3 x}+\frac{3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{128 a^{7/2}}+\frac{b \sqrt{a+b x} (A b-2 a B)}{16 a x^3}+\frac{(a+b x)^{3/2} (A b-2 a B)}{8 a x^4}-\frac{A (a+b x)^{5/2}}{5 a x^5} \]

[Out]

(b*(A*b - 2*a*B)*Sqrt[a + b*x])/(16*a*x^3) + (b^2*(A*b - 2*a*B)*Sqrt[a + b*x])/(64*a^2*x^2) - (3*b^3*(A*b - 2*
a*B)*Sqrt[a + b*x])/(128*a^3*x) + ((A*b - 2*a*B)*(a + b*x)^(3/2))/(8*a*x^4) - (A*(a + b*x)^(5/2))/(5*a*x^5) +
(3*b^4*(A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(128*a^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0804811, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {78, 47, 51, 63, 208} \[ \frac{b^2 \sqrt{a+b x} (A b-2 a B)}{64 a^2 x^2}-\frac{3 b^3 \sqrt{a+b x} (A b-2 a B)}{128 a^3 x}+\frac{3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{128 a^{7/2}}+\frac{b \sqrt{a+b x} (A b-2 a B)}{16 a x^3}+\frac{(a+b x)^{3/2} (A b-2 a B)}{8 a x^4}-\frac{A (a+b x)^{5/2}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(3/2)*(A + B*x))/x^6,x]

[Out]

(b*(A*b - 2*a*B)*Sqrt[a + b*x])/(16*a*x^3) + (b^2*(A*b - 2*a*B)*Sqrt[a + b*x])/(64*a^2*x^2) - (3*b^3*(A*b - 2*
a*B)*Sqrt[a + b*x])/(128*a^3*x) + ((A*b - 2*a*B)*(a + b*x)^(3/2))/(8*a*x^4) - (A*(a + b*x)^(5/2))/(5*a*x^5) +
(3*b^4*(A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/(128*a^(7/2))

Rule 78

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] || LtQ
[p, n]))))

Rule 47

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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{(a+b x)^{3/2} (A+B x)}{x^6} \, dx &=-\frac{A (a+b x)^{5/2}}{5 a x^5}+\frac{\left (-\frac{5 A b}{2}+5 a B\right ) \int \frac{(a+b x)^{3/2}}{x^5} \, dx}{5 a}\\ &=\frac{(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac{A (a+b x)^{5/2}}{5 a x^5}-\frac{(3 b (A b-2 a B)) \int \frac{\sqrt{a+b x}}{x^4} \, dx}{16 a}\\ &=\frac{b (A b-2 a B) \sqrt{a+b x}}{16 a x^3}+\frac{(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac{A (a+b x)^{5/2}}{5 a x^5}-\frac{\left (b^2 (A b-2 a B)\right ) \int \frac{1}{x^3 \sqrt{a+b x}} \, dx}{32 a}\\ &=\frac{b (A b-2 a B) \sqrt{a+b x}}{16 a x^3}+\frac{b^2 (A b-2 a B) \sqrt{a+b x}}{64 a^2 x^2}+\frac{(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac{A (a+b x)^{5/2}}{5 a x^5}+\frac{\left (3 b^3 (A b-2 a B)\right ) \int \frac{1}{x^2 \sqrt{a+b x}} \, dx}{128 a^2}\\ &=\frac{b (A b-2 a B) \sqrt{a+b x}}{16 a x^3}+\frac{b^2 (A b-2 a B) \sqrt{a+b x}}{64 a^2 x^2}-\frac{3 b^3 (A b-2 a B) \sqrt{a+b x}}{128 a^3 x}+\frac{(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac{A (a+b x)^{5/2}}{5 a x^5}-\frac{\left (3 b^4 (A b-2 a B)\right ) \int \frac{1}{x \sqrt{a+b x}} \, dx}{256 a^3}\\ &=\frac{b (A b-2 a B) \sqrt{a+b x}}{16 a x^3}+\frac{b^2 (A b-2 a B) \sqrt{a+b x}}{64 a^2 x^2}-\frac{3 b^3 (A b-2 a B) \sqrt{a+b x}}{128 a^3 x}+\frac{(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac{A (a+b x)^{5/2}}{5 a x^5}-\frac{\left (3 b^3 (A b-2 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{128 a^3}\\ &=\frac{b (A b-2 a B) \sqrt{a+b x}}{16 a x^3}+\frac{b^2 (A b-2 a B) \sqrt{a+b x}}{64 a^2 x^2}-\frac{3 b^3 (A b-2 a B) \sqrt{a+b x}}{128 a^3 x}+\frac{(A b-2 a B) (a+b x)^{3/2}}{8 a x^4}-\frac{A (a+b x)^{5/2}}{5 a x^5}+\frac{3 b^4 (A b-2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{128 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0210563, size = 57, normalized size = 0.33 \[ -\frac{(a+b x)^{5/2} \left (a^5 A+b^4 x^5 (2 a B-A b) \, _2F_1\left (\frac{5}{2},5;\frac{7}{2};\frac{b x}{a}+1\right )\right )}{5 a^6 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(3/2)*(A + B*x))/x^6,x]

[Out]

-((a + b*x)^(5/2)*(a^5*A + b^4*(-(A*b) + 2*a*B)*x^5*Hypergeometric2F1[5/2, 5, 7/2, 1 + (b*x)/a]))/(5*a^6*x^5)

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 129, normalized size = 0.8 \begin{align*} 2\,{b}^{4} \left ({\frac{1}{{b}^{5}{x}^{5}} \left ( -{\frac{ \left ( 3\,Ab-6\,Ba \right ) \left ( bx+a \right ) ^{9/2}}{256\,{a}^{3}}}+{\frac{ \left ( 7\,Ab-14\,Ba \right ) \left ( bx+a \right ) ^{7/2}}{128\,{a}^{2}}}-1/10\,{\frac{Ab \left ( bx+a \right ) ^{5/2}}{a}}+ \left ( -{\frac{7\,Ab}{128}}+{\frac{7\,Ba}{64}} \right ) \left ( bx+a \right ) ^{3/2}+{\frac{3\,a \left ( Ab-2\,Ba \right ) \sqrt{bx+a}}{256}} \right ) }+{\frac{3\,Ab-6\,Ba}{256\,{a}^{7/2}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(B*x+A)/x^6,x)

[Out]

2*b^4*((-3/256*(A*b-2*B*a)/a^3*(b*x+a)^(9/2)+7/128*(A*b-2*B*a)/a^2*(b*x+a)^(7/2)-1/10*A*b/a*(b*x+a)^(5/2)+(-7/
128*A*b+7/64*B*a)*(b*x+a)^(3/2)+3/256*a*(A*b-2*B*a)*(b*x+a)^(1/2))/b^5/x^5+3/256*(A*b-2*B*a)/a^(7/2)*arctanh((
b*x+a)^(1/2)/a^(1/2)))

________________________________________________________________________________________

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((b*x+a)^(3/2)*(B*x+A)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.27365, size = 695, normalized size = 4.04 \begin{align*} \left [-\frac{15 \,{\left (2 \, B a b^{4} - A b^{5}\right )} \sqrt{a} x^{5} \log \left (\frac{b x + 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (128 \, A a^{5} - 15 \,{\left (2 \, B a^{2} b^{3} - A a b^{4}\right )} x^{4} + 10 \,{\left (2 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{3} + 8 \,{\left (30 \, B a^{4} b + A a^{3} b^{2}\right )} x^{2} + 16 \,{\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x\right )} \sqrt{b x + a}}{1280 \, a^{4} x^{5}}, \frac{15 \,{\left (2 \, B a b^{4} - A b^{5}\right )} \sqrt{-a} x^{5} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) -{\left (128 \, A a^{5} - 15 \,{\left (2 \, B a^{2} b^{3} - A a b^{4}\right )} x^{4} + 10 \,{\left (2 \, B a^{3} b^{2} - A a^{2} b^{3}\right )} x^{3} + 8 \,{\left (30 \, B a^{4} b + A a^{3} b^{2}\right )} x^{2} + 16 \,{\left (10 \, B a^{5} + 11 \, A a^{4} b\right )} x\right )} \sqrt{b x + a}}{640 \, a^{4} x^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^6,x, algorithm="fricas")

[Out]

[-1/1280*(15*(2*B*a*b^4 - A*b^5)*sqrt(a)*x^5*log((b*x + 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(128*A*a^5 - 15*
(2*B*a^2*b^3 - A*a*b^4)*x^4 + 10*(2*B*a^3*b^2 - A*a^2*b^3)*x^3 + 8*(30*B*a^4*b + A*a^3*b^2)*x^2 + 16*(10*B*a^5
+ 11*A*a^4*b)*x)*sqrt(b*x + a))/(a^4*x^5), 1/640*(15*(2*B*a*b^4 - A*b^5)*sqrt(-a)*x^5*arctan(sqrt(b*x + a)*sq
rt(-a)/a) - (128*A*a^5 - 15*(2*B*a^2*b^3 - A*a*b^4)*x^4 + 10*(2*B*a^3*b^2 - A*a^2*b^3)*x^3 + 8*(30*B*a^4*b + A
*a^3*b^2)*x^2 + 16*(10*B*a^5 + 11*A*a^4*b)*x)*sqrt(b*x + a))/(a^4*x^5)]

________________________________________________________________________________________

Sympy [B]  time = 142.715, size = 1843, normalized size = 10.72 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)*(B*x+A)/x**6,x)

[Out]

-1930*A*a**6*b**5*sqrt(a + b*x)/(5120*a**10 + 6400*a**9*b*x - 12800*a**8*(a + b*x)**2 + 12800*a**7*(a + b*x)**
3 - 6400*a**6*(a + b*x)**4 + 1280*a**5*(a + b*x)**5) + 4740*A*a**5*b**5*(a + b*x)**(3/2)/(5120*a**10 + 6400*a*
*9*b*x - 12800*a**8*(a + b*x)**2 + 12800*a**7*(a + b*x)**3 - 6400*a**6*(a + b*x)**4 + 1280*a**5*(a + b*x)**5)
- 5376*A*a**4*b**5*(a + b*x)**(5/2)/(5120*a**10 + 6400*a**9*b*x - 12800*a**8*(a + b*x)**2 + 12800*a**7*(a + b*
x)**3 - 6400*a**6*(a + b*x)**4 + 1280*a**5*(a + b*x)**5) - 1116*A*a**4*b**5*sqrt(a + b*x)/(-1152*a**8 - 1536*a
**7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)**4) + 2940*A*a**3*b**5*(a + b*x
)**(7/2)/(5120*a**10 + 6400*a**9*b*x - 12800*a**8*(a + b*x)**2 + 12800*a**7*(a + b*x)**3 - 6400*a**6*(a + b*x)
**4 + 1280*a**5*(a + b*x)**5) + 2044*A*a**3*b**5*(a + b*x)**(3/2)/(-1152*a**8 - 1536*a**7*b*x + 2304*a**6*(a +
b*x)**2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)**4) - 630*A*a**2*b**5*(a + b*x)**(9/2)/(5120*a**10 + 64
00*a**9*b*x - 12800*a**8*(a + b*x)**2 + 12800*a**7*(a + b*x)**3 - 6400*a**6*(a + b*x)**4 + 1280*a**5*(a + b*x)
**5) - 1540*A*a**2*b**5*(a + b*x)**(5/2)/(-1152*a**8 - 1536*a**7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a +
b*x)**3 + 384*a**4*(a + b*x)**4) - 66*A*a**2*b**5*sqrt(a + b*x)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)*
*2 + 48*a**3*(a + b*x)**3) - 63*A*a**2*b**5*sqrt(a**(-11))*log(-a**6*sqrt(a**(-11)) + sqrt(a + b*x))/256 + 63*
A*a**2*b**5*sqrt(a**(-11))*log(a**6*sqrt(a**(-11)) + sqrt(a + b*x))/256 + 420*A*a*b**5*(a + b*x)**(7/2)/(-1152
*a**8 - 1536*a**7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)**4) + 80*A*a*b**5
*(a + b*x)**(3/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) + 35*A*a*b**5*sqrt(a
**(-9))*log(-a**5*sqrt(a**(-9)) + sqrt(a + b*x))/64 - 35*A*a*b**5*sqrt(a**(-9))*log(a**5*sqrt(a**(-9)) + sqrt(
a + b*x))/64 - 30*A*b**5*(a + b*x)**(5/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)*
*3) - 5*A*b**5*sqrt(a**(-7))*log(-a**4*sqrt(a**(-7)) + sqrt(a + b*x))/16 + 5*A*b**5*sqrt(a**(-7))*log(a**4*sqr
t(a**(-7)) + sqrt(a + b*x))/16 - 558*B*a**5*b**4*sqrt(a + b*x)/(-1152*a**8 - 1536*a**7*b*x + 2304*a**6*(a + b*
x)**2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)**4) + 1022*B*a**4*b**4*(a + b*x)**(3/2)/(-1152*a**8 - 1536
*a**7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)**4) - 770*B*a**3*b**4*(a + b*
x)**(5/2)/(-1152*a**8 - 1536*a**7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a + b*x)**3 + 384*a**4*(a + b*x)**
4) - 132*B*a**3*b**4*sqrt(a + b*x)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) + 2
10*B*a**2*b**4*(a + b*x)**(7/2)/(-1152*a**8 - 1536*a**7*b*x + 2304*a**6*(a + b*x)**2 - 1536*a**5*(a + b*x)**3
+ 384*a**4*(a + b*x)**4) + 160*B*a**2*b**4*(a + b*x)**(3/2)/(96*a**6 + 144*a**5*b*x - 144*a**4*(a + b*x)**2 +
48*a**3*(a + b*x)**3) + 35*B*a**2*b**4*sqrt(a**(-9))*log(-a**5*sqrt(a**(-9)) + sqrt(a + b*x))/128 - 35*B*a**2*
b**4*sqrt(a**(-9))*log(a**5*sqrt(a**(-9)) + sqrt(a + b*x))/128 - 60*B*a*b**4*(a + b*x)**(5/2)/(96*a**6 + 144*a
**5*b*x - 144*a**4*(a + b*x)**2 + 48*a**3*(a + b*x)**3) - 10*B*a*b**4*sqrt(a + b*x)/(-8*a**4 - 16*a**3*b*x + 8
*a**2*(a + b*x)**2) - 5*B*a*b**4*sqrt(a**(-7))*log(-a**4*sqrt(a**(-7)) + sqrt(a + b*x))/8 + 5*B*a*b**4*sqrt(a*
*(-7))*log(a**4*sqrt(a**(-7)) + sqrt(a + b*x))/8 + 6*B*b**4*(a + b*x)**(3/2)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(
a + b*x)**2) + 3*B*b**4*sqrt(a**(-5))*log(-a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 3*B*b**4*sqrt(a**(-5))*log(
a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8

________________________________________________________________________________________

Giac [A]  time = 1.2192, size = 259, normalized size = 1.51 \begin{align*} \frac{\frac{15 \,{\left (2 \, B a b^{5} - A b^{6}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} + \frac{30 \,{\left (b x + a\right )}^{\frac{9}{2}} B a b^{5} - 140 \,{\left (b x + a\right )}^{\frac{7}{2}} B a^{2} b^{5} + 140 \,{\left (b x + a\right )}^{\frac{3}{2}} B a^{4} b^{5} - 30 \, \sqrt{b x + a} B a^{5} b^{5} - 15 \,{\left (b x + a\right )}^{\frac{9}{2}} A b^{6} + 70 \,{\left (b x + a\right )}^{\frac{7}{2}} A a b^{6} - 128 \,{\left (b x + a\right )}^{\frac{5}{2}} A a^{2} b^{6} - 70 \,{\left (b x + a\right )}^{\frac{3}{2}} A a^{3} b^{6} + 15 \, \sqrt{b x + a} A a^{4} b^{6}}{a^{3} b^{5} x^{5}}}{640 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^6,x, algorithm="giac")

[Out]

1/640*(15*(2*B*a*b^5 - A*b^6)*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + (30*(b*x + a)^(9/2)*B*a*b^5 - 14
0*(b*x + a)^(7/2)*B*a^2*b^5 + 140*(b*x + a)^(3/2)*B*a^4*b^5 - 30*sqrt(b*x + a)*B*a^5*b^5 - 15*(b*x + a)^(9/2)*
A*b^6 + 70*(b*x + a)^(7/2)*A*a*b^6 - 128*(b*x + a)^(5/2)*A*a^2*b^6 - 70*(b*x + a)^(3/2)*A*a^3*b^6 + 15*sqrt(b*
x + a)*A*a^4*b^6)/(a^3*b^5*x^5))/b