3.679 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^{3/2}}{x^9} \, dx\)

Optimal. Leaf size=210 \[ -\frac{a^2 \sqrt{a^2+2 a b x+b^2 x^2} (a B+3 A b)}{7 x^7 (a+b x)}-\frac{a b \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{2 x^6 (a+b x)}-\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2} (3 a B+A b)}{5 x^5 (a+b x)}-\frac{a^3 A \sqrt{a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac{b^3 B \sqrt{a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)} \]

[Out]

-(a^3*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*x^8*(a + b*x)) - (a^2*(3*A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
7*x^7*(a + b*x)) - (a*b*(A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*x^6*(a + b*x)) - (b^2*(A*b + 3*a*B)*Sqrt
[a^2 + 2*a*b*x + b^2*x^2])/(5*x^5*(a + b*x)) - (b^3*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*x^4*(a + b*x))

________________________________________________________________________________________

Rubi [A]  time = 0.0807631, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {770, 76} \[ -\frac{a^2 \sqrt{a^2+2 a b x+b^2 x^2} (a B+3 A b)}{7 x^7 (a+b x)}-\frac{a b \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{2 x^6 (a+b x)}-\frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2} (3 a B+A b)}{5 x^5 (a+b x)}-\frac{a^3 A \sqrt{a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac{b^3 B \sqrt{a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^9,x]

[Out]

-(a^3*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*x^8*(a + b*x)) - (a^2*(3*A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(
7*x^7*(a + b*x)) - (a*b*(A*b + a*B)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*x^6*(a + b*x)) - (b^2*(A*b + 3*a*B)*Sqrt
[a^2 + 2*a*b*x + b^2*x^2])/(5*x^5*(a + b*x)) - (b^3*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*x^4*(a + b*x))

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^9} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right )^3 (A+B x)}{x^9} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (\frac{a^3 A b^3}{x^9}+\frac{a^2 b^3 (3 A b+a B)}{x^8}+\frac{3 a b^4 (A b+a B)}{x^7}+\frac{b^5 (A b+3 a B)}{x^6}+\frac{b^6 B}{x^5}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=-\frac{a^3 A \sqrt{a^2+2 a b x+b^2 x^2}}{8 x^8 (a+b x)}-\frac{a^2 (3 A b+a B) \sqrt{a^2+2 a b x+b^2 x^2}}{7 x^7 (a+b x)}-\frac{a b (A b+a B) \sqrt{a^2+2 a b x+b^2 x^2}}{2 x^6 (a+b x)}-\frac{b^2 (A b+3 a B) \sqrt{a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac{b^3 B \sqrt{a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0277857, size = 87, normalized size = 0.41 \[ -\frac{\sqrt{(a+b x)^2} \left (20 a^2 b x (6 A+7 B x)+5 a^3 (7 A+8 B x)+28 a b^2 x^2 (5 A+6 B x)+14 b^3 x^3 (4 A+5 B x)\right )}{280 x^8 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/x^9,x]

[Out]

-(Sqrt[(a + b*x)^2]*(14*b^3*x^3*(4*A + 5*B*x) + 28*a*b^2*x^2*(5*A + 6*B*x) + 20*a^2*b*x*(6*A + 7*B*x) + 5*a^3*
(7*A + 8*B*x)))/(280*x^8*(a + b*x))

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 92, normalized size = 0.4 \begin{align*} -{\frac{70\,B{x}^{4}{b}^{3}+56\,A{b}^{3}{x}^{3}+168\,B{x}^{3}a{b}^{2}+140\,A{x}^{2}a{b}^{2}+140\,B{x}^{2}{a}^{2}b+120\,A{a}^{2}bx+40\,{a}^{3}Bx+35\,A{a}^{3}}{280\,{x}^{8} \left ( bx+a \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^9,x)

[Out]

-1/280*(70*B*b^3*x^4+56*A*b^3*x^3+168*B*a*b^2*x^3+140*A*a*b^2*x^2+140*B*a^2*b*x^2+120*A*a^2*b*x+40*B*a^3*x+35*
A*a^3)*((b*x+a)^2)^(3/2)/x^8/(b*x+a)^3

________________________________________________________________________________________

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)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^9,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.28052, size = 167, normalized size = 0.8 \begin{align*} -\frac{70 \, B b^{3} x^{4} + 35 \, A a^{3} + 56 \,{\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 140 \,{\left (B a^{2} b + A a b^{2}\right )} x^{2} + 40 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} x}{280 \, x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^9,x, algorithm="fricas")

[Out]

-1/280*(70*B*b^3*x^4 + 35*A*a^3 + 56*(3*B*a*b^2 + A*b^3)*x^3 + 140*(B*a^2*b + A*a*b^2)*x^2 + 40*(B*a^3 + 3*A*a
^2*b)*x)/x^8

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}{x^{9}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2)/x**9,x)

[Out]

Integral((A + B*x)*((a + b*x)**2)**(3/2)/x**9, x)

________________________________________________________________________________________

Giac [A]  time = 1.20928, size = 201, normalized size = 0.96 \begin{align*} \frac{{\left (2 \, B a b^{7} - A b^{8}\right )} \mathrm{sgn}\left (b x + a\right )}{280 \, a^{5}} - \frac{70 \, B b^{3} x^{4} \mathrm{sgn}\left (b x + a\right ) + 168 \, B a b^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + 56 \, A b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + 140 \, B a^{2} b x^{2} \mathrm{sgn}\left (b x + a\right ) + 140 \, A a b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 40 \, B a^{3} x \mathrm{sgn}\left (b x + a\right ) + 120 \, A a^{2} b x \mathrm{sgn}\left (b x + a\right ) + 35 \, A a^{3} \mathrm{sgn}\left (b x + a\right )}{280 \, x^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^9,x, algorithm="giac")

[Out]

1/280*(2*B*a*b^7 - A*b^8)*sgn(b*x + a)/a^5 - 1/280*(70*B*b^3*x^4*sgn(b*x + a) + 168*B*a*b^2*x^3*sgn(b*x + a) +
56*A*b^3*x^3*sgn(b*x + a) + 140*B*a^2*b*x^2*sgn(b*x + a) + 140*A*a*b^2*x^2*sgn(b*x + a) + 40*B*a^3*x*sgn(b*x
+ a) + 120*A*a^2*b*x*sgn(b*x + a) + 35*A*a^3*sgn(b*x + a))/x^8