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

Optimal. Leaf size=77 \[ \frac{(a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (A b-a B)}{4 a^2 x^4}-\frac{A \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 a^2 x^5} \]

[Out]

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

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Rubi [A]  time = 0.0456049, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {769, 646, 37} \[ \frac{(a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (A b-a B)}{4 a^2 x^4}-\frac{A \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 a^2 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 769

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)^2), x] + Dist[(2*c*f -
b*g)/(2*c*d - b*e), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x]
&& EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 0]

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 37

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] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -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^6} \, dx &=-\frac{A \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 a^2 x^5}-\frac{\left (2 A b^2-2 a b B\right ) \int \frac{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{x^5} \, dx}{2 a b}\\ &=-\frac{A \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 a^2 x^5}-\frac{\left (\left (2 A b^2-2 a b B\right ) \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \frac{\left (a b+b^2 x\right )^3}{x^5} \, dx}{2 a b^3 \left (a b+b^2 x\right )}\\ &=\frac{(A b-a B) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{4 a^2 x^4}-\frac{A \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 a^2 x^5}\\ \end{align*}

Mathematica [A]  time = 0.0328356, size = 84, normalized size = 1.09 \[ -\frac{\sqrt{(a+b x)^2} \left (5 a^2 b x (3 A+4 B x)+a^3 (4 A+5 B x)+10 a b^2 x^2 (2 A+3 B x)+10 b^3 x^3 (A+2 B x)\right )}{20 x^5 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[(a + b*x)^2]*(10*b^3*x^3*(A + 2*B*x) + 10*a*b^2*x^2*(2*A + 3*B*x) + 5*a^2*b*x*(3*A + 4*B*x) + a^3*(4*A
+ 5*B*x)))/(20*x^5*(a + b*x))

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Maple [A]  time = 0.007, size = 92, normalized size = 1.2 \begin{align*} -{\frac{20\,B{x}^{4}{b}^{3}+10\,A{b}^{3}{x}^{3}+30\,B{x}^{3}a{b}^{2}+20\,A{x}^{2}a{b}^{2}+20\,B{x}^{2}{a}^{2}b+15\,A{a}^{2}bx+5\,{a}^{3}Bx+4\,A{a}^{3}}{20\,{x}^{5} \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^6,x)

[Out]

-1/20*(20*B*b^3*x^4+10*A*b^3*x^3+30*B*a*b^2*x^3+20*A*a*b^2*x^2+20*B*a^2*b*x^2+15*A*a^2*b*x+5*B*a^3*x+4*A*a^3)*
((b*x+a)^2)^(3/2)/x^5/(b*x+a)^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((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.41842, size = 162, normalized size = 2.1 \begin{align*} -\frac{20 \, B b^{3} x^{4} + 4 \, A a^{3} + 10 \,{\left (3 \, B a b^{2} + A b^{3}\right )} x^{3} + 20 \,{\left (B a^{2} b + A a b^{2}\right )} x^{2} + 5 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} x}{20 \, x^{5}} \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^6,x, algorithm="fricas")

[Out]

-1/20*(20*B*b^3*x^4 + 4*A*a^3 + 10*(3*B*a*b^2 + A*b^3)*x^3 + 20*(B*a^2*b + A*a*b^2)*x^2 + 5*(B*a^3 + 3*A*a^2*b
)*x)/x^5

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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^{6}}\, 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**6,x)

[Out]

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

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Giac [B]  time = 1.18778, size = 201, normalized size = 2.61 \begin{align*} -\frac{{\left (5 \, B a b^{4} - A b^{5}\right )} \mathrm{sgn}\left (b x + a\right )}{20 \, a^{2}} - \frac{20 \, B b^{3} x^{4} \mathrm{sgn}\left (b x + a\right ) + 30 \, B a b^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + 10 \, A b^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + 20 \, B a^{2} b x^{2} \mathrm{sgn}\left (b x + a\right ) + 20 \, A a b^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + 5 \, B a^{3} x \mathrm{sgn}\left (b x + a\right ) + 15 \, A a^{2} b x \mathrm{sgn}\left (b x + a\right ) + 4 \, A a^{3} \mathrm{sgn}\left (b x + a\right )}{20 \, x^{5}} \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^6,x, algorithm="giac")

[Out]

-1/20*(5*B*a*b^4 - A*b^5)*sgn(b*x + a)/a^2 - 1/20*(20*B*b^3*x^4*sgn(b*x + a) + 30*B*a*b^2*x^3*sgn(b*x + a) + 1
0*A*b^3*x^3*sgn(b*x + a) + 20*B*a^2*b*x^2*sgn(b*x + a) + 20*A*a*b^2*x^2*sgn(b*x + a) + 5*B*a^3*x*sgn(b*x + a)
+ 15*A*a^2*b*x*sgn(b*x + a) + 4*A*a^3*sgn(b*x + a))/x^5