3.127 \(\int \frac{A+B x}{x^3 (b x+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=128 \[ -\frac{16 c^2 (b+2 c x) (7 b B-8 A c)}{35 b^5 \sqrt{b x+c x^2}}+\frac{4 c (7 b B-8 A c)}{35 b^3 x \sqrt{b x+c x^2}}-\frac{2 (7 b B-8 A c)}{35 b^2 x^2 \sqrt{b x+c x^2}}-\frac{2 A}{7 b x^3 \sqrt{b x+c x^2}} \]

[Out]

(-2*A)/(7*b*x^3*Sqrt[b*x + c*x^2]) - (2*(7*b*B - 8*A*c))/(35*b^2*x^2*Sqrt[b*x + c*x^2]) + (4*c*(7*b*B - 8*A*c)
)/(35*b^3*x*Sqrt[b*x + c*x^2]) - (16*c^2*(7*b*B - 8*A*c)*(b + 2*c*x))/(35*b^5*Sqrt[b*x + c*x^2])

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Rubi [A]  time = 0.114742, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {792, 658, 613} \[ -\frac{16 c^2 (b+2 c x) (7 b B-8 A c)}{35 b^5 \sqrt{b x+c x^2}}+\frac{4 c (7 b B-8 A c)}{35 b^3 x \sqrt{b x+c x^2}}-\frac{2 (7 b B-8 A c)}{35 b^2 x^2 \sqrt{b x+c x^2}}-\frac{2 A}{7 b x^3 \sqrt{b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A)/(7*b*x^3*Sqrt[b*x + c*x^2]) - (2*(7*b*B - 8*A*c))/(35*b^2*x^2*Sqrt[b*x + c*x^2]) + (4*c*(7*b*B - 8*A*c)
)/(35*b^3*x*Sqrt[b*x + c*x^2]) - (16*c^2*(7*b*B - 8*A*c)*(b + 2*c*x))/(35*b^5*Sqrt[b*x + c*x^2])

Rule 792

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

Rule 658

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(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, m, p}, x] && NeQ[b^2 - 4*a*c
, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \frac{A+B x}{x^3 \left (b x+c x^2\right )^{3/2}} \, dx &=-\frac{2 A}{7 b x^3 \sqrt{b x+c x^2}}+\frac{\left (2 \left (\frac{1}{2} (b B-2 A c)-3 (-b B+A c)\right )\right ) \int \frac{1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx}{7 b}\\ &=-\frac{2 A}{7 b x^3 \sqrt{b x+c x^2}}-\frac{2 (7 b B-8 A c)}{35 b^2 x^2 \sqrt{b x+c x^2}}-\frac{(6 c (7 b B-8 A c)) \int \frac{1}{x \left (b x+c x^2\right )^{3/2}} \, dx}{35 b^2}\\ &=-\frac{2 A}{7 b x^3 \sqrt{b x+c x^2}}-\frac{2 (7 b B-8 A c)}{35 b^2 x^2 \sqrt{b x+c x^2}}+\frac{4 c (7 b B-8 A c)}{35 b^3 x \sqrt{b x+c x^2}}+\frac{\left (8 c^2 (7 b B-8 A c)\right ) \int \frac{1}{\left (b x+c x^2\right )^{3/2}} \, dx}{35 b^3}\\ &=-\frac{2 A}{7 b x^3 \sqrt{b x+c x^2}}-\frac{2 (7 b B-8 A c)}{35 b^2 x^2 \sqrt{b x+c x^2}}+\frac{4 c (7 b B-8 A c)}{35 b^3 x \sqrt{b x+c x^2}}-\frac{16 c^2 (7 b B-8 A c) (b+2 c x)}{35 b^5 \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0325647, size = 98, normalized size = 0.77 \[ -\frac{2 \left (A \left (16 b^2 c^2 x^2-8 b^3 c x+5 b^4-64 b c^3 x^3-128 c^4 x^4\right )+7 b B x \left (-2 b^2 c x+b^3+8 b c^2 x^2+16 c^3 x^3\right )\right )}{35 b^5 x^3 \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(7*b*B*x*(b^3 - 2*b^2*c*x + 8*b*c^2*x^2 + 16*c^3*x^3) + A*(5*b^4 - 8*b^3*c*x + 16*b^2*c^2*x^2 - 64*b*c^3*x
^3 - 128*c^4*x^4)))/(35*b^5*x^3*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.007, size = 110, normalized size = 0.9 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -128\,A{c}^{4}{x}^{4}+112\,Bb{c}^{3}{x}^{4}-64\,Ab{c}^{3}{x}^{3}+56\,B{b}^{2}{c}^{2}{x}^{3}+16\,A{b}^{2}{c}^{2}{x}^{2}-14\,B{b}^{3}c{x}^{2}-8\,A{b}^{3}cx+7\,{b}^{4}Bx+5\,A{b}^{4} \right ) }{35\,{x}^{2}{b}^{5}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(c*x+b)*(-128*A*c^4*x^4+112*B*b*c^3*x^4-64*A*b*c^3*x^3+56*B*b^2*c^2*x^3+16*A*b^2*c^2*x^2-14*B*b^3*c*x^2-
8*A*b^3*c*x+7*B*b^4*x+5*A*b^4)/x^2/b^5/(c*x^2+b*x)^(3/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.94433, size = 246, normalized size = 1.92 \begin{align*} -\frac{2 \,{\left (5 \, A b^{4} + 16 \,{\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{4} + 8 \,{\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{3} - 2 \,{\left (7 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )} x^{2} +{\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} x\right )} \sqrt{c x^{2} + b x}}{35 \,{\left (b^{5} c x^{5} + b^{6} x^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(5*A*b^4 + 16*(7*B*b*c^3 - 8*A*c^4)*x^4 + 8*(7*B*b^2*c^2 - 8*A*b*c^3)*x^3 - 2*(7*B*b^3*c - 8*A*b^2*c^2)*
x^2 + (7*B*b^4 - 8*A*b^3*c)*x)*sqrt(c*x^2 + b*x)/(b^5*c*x^5 + b^6*x^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x + A)/((c*x^2 + b*x)^(3/2)*x^3), x)