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]

-2/7*A/b/x^3/(c*x^2+b*x)^(1/2)-2/35*(-8*A*c+7*B*b)/b^2/x^2/(c*x^2+b*x)^(1/2)+4/35*c*(-8*A*c+7*B*b)/b^3/x/(c*x^
2+b*x)^(1/2)-16/35*c^2*(-8*A*c+7*B*b)*(2*c*x+b)/b^5/(c*x^2+b*x)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 128, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.03, size = 98, normalized size = 0.77 \[ -\frac {2 \left (A \left (5 b^4-8 b^3 c x+16 b^2 c^2 x^2-64 b c^3 x^3-128 c^4 x^4\right )+7 b B x \left (b^3-2 b^2 c x+8 b c^2 x^2+16 c^3 x^3\right )\right )}{35 b^5 x^3 \sqrt {x (b+c x)}} \]

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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fricas [A]  time = 0.87, size = 117, normalized size = 0.91 \[ -\frac {2 \, {\left (5 \, A b^{4} + 16 \, {\left (7 \, B b c^{3} - 8 \, A c^{4}\right )} x^{4} + 8 \, {\left (7 \, B b^{2} c^{2} - 8 \, A b c^{3}\right )} x^{3} - 2 \, {\left (7 \, B b^{3} c - 8 \, A b^{2} c^{2}\right )} x^{2} + {\left (7 \, B b^{4} - 8 \, A b^{3} c\right )} x\right )} \sqrt {c x^{2} + b x}}{35 \, {\left (b^{5} c x^{5} + b^{6} x^{4}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{3}}\,{d x} \]

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)

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maple [A]  time = 0.06, size = 110, normalized size = 0.86 \[ -\frac {2 \left (c x +b \right ) \left (-128 A \,c^{4} x^{4}+112 B b \,c^{3} x^{4}-64 A b \,c^{3} x^{3}+56 B \,b^{2} c^{2} x^{3}+16 A \,b^{2} c^{2} x^{2}-14 B \,b^{3} c \,x^{2}-8 A \,b^{3} c x +7 b^{4} B x +5 A \,b^{4}\right )}{35 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{5} x^{2}} \]

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 [A]  time = 0.92, size = 188, normalized size = 1.47 \[ -\frac {32 \, B c^{3} x}{5 \, \sqrt {c x^{2} + b x} b^{4}} + \frac {256 \, A c^{4} x}{35 \, \sqrt {c x^{2} + b x} b^{5}} - \frac {16 \, B c^{2}}{5 \, \sqrt {c x^{2} + b x} b^{3}} + \frac {128 \, A c^{3}}{35 \, \sqrt {c x^{2} + b x} b^{4}} + \frac {4 \, B c}{5 \, \sqrt {c x^{2} + b x} b^{2} x} - \frac {32 \, A c^{2}}{35 \, \sqrt {c x^{2} + b x} b^{3} x} - \frac {2 \, B}{5 \, \sqrt {c x^{2} + b x} b x^{2}} + \frac {16 \, A c}{35 \, \sqrt {c x^{2} + b x} b^{2} x^{2}} - \frac {2 \, A}{7 \, \sqrt {c x^{2} + b x} b x^{3}} \]

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]

-32/5*B*c^3*x/(sqrt(c*x^2 + b*x)*b^4) + 256/35*A*c^4*x/(sqrt(c*x^2 + b*x)*b^5) - 16/5*B*c^2/(sqrt(c*x^2 + b*x)
*b^3) + 128/35*A*c^3/(sqrt(c*x^2 + b*x)*b^4) + 4/5*B*c/(sqrt(c*x^2 + b*x)*b^2*x) - 32/35*A*c^2/(sqrt(c*x^2 + b
*x)*b^3*x) - 2/5*B/(sqrt(c*x^2 + b*x)*b*x^2) + 16/35*A*c/(sqrt(c*x^2 + b*x)*b^2*x^2) - 2/7*A/(sqrt(c*x^2 + b*x
)*b*x^3)

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mupad [B]  time = 1.29, size = 161, normalized size = 1.26 \[ -\frac {\left (14\,B\,b^2-26\,A\,b\,c\right )\,\sqrt {c\,x^2+b\,x}}{35\,b^4\,x^3}-\frac {2\,A\,\sqrt {c\,x^2+b\,x}}{7\,b^2\,x^4}-\frac {\sqrt {c\,x^2+b\,x}\,\left (x\,\left (\frac {116\,A\,c^4-84\,B\,b\,c^3}{35\,b^5}-\frac {4\,c^3\,\left (93\,A\,c-77\,B\,b\right )}{35\,b^5}\right )-\frac {2\,c^2\,\left (93\,A\,c-77\,B\,b\right )}{35\,b^4}\right )}{x\,\left (b+c\,x\right )}-\frac {2\,c\,\sqrt {c\,x^2+b\,x}\,\left (29\,A\,c-21\,B\,b\right )}{35\,b^4\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((14*B*b^2 - 26*A*b*c)*(b*x + c*x^2)^(1/2))/(35*b^4*x^3) - (2*A*(b*x + c*x^2)^(1/2))/(7*b^2*x^4) - ((b*x + c
*x^2)^(1/2)*(x*((116*A*c^4 - 84*B*b*c^3)/(35*b^5) - (4*c^3*(93*A*c - 77*B*b))/(35*b^5)) - (2*c^2*(93*A*c - 77*
B*b))/(35*b^4)))/(x*(b + c*x)) - (2*c*(b*x + c*x^2)^(1/2)*(29*A*c - 21*B*b))/(35*b^4*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x}{x^{3} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]

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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