3.1.74 \(\int \frac {(A+B x) \sqrt {b x+c x^2}}{x^4} \, dx\)

Optimal. Leaf size=57 \[ -\frac {2 \left (b x+c x^2\right )^{3/2} (5 b B-2 A c)}{15 b^2 x^3}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{5 b x^4} \]

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Rubi [A]  time = 0.05, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {792, 650} \begin {gather*} -\frac {2 \left (b x+c x^2\right )^{3/2} (5 b B-2 A c)}{15 b^2 x^3}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{5 b x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[b*x + c*x^2])/x^4,x]

[Out]

(-2*A*(b*x + c*x^2)^(3/2))/(5*b*x^4) - (2*(5*b*B - 2*A*c)*(b*x + c*x^2)^(3/2))/(15*b^2*x^3)

Rule 650

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))/((p + 1)*(2*c*d - b*e)), 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] && EqQ[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) \sqrt {b x+c x^2}}{x^4} \, dx &=-\frac {2 A \left (b x+c x^2\right )^{3/2}}{5 b x^4}+\frac {\left (2 \left (-4 (-b B+A c)+\frac {3}{2} (-b B+2 A c)\right )\right ) \int \frac {\sqrt {b x+c x^2}}{x^3} \, dx}{5 b}\\ &=-\frac {2 A \left (b x+c x^2\right )^{3/2}}{5 b x^4}-\frac {2 (5 b B-2 A c) \left (b x+c x^2\right )^{3/2}}{15 b^2 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 36, normalized size = 0.63 \begin {gather*} -\frac {2 (x (b+c x))^{3/2} (3 A b-2 A c x+5 b B x)}{15 b^2 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[b*x + c*x^2])/x^4,x]

[Out]

(-2*(x*(b + c*x))^(3/2)*(3*A*b + 5*b*B*x - 2*A*c*x))/(15*b^2*x^4)

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IntegrateAlgebraic [A]  time = 0.30, size = 60, normalized size = 1.05 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (-3 A b^2-A b c x+2 A c^2 x^2-5 b^2 B x-5 b B c x^2\right )}{15 b^2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*Sqrt[b*x + c*x^2])/x^4,x]

[Out]

(2*Sqrt[b*x + c*x^2]*(-3*A*b^2 - 5*b^2*B*x - A*b*c*x - 5*b*B*c*x^2 + 2*A*c^2*x^2))/(15*b^2*x^3)

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fricas [A]  time = 0.40, size = 55, normalized size = 0.96 \begin {gather*} -\frac {2 \, {\left (3 \, A b^{2} + {\left (5 \, B b c - 2 \, A c^{2}\right )} x^{2} + {\left (5 \, B b^{2} + A b c\right )} x\right )} \sqrt {c x^{2} + b x}}{15 \, b^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^4,x, algorithm="fricas")

[Out]

-2/15*(3*A*b^2 + (5*B*b*c - 2*A*c^2)*x^2 + (5*B*b^2 + A*b*c)*x)*sqrt(c*x^2 + b*x)/(b^2*x^3)

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giac [B]  time = 0.25, size = 191, normalized size = 3.35 \begin {gather*} \frac {2 \, {\left (15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{4} B c + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B b \sqrt {c} + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} A c^{\frac {3}{2}} + 5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b^{2} + 25 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A b c + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b^{2} \sqrt {c} + 3 \, A b^{3}\right )}}{15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^4,x, algorithm="giac")

[Out]

2/15*(15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*c + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b*sqrt(c) + 15*(sqrt(c
)*x - sqrt(c*x^2 + b*x))^3*A*c^(3/2) + 5*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^2 + 25*(sqrt(c)*x - sqrt(c*x^2
+ b*x))^2*A*b*c + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^2*sqrt(c) + 3*A*b^3)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^
5

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maple [A]  time = 0.05, size = 40, normalized size = 0.70 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-2 A c x +5 B b x +3 A b \right ) \sqrt {c \,x^{2}+b x}}{15 b^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(1/2)/x^4,x)

[Out]

-2/15*(c*x+b)*(-2*A*c*x+5*B*b*x+3*A*b)*(c*x^2+b*x)^(1/2)/b^2/x^3

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maxima [B]  time = 0.95, size = 100, normalized size = 1.75 \begin {gather*} -\frac {2 \, \sqrt {c x^{2} + b x} B c}{3 \, b x} + \frac {4 \, \sqrt {c x^{2} + b x} A c^{2}}{15 \, b^{2} x} - \frac {2 \, \sqrt {c x^{2} + b x} B}{3 \, x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} A c}{15 \, b x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} A}{5 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^4,x, algorithm="maxima")

[Out]

-2/3*sqrt(c*x^2 + b*x)*B*c/(b*x) + 4/15*sqrt(c*x^2 + b*x)*A*c^2/(b^2*x) - 2/3*sqrt(c*x^2 + b*x)*B/x^2 - 2/15*s
qrt(c*x^2 + b*x)*A*c/(b*x^2) - 2/5*sqrt(c*x^2 + b*x)*A/x^3

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mupad [B]  time = 1.44, size = 100, normalized size = 1.75 \begin {gather*} \frac {4\,A\,c^2\,\sqrt {c\,x^2+b\,x}}{15\,b^2\,x}-\frac {2\,B\,\sqrt {c\,x^2+b\,x}}{3\,x^2}-\frac {2\,A\,c\,\sqrt {c\,x^2+b\,x}}{15\,b\,x^2}-\frac {2\,B\,c\,\sqrt {c\,x^2+b\,x}}{3\,b\,x}-\frac {2\,A\,\sqrt {c\,x^2+b\,x}}{5\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^4,x)

[Out]

(4*A*c^2*(b*x + c*x^2)^(1/2))/(15*b^2*x) - (2*B*(b*x + c*x^2)^(1/2))/(3*x^2) - (2*A*c*(b*x + c*x^2)^(1/2))/(15
*b*x^2) - (2*B*c*(b*x + c*x^2)^(1/2))/(3*b*x) - (2*A*(b*x + c*x^2)^(1/2))/(5*x^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**(1/2)/x**4,x)

[Out]

Integral(sqrt(x*(b + c*x))*(A + B*x)/x**4, x)

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