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

Optimal. Leaf size=48 \[ \frac {4 c \left (b x+c x^2\right )^{3/2}}{15 b^2 x^3}-\frac {2 \left (b x+c x^2\right )^{3/2}}{5 b x^4} \]

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Rubi [A]  time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {658, 650} \begin {gather*} \frac {4 c \left (b x+c x^2\right )^{3/2}}{15 b^2 x^3}-\frac {2 \left (b x+c x^2\right )^{3/2}}{5 b x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x + c*x^2]/x^4,x]

[Out]

(-2*(b*x + c*x^2)^(3/2))/(5*b*x^4) + (4*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 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]

Rubi steps

\begin {align*} \int \frac {\sqrt {b x+c x^2}}{x^4} \, dx &=-\frac {2 \left (b x+c x^2\right )^{3/2}}{5 b x^4}-\frac {(2 c) \int \frac {\sqrt {b x+c x^2}}{x^3} \, dx}{5 b}\\ &=-\frac {2 \left (b x+c x^2\right )^{3/2}}{5 b x^4}+\frac {4 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 = 29, normalized size = 0.60 \begin {gather*} \frac {2 (x (b+c x))^{3/2} (2 c x-3 b)}{15 b^2 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[b*x + c*x^2]/x^4,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 42, normalized size = 0.88 \begin {gather*} \frac {2 \sqrt {b x+c x^2} \left (-3 b^2-b c x+2 c^2 x^2\right )}{15 b^2 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[b*x + c*x^2]/x^4,x]

[Out]

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

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fricas [A]  time = 0.39, size = 38, normalized size = 0.79 \begin {gather*} \frac {2 \, {\left (2 \, c^{2} x^{2} - b c x - 3 \, b^{2}\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((c*x^2+b*x)^(1/2)/x^4,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.23, size = 107, normalized size = 2.23 \begin {gather*} \frac {2 \, {\left (15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} c^{\frac {3}{2}} + 25 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} b c + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} b^{2} \sqrt {c} + 3 \, 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((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))^3*c^(3/2) + 25*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b*c + 15*(sqrt(c)*x
- sqrt(c*x^2 + b*x))*b^2*sqrt(c) + 3*b^3)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^5

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maple [A]  time = 0.05, size = 33, normalized size = 0.69 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-2 c x +3 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((c*x^2+b*x)^(1/2)/x^4,x)

[Out]

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

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maxima [A]  time = 0.96, size = 59, normalized size = 1.23 \begin {gather*} \frac {4 \, \sqrt {c x^{2} + b x} c^{2}}{15 \, b^{2} x} - \frac {2 \, \sqrt {c x^{2} + b x} c}{15 \, b x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x}}{5 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.34, size = 37, normalized size = 0.77 \begin {gather*} -\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (3\,b^2+b\,c\,x-2\,c^2\,x^2\right )}{15\,b^2\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^(1/2)/x^4,x)

[Out]

-(2*(b*x + c*x^2)^(1/2)*(3*b^2 - 2*c^2*x^2 + b*c*x))/(15*b^2*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 )}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x*(b + c*x))/x**4, x)

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