3.3.20 \(\int \frac {1}{x^3 \sqrt {a x^3+b x^4}} \, dx\)

Optimal. Leaf size=108 \[ \frac {32 b^3 \sqrt {a x^3+b x^4}}{35 a^4 x^2}-\frac {16 b^2 \sqrt {a x^3+b x^4}}{35 a^3 x^3}+\frac {12 b \sqrt {a x^3+b x^4}}{35 a^2 x^4}-\frac {2 \sqrt {a x^3+b x^4}}{7 a x^5} \]

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Rubi [A]  time = 0.13, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2016, 2000} \begin {gather*} \frac {32 b^3 \sqrt {a x^3+b x^4}}{35 a^4 x^2}-\frac {16 b^2 \sqrt {a x^3+b x^4}}{35 a^3 x^3}+\frac {12 b \sqrt {a x^3+b x^4}}{35 a^2 x^4}-\frac {2 \sqrt {a x^3+b x^4}}{7 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[a*x^3 + b*x^4]),x]

[Out]

(-2*Sqrt[a*x^3 + b*x^4])/(7*a*x^5) + (12*b*Sqrt[a*x^3 + b*x^4])/(35*a^2*x^4) - (16*b^2*Sqrt[a*x^3 + b*x^4])/(3
5*a^3*x^3) + (32*b^3*Sqrt[a*x^3 + b*x^4])/(35*a^4*x^2)

Rule 2000

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x
^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 0]

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps

\begin {align*} \int \frac {1}{x^3 \sqrt {a x^3+b x^4}} \, dx &=-\frac {2 \sqrt {a x^3+b x^4}}{7 a x^5}-\frac {(6 b) \int \frac {1}{x^2 \sqrt {a x^3+b x^4}} \, dx}{7 a}\\ &=-\frac {2 \sqrt {a x^3+b x^4}}{7 a x^5}+\frac {12 b \sqrt {a x^3+b x^4}}{35 a^2 x^4}+\frac {\left (24 b^2\right ) \int \frac {1}{x \sqrt {a x^3+b x^4}} \, dx}{35 a^2}\\ &=-\frac {2 \sqrt {a x^3+b x^4}}{7 a x^5}+\frac {12 b \sqrt {a x^3+b x^4}}{35 a^2 x^4}-\frac {16 b^2 \sqrt {a x^3+b x^4}}{35 a^3 x^3}-\frac {\left (16 b^3\right ) \int \frac {1}{\sqrt {a x^3+b x^4}} \, dx}{35 a^3}\\ &=-\frac {2 \sqrt {a x^3+b x^4}}{7 a x^5}+\frac {12 b \sqrt {a x^3+b x^4}}{35 a^2 x^4}-\frac {16 b^2 \sqrt {a x^3+b x^4}}{35 a^3 x^3}+\frac {32 b^3 \sqrt {a x^3+b x^4}}{35 a^4 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 53, normalized size = 0.49 \begin {gather*} \frac {2 \sqrt {x^3 (a+b x)} \left (-5 a^3+6 a^2 b x-8 a b^2 x^2+16 b^3 x^3\right )}{35 a^4 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[a*x^3 + b*x^4]),x]

[Out]

(2*Sqrt[x^3*(a + b*x)]*(-5*a^3 + 6*a^2*b*x - 8*a*b^2*x^2 + 16*b^3*x^3))/(35*a^4*x^5)

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IntegrateAlgebraic [A]  time = 0.18, size = 55, normalized size = 0.51 \begin {gather*} \frac {2 \left (-5 a^3+6 a^2 b x-8 a b^2 x^2+16 b^3 x^3\right ) \sqrt {a x^3+b x^4}}{35 a^4 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^3*Sqrt[a*x^3 + b*x^4]),x]

[Out]

(2*(-5*a^3 + 6*a^2*b*x - 8*a*b^2*x^2 + 16*b^3*x^3)*Sqrt[a*x^3 + b*x^4])/(35*a^4*x^5)

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fricas [A]  time = 0.40, size = 51, normalized size = 0.47 \begin {gather*} \frac {2 \, {\left (16 \, b^{3} x^{3} - 8 \, a b^{2} x^{2} + 6 \, a^{2} b x - 5 \, a^{3}\right )} \sqrt {b x^{4} + a x^{3}}}{35 \, a^{4} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a*x^3)^(1/2),x, algorithm="fricas")

[Out]

2/35*(16*b^3*x^3 - 8*a*b^2*x^2 + 6*a^2*b*x - 5*a^3)*sqrt(b*x^4 + a*x^3)/(a^4*x^5)

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giac [A]  time = 0.27, size = 57, normalized size = 0.53 \begin {gather*} -\frac {2 \, {\left (5 \, {\left (b + \frac {a}{x}\right )}^{\frac {7}{2}} - 21 \, {\left (b + \frac {a}{x}\right )}^{\frac {5}{2}} b + 35 \, {\left (b + \frac {a}{x}\right )}^{\frac {3}{2}} b^{2} - 35 \, \sqrt {b + \frac {a}{x}} b^{3}\right )}}{35 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a*x^3)^(1/2),x, algorithm="giac")

[Out]

-2/35*(5*(b + a/x)^(7/2) - 21*(b + a/x)^(5/2)*b + 35*(b + a/x)^(3/2)*b^2 - 35*sqrt(b + a/x)*b^3)/a^4

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maple [A]  time = 0.05, size = 57, normalized size = 0.53 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 \sqrt {b \,x^{4}+a \,x^{3}}\, a^{4} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^4+a*x^3)^(1/2),x)

[Out]

-2/35*(b*x+a)*(-16*b^3*x^3+8*a*b^2*x^2-6*a^2*b*x+5*a^3)/x^2/a^4/(b*x^4+a*x^3)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {b x^{4} + a x^{3}} x^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^4+a*x^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*x^4 + a*x^3)*x^3), x)

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mupad [B]  time = 5.14, size = 92, normalized size = 0.85 \begin {gather*} \frac {12\,b\,\sqrt {b\,x^4+a\,x^3}}{35\,a^2\,x^4}-\frac {2\,\sqrt {b\,x^4+a\,x^3}}{7\,a\,x^5}-\frac {16\,b^2\,\sqrt {b\,x^4+a\,x^3}}{35\,a^3\,x^3}+\frac {32\,b^3\,\sqrt {b\,x^4+a\,x^3}}{35\,a^4\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a*x^3 + b*x^4)^(1/2)),x)

[Out]

(12*b*(a*x^3 + b*x^4)^(1/2))/(35*a^2*x^4) - (2*(a*x^3 + b*x^4)^(1/2))/(7*a*x^5) - (16*b^2*(a*x^3 + b*x^4)^(1/2
))/(35*a^3*x^3) + (32*b^3*(a*x^3 + b*x^4)^(1/2))/(35*a^4*x^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \sqrt {x^{3} \left (a + b x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x**4+a*x**3)**(1/2),x)

[Out]

Integral(1/(x**3*sqrt(x**3*(a + b*x))), x)

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