3.4.17 \(\int \frac {1}{x^3 \sqrt {a x^3+b x^4}} \, dx\) [317]

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

[Out]

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

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Rubi [A]
time = 0.09, 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 = {2041, 2025} \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 2025

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 2041

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 = 0.01, 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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Maple [A]
time = 0.37, size = 72, normalized size = 0.67

method	result	size
trager	\(-\frac {2 \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right ) \sqrt {b \,x^{4}+a \,x^{3}}}{35 a^{4} x^{5}}\)	\(52\)
risch	\(-\frac {2 \left (b x +a \right ) \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 x^{2} \sqrt {x^{3} \left (b x +a \right )}\, a^{4}}\)	\(55\)
gosper	\(-\frac {2 \left (b x +a \right ) \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 x^{2} a^{4} \sqrt {b \,x^{4}+a \,x^{3}}}\)	\(57\)
default	\(-\frac {2 \sqrt {x \left (b x +a \right )}\, \sqrt {b \,x^{2}+a x}\, \left (-16 b^{3} x^{3}+8 a \,b^{2} x^{2}-6 a^{2} b x +5 a^{3}\right )}{35 x^{3} \sqrt {b \,x^{4}+a \,x^{3}}\, a^{4}}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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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Fricas [A]
time = 1.94, 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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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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Giac [A]
time = 0.57, size = 111, normalized size = 1.03 \begin {gather*} \frac {2 \, {\left (70 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{3} b^{\frac {3}{2}} + 84 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a b + 35 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a^{2} \sqrt {b} + 5 \, a^{3}\right )}}{35 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{7} \mathrm {sgn}\left (x\right )} \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*(70*(sqrt(b)*x - sqrt(b*x^2 + a*x))^3*b^(3/2) + 84*(sqrt(b)*x - sqrt(b*x^2 + a*x))^2*a*b + 35*(sqrt(b)*x
- sqrt(b*x^2 + a*x))*a^2*sqrt(b) + 5*a^3)/((sqrt(b)*x - sqrt(b*x^2 + a*x))^7*sgn(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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