3.4.11 \(\int \frac {x^3}{\sqrt {a x^3+b x^4}} \, dx\) [311]

Optimal. Leaf size=86 \[ \frac {\sqrt {a x^3+b x^4}}{2 b}-\frac {3 a \sqrt {a x^3+b x^4}}{4 b^2 x}+\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a x^3+b x^4}}\right )}{4 b^{5/2}} \]

[Out]

3/4*a^2*arctanh(x^2*b^(1/2)/(b*x^4+a*x^3)^(1/2))/b^(5/2)+1/2*(b*x^4+a*x^3)^(1/2)/b-3/4*a*(b*x^4+a*x^3)^(1/2)/b
^2/x

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Rubi [A]
time = 0.09, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2049, 2054, 212} \begin {gather*} \frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a x^3+b x^4}}\right )}{4 b^{5/2}}-\frac {3 a \sqrt {a x^3+b x^4}}{4 b^2 x}+\frac {\sqrt {a x^3+b x^4}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[a*x^3 + b*x^4],x]

[Out]

Sqrt[a*x^3 + b*x^4]/(2*b) - (3*a*Sqrt[a*x^3 + b*x^4])/(4*b^2*x) + (3*a^2*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a*x^3 + b*
x^4]])/(4*b^(5/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2049

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

Rule 2054

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rubi steps

\begin {align*} \int \frac {x^3}{\sqrt {a x^3+b x^4}} \, dx &=\frac {\sqrt {a x^3+b x^4}}{2 b}-\frac {(3 a) \int \frac {x^2}{\sqrt {a x^3+b x^4}} \, dx}{4 b}\\ &=\frac {\sqrt {a x^3+b x^4}}{2 b}-\frac {3 a \sqrt {a x^3+b x^4}}{4 b^2 x}+\frac {\left (3 a^2\right ) \int \frac {x}{\sqrt {a x^3+b x^4}} \, dx}{8 b^2}\\ &=\frac {\sqrt {a x^3+b x^4}}{2 b}-\frac {3 a \sqrt {a x^3+b x^4}}{4 b^2 x}+\frac {\left (3 a^2\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a x^3+b x^4}}\right )}{4 b^2}\\ &=\frac {\sqrt {a x^3+b x^4}}{2 b}-\frac {3 a \sqrt {a x^3+b x^4}}{4 b^2 x}+\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a x^3+b x^4}}\right )}{4 b^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 93, normalized size = 1.08 \begin {gather*} \frac {\sqrt {b} x^2 \left (-3 a^2-a b x+2 b^2 x^2\right )-3 a^2 x^{3/2} \sqrt {a+b x} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )}{4 b^{5/2} \sqrt {x^3 (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[a*x^3 + b*x^4],x]

[Out]

(Sqrt[b]*x^2*(-3*a^2 - a*b*x + 2*b^2*x^2) - 3*a^2*x^(3/2)*Sqrt[a + b*x]*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]
])/(4*b^(5/2)*Sqrt[x^3*(a + b*x)])

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Maple [A]
time = 0.38, size = 98, normalized size = 1.14

method	result	size
risch	\(-\frac {\left (-2 b x +3 a \right ) x^{2} \left (b x +a \right )}{4 b^{2} \sqrt {x^{3} \left (b x +a \right )}}+\frac {3 a^{2} \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right ) x \sqrt {x \left (b x +a \right )}}{8 b^{\frac {5}{2}} \sqrt {x^{3} \left (b x +a \right )}}\)	\(87\)
default	\(\frac {x \sqrt {x \left (b x +a \right )}\, \left (4 x \sqrt {b \,x^{2}+a x}\, b^{\frac {5}{2}}-6 b^{\frac {3}{2}} \sqrt {b \,x^{2}+a x}\, a +3 \ln \left (\frac {2 \sqrt {b \,x^{2}+a x}\, \sqrt {b}+2 b x +a}{2 \sqrt {b}}\right ) a^{2} b \right )}{8 \sqrt {b \,x^{4}+a \,x^{3}}\, b^{\frac {7}{2}}}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x*(x*(b*x+a))^(1/2)*(4*x*(b*x^2+a*x)^(1/2)*b^(5/2)-6*b^(3/2)*(b*x^2+a*x)^(1/2)*a+3*ln(1/2*(2*(b*x^2+a*x)^(
1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^2*b)/(b*x^4+a*x^3)^(1/2)/b^(7/2)

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

[Out]

integrate(x^3/sqrt(b*x^4 + a*x^3), x)

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Fricas [A]
time = 1.87, size = 150, normalized size = 1.74 \begin {gather*} \left [\frac {3 \, a^{2} \sqrt {b} x \log \left (\frac {2 \, b x^{2} + a x + 2 \, \sqrt {b x^{4} + a x^{3}} \sqrt {b}}{x}\right ) + 2 \, \sqrt {b x^{4} + a x^{3}} {\left (2 \, b^{2} x - 3 \, a b\right )}}{8 \, b^{3} x}, -\frac {3 \, a^{2} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{4} + a x^{3}} \sqrt {-b}}{b x^{2}}\right ) - \sqrt {b x^{4} + a x^{3}} {\left (2 \, b^{2} x - 3 \, a b\right )}}{4 \, b^{3} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*a^2*sqrt(b)*x*log((2*b*x^2 + a*x + 2*sqrt(b*x^4 + a*x^3)*sqrt(b))/x) + 2*sqrt(b*x^4 + a*x^3)*(2*b^2*x
- 3*a*b))/(b^3*x), -1/4*(3*a^2*sqrt(-b)*x*arctan(sqrt(b*x^4 + a*x^3)*sqrt(-b)/(b*x^2)) - sqrt(b*x^4 + a*x^3)*(
2*b^2*x - 3*a*b))/(b^3*x)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {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(x**3/(b*x**4+a*x**3)**(1/2),x)

[Out]

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

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Giac [A]
time = 0.67, size = 90, normalized size = 1.05 \begin {gather*} \frac {1}{4} \, \sqrt {b x^{2} + a x} {\left (\frac {2 \, x}{b \mathrm {sgn}\left (x\right )} - \frac {3 \, a}{b^{2} \mathrm {sgn}\left (x\right )}\right )} + \frac {3 \, a^{2} \log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{8 \, b^{\frac {5}{2}}} - \frac {3 \, a^{2} \log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right )}{8 \, b^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(b*x^2 + a*x)*(2*x/(b*sgn(x)) - 3*a/(b^2*sgn(x))) + 3/8*a^2*log(abs(a))*sgn(x)/b^(5/2) - 3/8*a^2*log(a
bs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a))/(b^(5/2)*sgn(x))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3}{\sqrt {b\,x^4+a\,x^3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a*x^3 + b*x^4)^(1/2),x)

[Out]

int(x^3/(a*x^3 + b*x^4)^(1/2), x)

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