3.3.59 \(\int \frac {1}{x^3 \sqrt {a x^2+b x^3}} \, dx\) [259]

Optimal. Leaf size=115 \[ -\frac {\sqrt {a x^2+b x^3}}{3 a x^4}+\frac {5 b \sqrt {a x^2+b x^3}}{12 a^2 x^3}-\frac {5 b^2 \sqrt {a x^2+b x^3}}{8 a^3 x^2}+\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{7/2}} \]

[Out]

5/8*b^3*arctanh(x*a^(1/2)/(b*x^3+a*x^2)^(1/2))/a^(7/2)-1/3*(b*x^3+a*x^2)^(1/2)/a/x^4+5/12*b*(b*x^3+a*x^2)^(1/2
)/a^2/x^3-5/8*b^2*(b*x^3+a*x^2)^(1/2)/a^3/x^2

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Rubi [A]
time = 0.09, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2050, 2033, 212} \begin {gather*} \frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{7/2}}-\frac {5 b^2 \sqrt {a x^2+b x^3}}{8 a^3 x^2}+\frac {5 b \sqrt {a x^2+b x^3}}{12 a^2 x^3}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[a*x^2 + b*x^3]),x]

[Out]

-1/3*Sqrt[a*x^2 + b*x^3]/(a*x^4) + (5*b*Sqrt[a*x^2 + b*x^3])/(12*a^2*x^3) - (5*b^2*Sqrt[a*x^2 + b*x^3])/(8*a^3
*x^2) + (5*b^3*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(8*a^(7/2))

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 2033

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

Rule 2050

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, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {1}{x^3 \sqrt {a x^2+b x^3}} \, dx &=-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}-\frac {(5 b) \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx}{6 a}\\ &=-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}+\frac {5 b \sqrt {a x^2+b x^3}}{12 a^2 x^3}+\frac {\left (5 b^2\right ) \int \frac {1}{x \sqrt {a x^2+b x^3}} \, dx}{8 a^2}\\ &=-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}+\frac {5 b \sqrt {a x^2+b x^3}}{12 a^2 x^3}-\frac {5 b^2 \sqrt {a x^2+b x^3}}{8 a^3 x^2}-\frac {\left (5 b^3\right ) \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx}{16 a^3}\\ &=-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}+\frac {5 b \sqrt {a x^2+b x^3}}{12 a^2 x^3}-\frac {5 b^2 \sqrt {a x^2+b x^3}}{8 a^3 x^2}+\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {x}{\sqrt {a x^2+b x^3}}\right )}{8 a^3}\\ &=-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}+\frac {5 b \sqrt {a x^2+b x^3}}{12 a^2 x^3}-\frac {5 b^2 \sqrt {a x^2+b x^3}}{8 a^3 x^2}+\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 96, normalized size = 0.83 \begin {gather*} \frac {-\sqrt {a} \left (8 a^3-2 a^2 b x+5 a b^2 x^2+15 b^3 x^3\right )+15 b^3 x^3 \sqrt {a+b x} \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{24 a^{7/2} x^2 \sqrt {x^2 (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[a*x^2 + b*x^3]),x]

[Out]

(-(Sqrt[a]*(8*a^3 - 2*a^2*b*x + 5*a*b^2*x^2 + 15*b^3*x^3)) + 15*b^3*x^3*Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/Sq
rt[a]])/(24*a^(7/2)*x^2*Sqrt[x^2*(a + b*x)])

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Maple [A]
time = 0.39, size = 95, normalized size = 0.83

method	result	size
risch	\(-\frac {\left (b x +a \right ) \left (15 b^{2} x^{2}-10 a b x +8 a^{2}\right )}{24 a^{3} x^{2} \sqrt {x^{2} \left (b x +a \right )}}+\frac {5 b^{3} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {b x +a}\, x}{8 a^{\frac {7}{2}} \sqrt {x^{2} \left (b x +a \right )}}\)	\(84\)
default	\(-\frac {\sqrt {b x +a}\, \left (15 \sqrt {b x +a}\, a^{\frac {3}{2}} b^{2} x^{2}-15 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{3} x^{3}-10 \sqrt {b x +a}\, a^{\frac {5}{2}} b x +8 \sqrt {b x +a}\, a^{\frac {7}{2}}\right )}{24 x^{2} \sqrt {b \,x^{3}+a \,x^{2}}\, a^{\frac {9}{2}}}\)	\(95\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24/x^2*(b*x+a)^(1/2)*(15*(b*x+a)^(1/2)*a^(3/2)*b^2*x^2-15*arctanh((b*x+a)^(1/2)/a^(1/2))*a*b^3*x^3-10*(b*x+
a)^(1/2)*a^(5/2)*b*x+8*(b*x+a)^(1/2)*a^(7/2))/(b*x^3+a*x^2)^(1/2)/a^(9/2)

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

[Out]

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

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Fricas [A]
time = 2.03, size = 175, normalized size = 1.52 \begin {gather*} \left [\frac {15 \, \sqrt {a} b^{3} x^{4} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, {\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, a^{4} x^{4}}, -\frac {15 \, \sqrt {-a} b^{3} x^{4} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + {\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, a^{4} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(15*sqrt(a)*b^3*x^4*log((b*x^2 + 2*a*x + 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) - 2*(15*a*b^2*x^2 - 10*a^2*
b*x + 8*a^3)*sqrt(b*x^3 + a*x^2))/(a^4*x^4), -1/24*(15*sqrt(-a)*b^3*x^4*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(a
*x)) + (15*a*b^2*x^2 - 10*a^2*b*x + 8*a^3)*sqrt(b*x^3 + a*x^2))/(a^4*x^4)]

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

[Out]

Integral(1/(x**3*sqrt(x**2*(a + b*x))), x)

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Giac [A]
time = 0.96, size = 88, normalized size = 0.77 \begin {gather*} -\frac {\frac {15 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {15 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} + 33 \, \sqrt {b x + a} a^{2} b^{4}}{a^{3} b^{3} x^{3}}}{24 \, b \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(15*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + (15*(b*x + a)^(5/2)*b^4 - 40*(b*x + a)^(3/2)*a*b
^4 + 33*sqrt(b*x + a)*a^2*b^4)/(a^3*b^3*x^3))/(b*sgn(x))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^3\,\sqrt {b\,x^3+a\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a*x^2 + b*x^3)^(1/2)),x)

[Out]

int(1/(x^3*(a*x^2 + b*x^3)^(1/2)), x)

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