3.2.69 \(\int \frac {(a x^2+b x^3)^{3/2}}{x^6} \, dx\)

Optimal. Leaf size=81 \[ -\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 \sqrt {a}}-\frac {3 b \sqrt {a x^2+b x^3}}{4 x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5} \]

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Rubi [A]  time = 0.09, antiderivative size = 81, 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 = {2020, 2008, 206} \begin {gather*} -\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 \sqrt {a}}-\frac {3 b \sqrt {a x^2+b x^3}}{4 x^2}-\frac {\left (a x^2+b x^3\right )^{3/2}}{2 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x^6,x]

[Out]

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

Rule 206

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

Rule 2008

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 2020

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 72, normalized size = 0.89 \begin {gather*} -\frac {2 a^2+3 b^2 x^2 \sqrt {\frac {b x}{a}+1} \tanh ^{-1}\left (\sqrt {\frac {b x}{a}+1}\right )+7 a b x+5 b^2 x^2}{4 x \sqrt {x^2 (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3)^(3/2)/x^6,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 15.71, size = 82, normalized size = 1.01 \begin {gather*} \frac {\left (x^2 (a+b x)\right )^{3/2} \left (-\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {\sqrt {a+b x} (5 (a+b x)-3 a)}{4 x^2}\right )}{x^3 (a+b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a*x^2 + b*x^3)^(3/2)/x^6,x]

[Out]

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

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fricas [A]  time = 0.43, size = 154, normalized size = 1.90 \begin {gather*} \left [\frac {3 \, \sqrt {a} b^{2} x^{3} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, \sqrt {b x^{3} + a x^{2}} {\left (5 \, a b x + 2 \, a^{2}\right )}}{8 \, a x^{3}}, \frac {3 \, \sqrt {-a} b^{2} x^{3} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) - \sqrt {b x^{3} + a x^{2}} {\left (5 \, a b x + 2 \, a^{2}\right )}}{4 \, a x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="fricas")

[Out]

[1/8*(3*sqrt(a)*b^2*x^3*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) - 2*sqrt(b*x^3 + a*x^2)*(5*a*
b*x + 2*a^2))/(a*x^3), 1/4*(3*sqrt(-a)*b^2*x^3*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(a*x)) - sqrt(b*x^3 + a*x^2
)*(5*a*b*x + 2*a^2))/(a*x^3)]

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giac [A]  time = 0.26, size = 70, normalized size = 0.86 \begin {gather*} \frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\relax (x)}{\sqrt {-a}} - \frac {5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3} \mathrm {sgn}\relax (x) - 3 \, \sqrt {b x + a} a b^{3} \mathrm {sgn}\relax (x)}{b^{2} x^{2}}}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 74, normalized size = 0.91 \begin {gather*} -\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (3 b^{2} x^{2} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-3 \sqrt {b x +a}\, a^{\frac {3}{2}}+5 \left (b x +a \right )^{\frac {3}{2}} \sqrt {a}\right )}{4 \left (b x +a \right )^{\frac {3}{2}} \sqrt {a}\, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a*x^2)^(3/2)/x^6,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}{x^{6}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="maxima")

[Out]

integrate((b*x^3 + a*x^2)^(3/2)/x^6, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^3+a\,x^2\right )}^{3/2}}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3)^(3/2)/x^6,x)

[Out]

int((a*x^2 + b*x^3)^(3/2)/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}{x^{6}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a*x**2)**(3/2)/x**6,x)

[Out]

Integral((x**2*(a + b*x))**(3/2)/x**6, x)

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