3.3.48 \(\int \frac {(a x^2+b x^3)^{3/2}}{x^6} \, dx\) [248]

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

[Out]

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

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Rubi [A]
time = 0.06, 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 = {2045, 2033, 212} \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 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 2045

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 ) \text {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 = 0.10, size = 82, normalized size = 1.01 \begin {gather*} -\frac {\sqrt {x^2 (a+b x)} \left (\sqrt {a} \sqrt {a+b x} (2 a+5 b x)+3 b^2 x^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{4 \sqrt {a} x^3 \sqrt {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*(Sqrt[x^2*(a + b*x)]*(Sqrt[a]*Sqrt[a + b*x]*(2*a + 5*b*x) + 3*b^2*x^2*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(S
qrt[a]*x^3*Sqrt[a + b*x])

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Maple [A]
time = 0.38, size = 74, normalized size = 0.91

method	result	size
risch	\(-\frac {\left (5 b x +2 a \right ) \sqrt {x^{2} \left (b x +a \right )}}{4 x^{3}}-\frac {3 b^{2} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{4 \sqrt {a}\, x \sqrt {b x +a}}\)	\(67\)
default	\(-\frac {\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (3 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{2} x^{2}+5 \left (b x +a \right )^{\frac {3}{2}} \sqrt {a}-3 \sqrt {b x +a}\, a^{\frac {3}{2}}\right )}{4 x^{5} \left (b x +a \right )^{\frac {3}{2}} \sqrt {a}}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b*x^3+a*x^2)^(3/2)*(3*arctanh((b*x+a)^(1/2)/a^(1/2))*b^2*x^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*} \text {Failed to integrate} \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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Fricas [A]
time = 1.10, 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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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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Giac [A]
time = 1.03, 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}\left (x\right )}{\sqrt {-a}} - \frac {5 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{3} \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {b x + a} a b^{3} \mathrm {sgn}\left (x\right )}{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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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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