3.3.39 \(\int \frac {\sqrt {a x^2+b x^3}}{x^5} \, dx\) [239]

Optimal. Leaf size=112 \[ -\frac {\sqrt {a x^2+b x^3}}{3 x^4}-\frac {b \sqrt {a x^2+b x^3}}{12 a x^3}+\frac {b^2 \sqrt {a x^2+b x^3}}{8 a^2 x^2}-\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2045, 2050, 2033, 212} \begin {gather*} -\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{5/2}}+\frac {b^2 \sqrt {a x^2+b x^3}}{8 a^2 x^2}-\frac {b \sqrt {a x^2+b x^3}}{12 a x^3}-\frac {\sqrt {a x^2+b x^3}}{3 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^2 + b*x^3]/x^5,x]

[Out]

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

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

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Mathematica [A]
time = 0.11, size = 93, normalized size = 0.83 \begin {gather*} -\frac {\sqrt {x^2 (a+b x)} \left (\sqrt {a} \sqrt {a+b x} \left (8 a^2+2 a b x-3 b^2 x^2\right )+3 b^3 x^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{24 a^{5/2} x^4 \sqrt {a+b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a*x^2 + b*x^3]/x^5,x]

[Out]

-1/24*(Sqrt[x^2*(a + b*x)]*(Sqrt[a]*Sqrt[a + b*x]*(8*a^2 + 2*a*b*x - 3*b^2*x^2) + 3*b^3*x^3*ArcTanh[Sqrt[a + b
*x]/Sqrt[a]]))/(a^(5/2)*x^4*Sqrt[a + b*x])

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Maple [A]
time = 0.39, size = 89, normalized size = 0.79

method	result	size
risch	\(-\frac {\left (-3 b^{2} x^{2}+2 a b x +8 a^{2}\right ) \sqrt {x^{2} \left (b x +a \right )}}{24 x^{4} a^{2}}-\frac {b^{3} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {x^{2} \left (b x +a \right )}}{8 a^{\frac {5}{2}} x \sqrt {b x +a}}\)	\(81\)
default	\(\frac {\sqrt {b \,x^{3}+a \,x^{2}}\, \left (3 \left (b x +a \right )^{\frac {5}{2}} a^{\frac {5}{2}}-8 \left (b x +a \right )^{\frac {3}{2}} a^{\frac {7}{2}}-3 \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a^{2} b^{3} x^{3}-3 \sqrt {b x +a}\, a^{\frac {9}{2}}\right )}{24 x^{4} \sqrt {b x +a}\, a^{\frac {9}{2}}}\)	\(89\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a*x^2)^(1/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

1/24*(b*x^3+a*x^2)^(1/2)*(3*(b*x+a)^(5/2)*a^(5/2)-8*(b*x+a)^(3/2)*a^(7/2)-3*arctanh((b*x+a)^(1/2)/a^(1/2))*a^2
*b^3*x^3-3*(b*x+a)^(1/2)*a^(9/2))/x^4/(b*x+a)^(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((b*x^3+a*x^2)^(1/2)/x^5,x, algorithm="maxima")

[Out]

integrate(sqrt(b*x^3 + a*x^2)/x^5, x)

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Fricas [A]
time = 1.33, size = 175, normalized size = 1.56 \begin {gather*} \left [\frac {3 \, \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 (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, a^{3} x^{4}}, \frac {3 \, \sqrt {-a} b^{3} x^{4} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + {\left (3 \, a b^{2} x^{2} - 2 \, a^{2} b x - 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, a^{3} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(1/2)/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{2} \left (a + b x\right )}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a*x**2)**(1/2)/x**5,x)

[Out]

Integral(sqrt(x**2*(a + b*x))/x**5, x)

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Giac [A]
time = 1.49, size = 92, normalized size = 0.82 \begin {gather*} \frac {\frac {3 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} \mathrm {sgn}\left (x\right ) - 8 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} \mathrm {sgn}\left (x\right ) - 3 \, \sqrt {b x + a} a^{2} b^{4} \mathrm {sgn}\left (x\right )}{a^{2} b^{3} x^{3}}}{24 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a*x^2)^(1/2)/x^5,x, algorithm="giac")

[Out]

1/24*(3*b^4*arctan(sqrt(b*x + a)/sqrt(-a))*sgn(x)/(sqrt(-a)*a^2) + (3*(b*x + a)^(5/2)*b^4*sgn(x) - 8*(b*x + a)
^(3/2)*a*b^4*sgn(x) - 3*sqrt(b*x + a)*a^2*b^4*sgn(x))/(a^2*b^3*x^3))/b

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {b\,x^3+a\,x^2}}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 + b*x^3)^(1/2)/x^5,x)

[Out]

int((a*x^2 + b*x^3)^(1/2)/x^5, x)

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