3.4.74 \(\int \frac {\sqrt {a+b x^3}}{x^7} \, dx\) [374]

Optimal. Leaf size=71 \[ -\frac {\sqrt {a+b x^3}}{6 x^6}-\frac {b \sqrt {a+b x^3}}{12 a x^3}+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{12 a^{3/2}} \]

[Out]

1/12*b^2*arctanh((b*x^3+a)^(1/2)/a^(1/2))/a^(3/2)-1/6*(b*x^3+a)^(1/2)/x^6-1/12*b*(b*x^3+a)^(1/2)/a/x^3

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Rubi [A]
time = 0.03, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 43, 44, 65, 214} \begin {gather*} \frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{12 a^{3/2}}-\frac {b \sqrt {a+b x^3}}{12 a x^3}-\frac {\sqrt {a+b x^3}}{6 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^3]/x^7,x]

[Out]

-1/6*Sqrt[a + b*x^3]/x^6 - (b*Sqrt[a + b*x^3])/(12*a*x^3) + (b^2*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(12*a^(3/2)
)

Rule 43

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

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x^3}}{x^7} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x^3} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {a+b x^3}}{6 x^6}+\frac {1}{12} b \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {a+b x^3}}{6 x^6}-\frac {b \sqrt {a+b x^3}}{12 a x^3}-\frac {b^2 \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^3\right )}{24 a}\\ &=-\frac {\sqrt {a+b x^3}}{6 x^6}-\frac {b \sqrt {a+b x^3}}{12 a x^3}-\frac {b \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^3}\right )}{12 a}\\ &=-\frac {\sqrt {a+b x^3}}{6 x^6}-\frac {b \sqrt {a+b x^3}}{12 a x^3}+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{12 a^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 62, normalized size = 0.87 \begin {gather*} \frac {\left (-2 a-b x^3\right ) \sqrt {a+b x^3}}{12 a x^6}+\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{12 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^3]/x^7,x]

[Out]

((-2*a - b*x^3)*Sqrt[a + b*x^3])/(12*a*x^6) + (b^2*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(12*a^(3/2))

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Maple [A]
time = 0.14, size = 56, normalized size = 0.79

method	result	size
risch	\(-\frac {\sqrt {b \,x^{3}+a}\, \left (b \,x^{3}+2 a \right )}{12 x^{6} a}+\frac {b^{2} \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{12 a^{\frac {3}{2}}}\)	\(50\)
default	\(\frac {b^{2} \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{12 a^{\frac {3}{2}}}-\frac {\sqrt {b \,x^{3}+a}}{6 x^{6}}-\frac {b \sqrt {b \,x^{3}+a}}{12 a \,x^{3}}\)	\(56\)
elliptic	\(\frac {b^{2} \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{12 a^{\frac {3}{2}}}-\frac {\sqrt {b \,x^{3}+a}}{6 x^{6}}-\frac {b \sqrt {b \,x^{3}+a}}{12 a \,x^{3}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^(1/2)/x^7,x,method=_RETURNVERBOSE)

[Out]

1/12*b^2*arctanh((b*x^3+a)^(1/2)/a^(1/2))/a^(3/2)-1/6*(b*x^3+a)^(1/2)/x^6-1/12*b*(b*x^3+a)^(1/2)/a/x^3

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Maxima [A]
time = 0.50, size = 100, normalized size = 1.41 \begin {gather*} -\frac {b^{2} \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{24 \, a^{\frac {3}{2}}} - \frac {{\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2} + \sqrt {b x^{3} + a} a b^{2}}{12 \, {\left ({\left (b x^{3} + a\right )}^{2} a - 2 \, {\left (b x^{3} + a\right )} a^{2} + a^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/2)/x^7,x, algorithm="maxima")

[Out]

-1/24*b^2*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^(3/2) - 1/12*((b*x^3 + a)^(3/2)*b^2 +
sqrt(b*x^3 + a)*a*b^2)/((b*x^3 + a)^2*a - 2*(b*x^3 + a)*a^2 + a^3)

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Fricas [A]
time = 0.36, size = 133, normalized size = 1.87 \begin {gather*} \left [\frac {\sqrt {a} b^{2} x^{6} \log \left (\frac {b x^{3} + 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) - 2 \, {\left (a b x^{3} + 2 \, a^{2}\right )} \sqrt {b x^{3} + a}}{24 \, a^{2} x^{6}}, -\frac {\sqrt {-a} b^{2} x^{6} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + {\left (a b x^{3} + 2 \, a^{2}\right )} \sqrt {b x^{3} + a}}{12 \, a^{2} x^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/2)/x^7,x, algorithm="fricas")

[Out]

[1/24*(sqrt(a)*b^2*x^6*log((b*x^3 + 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) - 2*(a*b*x^3 + 2*a^2)*sqrt(b*x^3 + a
))/(a^2*x^6), -1/12*(sqrt(-a)*b^2*x^6*arctan(sqrt(b*x^3 + a)*sqrt(-a)/a) + (a*b*x^3 + 2*a^2)*sqrt(b*x^3 + a))/
(a^2*x^6)]

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Sympy [A]
time = 2.06, size = 100, normalized size = 1.41 \begin {gather*} - \frac {a}{6 \sqrt {b} x^{\frac {15}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {\sqrt {b}}{4 x^{\frac {9}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {b^{\frac {3}{2}}}{12 a x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{12 a^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**(1/2)/x**7,x)

[Out]

-a/(6*sqrt(b)*x**(15/2)*sqrt(a/(b*x**3) + 1)) - sqrt(b)/(4*x**(9/2)*sqrt(a/(b*x**3) + 1)) - b**(3/2)/(12*a*x**
(3/2)*sqrt(a/(b*x**3) + 1)) + b**2*asinh(sqrt(a)/(sqrt(b)*x**(3/2)))/(12*a**(3/2))

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Giac [A]
time = 1.45, size = 72, normalized size = 1.01 \begin {gather*} -\frac {\frac {b^{3} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {{\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{3} + \sqrt {b x^{3} + a} a b^{3}}{a b^{2} x^{6}}}{12 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^(1/2)/x^7,x, algorithm="giac")

[Out]

-1/12*(b^3*arctan(sqrt(b*x^3 + a)/sqrt(-a))/(sqrt(-a)*a) + ((b*x^3 + a)^(3/2)*b^3 + sqrt(b*x^3 + a)*a*b^3)/(a*
b^2*x^6))/b

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Mupad [B]
time = 1.35, size = 76, normalized size = 1.07 \begin {gather*} \frac {b^2\,\ln \left (\frac {\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )\,{\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}^3}{x^6}\right )}{24\,a^{3/2}}-\frac {\sqrt {b\,x^3+a}}{6\,x^6}-\frac {b\,\sqrt {b\,x^3+a}}{12\,a\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(1/2)/x^7,x)

[Out]

(b^2*log((((a + b*x^3)^(1/2) - a^(1/2))*((a + b*x^3)^(1/2) + a^(1/2))^3)/x^6))/(24*a^(3/2)) - (a + b*x^3)^(1/2
)/(6*x^6) - (b*(a + b*x^3)^(1/2))/(12*a*x^3)

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