3.8.55 \(\int \frac {\sqrt {b+a x^3} (2 b+a x^3)}{x^7} \, dx\)

Optimal. Leaf size=59 \[ \frac {\left (-3 a x^3-2 b\right ) \sqrt {a x^3+b}}{6 x^6}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {a x^3+b}}{\sqrt {b}}\right )}{6 \sqrt {b}} \]

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Rubi [A]  time = 0.04, antiderivative size = 68, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {446, 78, 47, 63, 208} \begin {gather*} -\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {a x^3+b}}{\sqrt {b}}\right )}{6 \sqrt {b}}-\frac {a \sqrt {a x^3+b}}{6 x^3}-\frac {\left (a x^3+b\right )^{3/2}}{3 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[b + a*x^3]*(2*b + a*x^3))/x^7,x]

[Out]

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

Rule 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 208

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

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 75, normalized size = 1.27 \begin {gather*} -\frac {a^2 x^6 \sqrt {\frac {a x^3}{b}+1} \tanh ^{-1}\left (\sqrt {\frac {a x^3}{b}+1}\right )+3 a^2 x^6+5 a b x^3+2 b^2}{6 x^6 \sqrt {a x^3+b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[b + a*x^3]*(2*b + a*x^3))/x^7,x]

[Out]

-1/6*(2*b^2 + 5*a*b*x^3 + 3*a^2*x^6 + a^2*x^6*Sqrt[1 + (a*x^3)/b]*ArcTanh[Sqrt[1 + (a*x^3)/b]])/(x^6*Sqrt[b +
a*x^3])

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IntegrateAlgebraic [A]  time = 0.10, size = 59, normalized size = 1.00 \begin {gather*} \frac {\left (-2 b-3 a x^3\right ) \sqrt {b+a x^3}}{6 x^6}-\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b+a x^3}}{\sqrt {b}}\right )}{6 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[b + a*x^3]*(2*b + a*x^3))/x^7,x]

[Out]

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

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fricas [A]  time = 0.62, size = 136, normalized size = 2.31 \begin {gather*} \left [\frac {a^{2} \sqrt {b} x^{6} \log \left (\frac {a x^{3} - 2 \, \sqrt {a x^{3} + b} \sqrt {b} + 2 \, b}{x^{3}}\right ) - 2 \, {\left (3 \, a b x^{3} + 2 \, b^{2}\right )} \sqrt {a x^{3} + b}}{12 \, b x^{6}}, \frac {a^{2} \sqrt {-b} x^{6} \arctan \left (\frac {\sqrt {a x^{3} + b} \sqrt {-b}}{b}\right ) - {\left (3 \, a b x^{3} + 2 \, b^{2}\right )} \sqrt {a x^{3} + b}}{6 \, b x^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.28, size = 69, normalized size = 1.17 \begin {gather*} \frac {\frac {a^{3} \arctan \left (\frac {\sqrt {a x^{3} + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {3 \, {\left (a x^{3} + b\right )}^{\frac {3}{2}} a^{3} - \sqrt {a x^{3} + b} a^{3} b}{a^{2} x^{6}}}{6 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.11, size = 48, normalized size = 0.81

method	result	size
risch	\(-\frac {\sqrt {a \,x^{3}+b}\, \left (3 a \,x^{3}+2 b \right )}{6 x^{6}}-\frac {a^{2} \arctanh \left (\frac {\sqrt {a \,x^{3}+b}}{\sqrt {b}}\right )}{6 \sqrt {b}}\)	\(48\)
elliptic	\(-\frac {b \sqrt {a \,x^{3}+b}}{3 x^{6}}-\frac {a \sqrt {a \,x^{3}+b}}{2 x^{3}}-\frac {a^{2} \arctanh \left (\frac {\sqrt {a \,x^{3}+b}}{\sqrt {b}}\right )}{6 \sqrt {b}}\)	\(54\)
default	\(a \left (-\frac {\sqrt {a \,x^{3}+b}}{3 x^{3}}-\frac {a \arctanh \left (\frac {\sqrt {a \,x^{3}+b}}{\sqrt {b}}\right )}{3 \sqrt {b}}\right )+2 b \left (-\frac {\sqrt {a \,x^{3}+b}}{6 x^{6}}-\frac {a \sqrt {a \,x^{3}+b}}{12 b \,x^{3}}+\frac {a^{2} \arctanh \left (\frac {\sqrt {a \,x^{3}+b}}{\sqrt {b}}\right )}{12 b^{\frac {3}{2}}}\right )\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.43, size = 158, normalized size = 2.68 \begin {gather*} \frac {1}{6} \, {\left (\frac {a \log \left (\frac {\sqrt {a x^{3} + b} - \sqrt {b}}{\sqrt {a x^{3} + b} + \sqrt {b}}\right )}{\sqrt {b}} - \frac {2 \, \sqrt {a x^{3} + b}}{x^{3}}\right )} a - \frac {1}{12} \, {\left (\frac {a^{2} \log \left (\frac {\sqrt {a x^{3} + b} - \sqrt {b}}{\sqrt {a x^{3} + b} + \sqrt {b}}\right )}{b^{\frac {3}{2}}} + \frac {2 \, {\left ({\left (a x^{3} + b\right )}^{\frac {3}{2}} a^{2} + \sqrt {a x^{3} + b} a^{2} b\right )}}{{\left (a x^{3} + b\right )}^{2} b - 2 \, {\left (a x^{3} + b\right )} b^{2} + b^{3}}\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.93, size = 74, normalized size = 1.25 \begin {gather*} \frac {a^2\,\ln \left (\frac {{\left (\sqrt {a\,x^3+b}-\sqrt {b}\right )}^3\,\left (\sqrt {a\,x^3+b}+\sqrt {b}\right )}{x^6}\right )}{12\,\sqrt {b}}-\frac {b\,\sqrt {a\,x^3+b}}{3\,x^6}-\frac {a\,\sqrt {a\,x^3+b}}{2\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 68.22, size = 128, normalized size = 2.17 \begin {gather*} - \frac {a^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x^{3}}}}{3 x^{\frac {3}{2}}} - \frac {a^{\frac {3}{2}}}{6 x^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x^{3}}}} - \frac {\sqrt {a} b}{2 x^{\frac {9}{2}} \sqrt {1 + \frac {b}{a x^{3}}}} - \frac {a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{\frac {3}{2}}} \right )}}{6 \sqrt {b}} - \frac {b^{2}}{3 \sqrt {a} x^{\frac {15}{2}} \sqrt {1 + \frac {b}{a x^{3}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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