3.5.27 \(\int \frac {1}{x^7 (a+b x^3)^{3/2}} \, dx\) [427]

Optimal. Leaf size=95 \[ \frac {5 b^2}{4 a^3 \sqrt {a+b x^3}}-\frac {1}{6 a x^6 \sqrt {a+b x^3}}+\frac {5 b}{12 a^2 x^3 \sqrt {a+b x^3}}-\frac {5 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{4 a^{7/2}} \]

[Out]

-5/4*b^2*arctanh((b*x^3+a)^(1/2)/a^(1/2))/a^(7/2)+5/4*b^2/a^3/(b*x^3+a)^(1/2)-1/6/a/x^6/(b*x^3+a)^(1/2)+5/12*b
/a^2/x^3/(b*x^3+a)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 44, 53, 65, 214} \begin {gather*} -\frac {5 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{4 a^{7/2}}+\frac {5 b^2}{4 a^3 \sqrt {a+b x^3}}+\frac {5 b}{12 a^2 x^3 \sqrt {a+b x^3}}-\frac {1}{6 a x^6 \sqrt {a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(a + b*x^3)^(3/2)),x]

[Out]

(5*b^2)/(4*a^3*Sqrt[a + b*x^3]) - 1/(6*a*x^6*Sqrt[a + b*x^3]) + (5*b)/(12*a^2*x^3*Sqrt[a + b*x^3]) - (5*b^2*Ar
cTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(4*a^(7/2))

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 53

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 {1}{x^7 \left (a+b x^3\right )^{3/2}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {1}{x^3 (a+b x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac {2}{3 a x^6 \sqrt {a+b x^3}}+\frac {5 \text {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,x^3\right )}{3 a}\\ &=\frac {2}{3 a x^6 \sqrt {a+b x^3}}-\frac {5 \sqrt {a+b x^3}}{6 a^2 x^6}-\frac {(5 b) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^3\right )}{4 a^2}\\ &=\frac {2}{3 a x^6 \sqrt {a+b x^3}}-\frac {5 \sqrt {a+b x^3}}{6 a^2 x^6}+\frac {5 b \sqrt {a+b x^3}}{4 a^3 x^3}+\frac {\left (5 b^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^3\right )}{8 a^3}\\ &=\frac {2}{3 a x^6 \sqrt {a+b x^3}}-\frac {5 \sqrt {a+b x^3}}{6 a^2 x^6}+\frac {5 b \sqrt {a+b x^3}}{4 a^3 x^3}+\frac {(5 b) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^3}\right )}{4 a^3}\\ &=\frac {2}{3 a x^6 \sqrt {a+b x^3}}-\frac {5 \sqrt {a+b x^3}}{6 a^2 x^6}+\frac {5 b \sqrt {a+b x^3}}{4 a^3 x^3}-\frac {5 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{4 a^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 73, normalized size = 0.77 \begin {gather*} \frac {-2 a^2+5 a b x^3+15 b^2 x^6}{12 a^3 x^6 \sqrt {a+b x^3}}-\frac {5 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{4 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(a + b*x^3)^(3/2)),x]

[Out]

(-2*a^2 + 5*a*b*x^3 + 15*b^2*x^6)/(12*a^3*x^6*Sqrt[a + b*x^3]) - (5*b^2*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(4*a
^(7/2))

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Maple [A]
time = 0.15, size = 80, normalized size = 0.84

method	result	size
default	\(-\frac {\sqrt {b \,x^{3}+a}}{6 a^{2} x^{6}}+\frac {7 b \sqrt {b \,x^{3}+a}}{12 a^{3} x^{3}}+\frac {2 b^{2}}{3 a^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {5 b^{2} \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{4 a^{\frac {7}{2}}}\)	\(80\)
elliptic	\(-\frac {\sqrt {b \,x^{3}+a}}{6 a^{2} x^{6}}+\frac {7 b \sqrt {b \,x^{3}+a}}{12 a^{3} x^{3}}+\frac {2 b^{2}}{3 a^{3} \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {5 b^{2} \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{4 a^{\frac {7}{2}}}\)	\(80\)
risch	\(-\frac {\sqrt {b \,x^{3}+a}\, \left (-7 b \,x^{3}+2 a \right )}{12 a^{3} x^{6}}+\frac {b^{2} \left (-\frac {14}{3 \sqrt {b \,x^{3}+a}}+15 a \left (\frac {2}{3 a \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}}-\frac {2 \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {3}{2}}}\right )\right )}{8 a^{3}}\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/a^2*(b*x^3+a)^(1/2)/x^6+7/12/a^3*b*(b*x^3+a)^(1/2)/x^3+2/3*b^2/a^3/((x^3+a/b)*b)^(1/2)-5/4*b^2*arctanh((b
*x^3+a)^(1/2)/a^(1/2))/a^(7/2)

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Maxima [A]
time = 0.53, size = 122, normalized size = 1.28 \begin {gather*} \frac {15 \, {\left (b x^{3} + a\right )}^{2} b^{2} - 25 \, {\left (b x^{3} + a\right )} a b^{2} + 8 \, a^{2} b^{2}}{12 \, {\left ({\left (b x^{3} + a\right )}^{\frac {5}{2}} a^{3} - 2 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{4} + \sqrt {b x^{3} + a} a^{5}\right )}} + \frac {5 \, b^{2} \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{8 \, a^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(15*(b*x^3 + a)^2*b^2 - 25*(b*x^3 + a)*a*b^2 + 8*a^2*b^2)/((b*x^3 + a)^(5/2)*a^3 - 2*(b*x^3 + a)^(3/2)*a^
4 + sqrt(b*x^3 + a)*a^5) + 5/8*b^2*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^(7/2)

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Fricas [A]
time = 0.37, size = 203, normalized size = 2.14 \begin {gather*} \left [\frac {15 \, {\left (b^{3} x^{9} + a b^{2} x^{6}\right )} \sqrt {a} \log \left (\frac {b x^{3} - 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + 2 \, {\left (15 \, a b^{2} x^{6} + 5 \, a^{2} b x^{3} - 2 \, a^{3}\right )} \sqrt {b x^{3} + a}}{24 \, {\left (a^{4} b x^{9} + a^{5} x^{6}\right )}}, \frac {15 \, {\left (b^{3} x^{9} + a b^{2} x^{6}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + {\left (15 \, a b^{2} x^{6} + 5 \, a^{2} b x^{3} - 2 \, a^{3}\right )} \sqrt {b x^{3} + a}}{12 \, {\left (a^{4} b x^{9} + a^{5} x^{6}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(15*(b^3*x^9 + a*b^2*x^6)*sqrt(a)*log((b*x^3 - 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) + 2*(15*a*b^2*x^6 +
5*a^2*b*x^3 - 2*a^3)*sqrt(b*x^3 + a))/(a^4*b*x^9 + a^5*x^6), 1/12*(15*(b^3*x^9 + a*b^2*x^6)*sqrt(-a)*arctan(s
qrt(b*x^3 + a)*sqrt(-a)/a) + (15*a*b^2*x^6 + 5*a^2*b*x^3 - 2*a^3)*sqrt(b*x^3 + a))/(a^4*b*x^9 + a^5*x^6)]

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Sympy [A]
time = 4.09, size = 112, normalized size = 1.18 \begin {gather*} - \frac {1}{6 a \sqrt {b} x^{\frac {15}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {5 \sqrt {b}}{12 a^{2} x^{\frac {9}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} + \frac {5 b^{\frac {3}{2}}}{4 a^{3} x^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}} - \frac {5 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{4 a^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(6*a*sqrt(b)*x**(15/2)*sqrt(a/(b*x**3) + 1)) + 5*sqrt(b)/(12*a**2*x**(9/2)*sqrt(a/(b*x**3) + 1)) + 5*b**(3/
2)/(4*a**3*x**(3/2)*sqrt(a/(b*x**3) + 1)) - 5*b**2*asinh(sqrt(a)/(sqrt(b)*x**(3/2)))/(4*a**(7/2))

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Giac [A]
time = 1.22, size = 88, normalized size = 0.93 \begin {gather*} \frac {5 \, b^{2} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3}} + \frac {2 \, b^{2}}{3 \, \sqrt {b x^{3} + a} a^{3}} + \frac {7 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2} - 9 \, \sqrt {b x^{3} + a} a b^{2}}{12 \, a^{3} b^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4*b^2*arctan(sqrt(b*x^3 + a)/sqrt(-a))/(sqrt(-a)*a^3) + 2/3*b^2/(sqrt(b*x^3 + a)*a^3) + 1/12*(7*(b*x^3 + a)^
(3/2)*b^2 - 9*sqrt(b*x^3 + a)*a*b^2)/(a^3*b^2*x^6)

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Mupad [B]
time = 1.47, size = 96, normalized size = 1.01 \begin {gather*} \frac {2\,b^2}{3\,a^3\,\sqrt {b\,x^3+a}}-\frac {\sqrt {b\,x^3+a}}{6\,a^2\,x^6}+\frac {5\,b^2\,\ln \left (\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3\,\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}{x^6}\right )}{8\,a^{7/2}}+\frac {7\,b\,\sqrt {b\,x^3+a}}{12\,a^3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(a + b*x^3)^(3/2)),x)

[Out]

(2*b^2)/(3*a^3*(a + b*x^3)^(1/2)) - (a + b*x^3)^(1/2)/(6*a^2*x^6) + (5*b^2*log((((a + b*x^3)^(1/2) - a^(1/2))^
3*((a + b*x^3)^(1/2) + a^(1/2)))/x^6))/(8*a^(7/2)) + (7*b*(a + b*x^3)^(1/2))/(12*a^3*x^3)

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