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

Optimal. Leaf size=68 \[ -\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {b \sqrt {a+b x^3}}{4 x^3}-\frac {\left (a+b x^3\right )^{3/2}}{6 x^6} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 47, 63, 208} \[ -\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {b \sqrt {a+b x^3}}{4 x^3}-\frac {\left (a+b x^3\right )^{3/2}}{6 x^6} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^(3/2)/x^7,x]

[Out]

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

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 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 266

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

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Mathematica [A]  time = 0.04, size = 76, normalized size = 1.12 \[ -\frac {2 a^2+3 b^2 x^6 \sqrt {\frac {b x^3}{a}+1} \tanh ^{-1}\left (\sqrt {\frac {b x^3}{a}+1}\right )+7 a b x^3+5 b^2 x^6}{12 x^6 \sqrt {a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^(3/2)/x^7,x]

[Out]

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

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fricas [A]  time = 0.77, size = 138, normalized size = 2.03 \[ \left [\frac {3 \, \sqrt {a} b^{2} x^{6} \log \left (\frac {b x^{3} - 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) - 2 \, {\left (5 \, a b x^{3} + 2 \, a^{2}\right )} \sqrt {b x^{3} + a}}{24 \, a x^{6}}, \frac {3 \, \sqrt {-a} b^{2} x^{6} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) - {\left (5 \, a b x^{3} + 2 \, a^{2}\right )} \sqrt {b x^{3} + a}}{12 \, a x^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 70, normalized size = 1.03 \[ \frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{3} - 3 \, \sqrt {b x^{3} + a} a b^{3}}{b^{2} x^{6}}}{12 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 54, normalized size = 0.79 \[ -\frac {b^{2} \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {5 \sqrt {b \,x^{3}+a}\, b}{12 x^{3}}-\frac {\sqrt {b \,x^{3}+a}\, a}{6 x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.90, size = 98, normalized size = 1.44 \[ \frac {b^{2} \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{8 \, \sqrt {a}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2} - 3 \, \sqrt {b x^{3} + a} a b^{2}}{12 \, {\left ({\left (b x^{3} + a\right )}^{2} - 2 \, {\left (b x^{3} + a\right )} a + a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.44, size = 74, normalized size = 1.09 \[ \frac {b^2\,\ln \left (\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3\,\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}{x^6}\right )}{8\,\sqrt {a}}-\frac {5\,b\,\sqrt {b\,x^3+a}}{12\,x^3}-\frac {a\,\sqrt {b\,x^3+a}}{6\,x^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^(3/2)/x^7,x)

[Out]

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

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sympy [A]  time = 7.54, size = 78, normalized size = 1.15 \[ - \frac {a \sqrt {b} \sqrt {\frac {a}{b x^{3}} + 1}}{6 x^{\frac {9}{2}}} - \frac {5 b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}}{12 x^{\frac {3}{2}}} - \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{4 \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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