\(\int \frac {(a+b x^3)^{3/2}}{x^7} \, dx\) [393]

Optimal result

Integrand size = 15, antiderivative size = 68 \[ \int \frac {\left (a+b x^3\right )^{3/2}}{x^7} \, dx=-\frac {b \sqrt {a+b x^3}}{4 x^3}-\frac {\left (a+b x^3\right )^{3/2}}{6 x^6}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{4 \sqrt {a}} \]

[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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {272, 43, 65, 214} \[ \int \frac {\left (a+b x^3\right )^{3/2}}{x^7} \, dx=-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{4 \sqrt {a}}-\frac {b \sqrt {a+b x^3}}{4 x^3}-\frac {\left (a+b x^3\right )^{3/2}}{6 x^6} \]

[In]

Int[(a + b*x^3)^(3/2)/x^7,x]

[Out]

-1/4*(b*Sqrt[a + b*x^3])/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 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 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*} \text {integral}& = \frac {1}{3} \text {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 \text {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 \text {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 \text {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*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b x^3\right )^{3/2}}{x^7} \, dx=\frac {\left (-2 a-5 b x^3\right ) \sqrt {a+b x^3}}{12 x^6}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{4 \sqrt {a}} \]

[In]

Integrate[(a + b*x^3)^(3/2)/x^7,x]

[Out]

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

Maple [A] (verified)

Time = 3.91 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.71

method	result	size
risch	\(-\frac {\sqrt {b \,x^{3}+a}\, \left (5 b \,x^{3}+2 a \right )}{12 x^{6}}-\frac {b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{4 \sqrt {a}}\)	\(48\)
default	\(-\frac {a \sqrt {b \,x^{3}+a}}{6 x^{6}}-\frac {5 b \sqrt {b \,x^{3}+a}}{12 x^{3}}-\frac {b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{4 \sqrt {a}}\)	\(54\)
elliptic	\(-\frac {a \sqrt {b \,x^{3}+a}}{6 x^{6}}-\frac {5 b \sqrt {b \,x^{3}+a}}{12 x^{3}}-\frac {b^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{4 \sqrt {a}}\)	\(54\)
pseudoelliptic	\(\frac {-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right ) b^{2} x^{6}-5 b \,x^{3} \sqrt {b \,x^{3}+a}\, \sqrt {a}-2 a^{\frac {3}{2}} \sqrt {b \,x^{3}+a}}{12 x^{6} \sqrt {a}}\)	\(64\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.03 \[ \int \frac {\left (a+b x^3\right )^{3/2}}{x^7} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} x^{6} \log \left (\frac {b x^{3} - 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) - 2 \, {\left (5 \, a b x^{3} + 2 \, a^{2}\right )} \sqrt {b x^{3} + a}}{24 \, a x^{6}}, \frac {3 \, \sqrt {-a} b^{2} x^{6} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) - {\left (5 \, a b x^{3} + 2 \, a^{2}\right )} \sqrt {b x^{3} + a}}{12 \, a x^{6}}\right ] \]

[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)]

Sympy [A] (verification not implemented)

Time = 1.63 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x^3\right )^{3/2}}{x^7} \, dx=- \frac {a \sqrt {b} \sqrt {\frac {a}{b x^{3}} + 1}}{6 x^{\frac {9}{2}}} - \frac {5 b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{3}} + 1}}{12 x^{\frac {3}{2}}} - \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{4 \sqrt {a}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.44 \[ \int \frac {\left (a+b x^3\right )^{3/2}}{x^7} \, dx=\frac {b^{2} \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{8 \, \sqrt {a}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{2} - 3 \, \sqrt {b x^{3} + a} a b^{2}}{12 \, {\left ({\left (b x^{3} + a\right )}^{2} - 2 \, {\left (b x^{3} + a\right )} a + a^{2}\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^3\right )^{3/2}}{x^7} \, dx=\frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b^{3} - 3 \, \sqrt {b x^{3} + a} a b^{3}}{b^{2} x^{6}}}{12 \, b} \]

[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

Mupad [B] (verification not implemented)

Time = 6.23 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^3\right )^{3/2}}{x^7} \, dx=\frac {b^2\,\ln \left (\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3\,\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}{x^6}\right )}{8\,\sqrt {a}}-\frac {5\,b\,\sqrt {b\,x^3+a}}{12\,x^3}-\frac {a\,\sqrt {b\,x^3+a}}{6\,x^6} \]

[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)