3.4.76 \(\int \frac {(a+b x^2)^{3/2}}{x^7} \, dx\) [376]

Optimal. Leaf size=92 \[ -\frac {b \sqrt {a+b x^2}}{8 x^4}-\frac {b^2 \sqrt {a+b x^2}}{16 a x^2}-\frac {\left (a+b x^2\right )^{3/2}}{6 x^6}+\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{3/2}} \]

[Out]

-1/6*(b*x^2+a)^(3/2)/x^6+1/16*b^3*arctanh((b*x^2+a)^(1/2)/a^(1/2))/a^(3/2)-1/8*b*(b*x^2+a)^(1/2)/x^4-1/16*b^2*
(b*x^2+a)^(1/2)/a/x^2

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Rubi [A]
time = 0.04, antiderivative size = 92, 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, 43, 44, 65, 214} \begin {gather*} \frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{3/2}}-\frac {b^2 \sqrt {a+b x^2}}{16 a x^2}-\frac {\left (a+b x^2\right )^{3/2}}{6 x^6}-\frac {b \sqrt {a+b x^2}}{8 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/x^7,x]

[Out]

-1/8*(b*Sqrt[a + b*x^2])/x^4 - (b^2*Sqrt[a + b*x^2])/(16*a*x^2) - (a + b*x^2)^(3/2)/(6*x^6) + (b^3*ArcTanh[Sqr
t[a + b*x^2]/Sqrt[a]])/(16*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 {\left (a+b x^2\right )^{3/2}}{x^7} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2\right )^{3/2}}{6 x^6}+\frac {1}{4} b \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {b \sqrt {a+b x^2}}{8 x^4}-\frac {\left (a+b x^2\right )^{3/2}}{6 x^6}+\frac {1}{16} b^2 \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=-\frac {b \sqrt {a+b x^2}}{8 x^4}-\frac {b^2 \sqrt {a+b x^2}}{16 a x^2}-\frac {\left (a+b x^2\right )^{3/2}}{6 x^6}-\frac {b^3 \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{32 a}\\ &=-\frac {b \sqrt {a+b x^2}}{8 x^4}-\frac {b^2 \sqrt {a+b x^2}}{16 a x^2}-\frac {\left (a+b x^2\right )^{3/2}}{6 x^6}-\frac {b^2 \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{16 a}\\ &=-\frac {b \sqrt {a+b x^2}}{8 x^4}-\frac {b^2 \sqrt {a+b x^2}}{16 a x^2}-\frac {\left (a+b x^2\right )^{3/2}}{6 x^6}+\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 73, normalized size = 0.79 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-8 a^2-14 a b x^2-3 b^2 x^4\right )}{48 a x^6}+\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/x^7,x]

[Out]

(Sqrt[a + b*x^2]*(-8*a^2 - 14*a*b*x^2 - 3*b^2*x^4))/(48*a*x^6) + (b^3*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(16*a^
(3/2))

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Maple [A]
time = 0.06, size = 125, normalized size = 1.36

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (3 b^{2} x^{4}+14 a b \,x^{2}+8 a^{2}\right )}{48 x^{6} a}+\frac {b^{3} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{16 a^{\frac {3}{2}}}\)	\(71\)
default	\(-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 a \,x^{6}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{4 a \,x^{4}}+\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{2 a \,x^{2}}+\frac {3 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )}{2 a}\right )}{4 a}\right )}{6 a}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.32, size = 110, normalized size = 1.20 \begin {gather*} \frac {b^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, a^{\frac {3}{2}}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}}{48 \, a^{3}} - \frac {\sqrt {b x^{2} + a} b^{3}}{16 \, a^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}}{48 \, a^{3} x^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b}{24 \, a^{2} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}}}{6 \, a x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*b^3*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(3/2) - 1/48*(b*x^2 + a)^(3/2)*b^3/a^3 - 1/16*sqrt(b*x^2 + a)*b^3/a^2
+ 1/48*(b*x^2 + a)^(5/2)*b^2/(a^3*x^2) + 1/24*(b*x^2 + a)^(5/2)*b/(a^2*x^4) - 1/6*(b*x^2 + a)^(5/2)/(a*x^6)

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Fricas [A]
time = 1.32, size = 157, normalized size = 1.71 \begin {gather*} \left [\frac {3 \, \sqrt {a} b^{3} x^{6} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (3 \, a b^{2} x^{4} + 14 \, a^{2} b x^{2} + 8 \, a^{3}\right )} \sqrt {b x^{2} + a}}{96 \, a^{2} x^{6}}, -\frac {3 \, \sqrt {-a} b^{3} x^{6} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, a b^{2} x^{4} + 14 \, a^{2} b x^{2} + 8 \, a^{3}\right )} \sqrt {b x^{2} + a}}{48 \, a^{2} x^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*sqrt(a)*b^3*x^6*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(3*a*b^2*x^4 + 14*a^2*b*x^2 +
8*a^3)*sqrt(b*x^2 + a))/(a^2*x^6), -1/48*(3*sqrt(-a)*b^3*x^6*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (3*a*b^2*x^4
+ 14*a^2*b*x^2 + 8*a^3)*sqrt(b*x^2 + a))/(a^2*x^6)]

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Sympy [A]
time = 3.87, size = 119, normalized size = 1.29 \begin {gather*} - \frac {a^{2}}{6 \sqrt {b} x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {11 a \sqrt {b}}{24 x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {17 b^{\frac {3}{2}}}{48 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {b^{\frac {5}{2}}}{16 a x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{16 a^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2/(6*sqrt(b)*x**7*sqrt(a/(b*x**2) + 1)) - 11*a*sqrt(b)/(24*x**5*sqrt(a/(b*x**2) + 1)) - 17*b**(3/2)/(48*x*
*3*sqrt(a/(b*x**2) + 1)) - b**(5/2)/(16*a*x*sqrt(a/(b*x**2) + 1)) + b**3*asinh(sqrt(a)/(sqrt(b)*x))/(16*a**(3/
2))

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Giac [A]
time = 0.87, size = 92, normalized size = 1.00 \begin {gather*} -\frac {\frac {3 \, b^{4} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{4} + 8 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{4} - 3 \, \sqrt {b x^{2} + a} a^{2} b^{4}}{a b^{3} x^{6}}}{48 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(3*b^4*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a) + (3*(b*x^2 + a)^(5/2)*b^4 + 8*(b*x^2 + a)^(3/2)*a*
b^4 - 3*sqrt(b*x^2 + a)*a^2*b^4)/(a*b^3*x^6))/b

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Mupad [B]
time = 4.94, size = 72, normalized size = 0.78 \begin {gather*} \frac {a\,\sqrt {b\,x^2+a}}{16\,x^6}-\frac {{\left (b\,x^2+a\right )}^{3/2}}{6\,x^6}-\frac {{\left (b\,x^2+a\right )}^{5/2}}{16\,a\,x^6}-\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{16\,a^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2)/x^7,x)

[Out]

(a*(a + b*x^2)^(1/2))/(16*x^6) - (b^3*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*1i)/(16*a^(3/2)) - (a + b*x^2)^(3/2
)/(6*x^6) - (a + b*x^2)^(5/2)/(16*a*x^6)

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