3.4.93 \(\int \frac {(a+b x^2)^{5/2}}{x^5} \, dx\) [393]

Optimal. Leaf size=86 \[ \frac {15}{8} b^2 \sqrt {a+b x^2}-\frac {5 b \left (a+b x^2\right )^{3/2}}{8 x^2}-\frac {\left (a+b x^2\right )^{5/2}}{4 x^4}-\frac {15}{8} \sqrt {a} b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \]

[Out]

-5/8*b*(b*x^2+a)^(3/2)/x^2-1/4*(b*x^2+a)^(5/2)/x^4-15/8*b^2*arctanh((b*x^2+a)^(1/2)/a^(1/2))*a^(1/2)+15/8*b^2*
(b*x^2+a)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 86, 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, 52, 65, 214} \begin {gather*} \frac {15}{8} b^2 \sqrt {a+b x^2}-\frac {15}{8} \sqrt {a} b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )-\frac {5 b \left (a+b x^2\right )^{3/2}}{8 x^2}-\frac {\left (a+b x^2\right )^{5/2}}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/x^5,x]

[Out]

(15*b^2*Sqrt[a + b*x^2])/8 - (5*b*(a + b*x^2)^(3/2))/(8*x^2) - (a + b*x^2)^(5/2)/(4*x^4) - (15*Sqrt[a]*b^2*Arc
Tanh[Sqrt[a + b*x^2]/Sqrt[a]])/8

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 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 {\left (a+b x^2\right )^{5/2}}{x^5} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(a+b x)^{5/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2\right )^{5/2}}{4 x^4}+\frac {1}{8} (5 b) \text {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {5 b \left (a+b x^2\right )^{3/2}}{8 x^2}-\frac {\left (a+b x^2\right )^{5/2}}{4 x^4}+\frac {1}{16} \left (15 b^2\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac {15}{8} b^2 \sqrt {a+b x^2}-\frac {5 b \left (a+b x^2\right )^{3/2}}{8 x^2}-\frac {\left (a+b x^2\right )^{5/2}}{4 x^4}+\frac {1}{16} \left (15 a b^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=\frac {15}{8} b^2 \sqrt {a+b x^2}-\frac {5 b \left (a+b x^2\right )^{3/2}}{8 x^2}-\frac {\left (a+b x^2\right )^{5/2}}{4 x^4}+\frac {1}{8} (15 a b) \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )\\ &=\frac {15}{8} b^2 \sqrt {a+b x^2}-\frac {5 b \left (a+b x^2\right )^{3/2}}{8 x^2}-\frac {\left (a+b x^2\right )^{5/2}}{4 x^4}-\frac {15}{8} \sqrt {a} b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 70, normalized size = 0.81 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-2 a^2-9 a b x^2+8 b^2 x^4\right )}{8 x^4}-\frac {15}{8} \sqrt {a} b^2 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x^5,x]

[Out]

(Sqrt[a + b*x^2]*(-2*a^2 - 9*a*b*x^2 + 8*b^2*x^4))/(8*x^4) - (15*Sqrt[a]*b^2*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])
/8

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Maple [A]
time = 0.06, size = 115, normalized size = 1.34

method	result	size
risch	\(-\frac {a \sqrt {b \,x^{2}+a}\, \left (9 b \,x^{2}+2 a \right )}{8 x^{4}}-\frac {15 \sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) b^{2}}{8}+b^{2} \sqrt {b \,x^{2}+a}\)	\(71\)
default	\(-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{4 a \,x^{4}}+\frac {3 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{2 a \,x^{2}}+\frac {5 b \left (\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5}+a \left (\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{3}+a \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )\right )\right )}{2 a}\right )}{4 a}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/4/a/x^4*(b*x^2+a)^(7/2)+3/4*b/a*(-1/2/a/x^2*(b*x^2+a)^(7/2)+5/2*b/a*(1/5*(b*x^2+a)^(5/2)+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 = 0.27, size = 104, normalized size = 1.21 \begin {gather*} -\frac {15}{8} \, \sqrt {a} b^{2} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) + \frac {15}{8} \, \sqrt {b x^{2} + a} b^{2} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}}{8 \, a^{2}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}{8 \, a} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} b}{8 \, a^{2} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}}}{4 \, a x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^5,x, algorithm="maxima")

[Out]

-15/8*sqrt(a)*b^2*arcsinh(a/(sqrt(a*b)*abs(x))) + 15/8*sqrt(b*x^2 + a)*b^2 + 3/8*(b*x^2 + a)^(5/2)*b^2/a^2 + 5
/8*(b*x^2 + a)^(3/2)*b^2/a - 3/8*(b*x^2 + a)^(7/2)*b/(a^2*x^2) - 1/4*(b*x^2 + a)^(7/2)/(a*x^4)

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Fricas [A]
time = 1.41, size = 145, normalized size = 1.69 \begin {gather*} \left [\frac {15 \, \sqrt {a} b^{2} x^{4} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (8 \, b^{2} x^{4} - 9 \, a b x^{2} - 2 \, a^{2}\right )} \sqrt {b x^{2} + a}}{16 \, x^{4}}, \frac {15 \, \sqrt {-a} b^{2} x^{4} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, b^{2} x^{4} - 9 \, a b x^{2} - 2 \, a^{2}\right )} \sqrt {b x^{2} + a}}{8 \, x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^5,x, algorithm="fricas")

[Out]

[1/16*(15*sqrt(a)*b^2*x^4*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(8*b^2*x^4 - 9*a*b*x^2 - 2*a
^2)*sqrt(b*x^2 + a))/x^4, 1/8*(15*sqrt(-a)*b^2*x^4*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (8*b^2*x^4 - 9*a*b*x^2 -
2*a^2)*sqrt(b*x^2 + a))/x^4]

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Sympy [A]
time = 2.39, size = 117, normalized size = 1.36 \begin {gather*} - \frac {15 \sqrt {a} b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{8} - \frac {a^{3}}{4 \sqrt {b} x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {11 a^{2} \sqrt {b}}{8 x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {a b^{\frac {3}{2}}}{8 x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {b^{\frac {5}{2}} x}{\sqrt {\frac {a}{b x^{2}} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)/x**5,x)

[Out]

-15*sqrt(a)*b**2*asinh(sqrt(a)/(sqrt(b)*x))/8 - a**3/(4*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)) - 11*a**2*sqrt(b)/(
8*x**3*sqrt(a/(b*x**2) + 1)) - a*b**(3/2)/(8*x*sqrt(a/(b*x**2) + 1)) + b**(5/2)*x/sqrt(a/(b*x**2) + 1)

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Giac [A]
time = 0.66, size = 88, normalized size = 1.02 \begin {gather*} \frac {\frac {15 \, a b^{3} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 8 \, \sqrt {b x^{2} + a} b^{3} - \frac {9 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{3} - 7 \, \sqrt {b x^{2} + a} a^{2} b^{3}}{b^{2} x^{4}}}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^5,x, algorithm="giac")

[Out]

1/8*(15*a*b^3*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 8*sqrt(b*x^2 + a)*b^3 - (9*(b*x^2 + a)^(3/2)*a*b^3 -
7*sqrt(b*x^2 + a)*a^2*b^3)/(b^2*x^4))/b

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Mupad [B]
time = 5.01, size = 71, normalized size = 0.83 \begin {gather*} b^2\,\sqrt {b\,x^2+a}-\frac {9\,a\,{\left (b\,x^2+a\right )}^{3/2}}{8\,x^4}+\frac {7\,a^2\,\sqrt {b\,x^2+a}}{8\,x^4}+\frac {\sqrt {a}\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,15{}\mathrm {i}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2)/x^5,x)

[Out]

b^2*(a + b*x^2)^(1/2) + (a^(1/2)*b^2*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*15i)/8 - (9*a*(a + b*x^2)^(3/2))/(8*
x^4) + (7*a^2*(a + b*x^2)^(1/2))/(8*x^4)

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