3.4.85 \(\int \frac {(a+b x^2)^{5/2}}{x^7} \, dx\)

Optimal. Leaf size=89 \[ -\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}}-\frac {5 b^2 \sqrt {a+b x^2}}{16 x^2}-\frac {\left (a+b x^2\right )^{5/2}}{6 x^6}-\frac {5 b \left (a+b x^2\right )^{3/2}}{24 x^4} \]

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Rubi [A]  time = 0.05, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 47, 63, 208} \begin {gather*} -\frac {5 b^2 \sqrt {a+b x^2}}{16 x^2}-\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}}-\frac {5 b \left (a+b x^2\right )^{3/2}}{24 x^4}-\frac {\left (a+b x^2\right )^{5/2}}{6 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2)/x^7,x]

[Out]

(-5*b^2*Sqrt[a + b*x^2])/(16*x^2) - (5*b*(a + b*x^2)^(3/2))/(24*x^4) - (a + b*x^2)^(5/2)/(6*x^6) - (5*b^3*ArcT
anh[Sqrt[a + b*x^2]/Sqrt[a]])/(16*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^2\right )^{5/2}}{x^7} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(a+b x)^{5/2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac {\left (a+b x^2\right )^{5/2}}{6 x^6}+\frac {1}{12} (5 b) \operatorname {Subst}\left (\int \frac {(a+b x)^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {5 b \left (a+b x^2\right )^{3/2}}{24 x^4}-\frac {\left (a+b x^2\right )^{5/2}}{6 x^6}+\frac {1}{16} \left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {5 b^2 \sqrt {a+b x^2}}{16 x^2}-\frac {5 b \left (a+b x^2\right )^{3/2}}{24 x^4}-\frac {\left (a+b x^2\right )^{5/2}}{6 x^6}+\frac {1}{32} \left (5 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=-\frac {5 b^2 \sqrt {a+b x^2}}{16 x^2}-\frac {5 b \left (a+b x^2\right )^{3/2}}{24 x^4}-\frac {\left (a+b x^2\right )^{5/2}}{6 x^6}+\frac {1}{16} \left (5 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )\\ &=-\frac {5 b^2 \sqrt {a+b x^2}}{16 x^2}-\frac {5 b \left (a+b x^2\right )^{3/2}}{24 x^4}-\frac {\left (a+b x^2\right )^{5/2}}{6 x^6}-\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 87, normalized size = 0.98 \begin {gather*} -\frac {8 a^3+34 a^2 b x^2+15 b^3 x^6 \sqrt {\frac {b x^2}{a}+1} \tanh ^{-1}\left (\sqrt {\frac {b x^2}{a}+1}\right )+59 a b^2 x^4+33 b^3 x^6}{48 x^6 \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x^7,x]

[Out]

-1/48*(8*a^3 + 34*a^2*b*x^2 + 59*a*b^2*x^4 + 33*b^3*x^6 + 15*b^3*x^6*Sqrt[1 + (b*x^2)/a]*ArcTanh[Sqrt[1 + (b*x
^2)/a]])/(x^6*Sqrt[a + b*x^2])

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IntegrateAlgebraic [A]  time = 0.12, size = 70, normalized size = 0.79 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-8 a^2-26 a b x^2-33 b^2 x^4\right )}{48 x^6}-\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{16 \sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x^2)^(5/2)/x^7,x]

[Out]

(Sqrt[a + b*x^2]*(-8*a^2 - 26*a*b*x^2 - 33*b^2*x^4))/(48*x^6) - (5*b^3*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(16*S
qrt[a])

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fricas [A]  time = 0.99, size = 158, normalized size = 1.78 \begin {gather*} \left [\frac {15 \, \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 (33 \, a b^{2} x^{4} + 26 \, a^{2} b x^{2} + 8 \, a^{3}\right )} \sqrt {b x^{2} + a}}{96 \, a x^{6}}, \frac {15 \, \sqrt {-a} b^{3} x^{6} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) - {\left (33 \, a b^{2} x^{4} + 26 \, a^{2} b x^{2} + 8 \, a^{3}\right )} \sqrt {b x^{2} + a}}{48 \, a x^{6}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^7,x, algorithm="fricas")

[Out]

[1/96*(15*sqrt(a)*b^3*x^6*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(33*a*b^2*x^4 + 26*a^2*b*x^2
+ 8*a^3)*sqrt(b*x^2 + a))/(a*x^6), 1/48*(15*sqrt(-a)*b^3*x^6*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (33*a*b^2*x^4
+ 26*a^2*b*x^2 + 8*a^3)*sqrt(b*x^2 + a))/(a*x^6)]

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giac [A]  time = 1.22, size = 87, normalized size = 0.98 \begin {gather*} \frac {\frac {15 \, b^{4} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} - \frac {33 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{4} - 40 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{4} + 15 \, \sqrt {b x^{2} + a} a^{2} b^{4}}{b^{3} x^{6}}}{48 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^7,x, algorithm="giac")

[Out]

1/48*(15*b^4*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) - (33*(b*x^2 + a)^(5/2)*b^4 - 40*(b*x^2 + a)^(3/2)*a*b^
4 + 15*sqrt(b*x^2 + a)*a^2*b^4)/(b^3*x^6))/b

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maple [A]  time = 0.01, size = 139, normalized size = 1.56 \begin {gather*} -\frac {5 b^{3} \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{16 \sqrt {a}}+\frac {5 \sqrt {b \,x^{2}+a}\, b^{3}}{16 a}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{3}}{48 a^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} b^{3}}{16 a^{3}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} b^{2}}{16 a^{3} x^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} b}{24 a^{2} x^{4}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{6 a \,x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)/x^7,x)

[Out]

-1/6/a/x^6*(b*x^2+a)^(7/2)-1/24/a^2*b/x^4*(b*x^2+a)^(7/2)-1/16/a^3*b^2/x^2*(b*x^2+a)^(7/2)+1/16/a^3*b^3*(b*x^2
+a)^(5/2)+5/48/a^2*b^3*(b*x^2+a)^(3/2)-5/16/a^(1/2)*b^3*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)+5/16/a*b^3*(b*x^
2+a)^(1/2)

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maxima [A]  time = 1.47, size = 127, normalized size = 1.43 \begin {gather*} -\frac {5 \, b^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, \sqrt {a}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{3}}{16 \, a^{3}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{3}}{48 \, a^{2}} + \frac {5 \, \sqrt {b x^{2} + a} b^{3}}{16 \, a} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} b^{2}}{16 \, a^{3} x^{2}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} b}{24 \, a^{2} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}}}{6 \, a x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)/x^7,x, algorithm="maxima")

[Out]

-5/16*b^3*arcsinh(a/(sqrt(a*b)*abs(x)))/sqrt(a) + 1/16*(b*x^2 + a)^(5/2)*b^3/a^3 + 5/48*(b*x^2 + a)^(3/2)*b^3/
a^2 + 5/16*sqrt(b*x^2 + a)*b^3/a - 1/16*(b*x^2 + a)^(7/2)*b^2/(a^3*x^2) - 1/24*(b*x^2 + a)^(7/2)*b/(a^2*x^4) -
1/6*(b*x^2 + a)^(7/2)/(a*x^6)

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mupad [B]  time = 5.20, size = 72, normalized size = 0.81 \begin {gather*} \frac {5\,a\,{\left (b\,x^2+a\right )}^{3/2}}{6\,x^6}-\frac {11\,{\left (b\,x^2+a\right )}^{5/2}}{16\,x^6}-\frac {5\,a^2\,\sqrt {b\,x^2+a}}{16\,x^6}+\frac {b^3\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{16\,\sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2)/x^7,x)

[Out]

(b^3*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*5i)/(16*a^(1/2)) - (11*(a + b*x^2)^(5/2))/(16*x^6) + (5*a*(a + b*x^2
)^(3/2))/(6*x^6) - (5*a^2*(a + b*x^2)^(1/2))/(16*x^6)

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sympy [A]  time = 4.53, size = 99, normalized size = 1.11 \begin {gather*} - \frac {a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{6 x^{5}} - \frac {13 a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{24 x^{3}} - \frac {11 b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{16 x} - \frac {5 b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{16 \sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)/x**7,x)

[Out]

-a**2*sqrt(b)*sqrt(a/(b*x**2) + 1)/(6*x**5) - 13*a*b**(3/2)*sqrt(a/(b*x**2) + 1)/(24*x**3) - 11*b**(5/2)*sqrt(
a/(b*x**2) + 1)/(16*x) - 5*b**3*asinh(sqrt(a)/(sqrt(b)*x))/(16*sqrt(a))

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