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

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

[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])

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Rubi [A]  time = 0.0525926, antiderivative size = 89, normalized size of antiderivative = 1., 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} \[ -\frac{5 b^2 \sqrt{a+b x^2}}{16 x^2}-\frac{5 b^3 \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{16 \sqrt{a}}-\frac{5 b \left (a+b x^2\right )^{3/2}}{24 x^4}-\frac{\left (a+b x^2\right )^{5/2}}{6 x^6} \]

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 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]]

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]

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*}

Mathematica [A]  time = 0.0398797, size = 87, normalized size = 0.98 \[ -\frac{34 a^2 b x^2+8 a^3+59 a b^2 x^4+15 b^3 x^6 \sqrt{\frac{b x^2}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{b x^2}{a}+1}\right )+33 b^3 x^6}{48 x^6 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2)/x^7,x]

[Out]

-(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
]])/(48*x^6*Sqrt[a + b*x^2])

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Maple [A]  time = 0.008, size = 139, normalized size = 1.6 \begin{align*} -{\frac{1}{6\,a{x}^{6}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{b}{24\,{a}^{2}{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{{b}^{2}}{16\,{a}^{3}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{{b}^{3}}{16\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,{b}^{3}}{48\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,{b}^{3}}{16}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{5\,{b}^{3}}{16\,a}\sqrt{b{x}^{2}+a}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61243, size = 370, normalized size = 4.16 \begin{align*} \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{align*}

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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Sympy [A]  time = 4.75573, size = 99, normalized size = 1.11 \begin{align*} - \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{align*}

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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Giac [A]  time = 1.72511, size = 101, normalized size = 1.13 \begin{align*} \frac{1}{48} \, b^{3}{\left (\frac{15 \, \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{33 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} - 40 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a + 15 \, \sqrt{b x^{2} + a} a^{2}}{b^{3} x^{6}}\right )} \end{align*}

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*b^3*(15*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) - (33*(b*x^2 + a)^(5/2) - 40*(b*x^2 + a)^(3/2)*a + 15*s
qrt(b*x^2 + a)*a^2)/(b^3*x^6))