3.535 \(\int \frac{(a+b x^2)^{3/2} (A+B x^2)}{x^9} \, dx\)

Optimal. Leaf size=156 \[ \frac{b^2 \sqrt{a+b x^2} (3 A b-8 a B)}{128 a^2 x^2}-\frac{b^3 (3 A b-8 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{128 a^{5/2}}+\frac{b \sqrt{a+b x^2} (3 A b-8 a B)}{64 a x^4}+\frac{\left (a+b x^2\right )^{3/2} (3 A b-8 a B)}{48 a x^6}-\frac{A \left (a+b x^2\right )^{5/2}}{8 a x^8} \]

[Out]

(b*(3*A*b - 8*a*B)*Sqrt[a + b*x^2])/(64*a*x^4) + (b^2*(3*A*b - 8*a*B)*Sqrt[a + b*x^2])/(128*a^2*x^2) + ((3*A*b
- 8*a*B)*(a + b*x^2)^(3/2))/(48*a*x^6) - (A*(a + b*x^2)^(5/2))/(8*a*x^8) - (b^3*(3*A*b - 8*a*B)*ArcTanh[Sqrt[
a + b*x^2]/Sqrt[a]])/(128*a^(5/2))

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Rubi [A]  time = 0.123087, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {446, 78, 47, 51, 63, 208} \[ \frac{b^2 \sqrt{a+b x^2} (3 A b-8 a B)}{128 a^2 x^2}-\frac{b^3 (3 A b-8 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{128 a^{5/2}}+\frac{b \sqrt{a+b x^2} (3 A b-8 a B)}{64 a x^4}+\frac{\left (a+b x^2\right )^{3/2} (3 A b-8 a B)}{48 a x^6}-\frac{A \left (a+b x^2\right )^{5/2}}{8 a x^8} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(3/2)*(A + B*x^2))/x^9,x]

[Out]

(b*(3*A*b - 8*a*B)*Sqrt[a + b*x^2])/(64*a*x^4) + (b^2*(3*A*b - 8*a*B)*Sqrt[a + b*x^2])/(128*a^2*x^2) + ((3*A*b
- 8*a*B)*(a + b*x^2)^(3/2))/(48*a*x^6) - (A*(a + b*x^2)^(5/2))/(8*a*x^8) - (b^3*(3*A*b - 8*a*B)*ArcTanh[Sqrt[
a + b*x^2]/Sqrt[a]])/(128*a^(5/2))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, 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 51

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 )^{3/2} \left (A+B x^2\right )}{x^9} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2} (A+B x)}{x^5} \, dx,x,x^2\right )\\ &=-\frac{A \left (a+b x^2\right )^{5/2}}{8 a x^8}+\frac{\left (-\frac{3 A b}{2}+4 a B\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x^4} \, dx,x,x^2\right )}{8 a}\\ &=\frac{(3 A b-8 a B) \left (a+b x^2\right )^{3/2}}{48 a x^6}-\frac{A \left (a+b x^2\right )^{5/2}}{8 a x^8}-\frac{(b (3 A b-8 a B)) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^3} \, dx,x,x^2\right )}{32 a}\\ &=\frac{b (3 A b-8 a B) \sqrt{a+b x^2}}{64 a x^4}+\frac{(3 A b-8 a B) \left (a+b x^2\right )^{3/2}}{48 a x^6}-\frac{A \left (a+b x^2\right )^{5/2}}{8 a x^8}-\frac{\left (b^2 (3 A b-8 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,x^2\right )}{128 a}\\ &=\frac{b (3 A b-8 a B) \sqrt{a+b x^2}}{64 a x^4}+\frac{b^2 (3 A b-8 a B) \sqrt{a+b x^2}}{128 a^2 x^2}+\frac{(3 A b-8 a B) \left (a+b x^2\right )^{3/2}}{48 a x^6}-\frac{A \left (a+b x^2\right )^{5/2}}{8 a x^8}+\frac{\left (b^3 (3 A b-8 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{256 a^2}\\ &=\frac{b (3 A b-8 a B) \sqrt{a+b x^2}}{64 a x^4}+\frac{b^2 (3 A b-8 a B) \sqrt{a+b x^2}}{128 a^2 x^2}+\frac{(3 A b-8 a B) \left (a+b x^2\right )^{3/2}}{48 a x^6}-\frac{A \left (a+b x^2\right )^{5/2}}{8 a x^8}+\frac{\left (b^2 (3 A b-8 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{128 a^2}\\ &=\frac{b (3 A b-8 a B) \sqrt{a+b x^2}}{64 a x^4}+\frac{b^2 (3 A b-8 a B) \sqrt{a+b x^2}}{128 a^2 x^2}+\frac{(3 A b-8 a B) \left (a+b x^2\right )^{3/2}}{48 a x^6}-\frac{A \left (a+b x^2\right )^{5/2}}{8 a x^8}-\frac{b^3 (3 A b-8 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{128 a^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0244022, size = 62, normalized size = 0.4 \[ -\frac{\left (a+b x^2\right )^{5/2} \left (5 a^4 A+b^3 x^8 (3 A b-8 a B) \, _2F_1\left (\frac{5}{2},4;\frac{7}{2};\frac{b x^2}{a}+1\right )\right )}{40 a^5 x^8} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/x^9,x]

[Out]

-((a + b*x^2)^(5/2)*(5*a^4*A + b^3*(3*A*b - 8*a*B)*x^8*Hypergeometric2F1[5/2, 4, 7/2, 1 + (b*x^2)/a]))/(40*a^5
*x^8)

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Maple [B]  time = 0.017, size = 275, normalized size = 1.8 \begin{align*} -{\frac{A}{8\,a{x}^{8}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{Ab}{16\,{a}^{2}{x}^{6}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{A{b}^{2}}{64\,{a}^{3}{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{A{b}^{3}}{128\,{a}^{4}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{A{b}^{4}}{128\,{a}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{3\,A{b}^{4}}{128}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}+{\frac{3\,A{b}^{4}}{128\,{a}^{3}}\sqrt{b{x}^{2}+a}}-{\frac{B}{6\,a{x}^{6}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{Bb}{24\,{a}^{2}{x}^{4}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{B{b}^{2}}{48\,{a}^{3}{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{B{b}^{3}}{48\,{a}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{B{b}^{3}}{16}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{B{b}^{3}}{16\,{a}^{2}}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)*(B*x^2+A)/x^9,x)

[Out]

-1/8*A*(b*x^2+a)^(5/2)/a/x^8+1/16*A*b/a^2/x^6*(b*x^2+a)^(5/2)-1/64*A*b^2/a^3/x^4*(b*x^2+a)^(5/2)-1/128*A*b^3/a
^4/x^2*(b*x^2+a)^(5/2)+1/128*A*b^4/a^4*(b*x^2+a)^(3/2)-3/128*A*b^4/a^(5/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/
x)+3/128*A*b^4/a^3*(b*x^2+a)^(1/2)-1/6*B/a/x^6*(b*x^2+a)^(5/2)+1/24*B*b/a^2/x^4*(b*x^2+a)^(5/2)+1/48*B*b^2/a^3
/x^2*(b*x^2+a)^(5/2)-1/48*B*b^3/a^3*(b*x^2+a)^(3/2)+1/16*B*b^3/a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)-1
/16*B*b^3/a^2*(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)^(3/2)*(B*x^2+A)/x^9,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.7552, size = 621, normalized size = 3.98 \begin{align*} \left [-\frac{3 \,{\left (8 \, B a b^{3} - 3 \, A b^{4}\right )} \sqrt{a} x^{8} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (3 \,{\left (8 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{6} + 48 \, A a^{4} + 2 \,{\left (56 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{4} + 8 \,{\left (8 \, B a^{4} + 9 \, A a^{3} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{768 \, a^{3} x^{8}}, -\frac{3 \,{\left (8 \, B a b^{3} - 3 \, A b^{4}\right )} \sqrt{-a} x^{8} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (3 \,{\left (8 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{6} + 48 \, A a^{4} + 2 \,{\left (56 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{4} + 8 \,{\left (8 \, B a^{4} + 9 \, A a^{3} b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{384 \, a^{3} x^{8}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^9,x, algorithm="fricas")

[Out]

[-1/768*(3*(8*B*a*b^3 - 3*A*b^4)*sqrt(a)*x^8*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(3*(8*B*a
^2*b^2 - 3*A*a*b^3)*x^6 + 48*A*a^4 + 2*(56*B*a^3*b + 3*A*a^2*b^2)*x^4 + 8*(8*B*a^4 + 9*A*a^3*b)*x^2)*sqrt(b*x^
2 + a))/(a^3*x^8), -1/384*(3*(8*B*a*b^3 - 3*A*b^4)*sqrt(-a)*x^8*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (3*(8*B*a^2
*b^2 - 3*A*a*b^3)*x^6 + 48*A*a^4 + 2*(56*B*a^3*b + 3*A*a^2*b^2)*x^4 + 8*(8*B*a^4 + 9*A*a^3*b)*x^2)*sqrt(b*x^2
+ a))/(a^3*x^8)]

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Sympy [B]  time = 136.73, size = 287, normalized size = 1.84 \begin{align*} - \frac{A a^{2}}{8 \sqrt{b} x^{9} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{5 A a \sqrt{b}}{16 x^{7} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{13 A b^{\frac{3}{2}}}{64 x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{A b^{\frac{5}{2}}}{128 a x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{3 A b^{\frac{7}{2}}}{128 a^{2} x \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{3 A b^{4} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{128 a^{\frac{5}{2}}} - \frac{B a^{2}}{6 \sqrt{b} x^{7} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{11 B a \sqrt{b}}{24 x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{17 B b^{\frac{3}{2}}}{48 x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{B b^{\frac{5}{2}}}{16 a x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{B b^{3} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{16 a^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)*(B*x**2+A)/x**9,x)

[Out]

-A*a**2/(8*sqrt(b)*x**9*sqrt(a/(b*x**2) + 1)) - 5*A*a*sqrt(b)/(16*x**7*sqrt(a/(b*x**2) + 1)) - 13*A*b**(3/2)/(
64*x**5*sqrt(a/(b*x**2) + 1)) + A*b**(5/2)/(128*a*x**3*sqrt(a/(b*x**2) + 1)) + 3*A*b**(7/2)/(128*a**2*x*sqrt(a
/(b*x**2) + 1)) - 3*A*b**4*asinh(sqrt(a)/(sqrt(b)*x))/(128*a**(5/2)) - B*a**2/(6*sqrt(b)*x**7*sqrt(a/(b*x**2)
+ 1)) - 11*B*a*sqrt(b)/(24*x**5*sqrt(a/(b*x**2) + 1)) - 17*B*b**(3/2)/(48*x**3*sqrt(a/(b*x**2) + 1)) - B*b**(5
/2)/(16*a*x*sqrt(a/(b*x**2) + 1)) + B*b**3*asinh(sqrt(a)/(sqrt(b)*x))/(16*a**(3/2))

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Giac [A]  time = 1.14682, size = 262, normalized size = 1.68 \begin{align*} -\frac{\frac{3 \,{\left (8 \, B a b^{4} - 3 \, A b^{5}\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{24 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} B a b^{4} + 40 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} B a^{2} b^{4} - 88 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B a^{3} b^{4} + 24 \, \sqrt{b x^{2} + a} B a^{4} b^{4} - 9 \,{\left (b x^{2} + a\right )}^{\frac{7}{2}} A b^{5} + 33 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} A a b^{5} + 33 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A a^{2} b^{5} - 9 \, \sqrt{b x^{2} + a} A a^{3} b^{5}}{a^{2} b^{4} x^{8}}}{384 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^9,x, algorithm="giac")

[Out]

-1/384*(3*(8*B*a*b^4 - 3*A*b^5)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a^2) + (24*(b*x^2 + a)^(7/2)*B*a*b^
4 + 40*(b*x^2 + a)^(5/2)*B*a^2*b^4 - 88*(b*x^2 + a)^(3/2)*B*a^3*b^4 + 24*sqrt(b*x^2 + a)*B*a^4*b^4 - 9*(b*x^2
+ a)^(7/2)*A*b^5 + 33*(b*x^2 + a)^(5/2)*A*a*b^5 + 33*(b*x^2 + a)^(3/2)*A*a^2*b^5 - 9*sqrt(b*x^2 + a)*A*a^3*b^5
)/(a^2*b^4*x^8))/b