3.532 \(\int \frac{(a+b x^2)^{3/2} (A+B x^2)}{x^6} \, dx\)

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

[Out]

-((b*B*Sqrt[a + b*x^2])/x) - (B*(a + b*x^2)^(3/2))/(3*x^3) - (A*(a + b*x^2)^(5/2))/(5*a*x^5) + b^(3/2)*B*ArcTa
nh[(Sqrt[b]*x)/Sqrt[a + b*x^2]]

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Rubi [A]  time = 0.0346065, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {451, 277, 217, 206} \[ -\frac{A \left (a+b x^2\right )^{5/2}}{5 a x^5}+b^{3/2} B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )-\frac{B \left (a+b x^2\right )^{3/2}}{3 x^3}-\frac{b B \sqrt{a+b x^2}}{x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(3/2)*(A + B*x^2))/x^6,x]

[Out]

-((b*B*Sqrt[a + b*x^2])/x) - (B*(a + b*x^2)^(3/2))/(3*x^3) - (A*(a + b*x^2)^(5/2))/(5*a*x^5) + b^(3/2)*B*ArcTa
nh[(Sqrt[b]*x)/Sqrt[a + b*x^2]]

Rule 451

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

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^6} \, dx &=-\frac{A \left (a+b x^2\right )^{5/2}}{5 a x^5}+B \int \frac{\left (a+b x^2\right )^{3/2}}{x^4} \, dx\\ &=-\frac{B \left (a+b x^2\right )^{3/2}}{3 x^3}-\frac{A \left (a+b x^2\right )^{5/2}}{5 a x^5}+(b B) \int \frac{\sqrt{a+b x^2}}{x^2} \, dx\\ &=-\frac{b B \sqrt{a+b x^2}}{x}-\frac{B \left (a+b x^2\right )^{3/2}}{3 x^3}-\frac{A \left (a+b x^2\right )^{5/2}}{5 a x^5}+\left (b^2 B\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx\\ &=-\frac{b B \sqrt{a+b x^2}}{x}-\frac{B \left (a+b x^2\right )^{3/2}}{3 x^3}-\frac{A \left (a+b x^2\right )^{5/2}}{5 a x^5}+\left (b^2 B\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )\\ &=-\frac{b B \sqrt{a+b x^2}}{x}-\frac{B \left (a+b x^2\right )^{3/2}}{3 x^3}-\frac{A \left (a+b x^2\right )^{5/2}}{5 a x^5}+b^{3/2} B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0355824, size = 76, normalized size = 0.88 \[ -\frac{A \left (a+b x^2\right )^{5/2}}{5 a x^5}-\frac{a B \sqrt{a+b x^2} \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};-\frac{b x^2}{a}\right )}{3 x^3 \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/x^6,x]

[Out]

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

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Maple [A]  time = 0.01, size = 115, normalized size = 1.3 \begin{align*} -{\frac{B}{3\,a{x}^{3}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{2\,Bb}{3\,{a}^{2}x} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{2\,{b}^{2}Bx}{3\,{a}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{{b}^{2}Bx}{a}\sqrt{b{x}^{2}+a}}+B{b}^{{\frac{3}{2}}}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) -{\frac{A}{5\,a{x}^{5}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}} \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^6,x)

[Out]

-1/3*B/a/x^3*(b*x^2+a)^(5/2)-2/3*B*b/a^2/x*(b*x^2+a)^(5/2)+2/3*B*b^2/a^2*x*(b*x^2+a)^(3/2)+B*b^2/a*x*(b*x^2+a)
^(1/2)+B*b^(3/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))-1/5*A*(b*x^2+a)^(5/2)/a/x^5

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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^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.63017, size = 439, normalized size = 5.1 \begin{align*} \left [\frac{15 \, B a b^{\frac{3}{2}} x^{5} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) - 2 \,{\left ({\left (20 \, B a b + 3 \, A b^{2}\right )} x^{4} + 3 \, A a^{2} +{\left (5 \, B a^{2} + 6 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{30 \, a x^{5}}, -\frac{15 \, B a \sqrt{-b} b x^{5} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) +{\left ({\left (20 \, B a b + 3 \, A b^{2}\right )} x^{4} + 3 \, A a^{2} +{\left (5 \, B a^{2} + 6 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{15 \, a x^{5}}\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^6,x, algorithm="fricas")

[Out]

[1/30*(15*B*a*b^(3/2)*x^5*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*((20*B*a*b + 3*A*b^2)*x^4 + 3*A*
a^2 + (5*B*a^2 + 6*A*a*b)*x^2)*sqrt(b*x^2 + a))/(a*x^5), -1/15*(15*B*a*sqrt(-b)*b*x^5*arctan(sqrt(-b)*x/sqrt(b
*x^2 + a)) + ((20*B*a*b + 3*A*b^2)*x^4 + 3*A*a^2 + (5*B*a^2 + 6*A*a*b)*x^2)*sqrt(b*x^2 + a))/(a*x^5)]

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Sympy [B]  time = 4.28487, size = 184, normalized size = 2.14 \begin{align*} - \frac{A a \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{5 x^{4}} - \frac{2 A b^{\frac{3}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{5 x^{2}} - \frac{A b^{\frac{5}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{5 a} - \frac{B \sqrt{a} b}{x \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{B a \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{3 x^{2}} - \frac{B b^{\frac{3}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{3} + B b^{\frac{3}{2}} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )} - \frac{B b^{2} x}{\sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \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**6,x)

[Out]

-A*a*sqrt(b)*sqrt(a/(b*x**2) + 1)/(5*x**4) - 2*A*b**(3/2)*sqrt(a/(b*x**2) + 1)/(5*x**2) - A*b**(5/2)*sqrt(a/(b
*x**2) + 1)/(5*a) - B*sqrt(a)*b/(x*sqrt(1 + b*x**2/a)) - B*a*sqrt(b)*sqrt(a/(b*x**2) + 1)/(3*x**2) - B*b**(3/2
)*sqrt(a/(b*x**2) + 1)/3 + B*b**(3/2)*asinh(sqrt(b)*x/sqrt(a)) - B*b**2*x/(sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [B]  time = 1.14851, size = 319, normalized size = 3.71 \begin{align*} -\frac{1}{2} \, B b^{\frac{3}{2}} \log \left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2}\right ) + \frac{2 \,{\left (30 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{8} B a b^{\frac{3}{2}} + 15 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{8} A b^{\frac{5}{2}} - 90 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{6} B a^{2} b^{\frac{3}{2}} + 110 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} B a^{3} b^{\frac{3}{2}} + 30 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} A a^{2} b^{\frac{5}{2}} - 70 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} B a^{4} b^{\frac{3}{2}} + 20 \, B a^{5} b^{\frac{3}{2}} + 3 \, A a^{4} b^{\frac{5}{2}}\right )}}{15 \,{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )}^{5}} \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^6,x, algorithm="giac")

[Out]

-1/2*B*b^(3/2)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2/15*(30*(sqrt(b)*x - sqrt(b*x^2 + a))^8*B*a*b^(3/2) + 1
5*(sqrt(b)*x - sqrt(b*x^2 + a))^8*A*b^(5/2) - 90*(sqrt(b)*x - sqrt(b*x^2 + a))^6*B*a^2*b^(3/2) + 110*(sqrt(b)*
x - sqrt(b*x^2 + a))^4*B*a^3*b^(3/2) + 30*(sqrt(b)*x - sqrt(b*x^2 + a))^4*A*a^2*b^(5/2) - 70*(sqrt(b)*x - sqrt
(b*x^2 + a))^2*B*a^4*b^(3/2) + 20*B*a^5*b^(3/2) + 3*A*a^4*b^(5/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^5