3.161 \(\int \frac{A+B x^3}{x^{5/2} (a+b x^3)} \, dx\)

Optimal. Leaf size=53 \[ -\frac{2 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{3 a^{3/2} \sqrt{b}}-\frac{2 A}{3 a x^{3/2}} \]

[Out]

(-2*A)/(3*a*x^(3/2)) - (2*(A*b - a*B)*ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]])/(3*a^(3/2)*Sqrt[b])

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Rubi [A]  time = 0.0370199, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {453, 329, 275, 205} \[ -\frac{2 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{3 a^{3/2} \sqrt{b}}-\frac{2 A}{3 a x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^(5/2)*(a + b*x^3)),x]

[Out]

(-2*A)/(3*a*x^(3/2)) - (2*(A*b - a*B)*ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]])/(3*a^(3/2)*Sqrt[b])

Rule 453

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[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0284999, size = 53, normalized size = 1. \[ \frac{2 (a B-A b) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{3 a^{3/2} \sqrt{b}}-\frac{2 A}{3 a x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/(x^(5/2)*(a + b*x^3)),x]

[Out]

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

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Maple [A]  time = 0.01, size = 53, normalized size = 1. \begin{align*} -{\frac{2\,Ab}{3\,a}\arctan \left ({b{x}^{{\frac{3}{2}}}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{2\,B}{3}\arctan \left ({b{x}^{{\frac{3}{2}}}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{2\,A}{3\,a}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/x^(5/2)/(b*x^3+a),x)

[Out]

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

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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^3+A)/x^(5/2)/(b*x^3+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.08949, size = 282, normalized size = 5.32 \begin{align*} \left [\frac{{\left (B a - A b\right )} \sqrt{-a b} x^{2} \log \left (\frac{b x^{3} + 2 \, \sqrt{-a b} x^{\frac{3}{2}} - a}{b x^{3} + a}\right ) - 2 \, A a b \sqrt{x}}{3 \, a^{2} b x^{2}}, \frac{2 \,{\left ({\left (B a - A b\right )} \sqrt{a b} x^{2} \arctan \left (\frac{\sqrt{a b} x^{\frac{3}{2}}}{a}\right ) - A a b \sqrt{x}\right )}}{3 \, a^{2} b x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^(5/2)/(b*x^3+a),x, algorithm="fricas")

[Out]

[1/3*((B*a - A*b)*sqrt(-a*b)*x^2*log((b*x^3 + 2*sqrt(-a*b)*x^(3/2) - a)/(b*x^3 + a)) - 2*A*a*b*sqrt(x))/(a^2*b
*x^2), 2/3*((B*a - A*b)*sqrt(a*b)*x^2*arctan(sqrt(a*b)*x^(3/2)/a) - A*a*b*sqrt(x))/(a^2*b*x^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)/x**(5/2)/(b*x**3+a),x)

[Out]

Timed out

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Giac [A]  time = 1.09644, size = 53, normalized size = 1. \begin{align*} \frac{2 \,{\left (B a - A b\right )} \arctan \left (\frac{b x^{\frac{3}{2}}}{\sqrt{a b}}\right )}{3 \, \sqrt{a b} a} - \frac{2 \, A}{3 \, a x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^(5/2)/(b*x^3+a),x, algorithm="giac")

[Out]

2/3*(B*a - A*b)*arctan(b*x^(3/2)/sqrt(a*b))/(sqrt(a*b)*a) - 2/3*A/(a*x^(3/2))