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

Optimal. Leaf size=97 \[ -\frac{3 A b-a B}{3 a^2 b x^{3/2}}-\frac{(3 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{3 a^{5/2} \sqrt{b}}+\frac{A b-a B}{3 a b x^{3/2} \left (a+b x^3\right )} \]

[Out]

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

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Rubi [A]  time = 0.0571234, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {457, 325, 329, 275, 205} \[ -\frac{3 A b-a B}{3 a^2 b x^{3/2}}-\frac{(3 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{3 a^{5/2} \sqrt{b}}+\frac{A b-a B}{3 a b x^{3/2} \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 325

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

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 )^2} \, dx &=\frac{A b-a B}{3 a b x^{3/2} \left (a+b x^3\right )}+\frac{\left (\frac{9 A b}{2}-\frac{3 a B}{2}\right ) \int \frac{1}{x^{5/2} \left (a+b x^3\right )} \, dx}{3 a b}\\ &=-\frac{3 A b-a B}{3 a^2 b x^{3/2}}+\frac{A b-a B}{3 a b x^{3/2} \left (a+b x^3\right )}-\frac{(3 A b-a B) \int \frac{\sqrt{x}}{a+b x^3} \, dx}{2 a^2}\\ &=-\frac{3 A b-a B}{3 a^2 b x^{3/2}}+\frac{A b-a B}{3 a b x^{3/2} \left (a+b x^3\right )}-\frac{(3 A b-a B) \operatorname{Subst}\left (\int \frac{x^2}{a+b x^6} \, dx,x,\sqrt{x}\right )}{a^2}\\ &=-\frac{3 A b-a B}{3 a^2 b x^{3/2}}+\frac{A b-a B}{3 a b x^{3/2} \left (a+b x^3\right )}-\frac{(3 A b-a B) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,x^{3/2}\right )}{3 a^2}\\ &=-\frac{3 A b-a B}{3 a^2 b x^{3/2}}+\frac{A b-a B}{3 a b x^{3/2} \left (a+b x^3\right )}-\frac{(3 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{3 a^{5/2} \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.116151, size = 79, normalized size = 0.81 \[ \frac{\frac{\sqrt{a} \left (-2 a A+a B x^3-3 A b x^3\right )}{x^{3/2} \left (a+b x^3\right )}+\frac{(a B-3 A b) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{\sqrt{b}}}{3 a^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[a]*(-2*a*A - 3*A*b*x^3 + a*B*x^3))/(x^(3/2)*(a + b*x^3)) + ((-3*A*b + a*B)*ArcTan[(Sqrt[b]*x^(3/2))/Sqr
t[a]])/Sqrt[b])/(3*a^(5/2))

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Maple [A]  time = 0.019, size = 93, normalized size = 1. \begin{align*} -{\frac{Ab}{3\,{a}^{2} \left ( b{x}^{3}+a \right ) }{x}^{{\frac{3}{2}}}}+{\frac{B}{3\,a \left ( b{x}^{3}+a \right ) }{x}^{{\frac{3}{2}}}}-{\frac{Ab}{{a}^{2}}\arctan \left ({b{x}^{{\frac{3}{2}}}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{B}{3\,a}\arctan \left ({b{x}^{{\frac{3}{2}}}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{2\,A}{3\,{a}^{2}}{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)^2,x)

[Out]

-1/3/a^2*x^(3/2)/(b*x^3+a)*A*b+1/3/a*x^(3/2)/(b*x^3+a)*B-1/a^2/(a*b)^(1/2)*arctan(b*x^(3/2)/(a*b)^(1/2))*A*b+1
/3/a/(a*b)^(1/2)*arctan(b*x^(3/2)/(a*b)^(1/2))*B-2/3*A/a^2/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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.92989, size = 498, normalized size = 5.13 \begin{align*} \left [\frac{{\left ({\left (B a b - 3 \, A b^{2}\right )} x^{5} +{\left (B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt{-a b} \log \left (\frac{b x^{3} + 2 \, \sqrt{-a b} x^{\frac{3}{2}} - a}{b x^{3} + a}\right ) - 2 \,{\left (2 \, A a^{2} b -{\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{3}\right )} \sqrt{x}}{6 \,{\left (a^{3} b^{2} x^{5} + a^{4} b x^{2}\right )}}, \frac{{\left ({\left (B a b - 3 \, A b^{2}\right )} x^{5} +{\left (B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x^{\frac{3}{2}}}{a}\right ) -{\left (2 \, A a^{2} b -{\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{3}\right )} \sqrt{x}}{3 \,{\left (a^{3} b^{2} x^{5} + a^{4} b x^{2}\right )}}\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)^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.10318, size = 89, normalized size = 0.92 \begin{align*} \frac{{\left (B a - 3 \, A b\right )} \arctan \left (\frac{b x^{\frac{3}{2}}}{\sqrt{a b}}\right )}{3 \, \sqrt{a b} a^{2}} + \frac{B a x^{3} - 3 \, A b x^{3} - 2 \, A a}{3 \,{\left (b x^{\frac{9}{2}} + a x^{\frac{3}{2}}\right )} a^{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)^2,x, algorithm="giac")

[Out]

1/3*(B*a - 3*A*b)*arctan(b*x^(3/2)/sqrt(a*b))/(sqrt(a*b)*a^2) + 1/3*(B*a*x^3 - 3*A*b*x^3 - 2*A*a)/((b*x^(9/2)
+ a*x^(3/2))*a^2)