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

Optimal. Leaf size=58 \[ \frac{2 (A b-a B)}{3 a b \sqrt{a+b x^3}}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{3 a^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0397789, antiderivative size = 58, 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 = {446, 78, 63, 208} \[ \frac{2 (A b-a B)}{3 a b \sqrt{a+b x^3}}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{3 a^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0521013, size = 58, normalized size = 1. \[ \frac{1}{3} \left (-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{a^{3/2}}-\frac{2 (a B-A b)}{a b \sqrt{a+b x^3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.026, size = 57, normalized size = 1. \begin{align*} -{\frac{2\,B}{3\,b}{\frac{1}{\sqrt{b{x}^{3}+a}}}}+A \left ({\frac{2}{3\,a}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{a}{b}} \right ) b}}}}-{\frac{2}{3}{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{3}{2}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.81584, size = 371, normalized size = 6.4 \begin{align*} \left [\frac{{\left (A b^{2} x^{3} + A a b\right )} \sqrt{a} \log \left (\frac{b x^{3} - 2 \, \sqrt{b x^{3} + a} \sqrt{a} + 2 \, a}{x^{3}}\right ) - 2 \, \sqrt{b x^{3} + a}{\left (B a^{2} - A a b\right )}}{3 \,{\left (a^{2} b^{2} x^{3} + a^{3} b\right )}}, \frac{2 \,{\left ({\left (A b^{2} x^{3} + A a b\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x^{3} + a} \sqrt{-a}}{a}\right ) - \sqrt{b x^{3} + a}{\left (B a^{2} - A a b\right )}\right )}}{3 \,{\left (a^{2} b^{2} x^{3} + a^{3} b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 9.94424, size = 56, normalized size = 0.97 \begin{align*} \frac{2 A \operatorname{atan}{\left (\frac{\sqrt{a + b x^{3}}}{\sqrt{- a}} \right )}}{3 a \sqrt{- a}} - \frac{2 \left (- A b + B a\right )}{3 a b \sqrt{a + b x^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*atan(sqrt(a + b*x**3)/sqrt(-a))/(3*a*sqrt(-a)) - 2*(-A*b + B*a)/(3*a*b*sqrt(a + b*x**3))

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Giac [A]  time = 1.09715, size = 72, normalized size = 1.24 \begin{align*} \frac{2 \, A \arctan \left (\frac{\sqrt{b x^{3} + a}}{\sqrt{-a}}\right )}{3 \, \sqrt{-a} a} - \frac{2 \,{\left (B a - A b\right )}}{3 \, \sqrt{b x^{3} + a} a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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