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

Optimal. Leaf size=81 \[ -\frac{2}{3} a^{3/2} A \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )+\frac{2}{9} A \left (a+b x^3\right )^{3/2}+\frac{2}{3} a A \sqrt{a+b x^3}+\frac{2 B \left (a+b x^3\right )^{5/2}}{15 b} \]

[Out]

(2*a*A*Sqrt[a + b*x^3])/3 + (2*A*(a + b*x^3)^(3/2))/9 + (2*B*(a + b*x^3)^(5/2))/(15*b) - (2*a^(3/2)*A*ArcTanh[
Sqrt[a + b*x^3]/Sqrt[a]])/3

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Rubi [A]  time = 0.0566899, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {446, 80, 50, 63, 208} \[ -\frac{2}{3} a^{3/2} A \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )+\frac{2}{9} A \left (a+b x^3\right )^{3/2}+\frac{2}{3} a A \sqrt{a+b x^3}+\frac{2 B \left (a+b x^3\right )^{5/2}}{15 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*A*Sqrt[a + b*x^3])/3 + (2*A*(a + b*x^3)^(3/2))/9 + (2*B*(a + b*x^3)^(5/2))/(15*b) - (2*a^(3/2)*A*ArcTanh[
Sqrt[a + b*x^3]/Sqrt[a]])/3

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 80

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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^3\right )^{3/2} \left (A+B x^3\right )}{x} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2} (A+B x)}{x} \, dx,x,x^3\right )\\ &=\frac{2 B \left (a+b x^3\right )^{5/2}}{15 b}+\frac{1}{3} A \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,x^3\right )\\ &=\frac{2}{9} A \left (a+b x^3\right )^{3/2}+\frac{2 B \left (a+b x^3\right )^{5/2}}{15 b}+\frac{1}{3} (a A) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^3\right )\\ &=\frac{2}{3} a A \sqrt{a+b x^3}+\frac{2}{9} A \left (a+b x^3\right )^{3/2}+\frac{2 B \left (a+b x^3\right )^{5/2}}{15 b}+\frac{1}{3} \left (a^2 A\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^3\right )\\ &=\frac{2}{3} a A \sqrt{a+b x^3}+\frac{2}{9} A \left (a+b x^3\right )^{3/2}+\frac{2 B \left (a+b x^3\right )^{5/2}}{15 b}+\frac{\left (2 a^2 A\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^3}\right )}{3 b}\\ &=\frac{2}{3} a A \sqrt{a+b x^3}+\frac{2}{9} A \left (a+b x^3\right )^{3/2}+\frac{2 B \left (a+b x^3\right )^{5/2}}{15 b}-\frac{2}{3} a^{3/2} A \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0671314, size = 80, normalized size = 0.99 \[ \frac{2 \left (-15 a^{3/2} A b \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )+5 A b \left (a+b x^3\right )^{3/2}+15 a A b \sqrt{a+b x^3}+3 B \left (a+b x^3\right )^{5/2}\right )}{45 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(15*a*A*b*Sqrt[a + b*x^3] + 5*A*b*(a + b*x^3)^(3/2) + 3*B*(a + b*x^3)^(5/2) - 15*a^(3/2)*A*b*ArcTanh[Sqrt[a
+ b*x^3]/Sqrt[a]]))/(45*b)

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Maple [A]  time = 0.017, size = 66, normalized size = 0.8 \begin{align*}{\frac{2\,B}{15\,b} \left ( b{x}^{3}+a \right ) ^{{\frac{5}{2}}}}+A \left ({\frac{2\,b{x}^{3}}{9}\sqrt{b{x}^{3}+a}}+{\frac{8\,a}{9}\sqrt{b{x}^{3}+a}}-{\frac{2}{3}{a}^{{\frac{3}{2}}}{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*B*(b*x^3+a)^(5/2)/b+A*(2/9*b*x^3*(b*x^3+a)^(1/2)+8/9*a*(b*x^3+a)^(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)^(3/2)*(B*x^3+A)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.74219, size = 409, normalized size = 5.05 \begin{align*} \left [\frac{15 \, A a^{\frac{3}{2}} b \log \left (\frac{b x^{3} - 2 \, \sqrt{b x^{3} + a} \sqrt{a} + 2 \, a}{x^{3}}\right ) + 2 \,{\left (3 \, B b^{2} x^{6} +{\left (6 \, B a b + 5 \, A b^{2}\right )} x^{3} + 3 \, B a^{2} + 20 \, A a b\right )} \sqrt{b x^{3} + a}}{45 \, b}, \frac{2 \,{\left (15 \, A \sqrt{-a} a b \arctan \left (\frac{\sqrt{b x^{3} + a} \sqrt{-a}}{a}\right ) +{\left (3 \, B b^{2} x^{6} +{\left (6 \, B a b + 5 \, A b^{2}\right )} x^{3} + 3 \, B a^{2} + 20 \, A a b\right )} \sqrt{b x^{3} + a}\right )}}{45 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/45*(15*A*a^(3/2)*b*log((b*x^3 - 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) + 2*(3*B*b^2*x^6 + (6*B*a*b + 5*A*b^2
)*x^3 + 3*B*a^2 + 20*A*a*b)*sqrt(b*x^3 + a))/b, 2/45*(15*A*sqrt(-a)*a*b*arctan(sqrt(b*x^3 + a)*sqrt(-a)/a) + (
3*B*b^2*x^6 + (6*B*a*b + 5*A*b^2)*x^3 + 3*B*a^2 + 20*A*a*b)*sqrt(b*x^3 + a))/b]

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Sympy [A]  time = 39.0241, size = 82, normalized size = 1.01 \begin{align*} \frac{2 A a^{2} \operatorname{atan}{\left (\frac{\sqrt{a + b x^{3}}}{\sqrt{- a}} \right )}}{3 \sqrt{- a}} + \frac{2 A a \sqrt{a + b x^{3}}}{3} + \frac{2 A \left (a + b x^{3}\right )^{\frac{3}{2}}}{9} + \frac{2 B \left (a + b x^{3}\right )^{\frac{5}{2}}}{15 b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a**2*atan(sqrt(a + b*x**3)/sqrt(-a))/(3*sqrt(-a)) + 2*A*a*sqrt(a + b*x**3)/3 + 2*A*(a + b*x**3)**(3/2)/9 +
2*B*(a + b*x**3)**(5/2)/(15*b)

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Giac [A]  time = 1.26816, size = 108, normalized size = 1.33 \begin{align*} \frac{2 \, A a^{2} \arctan \left (\frac{\sqrt{b x^{3} + a}}{\sqrt{-a}}\right )}{3 \, \sqrt{-a}} + \frac{2 \,{\left (3 \,{\left (b x^{3} + a\right )}^{\frac{5}{2}} B b^{4} + 5 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} A b^{5} + 15 \, \sqrt{b x^{3} + a} A a b^{5}\right )}}{45 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*A*a^2*arctan(sqrt(b*x^3 + a)/sqrt(-a))/sqrt(-a) + 2/45*(3*(b*x^3 + a)^(5/2)*B*b^4 + 5*(b*x^3 + a)^(3/2)*A*
b^5 + 15*sqrt(b*x^3 + a)*A*a*b^5)/b^5