3.183 \(\int \frac{\sqrt{a+b x^3} (A+B x^3)}{x^4} \, dx\)

Optimal. Leaf size=84 \[ \frac{\sqrt{a+b x^3} (2 a B+A b)}{3 a}-\frac{(2 a B+A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{3 \sqrt{a}}-\frac{A \left (a+b x^3\right )^{3/2}}{3 a x^3} \]

[Out]

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

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Rubi [A]  time = 0.063156, antiderivative size = 84, 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 = {446, 78, 50, 63, 208} \[ \frac{\sqrt{a+b x^3} (2 a B+A b)}{3 a}-\frac{(2 a B+A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{3 \sqrt{a}}-\frac{A \left (a+b x^3\right )^{3/2}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a + b*x^3]*(A + B*x^3))/x^4,x]

[Out]

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

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

Mathematica [A]  time = 0.0397215, size = 63, normalized size = 0.75 \[ \frac{1}{3} \left (\frac{\sqrt{a+b x^3} \left (2 B x^3-A\right )}{x^3}-\frac{(2 a B+A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a + b*x^3]*(A + B*x^3))/x^4,x]

[Out]

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

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Maple [A]  time = 0.021, size = 72, normalized size = 0.9 \begin{align*} A \left ( -{\frac{1}{3\,{x}^{3}}\sqrt{b{x}^{3}+a}}-{\frac{b}{3}{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ){\frac{1}{\sqrt{a}}}} \right ) +B \left ({\frac{2}{3}\sqrt{b{x}^{3}+a}}-{\frac{2}{3}\sqrt{a}{\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)*(b*x^3+a)^(1/2)/x^4,x)

[Out]

A*(-1/3*(b*x^3+a)^(1/2)/x^3-1/3*b*arctanh((b*x^3+a)^(1/2)/a^(1/2))/a^(1/2))+B*(2/3*(b*x^3+a)^(1/2)-2/3*a^(1/2)
*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)*(b*x^3+a)^(1/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.82532, size = 333, normalized size = 3.96 \begin{align*} \left [\frac{{\left (2 \, B a + A b\right )} \sqrt{a} x^{3} \log \left (\frac{b x^{3} - 2 \, \sqrt{b x^{3} + a} \sqrt{a} + 2 \, a}{x^{3}}\right ) + 2 \,{\left (2 \, B a x^{3} - A a\right )} \sqrt{b x^{3} + a}}{6 \, a x^{3}}, \frac{{\left (2 \, B a + A b\right )} \sqrt{-a} x^{3} \arctan \left (\frac{\sqrt{b x^{3} + a} \sqrt{-a}}{a}\right ) +{\left (2 \, B a x^{3} - A a\right )} \sqrt{b x^{3} + a}}{3 \, a x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*(b*x^3+a)^(1/2)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 23.2387, size = 134, normalized size = 1.6 \begin{align*} - \frac{A \sqrt{b} \sqrt{\frac{a}{b x^{3}} + 1}}{3 x^{\frac{3}{2}}} - \frac{A b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{\frac{3}{2}}} \right )}}{3 \sqrt{a}} - \frac{2 B \sqrt{a} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{\frac{3}{2}}} \right )}}{3} + \frac{2 B a}{3 \sqrt{b} x^{\frac{3}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} + \frac{2 B \sqrt{b} x^{\frac{3}{2}}}{3 \sqrt{\frac{a}{b x^{3}} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)*(b*x**3+a)**(1/2)/x**4,x)

[Out]

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

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Giac [A]  time = 1.13968, size = 92, normalized size = 1.1 \begin{align*} \frac{2 \, \sqrt{b x^{3} + a} B b + \frac{{\left (2 \, B a b + A b^{2}\right )} \arctan \left (\frac{\sqrt{b x^{3} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{\sqrt{b x^{3} + a} A b}{x^{3}}}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)*(b*x^3+a)^(1/2)/x^4,x, algorithm="giac")

[Out]

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