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

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

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Rubi [A]  time = 0.05, antiderivative size = 58, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {(A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2}}-\frac {A \sqrt {a+b x^3}}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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]]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 57, normalized size = 0.98 \begin {gather*} \frac {1}{3} \left (\frac {(A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {A \sqrt {a+b x^3}}{a x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.07, size = 58, normalized size = 1.00 \begin {gather*} \frac {(A b-2 a B) \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2}}-\frac {A \sqrt {a+b x^3}}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(A*Sqrt[a + b*x^3])/(a*x^3) + ((A*b - 2*a*B)*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(3*a^(3/2))

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fricas [A]  time = 1.51, size = 126, normalized size = 2.17 \begin {gather*} \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 \, \sqrt {b x^{3} + a} A a}{6 \, a^{2} 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 ) - \sqrt {b x^{3} + a} A a}{3 \, a^{2} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^4/(b*x^3+a)^(1/2),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*sqrt(b*x^3 + a)*A*a)/(
a^2*x^3), 1/3*((2*B*a - A*b)*sqrt(-a)*x^3*arctan(sqrt(b*x^3 + a)*sqrt(-a)/a) - sqrt(b*x^3 + a)*A*a)/(a^2*x^3)]

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giac [A]  time = 0.16, size = 62, normalized size = 1.07 \begin {gather*} \frac {\frac {{\left (2 \, B a b - A b^{2}\right )} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {\sqrt {b x^{3} + a} A b}{a x^{3}}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 62, normalized size = 1.07 \begin {gather*} -\frac {2 B \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 \sqrt {a}}+\left (\frac {b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {3}{2}}}-\frac {\sqrt {b \,x^{3}+a}}{3 a \,x^{3}}\right ) A \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.17, size = 109, normalized size = 1.88 \begin {gather*} -\frac {1}{6} \, A {\left (\frac {2 \, \sqrt {b x^{3} + a} b}{{\left (b x^{3} + a\right )} a - a^{2}} + \frac {b \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}}\right )} + \frac {B \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{3 \, \sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*A*(2*sqrt(b*x^3 + a)*b/((b*x^3 + a)*a - a^2) + b*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(
a)))/a^(3/2)) + 1/3*B*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/sqrt(a)

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mupad [B]  time = 2.89, size = 67, normalized size = 1.16 \begin {gather*} \frac {\ln \left (\frac {\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )\,{\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}^3}{x^6}\right )\,\left (A\,b-2\,B\,a\right )}{6\,a^{3/2}}-\frac {A\,\sqrt {b\,x^3+a}}{3\,a\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((((a + b*x^3)^(1/2) - a^(1/2))*((a + b*x^3)^(1/2) + a^(1/2))^3)/x^6)*(A*b - 2*B*a))/(6*a^(3/2)) - (A*(a +
b*x^3)^(1/2))/(3*a*x^3)

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sympy [A]  time = 31.83, size = 80, normalized size = 1.38 \begin {gather*} - \frac {A \sqrt {b} \sqrt {\frac {a}{b x^{3}} + 1}}{3 a x^{\frac {3}{2}}} + \frac {A b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{3 a^{\frac {3}{2}}} - \frac {2 B \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x^{\frac {3}{2}}} \right )}}{3 \sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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