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

Optimal. Leaf size=77 \[ -\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{5/2}}+\frac {2 A}{3 a^2 \sqrt {a+b x^3}}+\frac {2 (A b-a B)}{9 a b \left (a+b x^3\right )^{3/2}} \]

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Rubi [A]  time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, 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, 51, 63, 208} \begin {gather*} \frac {2 A}{3 a^2 \sqrt {a+b x^3}}-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{5/2}}+\frac {2 (A b-a B)}{9 a b \left (a+b x^3\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 51

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

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Mathematica [C]  time = 0.05, size = 62, normalized size = 0.81 \begin {gather*} \frac {2 a (A b-a B)+6 A b \left (a+b x^3\right ) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b x^3}{a}+1\right )}{9 a^2 b \left (a+b x^3\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(A*b - a*B) + 6*A*b*(a + b*x^3)*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*x^3)/a])/(9*a^2*b*(a + b*x^3)^(3/2
))

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IntegrateAlgebraic [A]  time = 0.09, size = 70, normalized size = 0.91 \begin {gather*} -\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{5/2}}-\frac {2 \left (a^2 B-4 a A b-3 A b^2 x^3\right )}{9 a^2 b \left (a+b x^3\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-4*a*A*b + a^2*B - 3*A*b^2*x^3))/(9*a^2*b*(a + b*x^3)^(3/2)) - (2*A*ArcTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(3*
a^(5/2))

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fricas [A]  time = 0.56, size = 243, normalized size = 3.16 \begin {gather*} \left [\frac {3 \, {\left (A b^{3} x^{6} + 2 \, A a b^{2} x^{3} + A a^{2} b\right )} \sqrt {a} \log \left (\frac {b x^{3} - 2 \, \sqrt {b x^{3} + a} \sqrt {a} + 2 \, a}{x^{3}}\right ) + 2 \, {\left (3 \, A a b^{2} x^{3} - B a^{3} + 4 \, A a^{2} b\right )} \sqrt {b x^{3} + a}}{9 \, {\left (a^{3} b^{3} x^{6} + 2 \, a^{4} b^{2} x^{3} + a^{5} b\right )}}, \frac {2 \, {\left (3 \, {\left (A b^{3} x^{6} + 2 \, A a b^{2} x^{3} + A a^{2} b\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a} \sqrt {-a}}{a}\right ) + {\left (3 \, A a b^{2} x^{3} - B a^{3} + 4 \, A a^{2} b\right )} \sqrt {b x^{3} + a}\right )}}{9 \, {\left (a^{3} b^{3} x^{6} + 2 \, a^{4} b^{2} x^{3} + a^{5} b\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x/(b*x^3+a)^(5/2),x, algorithm="fricas")

[Out]

[1/9*(3*(A*b^3*x^6 + 2*A*a*b^2*x^3 + A*a^2*b)*sqrt(a)*log((b*x^3 - 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) + 2*(
3*A*a*b^2*x^3 - B*a^3 + 4*A*a^2*b)*sqrt(b*x^3 + a))/(a^3*b^3*x^6 + 2*a^4*b^2*x^3 + a^5*b), 2/9*(3*(A*b^3*x^6 +
2*A*a*b^2*x^3 + A*a^2*b)*sqrt(-a)*arctan(sqrt(b*x^3 + a)*sqrt(-a)/a) + (3*A*a*b^2*x^3 - B*a^3 + 4*A*a^2*b)*sq
rt(b*x^3 + a))/(a^3*b^3*x^6 + 2*a^4*b^2*x^3 + a^5*b)]

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giac [A]  time = 0.17, size = 67, normalized size = 0.87 \begin {gather*} \frac {2 \, A \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{3 \, \sqrt {-a} a^{2}} - \frac {2 \, {\left (B a^{2} - 3 \, {\left (b x^{3} + a\right )} A b - A a b\right )}}{9 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x/(b*x^3+a)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.09, size = 85, normalized size = 1.10 \begin {gather*} \left (-\frac {2 \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right )}{3 a^{\frac {5}{2}}}+\frac {2}{3 \sqrt {\left (x^{3}+\frac {a}{b}\right ) b}\, a^{2}}+\frac {2 \sqrt {b \,x^{3}+a}}{9 \left (x^{3}+\frac {a}{b}\right )^{2} a \,b^{2}}\right ) A -\frac {2 B}{9 \left (b \,x^{3}+a \right )^{\frac {3}{2}} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/x/(b*x^3+a)^(5/2),x)

[Out]

-2/9*B/b/(b*x^3+a)^(3/2)+A*(2/9/a/b^2*(b*x^3+a)^(1/2)/(x^3+a/b)^2+2/3/a^2/((x^3+a/b)*b)^(1/2)-2/3/a^(5/2)*arct
anh((b*x^3+a)^(1/2)/a^(1/2)))

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maxima [A]  time = 1.17, size = 81, normalized size = 1.05 \begin {gather*} \frac {1}{9} \, A {\left (\frac {3 \, \log \left (\frac {\sqrt {b x^{3} + a} - \sqrt {a}}{\sqrt {b x^{3} + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, b x^{3} + 4 \, a\right )}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}} a^{2}}\right )} - \frac {2 \, B}{9 \, {\left (b x^{3} + a\right )}^{\frac {3}{2}} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x/(b*x^3+a)^(5/2),x, algorithm="maxima")

[Out]

1/9*A*(3*log((sqrt(b*x^3 + a) - sqrt(a))/(sqrt(b*x^3 + a) + sqrt(a)))/a^(5/2) + 2*(3*b*x^3 + 4*a)/((b*x^3 + a)
^(3/2)*a^2)) - 2/9*B/((b*x^3 + a)^(3/2)*b)

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mupad [B]  time = 2.78, size = 80, normalized size = 1.04 \begin {gather*} \frac {\frac {2\,A}{9\,a}-\frac {2\,B}{9\,b}}{{\left (b\,x^3+a\right )}^{3/2}}+\frac {2\,A}{3\,a^2\,\sqrt {b\,x^3+a}}+\frac {A\,\ln \left (\frac {{\left (\sqrt {b\,x^3+a}-\sqrt {a}\right )}^3\,\left (\sqrt {b\,x^3+a}+\sqrt {a}\right )}{x^6}\right )}{3\,a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^3)/(x*(a + b*x^3)^(5/2)),x)

[Out]

((2*A)/(9*a) - (2*B)/(9*b))/(a + b*x^3)^(3/2) + (2*A)/(3*a^2*(a + b*x^3)^(1/2)) + (A*log((((a + b*x^3)^(1/2) -
a^(1/2))^3*((a + b*x^3)^(1/2) + a^(1/2)))/x^6))/(3*a^(5/2))

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sympy [A]  time = 38.43, size = 76, normalized size = 0.99 \begin {gather*} \frac {2 A}{3 a^{2} \sqrt {a + b x^{3}}} + \frac {2 A \operatorname {atan}{\left (\frac {\sqrt {a + b x^{3}}}{\sqrt {- a}} \right )}}{3 a^{2} \sqrt {- a}} - \frac {2 \left (- A b + B a\right )}{9 a b \left (a + b x^{3}\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)/x/(b*x**3+a)**(5/2),x)

[Out]

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

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