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

Optimal. Leaf size=96 \[ -\frac {(3 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 a^{5/2} \sqrt {b}}+\frac {a B-3 A b}{3 a^2 b x^{3/2}}+\frac {A b-a B}{3 a b x^{3/2} \left (a+b x^3\right )} \]

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Rubi [A]  time = 0.06, antiderivative size = 97, normalized size of antiderivative = 1.01, 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 = {457, 325, 329, 275, 205} \begin {gather*} -\frac {3 A b-a B}{3 a^2 b x^{3/2}}-\frac {(3 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 a^{5/2} \sqrt {b}}+\frac {A b-a B}{3 a b x^{3/2} \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rubi steps

\begin {align*} \int \frac {A+B x^3}{x^{5/2} \left (a+b x^3\right )^2} \, dx &=\frac {A b-a B}{3 a b x^{3/2} \left (a+b x^3\right )}+\frac {\left (\frac {9 A b}{2}-\frac {3 a B}{2}\right ) \int \frac {1}{x^{5/2} \left (a+b x^3\right )} \, dx}{3 a b}\\ &=-\frac {3 A b-a B}{3 a^2 b x^{3/2}}+\frac {A b-a B}{3 a b x^{3/2} \left (a+b x^3\right )}-\frac {(3 A b-a B) \int \frac {\sqrt {x}}{a+b x^3} \, dx}{2 a^2}\\ &=-\frac {3 A b-a B}{3 a^2 b x^{3/2}}+\frac {A b-a B}{3 a b x^{3/2} \left (a+b x^3\right )}-\frac {(3 A b-a B) \operatorname {Subst}\left (\int \frac {x^2}{a+b x^6} \, dx,x,\sqrt {x}\right )}{a^2}\\ &=-\frac {3 A b-a B}{3 a^2 b x^{3/2}}+\frac {A b-a B}{3 a b x^{3/2} \left (a+b x^3\right )}-\frac {(3 A b-a B) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^{3/2}\right )}{3 a^2}\\ &=-\frac {3 A b-a B}{3 a^2 b x^{3/2}}+\frac {A b-a B}{3 a b x^{3/2} \left (a+b x^3\right )}-\frac {(3 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 a^{5/2} \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 79, normalized size = 0.82 \begin {gather*} \frac {\frac {(a B-3 A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{\sqrt {b}}+\frac {\sqrt {a} \left (-2 a A+a B x^3-3 A b x^3\right )}{x^{3/2} \left (a+b x^3\right )}}{3 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[a]*(-2*a*A - 3*A*b*x^3 + a*B*x^3))/(x^(3/2)*(a + b*x^3)) + ((-3*A*b + a*B)*ArcTan[(Sqrt[b]*x^(3/2))/Sqr
t[a]])/Sqrt[b])/(3*a^(5/2))

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IntegrateAlgebraic [A]  time = 0.11, size = 79, normalized size = 0.82 \begin {gather*} \frac {(a B-3 A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 a^{5/2} \sqrt {b}}+\frac {-2 a A+a B x^3-3 A b x^3}{3 a^2 x^{3/2} \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.82, size = 232, normalized size = 2.42 \begin {gather*} \left [\frac {{\left ({\left (B a b - 3 \, A b^{2}\right )} x^{5} + {\left (B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{3} + 2 \, \sqrt {-a b} x^{\frac {3}{2}} - a}{b x^{3} + a}\right ) - 2 \, {\left (2 \, A a^{2} b - {\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{3}\right )} \sqrt {x}}{6 \, {\left (a^{3} b^{2} x^{5} + a^{4} b x^{2}\right )}}, \frac {{\left ({\left (B a b - 3 \, A b^{2}\right )} x^{5} + {\left (B a^{2} - 3 \, A a b\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x^{\frac {3}{2}}}{a}\right ) - {\left (2 \, A a^{2} b - {\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{3}\right )} \sqrt {x}}{3 \, {\left (a^{3} b^{2} x^{5} + a^{4} b x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 66, normalized size = 0.69 \begin {gather*} \frac {{\left (B a - 3 \, A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} a^{2}} + \frac {B a x^{3} - 3 \, A b x^{3} - 2 \, A a}{3 \, {\left (b x^{\frac {9}{2}} + a x^{\frac {3}{2}}\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 93, normalized size = 0.97 \begin {gather*} -\frac {A b \,x^{\frac {3}{2}}}{3 \left (b \,x^{3}+a \right ) a^{2}}+\frac {B \,x^{\frac {3}{2}}}{3 \left (b \,x^{3}+a \right ) a}-\frac {A b \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{2}}+\frac {B \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \sqrt {a b}\, a}-\frac {2 A}{3 a^{2} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.19, size = 67, normalized size = 0.70 \begin {gather*} \frac {{\left (B a - 3 \, A b\right )} x^{3} - 2 \, A a}{3 \, {\left (a^{2} b x^{\frac {9}{2}} + a^{3} x^{\frac {3}{2}}\right )}} + \frac {{\left (B a - 3 \, A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.15, size = 139, normalized size = 1.45 \begin {gather*} -\frac {2\,A\,a^{3/2}\,\sqrt {b}-B\,a^2\,x^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x^{3/2}}{\sqrt {a}}\right )+3\,A\,b^2\,x^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x^{3/2}}{\sqrt {a}}\right )+3\,A\,\sqrt {a}\,b^{3/2}\,x^3-B\,a^{3/2}\,\sqrt {b}\,x^3+3\,A\,a\,b\,x^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x^{3/2}}{\sqrt {a}}\right )-B\,a\,b\,x^{9/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x^{3/2}}{\sqrt {a}}\right )}{3\,a^{7/2}\,\sqrt {b}\,x^{3/2}+3\,a^{5/2}\,b^{3/2}\,x^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*A*a^(3/2)*b^(1/2) - B*a^2*x^(3/2)*atan((b^(1/2)*x^(3/2))/a^(1/2)) + 3*A*b^2*x^(9/2)*atan((b^(1/2)*x^(3/2))
/a^(1/2)) + 3*A*a^(1/2)*b^(3/2)*x^3 - B*a^(3/2)*b^(1/2)*x^3 + 3*A*a*b*x^(3/2)*atan((b^(1/2)*x^(3/2))/a^(1/2))
- B*a*b*x^(9/2)*atan((b^(1/2)*x^(3/2))/a^(1/2)))/(3*a^(7/2)*b^(1/2)*x^(3/2) + 3*a^(5/2)*b^(3/2)*x^(9/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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