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

Optimal. Leaf size=71 \[ \frac {(a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 a^{3/2} b^{3/2}}+\frac {x^{3/2} (A b-a B)}{3 a b \left (a+b x^3\right )} \]

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Rubi [A]  time = 0.04, antiderivative size = 71, 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 = {457, 329, 275, 205} \begin {gather*} \frac {(a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 a^{3/2} b^{3/2}}+\frac {x^{3/2} (A b-a B)}{3 a b \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.06, size = 71, normalized size = 1.00 \begin {gather*} \frac {(a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 a^{3/2} b^{3/2}}+\frac {x^{3/2} (A b-a B)}{3 a b \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.10, size = 71, normalized size = 1.00 \begin {gather*} \frac {(a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a}}\right )}{3 a^{3/2} b^{3/2}}-\frac {x^{3/2} (a B-A b)}{3 a b \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 190, normalized size = 2.68 \begin {gather*} \left [-\frac {2 \, {\left (B a^{2} b - A a b^{2}\right )} x^{\frac {3}{2}} + {\left ({\left (B a b + A b^{2}\right )} x^{3} + B a^{2} + A a b\right )} \sqrt {-a b} \log \left (\frac {b x^{3} - 2 \, \sqrt {-a b} x^{\frac {3}{2}} - a}{b x^{3} + a}\right )}{6 \, {\left (a^{2} b^{3} x^{3} + a^{3} b^{2}\right )}}, -\frac {{\left (B a^{2} b - A a b^{2}\right )} x^{\frac {3}{2}} - {\left ({\left (B a b + A b^{2}\right )} x^{3} + B a^{2} + A a b\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x^{\frac {3}{2}}}{a}\right )}{3 \, {\left (a^{2} b^{3} x^{3} + a^{3} b^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.28, size = 63, normalized size = 0.89 \begin {gather*} \frac {{\left (B a + A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} a b} - \frac {B a x^{\frac {3}{2}} - A b x^{\frac {3}{2}}}{3 \, {\left (b x^{3} + a\right )} a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 74, normalized size = 1.04 \begin {gather*} \frac {A \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \sqrt {a b}\, a}+\frac {B \arctan \left (\frac {b \,x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \sqrt {a b}\, b}+\frac {\left (A b -B a \right ) x^{\frac {3}{2}}}{3 \left (b \,x^{3}+a \right ) a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.40, size = 61, normalized size = 0.86 \begin {gather*} -\frac {{\left (B a - A b\right )} x^{\frac {3}{2}}}{3 \, {\left (a b^{2} x^{3} + a^{2} b\right )}} + \frac {{\left (B a + A b\right )} \arctan \left (\frac {b x^{\frac {3}{2}}}{\sqrt {a b}}\right )}{3 \, \sqrt {a b} a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.14, size = 115, normalized size = 1.62 \begin {gather*} \frac {B\,a^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,x^{3/2}}{\sqrt {a}}\right )+A\,b^2\,x^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,x^{3/2}}{\sqrt {a}}\right )+A\,a\,b\,\mathrm {atan}\left (\frac {\sqrt {b}\,x^{3/2}}{\sqrt {a}}\right )+A\,\sqrt {a}\,b^{3/2}\,x^{3/2}-B\,a^{3/2}\,\sqrt {b}\,x^{3/2}+B\,a\,b\,x^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,x^{3/2}}{\sqrt {a}}\right )}{3\,a^{5/2}\,b^{3/2}+3\,a^{3/2}\,b^{5/2}\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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**(1/2)/(b*x**3+a)**2,x)

[Out]

Timed out

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