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

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

[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)
)

________________________________________________________________________________________

Rubi [A]  time = 0.0418958, antiderivative size = 71, normalized size of antiderivative = 1., 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} \[ \frac{(a B+A b) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{3 a^{3/2} b^{3/2}}+\frac{x^{3/2} (A b-a B)}{3 a b \left (a+b x^3\right )} \]

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 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))]))

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 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 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]

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*}

Mathematica [A]  time = 0.0515127, size = 71, normalized size = 1. \[ \frac{(a B+A b) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{3 a^{3/2} b^{3/2}}+\frac{x^{3/2} (A b-a B)}{3 a b \left (a+b x^3\right )} \]

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)
)

________________________________________________________________________________________

Maple [A]  time = 0.015, size = 74, normalized size = 1. \begin{align*}{\frac{Ab-Ba}{3\,ab \left ( b{x}^{3}+a \right ) }{x}^{{\frac{3}{2}}}}+{\frac{A}{3\,a}\arctan \left ({b{x}^{{\frac{3}{2}}}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{B}{3\,b}\arctan \left ({b{x}^{{\frac{3}{2}}}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.82273, size = 414, normalized size = 5.83 \begin{align*} \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{align*}

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)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

________________________________________________________________________________________

Giac [A]  time = 1.63099, size = 85, normalized size = 1.2 \begin{align*} \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{align*}

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
)