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

Optimal. Leaf size=104 \[ \frac{(a B+3 A b) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{12 a^{5/2} b^{3/2}}+\frac{x^{3/2} (a B+3 A b)}{12 a^2 b \left (a+b x^3\right )}+\frac{x^{3/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2} \]

[Out]

((A*b - a*B)*x^(3/2))/(6*a*b*(a + b*x^3)^2) + ((3*A*b + a*B)*x^(3/2))/(12*a^2*b*(a + b*x^3)) + ((3*A*b + a*B)*
ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]])/(12*a^(5/2)*b^(3/2))

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Rubi [A]  time = 0.0595053, antiderivative size = 104, normalized size of antiderivative = 1., 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, 290, 329, 275, 205} \[ \frac{(a B+3 A b) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{12 a^{5/2} b^{3/2}}+\frac{x^{3/2} (a B+3 A b)}{12 a^2 b \left (a+b x^3\right )}+\frac{x^{3/2} (A b-a B)}{6 a b \left (a+b x^3\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*x^(3/2))/(6*a*b*(a + b*x^3)^2) + ((3*A*b + a*B)*x^(3/2))/(12*a^2*b*(a + b*x^3)) + ((3*A*b + a*B)*
ArcTan[(Sqrt[b]*x^(3/2))/Sqrt[a]])/(12*a^(5/2)*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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -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 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 )^3} \, dx &=\frac{(A b-a B) x^{3/2}}{6 a b \left (a+b x^3\right )^2}+\frac{\left (\frac{9 A b}{2}+\frac{3 a B}{2}\right ) \int \frac{\sqrt{x}}{\left (a+b x^3\right )^2} \, dx}{6 a b}\\ &=\frac{(A b-a B) x^{3/2}}{6 a b \left (a+b x^3\right )^2}+\frac{(3 A b+a B) x^{3/2}}{12 a^2 b \left (a+b x^3\right )}+\frac{(3 A b+a B) \int \frac{\sqrt{x}}{a+b x^3} \, dx}{8 a^2 b}\\ &=\frac{(A b-a B) x^{3/2}}{6 a b \left (a+b x^3\right )^2}+\frac{(3 A b+a B) x^{3/2}}{12 a^2 b \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 )}{4 a^2 b}\\ &=\frac{(A b-a B) x^{3/2}}{6 a b \left (a+b x^3\right )^2}+\frac{(3 A b+a B) x^{3/2}}{12 a^2 b \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 )}{12 a^2 b}\\ &=\frac{(A b-a B) x^{3/2}}{6 a b \left (a+b x^3\right )^2}+\frac{(3 A b+a B) x^{3/2}}{12 a^2 b \left (a+b x^3\right )}+\frac{(3 A b+a B) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{12 a^{5/2} b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0730823, size = 94, normalized size = 0.9 \[ \frac{\frac{\sqrt{a} \sqrt{b} x^{3/2} \left (-a^2 B+a b \left (5 A+B x^3\right )+3 A b^2 x^3\right )}{\left (a+b x^3\right )^2}+(a B+3 A b) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{12 a^{5/2} b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[a]*Sqrt[b]*x^(3/2)*(-(a^2*B) + 3*A*b^2*x^3 + a*b*(5*A + B*x^3)))/(a + b*x^3)^2 + (3*A*b + a*B)*ArcTan[(
Sqrt[b]*x^(3/2))/Sqrt[a]])/(12*a^(5/2)*b^(3/2))

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Maple [A]  time = 0.014, size = 97, normalized size = 0.9 \begin{align*}{\frac{2}{3\, \left ( b{x}^{3}+a \right ) ^{2}} \left ({\frac{3\,Ab+Ba}{8\,{a}^{2}}{x}^{{\frac{9}{2}}}}+{\frac{5\,Ab-Ba}{8\,ab}{x}^{{\frac{3}{2}}}} \right ) }+{\frac{A}{4\,{a}^{2}}\arctan \left ({b{x}^{{\frac{3}{2}}}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{B}{12\,ab}\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)^3,x)

[Out]

2/3*(1/8*(3*A*b+B*a)/a^2*x^(9/2)+1/8*(5*A*b-B*a)/a/b*x^(3/2))/(b*x^3+a)^2+1/4/a^2/(a*b)^(1/2)*arctan(b*x^(3/2)
/(a*b)^(1/2))*A+1/12/a/b/(a*b)^(1/2)*arctan(b*x^(3/2)/(a*b)^(1/2))*B

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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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.11275, size = 664, normalized size = 6.38 \begin{align*} \left [-\frac{{\left ({\left (B a b^{2} + 3 \, A b^{3}\right )} x^{6} + B a^{3} + 3 \, A a^{2} b + 2 \,{\left (B a^{2} b + 3 \, A a b^{2}\right )} x^{3}\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 ({\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{4} -{\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt{x}}{24 \,{\left (a^{3} b^{4} x^{6} + 2 \, a^{4} b^{3} x^{3} + a^{5} b^{2}\right )}}, \frac{{\left ({\left (B a b^{2} + 3 \, A b^{3}\right )} x^{6} + B a^{3} + 3 \, A a^{2} b + 2 \,{\left (B a^{2} b + 3 \, A a b^{2}\right )} x^{3}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x^{\frac{3}{2}}}{a}\right ) +{\left ({\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{4} -{\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt{x}}{12 \,{\left (a^{3} b^{4} x^{6} + 2 \, a^{4} b^{3} x^{3} + a^{5} 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)^3,x, algorithm="fricas")

[Out]

[-1/24*(((B*a*b^2 + 3*A*b^3)*x^6 + B*a^3 + 3*A*a^2*b + 2*(B*a^2*b + 3*A*a*b^2)*x^3)*sqrt(-a*b)*log((b*x^3 - 2*
sqrt(-a*b)*x^(3/2) - a)/(b*x^3 + a)) - 2*((B*a^2*b^2 + 3*A*a*b^3)*x^4 - (B*a^3*b - 5*A*a^2*b^2)*x)*sqrt(x))/(a
^3*b^4*x^6 + 2*a^4*b^3*x^3 + a^5*b^2), 1/12*(((B*a*b^2 + 3*A*b^3)*x^6 + B*a^3 + 3*A*a^2*b + 2*(B*a^2*b + 3*A*a
*b^2)*x^3)*sqrt(a*b)*arctan(sqrt(a*b)*x^(3/2)/a) + ((B*a^2*b^2 + 3*A*a*b^3)*x^4 - (B*a^3*b - 5*A*a^2*b^2)*x)*s
qrt(x))/(a^3*b^4*x^6 + 2*a^4*b^3*x^3 + a^5*b^2)]

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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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.13459, size = 113, normalized size = 1.09 \begin{align*} \frac{{\left (B a + 3 \, A b\right )} \arctan \left (\frac{b x^{\frac{3}{2}}}{\sqrt{a b}}\right )}{12 \, \sqrt{a b} a^{2} b} + \frac{B a b x^{\frac{9}{2}} + 3 \, A b^{2} x^{\frac{9}{2}} - B a^{2} x^{\frac{3}{2}} + 5 \, A a b x^{\frac{3}{2}}}{12 \,{\left (b x^{3} + a\right )}^{2} a^{2} 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)^3,x, algorithm="giac")

[Out]

1/12*(B*a + 3*A*b)*arctan(b*x^(3/2)/sqrt(a*b))/(sqrt(a*b)*a^2*b) + 1/12*(B*a*b*x^(9/2) + 3*A*b^2*x^(9/2) - B*a
^2*x^(3/2) + 5*A*a*b*x^(3/2))/((b*x^3 + a)^2*a^2*b)