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

Optimal. Leaf size=53 \[ \frac{2 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{3 \sqrt{a} b^{3/2}}+\frac{2 B x^{3/2}}{3 b} \]

[Out]

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

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Rubi [A]  time = 0.0357264, antiderivative size = 53, 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 = {459, 329, 275, 205} \[ \frac{2 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{3 \sqrt{a} b^{3/2}}+\frac{2 B x^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

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 )}{a+b x^3} \, dx &=\frac{2 B x^{3/2}}{3 b}-\frac{\left (2 \left (-\frac{3 A b}{2}+\frac{3 a B}{2}\right )\right ) \int \frac{\sqrt{x}}{a+b x^3} \, dx}{3 b}\\ &=\frac{2 B x^{3/2}}{3 b}-\frac{\left (4 \left (-\frac{3 A b}{2}+\frac{3 a B}{2}\right )\right ) \operatorname{Subst}\left (\int \frac{x^2}{a+b x^6} \, dx,x,\sqrt{x}\right )}{3 b}\\ &=\frac{2 B x^{3/2}}{3 b}+\frac{(2 (A b-a B)) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,x^{3/2}\right )}{3 b}\\ &=\frac{2 B x^{3/2}}{3 b}+\frac{2 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{3 \sqrt{a} b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0450268, size = 52, normalized size = 0.98 \[ \frac{2}{3} \left (\frac{B x^{3/2}}{b}-\frac{(a B-A b) \tan ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a}}\right )}{\sqrt{a} b^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 53, normalized size = 1. \begin{align*}{\frac{2\,B}{3\,b}{x}^{{\frac{3}{2}}}}+{\frac{2\,A}{3}\arctan \left ({b{x}^{{\frac{3}{2}}}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{2\,Ba}{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),x)

[Out]

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

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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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.76878, size = 261, normalized size = 4.92 \begin{align*} \left [\frac{2 \, B a b x^{\frac{3}{2}} +{\left (B 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 )}{3 \, a b^{2}}, \frac{2 \,{\left (B a b x^{\frac{3}{2}} -{\left (B a - A b\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x^{\frac{3}{2}}}{a}\right )\right )}}{3 \, a b^{2}}\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),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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),x)

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.10648, size = 53, normalized size = 1. \begin{align*} \frac{2 \, B x^{\frac{3}{2}}}{3 \, b} - \frac{2 \,{\left (B a - A b\right )} \arctan \left (\frac{b x^{\frac{3}{2}}}{\sqrt{a b}}\right )}{3 \, \sqrt{a b} 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),x, algorithm="giac")

[Out]

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