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

Optimal. Leaf size=93 \[ \frac{x^{m+1} (a B (m+1)+A b (2-m)) \, _2F_1\left (1,\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )}{3 a^2 b (m+1)}+\frac{x^{m+1} (A b-a B)}{3 a b \left (a+b x^3\right )} \]

[Out]

((A*b - a*B)*x^(1 + m))/(3*a*b*(a + b*x^3)) + ((A*b*(2 - m) + a*B*(1 + m))*x^(1 + m)*Hypergeometric2F1[1, (1 +
m)/3, (4 + m)/3, -((b*x^3)/a)])/(3*a^2*b*(1 + m))

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Rubi [A]  time = 0.0415492, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {457, 364} \[ \frac{x^{m+1} (a B (m+1)+A b (2-m)) \, _2F_1\left (1,\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )}{3 a^2 b (m+1)}+\frac{x^{m+1} (A b-a B)}{3 a b \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*x^(1 + m))/(3*a*b*(a + b*x^3)) + ((A*b*(2 - m) + a*B*(1 + m))*x^(1 + m)*Hypergeometric2F1[1, (1 +
m)/3, (4 + m)/3, -((b*x^3)/a)])/(3*a^2*b*(1 + m))

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{x^m \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx &=\frac{(A b-a B) x^{1+m}}{3 a b \left (a+b x^3\right )}+\frac{(-A b (-2+m)+a B (1+m)) \int \frac{x^m}{a+b x^3} \, dx}{3 a b}\\ &=\frac{(A b-a B) x^{1+m}}{3 a b \left (a+b x^3\right )}+\frac{(A b (2-m)+a B (1+m)) x^{1+m} \, _2F_1\left (1,\frac{1+m}{3};\frac{4+m}{3};-\frac{b x^3}{a}\right )}{3 a^2 b (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0575832, size = 80, normalized size = 0.86 \[ \frac{x^{m+1} \left ((A b-a B) \, _2F_1\left (2,\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )+a B \, _2F_1\left (1,\frac{m+1}{3};\frac{m+4}{3};-\frac{b x^3}{a}\right )\right )}{a^2 b (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*(a*B*Hypergeometric2F1[1, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)] + (A*b - a*B)*Hypergeometric2F1[2, (1
+ m)/3, (4 + m)/3, -((b*x^3)/a)]))/(a^2*b*(1 + m))

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Maple [F]  time = 0.044, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{m} \left ( B{x}^{3}+A \right ) }{ \left ( b{x}^{3}+a \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{3} + A\right )} x^{m}}{{\left (b x^{3} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x^{3} + A\right )} x^{m}}{b^{2} x^{6} + 2 \, a b x^{3} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x^3 + A)*x^m/(b^2*x^6 + 2*a*b*x^3 + a^2), x)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{3} + A\right )} x^{m}}{{\left (b x^{3} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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