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

Optimal. Leaf size=66 \[ \frac {x^{m+1} (A b-a B) \, _2F_1\left (1,\frac {m+1}{3};\frac {m+4}{3};-\frac {b x^3}{a}\right )}{a b (m+1)}+\frac {B x^{m+1}}{b (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {459, 364} \[ \frac {x^{m+1} (A b-a B) \, _2F_1\left (1,\frac {m+1}{3};\frac {m+4}{3};-\frac {b x^3}{a}\right )}{a b (m+1)}+\frac {B x^{m+1}}{b (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(x^m*(A + B*x^3))/(a + b*x^3),x]

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 55, normalized size = 0.83 \[ \frac {x^{m+1} \left ((A b-a B) \, _2F_1\left (1,\frac {m+1}{3};\frac {m+4}{3};-\frac {b x^3}{a}\right )+a B\right )}{a b (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{3} + A\right )} x^{m}}{b x^{3} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{3} + A\right )} x^{m}}{b x^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.43, size = 0, normalized size = 0.00 \[ \int \frac {\left (B \,x^{3}+A \right ) x^{m}}{b \,x^{3}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{3} + A\right )} x^{m}}{b x^{3} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^m\,\left (B\,x^3+A\right )}{b\,x^3+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^m*(A + B*x^3))/(a + b*x^3),x)

[Out]

int((x^m*(A + B*x^3))/(a + b*x^3), x)

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sympy [C]  time = 22.06, size = 190, normalized size = 2.88 \[ \frac {A m x x^{m} \Phi \left (\frac {b x^{3} e^{i \pi }}{a}, 1, \frac {m}{3} + \frac {1}{3}\right ) \Gamma \left (\frac {m}{3} + \frac {1}{3}\right )}{9 a \Gamma \left (\frac {m}{3} + \frac {4}{3}\right )} + \frac {A x x^{m} \Phi \left (\frac {b x^{3} e^{i \pi }}{a}, 1, \frac {m}{3} + \frac {1}{3}\right ) \Gamma \left (\frac {m}{3} + \frac {1}{3}\right )}{9 a \Gamma \left (\frac {m}{3} + \frac {4}{3}\right )} + \frac {B m x^{4} x^{m} \Phi \left (\frac {b x^{3} e^{i \pi }}{a}, 1, \frac {m}{3} + \frac {4}{3}\right ) \Gamma \left (\frac {m}{3} + \frac {4}{3}\right )}{9 a \Gamma \left (\frac {m}{3} + \frac {7}{3}\right )} + \frac {4 B x^{4} x^{m} \Phi \left (\frac {b x^{3} e^{i \pi }}{a}, 1, \frac {m}{3} + \frac {4}{3}\right ) \Gamma \left (\frac {m}{3} + \frac {4}{3}\right )}{9 a \Gamma \left (\frac {m}{3} + \frac {7}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*m*x*x**m*lerchphi(b*x**3*exp_polar(I*pi)/a, 1, m/3 + 1/3)*gamma(m/3 + 1/3)/(9*a*gamma(m/3 + 4/3)) + A*x*x**m
*lerchphi(b*x**3*exp_polar(I*pi)/a, 1, m/3 + 1/3)*gamma(m/3 + 1/3)/(9*a*gamma(m/3 + 4/3)) + B*m*x**4*x**m*lerc
hphi(b*x**3*exp_polar(I*pi)/a, 1, m/3 + 4/3)*gamma(m/3 + 4/3)/(9*a*gamma(m/3 + 7/3)) + 4*B*x**4*x**m*lerchphi(
b*x**3*exp_polar(I*pi)/a, 1, m/3 + 4/3)*gamma(m/3 + 4/3)/(9*a*gamma(m/3 + 7/3))

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