3.4.22 \(\int \frac {x^m (A+B x^2)}{a+b x^2} \, dx\) [322]

Optimal. Leaf size=66 \[ \frac {B x^{1+m}}{b (1+m)}+\frac {(A b-a B) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{a b (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.02, 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 = {470, 371} \begin {gather*} \frac {x^{m+1} (A b-a B) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a b (m+1)}+\frac {B x^{m+1}}{b (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^m*(A + B*x^2))/(a + b*x^2),x]

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 470

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^2\right )}{a+b x^2} \, 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^2} \, 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}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{a b (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 55, normalized size = 0.83 \begin {gather*} \frac {x^{1+m} \left (a B+(A b-a B) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )\right )}{a b (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{m} \left (B \,x^{2}+A \right )}{b \,x^{2}+a}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 1.97, size = 190, normalized size = 2.88 \begin {gather*} \frac {A m x x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A x x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {B m x^{3} x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 B x^{3} x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*m*x*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(4*a*gamma(m/2 + 3/2)) + A*x*x**m
*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(4*a*gamma(m/2 + 3/2)) + B*m*x**3*x**m*lerc
hphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*a*gamma(m/2 + 5/2)) + 3*B*x**3*x**m*lerchphi(
b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*a*gamma(m/2 + 5/2))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^m\,\left (B\,x^2+A\right )}{b\,x^2+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^m*(A + B*x^2))/(a + b*x^2),x)

[Out]

int((x^m*(A + B*x^2))/(a + b*x^2), x)

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