3.4.32 \(\int \frac {x^m (c+d x^2)^2}{a+b x^2} \, dx\) [332]

Optimal. Leaf size=94 \[ \frac {d (2 b c-a d) x^{1+m}}{b^2 (1+m)}+\frac {d^2 x^{3+m}}{b (3+m)}+\frac {(b c-a d)^2 x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{a b^2 (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {472, 371} \begin {gather*} \frac {x^{m+1} (b c-a d)^2 \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a b^2 (m+1)}+\frac {d x^{m+1} (2 b c-a d)}{b^2 (m+1)}+\frac {d^2 x^{m+3}}{b (m+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^m*(c + d*x^2)^2)/(a + b*x^2),x]

[Out]

(d*(2*b*c - a*d)*x^(1 + m))/(b^2*(1 + m)) + (d^2*x^(3 + m))/(b*(3 + m)) + ((b*c - a*d)^2*x^(1 + m)*Hypergeomet
ric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*b^2*(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 472

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[(e*x)^m*((a + b*x^n)^p/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
time = 0.37, size = 85, normalized size = 0.90 \begin {gather*} \frac {x^{1+m} \left (c^2 \Phi \left (-\frac {b x^2}{a},1,\frac {1+m}{2}\right )+d x^2 \left (2 c \Phi \left (-\frac {b x^2}{a},1,\frac {3+m}{2}\right )+d x^2 \Phi \left (-\frac {b x^2}{a},1,\frac {5+m}{2}\right )\right )\right )}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^m*(c + d*x^2)^2)/(a + b*x^2),x]

[Out]

(x^(1 + m)*(c^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (1 + m)/2] + d*x^2*(2*c*HurwitzLerchPhi[-((b*x^2)/a), 1, (3 +
m)/2] + d*x^2*HurwitzLerchPhi[-((b*x^2)/a), 1, (5 + m)/2])))/(2*a)

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {x^{m} \left (d \,x^{2}+c \right )^{2}}{b \,x^{2}+a}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(d*x^2+c)^2/(b*x^2+a),x)

[Out]

int(x^m*(d*x^2+c)^2/(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*(d*x^2+c)^2/(b*x^2+a),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)^2*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*(d*x^2+c)^2/(b*x^2+a),x, algorithm="fricas")

[Out]

integral((d^2*x^4 + 2*c*d*x^2 + c^2)*x^m/(b*x^2 + a), x)

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Sympy [C] Result contains complex when optimal does not.
time = 3.13, size = 299, normalized size = 3.18 \begin {gather*} \frac {c^{2} 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 {c^{2} 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 {c d 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 )}{2 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 c d 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 )}{2 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {d^{2} m x^{5} x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {5 d^{2} x^{5} x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(d*x**2+c)**2/(b*x**2+a),x)

[Out]

c**2*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)) + c**2*
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)) + c*d*m*x**3*x
**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(2*a*gamma(m/2 + 5/2)) + 3*c*d*x**3*x**m
*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(2*a*gamma(m/2 + 5/2)) + d**2*m*x**5*x**m*l
erchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(4*a*gamma(m/2 + 7/2)) + 5*d**2*x**5*x**m*ler
chphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(4*a*gamma(m/2 + 7/2))

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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*(d*x^2+c)^2/(b*x^2+a),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)^2*x^m/(b*x^2 + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^m\,{\left (d\,x^2+c\right )}^2}{b\,x^2+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^m*(c + d*x^2)^2)/(a + b*x^2),x)

[Out]

int((x^m*(c + d*x^2)^2)/(a + b*x^2), x)

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