3.332 \(\int \frac{x^m (c+d x^2)^2}{a+b x^2} \, dx\)

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

[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))

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Rubi [A]  time = 0.0576673, antiderivative size = 94, normalized size of antiderivative = 1., 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 = {461, 364} \[ \frac{x^{m+1} (b c-a d)^2 \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a b^2 (m+1)}+\frac{d x^{m+1} (2 b c-a d)}{b^2 (m+1)}+\frac{d^2 x^{m+3}}{b (m+3)} \]

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 461

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])

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 (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*}

Mathematica [C]  time = 0.414335, size = 85, normalized size = 0.9 \[ \frac{x^{m+1} \left (c^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+1}{2}\right )+d x^2 \left (2 c \Phi \left (-\frac{b x^2}{a},1,\frac{m+3}{2}\right )+d x^2 \Phi \left (-\frac{b x^2}{a},1,\frac{m+5}{2}\right )\right )\right )}{2 a} \]

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 = 0.043, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d{x}^{2}+c \right ) ^{2}{x}^{m}}{b{x}^{2}+a}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )}^{2} x^{m}}{b x^{2} + a}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (d^{2} x^{4} + 2 \, c d x^{2} + c^{2}\right )} x^{m}}{b x^{2} + a}, x\right ) \end{align*}

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]  time = 9.32418, size = 299, normalized size = 3.18 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )}^{2} x^{m}}{b x^{2} + a}\,{d x} \end{align*}

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)