3.4.28 \(\int \frac {x^m (a+b x^2)^2}{c+d x^2} \, dx\) [328]

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

[Out]

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

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Rubi [A]
time = 0.05, 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 {d x^2}{c}\right )}{c d^2 (m+1)}-\frac {b x^{m+1} (b c-2 a d)}{d^2 (m+1)}+\frac {b^2 x^{m+3}}{d (m+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(b*c - 2*a*d)*x^(1 + m))/(d^2*(1 + m))) + (b^2*x^(3 + m))/(d*(3 + m)) + ((b*c - a*d)^2*x^(1 + m)*Hypergeo
metric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*d^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 (a+b x^2\right )^2}{c+d x^2} \, dx &=\int \left (-\frac {b (b c-2 a d) x^m}{d^2}+\frac {b^2 x^{2+m}}{d}+\frac {\left (b^2 c^2-2 a b c d+a^2 d^2\right ) x^m}{d^2 \left (c+d x^2\right )}\right ) \, dx\\ &=-\frac {b (b c-2 a d) x^{1+m}}{d^2 (1+m)}+\frac {b^2 x^{3+m}}{d (3+m)}+\frac {(b c-a d)^2 \int \frac {x^m}{c+d x^2} \, dx}{d^2}\\ &=-\frac {b (b c-2 a d) x^{1+m}}{d^2 (1+m)}+\frac {b^2 x^{3+m}}{d (3+m)}+\frac {(b c-a d)^2 x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{c d^2 (1+m)}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 118, normalized size = 1.26 \begin {gather*} \frac {x^{1+m} \left (\frac {a^2 \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{1+m}+b x^2 \left (\frac {2 a \, _2F_1\left (1,\frac {3+m}{2};\frac {5+m}{2};-\frac {d x^2}{c}\right )}{3+m}+\frac {b x^2 \, _2F_1\left (1,\frac {5+m}{2};\frac {7+m}{2};-\frac {d x^2}{c}\right )}{5+m}\right )\right )}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*((a^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(1 + m) + b*x^2*((2*a*Hypergeometri
c2F1[1, (3 + m)/2, (5 + m)/2, -((d*x^2)/c)])/(3 + m) + (b*x^2*Hypergeometric2F1[1, (5 + m)/2, (7 + m)/2, -((d*
x^2)/c)])/(5 + m))))/c

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {x^{m} \left (b \,x^{2}+a \right )^{2}}{d \,x^{2}+c}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 3.00, size = 299, normalized size = 3.18 \begin {gather*} \frac {a^{2} m x x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {a^{2} x x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {a b m x^{3} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{2 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 a b x^{3} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{2 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {b^{2} m x^{5} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {5 b^{2} x^{5} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 c \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*(b*x**2+a)**2/(d*x**2+c),x)

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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