3.4.34 \(\int \frac {x^m}{(a+b x^2) (c+d x^2)} \, dx\) [334]

Optimal. Leaf size=102 \[ \frac {b x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{a (b c-a d) (1+m)}-\frac {d x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{c (b c-a d) (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 102, 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 = {493, 371} \begin {gather*} \frac {b x^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a (m+1) (b c-a d)}-\frac {d x^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )}{c (m+1) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^m/((a + b*x^2)*(c + d*x^2)),x]

[Out]

(b*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*(b*c - a*d)*(1 + m)) - (d*x^(1 + m)*
Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*(b*c - a*d)*(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 493

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[b/(b*c - a*d), I
nt[(e*x)^m/(a + b*x^n), x], x] - Dist[d/(b*c - a*d), Int[(e*x)^m/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 85, normalized size = 0.83 \begin {gather*} \frac {x^{1+m} \left (-b c \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+a d \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )\right )}{a c (-b c+a d) (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {x^{m}}{\left (b \,x^{2}+a \right ) \left (d \,x^{2}+c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 3.55, size = 354, normalized size = 3.47 \begin {gather*} \frac {a m x^{m} \Phi \left (\frac {a e^{i \pi }}{b x^{2}}, 1, \frac {3}{2} - \frac {m}{2}\right ) \Gamma ^{2}\left (\frac {3}{2} - \frac {m}{2}\right )}{x^{3} \cdot \left (4 a b d \Gamma \left (\frac {3}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {5}{2} - \frac {m}{2}\right ) - 4 b^{2} c \Gamma \left (\frac {3}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {5}{2} - \frac {m}{2}\right )\right )} - \frac {3 a x^{m} \Phi \left (\frac {a e^{i \pi }}{b x^{2}}, 1, \frac {3}{2} - \frac {m}{2}\right ) \Gamma ^{2}\left (\frac {3}{2} - \frac {m}{2}\right )}{x^{3} \cdot \left (4 a b d \Gamma \left (\frac {3}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {5}{2} - \frac {m}{2}\right ) - 4 b^{2} c \Gamma \left (\frac {3}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {5}{2} - \frac {m}{2}\right )\right )} + \frac {b m x^{m} \Phi \left (\frac {c e^{i \pi }}{d x^{2}}, 1, \frac {1}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {1}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {5}{2} - \frac {m}{2}\right )}{x \left (4 a b d \Gamma \left (\frac {3}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {5}{2} - \frac {m}{2}\right ) - 4 b^{2} c \Gamma \left (\frac {3}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {5}{2} - \frac {m}{2}\right )\right )} - \frac {b x^{m} \Phi \left (\frac {c e^{i \pi }}{d x^{2}}, 1, \frac {1}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {1}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {5}{2} - \frac {m}{2}\right )}{x \left (4 a b d \Gamma \left (\frac {3}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {5}{2} - \frac {m}{2}\right ) - 4 b^{2} c \Gamma \left (\frac {3}{2} - \frac {m}{2}\right ) \Gamma \left (\frac {5}{2} - \frac {m}{2}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(x^m/((b*x^2 + a)*(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 )\,\left (d\,x^2+c\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/((a + b*x^2)*(c + d*x^2)),x)

[Out]

int(x^m/((a + b*x^2)*(c + d*x^2)), x)

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