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

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

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.0437234, antiderivative size = 102, 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 = {482, 364} \[ \frac{b x^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a (m+1) (b c-a d)}-\frac{d x^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c (m+1) (b c-a d)} \]

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 482

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]

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 (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*}

Mathematica [A]  time = 0.0535, size = 85, normalized size = 0.83 \[ \frac{x^{m+1} \left (a d \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )-b c \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )\right )}{a c (m+1) (a d-b c)} \]

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))

________________________________________________________________________________________

Maple [F]  time = 0.053, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{m}}{ \left ( b{x}^{2}+a \right ) \left ( d{x}^{2}+c \right ) }}\, dx \end{align*}

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

Sympy [C]  time = 12.0885, size = 354, normalized size = 3.47 \begin{align*} \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} \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} \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{align*}

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)))

________________________________________________________________________________________

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

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)