3.25 \(\int \frac {(e x)^m (a+b x^2) (A+B x^2)}{c+d x^2} \, dx\)

Optimal. Leaf size=120 \[ \frac {(e x)^{m+1} (b c-a d) (B c-A d) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )}{c d^2 e (m+1)}-\frac {(e x)^{m+1} (-a B d-A b d+b B c)}{d^2 e (m+1)}+\frac {b B (e x)^{m+3}}{d e^3 (m+3)} \]

[Out]

-(-A*b*d-B*a*d+B*b*c)*(e*x)^(1+m)/d^2/e/(1+m)+b*B*(e*x)^(3+m)/d/e^3/(3+m)+(-a*d+b*c)*(-A*d+B*c)*(e*x)^(1+m)*hy
pergeom([1, 1/2+1/2*m],[3/2+1/2*m],-d*x^2/c)/c/d^2/e/(1+m)

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Rubi [A]  time = 0.10, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {570, 364} \[ \frac {(e x)^{m+1} (b c-a d) (B c-A d) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )}{c d^2 e (m+1)}-\frac {(e x)^{m+1} (-a B d-A b d+b B c)}{d^2 e (m+1)}+\frac {b B (e x)^{m+3}}{d e^3 (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 570

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^
(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a,
b, c, d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 93, normalized size = 0.78 \[ \frac {x (e x)^m \left (\frac {(b c-a d) (B c-A d) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )}{c (m+1)}+\frac {a B d+A b d-b B c}{m+1}+\frac {b B d x^2}{m+3}\right )}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*((-(b*B*c) + A*b*d + a*B*d)/(1 + m) + (b*B*d*x^2)/(3 + m) + ((b*c - a*d)*(B*c - A*d)*Hypergeometric
2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*(1 + m))))/d^2

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B b x^{4} + {\left (B a + A b\right )} x^{2} + A a\right )} \left (e x\right )^{m}}{d x^{2} + c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(b*x^2+a)*(B*x^2+A)/(d*x^2+c),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right ) \left (B \,x^{2}+A \right ) \left (e x \right )^{m}}{d \,x^{2}+c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(b*x^2+a)*(B*x^2+A)/(d*x^2+c),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,\left (b\,x^2+a\right )}{d\,x^2+c} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 13.66, size = 428, normalized size = 3.57 \[ \frac {A a e^{m} m x x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A a e^{m} x x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A b e^{m} m x^{3} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 A b e^{m} x^{3} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {B a e^{m} m x^{3} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 B a e^{m} x^{3} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {B b e^{m} m x^{5} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {5 B b e^{m} x^{5} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a*e**m*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
*a*e**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*b*
e**m*m*x**3*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*c*gamma(m/2 + 5/2)) + 3*
A*b*e**m*x**3*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*c*gamma(m/2 + 5/2)) +
B*a*e**m*m*x**3*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*c*gamma(m/2 + 5/2))
+ 3*B*a*e**m*x**3*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*c*gamma(m/2 + 5/2)
) + B*b*e**m*m*x**5*x**m*lerchphi(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*b*e**m*x**5*x**m*lerchphi(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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