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

Optimal. Leaf size=118 \[ \frac {(e x)^{m+1} (A b-a B) (b c-a d) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a b^2 e (m+1)}+\frac {(e x)^{m+1} (-a B d+A b d+b B c)}{b^2 e (m+1)}+\frac {B d (e x)^{m+3}}{b e^3 (m+3)} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 118, 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} \[ \frac {(e x)^{m+1} (A b-a B) (b c-a d) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a b^2 e (m+1)}+\frac {(e x)^{m+1} (-a B d+A b d+b B c)}{b^2 e (m+1)}+\frac {B d (e x)^{m+3}}{b e^3 (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B*c + A*b*d - a*B*d)*(e*x)^(1 + m))/(b^2*e*(1 + m)) + (B*d*(e*x)^(3 + m))/(b*e^3*(3 + m)) + ((A*b - a*B)*(
b*c - a*d)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*b^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 (c+d x^2\right )}{a+b x^2} \, dx &=\int \left (\frac {(b B c+A b d-a B d) (e x)^m}{b^2}+\frac {B d (e x)^{2+m}}{b e^2}+\frac {\left (A b^2 c-a b B c-a A b d+a^2 B d\right ) (e x)^m}{b^2 \left (a+b x^2\right )}\right ) \, dx\\ &=\frac {(b B c+A b d-a B d) (e x)^{1+m}}{b^2 e (1+m)}+\frac {B d (e x)^{3+m}}{b e^3 (3+m)}+\frac {((A b-a B) (b c-a d)) \int \frac {(e x)^m}{a+b x^2} \, dx}{b^2}\\ &=\frac {(b B c+A b d-a B d) (e x)^{1+m}}{b^2 e (1+m)}+\frac {B d (e x)^{3+m}}{b e^3 (3+m)}+\frac {(A b-a B) (b c-a d) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{a b^2 e (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 93, normalized size = 0.79 \[ \frac {x (e x)^m \left (\frac {(a B-A b) (a d-b c) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a (m+1)}+\frac {-a B d+A b d+b B c}{m+1}+\frac {b B d x^2}{m+3}\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^m*(A + B*x^2)*(c + d*x^2))/(a + b*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) + ((-(A*b) + a*B)*(-(b*c) + a*d)*Hypergeomet
ric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*(1 + m))))/b^2

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fricas [F]  time = 1.13, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B d x^{4} + {\left (B c + A d\right )} x^{2} + A c\right )} \left (e x\right )^{m}}{b x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {\left (B \,x^{2}+A \right ) \left (d \,x^{2}+c \right ) \left (e x \right )^{m}}{b \,x^{2}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 11.54, size = 428, normalized size = 3.63 \[ \frac {A c e^{m} m x x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A c e^{m} x x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A d e^{m} m x^{3} x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 A d e^{m} x^{3} x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {B c e^{m} m x^{3} x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 B c e^{m} x^{3} x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {B d e^{m} m x^{5} x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {5 B d e^{m} x^{5} x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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