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

Optimal. Leaf size=178 \[ \frac {(e x)^{m+1} \left (a^2 B d^2-a b d (A d+2 B c)+b^2 c (2 A d+B c)\right )}{b^3 e (m+1)}+\frac {(e x)^{m+1} (A b-a B) (b c-a d)^2 \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a b^3 e (m+1)}+\frac {d (e x)^{m+3} (-a B d+A b d+2 b B c)}{b^2 e^3 (m+3)}+\frac {B d^2 (e x)^{m+5}}{b e^5 (m+5)} \]

[Out]

(a^2*B*d^2-a*b*d*(A*d+2*B*c)+b^2*c*(2*A*d+B*c))*(e*x)^(1+m)/b^3/e/(1+m)+d*(A*b*d-B*a*d+2*B*b*c)*(e*x)^(3+m)/b^
2/e^3/(3+m)+B*d^2*(e*x)^(5+m)/b/e^5/(5+m)+(A*b-B*a)*(-a*d+b*c)^2*(e*x)^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2
*m],-b*x^2/a)/a/b^3/e/(1+m)

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {570, 364} \[ \frac {(e x)^{m+1} \left (a^2 B d^2-a b d (A d+2 B c)+b^2 c (2 A d+B c)\right )}{b^3 e (m+1)}+\frac {d (e x)^{m+3} (-a B d+A b d+2 b B c)}{b^2 e^3 (m+3)}+\frac {(e x)^{m+1} (A b-a B) (b c-a d)^2 \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a b^3 e (m+1)}+\frac {B d^2 (e x)^{m+5}}{b e^5 (m+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2*B*d^2 - a*b*d*(2*B*c + A*d) + b^2*c*(B*c + 2*A*d))*(e*x)^(1 + m))/(b^3*e*(1 + m)) + (d*(2*b*B*c + A*b*d
- a*B*d)*(e*x)^(3 + m))/(b^2*e^3*(3 + m)) + (B*d^2*(e*x)^(5 + m))/(b*e^5*(5 + m)) + ((A*b - a*B)*(b*c - a*d)^2
*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*b^3*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 )^2}{a+b x^2} \, dx &=\int \left (\frac {\left (a^2 B d^2-a b d (2 B c+A d)+b^2 c (B c+2 A d)\right ) (e x)^m}{b^3}+\frac {d (2 b B c+A b d-a B d) (e x)^{2+m}}{b^2 e^2}+\frac {B d^2 (e x)^{4+m}}{b e^4}+\frac {\left (A b^3 c^2-a b^2 B c^2-2 a A b^2 c d+2 a^2 b B c d+a^2 A b d^2-a^3 B d^2\right ) (e x)^m}{b^3 \left (a+b x^2\right )}\right ) \, dx\\ &=\frac {\left (a^2 B d^2-a b d (2 B c+A d)+b^2 c (B c+2 A d)\right ) (e x)^{1+m}}{b^3 e (1+m)}+\frac {d (2 b B c+A b d-a B d) (e x)^{3+m}}{b^2 e^3 (3+m)}+\frac {B d^2 (e x)^{5+m}}{b e^5 (5+m)}+\frac {\left ((A b-a B) (b c-a d)^2\right ) \int \frac {(e x)^m}{a+b x^2} \, dx}{b^3}\\ &=\frac {\left (a^2 B d^2-a b d (2 B c+A d)+b^2 c (B c+2 A d)\right ) (e x)^{1+m}}{b^3 e (1+m)}+\frac {d (2 b B c+A b d-a B d) (e x)^{3+m}}{b^2 e^3 (3+m)}+\frac {B d^2 (e x)^{5+m}}{b e^5 (5+m)}+\frac {(A b-a B) (b c-a d)^2 (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{a b^3 e (1+m)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.26, size = 146, normalized size = 0.82 \[ \frac {x (e x)^m \left (\frac {a^2 B d^2-a b d (A d+2 B c)+b^2 c (2 A d+B c)}{m+1}+\frac {(A b-a B) (b c-a d)^2 \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a (m+1)}+\frac {b d x^2 (-a B d+A b d+2 b B c)}{m+3}+\frac {b^2 B d^2 x^4}{m+5}\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*((a^2*B*d^2 - a*b*d*(2*B*c + A*d) + b^2*c*(B*c + 2*A*d))/(1 + m) + (b*d*(2*b*B*c + A*b*d - a*B*d)*x
^2)/(3 + m) + (b^2*B*d^2*x^4)/(5 + m) + ((A*b - a*B)*(b*c - a*d)^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2,
-((b*x^2)/a)])/(a*(1 + m))))/b^3

________________________________________________________________________________________

fricas [F]  time = 1.13, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B d^{2} x^{6} + {\left (2 \, B c d + A d^{2}\right )} x^{4} + A c^{2} + {\left (B c^{2} + 2 \, A c d\right )} x^{2}\right )} \left (e x\right )^{m}}{b x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*d^2*x^6 + (2*B*c*d + A*d^2)*x^4 + A*c^2 + (B*c^2 + 2*A*c*d)*x^2)*(e*x)^m/(b*x^2 + a), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 20.59, size = 666, normalized size = 3.74 \[ \frac {A c^{2} 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^{2} 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 c 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 )}{2 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 A c 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 )}{2 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {A d^{2} 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 A d^{2} 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 )} + \frac {B c^{2} 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^{2} 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 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 )}{2 a \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {5 B c 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 )}{2 a \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {B d^{2} e^{m} m x^{7} x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {7}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {9}{2}\right )} + \frac {7 B d^{2} e^{m} x^{7} x^{m} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {7}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {9}{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)**2/(b*x**2+a),x)

[Out]

A*c**2*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**2*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*c*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)/(2*a*gamma(m/2 + 5/
2)) + 3*A*c*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)/(2*a*gamma(m/2
+ 5/2)) + A*d**2*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*A*d**2*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)) + B*c**2*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**2*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*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)/(2*a*gamma(m/2 + 7/2)) + 5*B*c*d*e**m*x**5*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 5/2)*gamm
a(m/2 + 5/2)/(2*a*gamma(m/2 + 7/2)) + B*d**2*e**m*m*x**7*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 7/2)
*gamma(m/2 + 7/2)/(4*a*gamma(m/2 + 9/2)) + 7*B*d**2*e**m*x**7*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 +
7/2)*gamma(m/2 + 7/2)/(4*a*gamma(m/2 + 9/2))

________________________________________________________________________________________