3.1.31 \(\int \frac {(e x)^m (a+b x^2)^2 (A+B x^2)}{(c+d x^2)^2} \, dx\) [31]

Optimal. Leaf size=246 \[ -\frac {b (2 a d (A d (1+m)-B c (3+m))-b c (A d (3+m)-B c (5+m))) (e x)^{1+m}}{2 c d^3 e (1+m)}-\frac {b^2 (A d (3+m)-B c (5+m)) (e x)^{3+m}}{2 c d^2 e^3 (3+m)}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{2 c d e \left (c+d x^2\right )}-\frac {(b c-a d) (a d (A d (1-m)+B c (1+m))+b c (A d (3+m)-B c (5+m))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{2 c^2 d^3 e (1+m)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.27, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {591, 584, 371} \begin {gather*} -\frac {(e x)^{m+1} (b c-a d) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right ) (a d (A d (1-m)+B c (m+1))+b c (A d (m+3)-B c (m+5)))}{2 c^2 d^3 e (m+1)}-\frac {b (e x)^{m+1} (2 a d (A d (m+1)-B c (m+3))-b c (A d (m+3)-B c (m+5)))}{2 c d^3 e (m+1)}-\frac {\left (a+b x^2\right )^2 (e x)^{m+1} (B c-A d)}{2 c d e \left (c+d x^2\right )}-\frac {b^2 (e x)^{m+3} (A d (m+3)-B c (m+5))}{2 c d^2 e^3 (m+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*(2*a*d*(A*d*(1 + m) - B*c*(3 + m)) - b*c*(A*d*(3 + m) - B*c*(5 + m)))*(e*x)^(1 + m))/(c*d^3*e*(1 + m))
- (b^2*(A*d*(3 + m) - B*c*(5 + m))*(e*x)^(3 + m))/(2*c*d^2*e^3*(3 + m)) - ((B*c - A*d)*(e*x)^(1 + m)*(a + b*x
^2)^2)/(2*c*d*e*(c + d*x^2)) - ((b*c - a*d)*(a*d*(A*d*(1 - m) + B*c*(1 + m)) + b*c*(A*d*(3 + m) - B*c*(5 + m))
)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(2*c^2*d^3*e*(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 584

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]

Rule 591

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*b*g*n*(p + 1))), x] + Dis
t[1/(a*b*n*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + (b*e - a*f)*(
m + 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m + n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])

Rubi steps

\begin {align*} \int \frac {(e x)^m \left (a+b x^2\right )^2 \left (A+B x^2\right )}{\left (c+d x^2\right )^2} \, dx &=-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{2 c d e \left (c+d x^2\right )}-\frac {\int \frac {(e x)^m \left (a+b x^2\right ) \left (-a (A d (1-m)+B c (1+m))+b (A d (3+m)-B c (5+m)) x^2\right )}{c+d x^2} \, dx}{2 c d}\\ &=-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{2 c d e \left (c+d x^2\right )}-\frac {\int \left (\frac {b (2 a d (A d (1+m)-B c (3+m))-b c (A d (3+m)-B c (5+m))) (e x)^m}{d^2}+\frac {b^2 (A d (3+m)-B c (5+m)) (e x)^{2+m}}{d e^2}+\frac {\left (-5 b^2 B c^3+3 A b^2 c^2 d+6 a b B c^2 d-2 a A b c d^2-a^2 B c d^2-a^2 A d^3-b^2 B c^3 m+A b^2 c^2 d m+2 a b B c^2 d m-2 a A b c d^2 m-a^2 B c d^2 m+a^2 A d^3 m\right ) (e x)^m}{d^2 \left (c+d x^2\right )}\right ) \, dx}{2 c d}\\ &=-\frac {b (2 a d (A d (1+m)-B c (3+m))-b c (A d (3+m)-B c (5+m))) (e x)^{1+m}}{2 c d^3 e (1+m)}-\frac {b^2 (A d (3+m)-B c (5+m)) (e x)^{3+m}}{2 c d^2 e^3 (3+m)}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{2 c d e \left (c+d x^2\right )}-\frac {((b c-a d) (a d (A d (1-m)+B c (1+m))+b c (A d (3+m)-B c (5+m)))) \int \frac {(e x)^m}{c+d x^2} \, dx}{2 c d^3}\\ &=-\frac {b (2 a d (A d (1+m)-B c (3+m))-b c (A d (3+m)-B c (5+m))) (e x)^{1+m}}{2 c d^3 e (1+m)}-\frac {b^2 (A d (3+m)-B c (5+m)) (e x)^{3+m}}{2 c d^2 e^3 (3+m)}-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^2}{2 c d e \left (c+d x^2\right )}-\frac {(b c-a d) (a d (A d (1-m)+B c (1+m))+b c (A d (3+m)-B c (5+m))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{2 c^2 d^3 e (1+m)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.81, size = 165, normalized size = 0.67 \begin {gather*} \frac {x (e x)^m \left (b c^2 \left (2 a B d (3+m)+b \left (-2 B c (3+m)+A d (3+m)+B d (1+m) x^2\right )\right )+c (b c-a d) (3 b B c-2 A b d-a B d) (3+m) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )-(b c-a d)^2 (B c-A d) (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )\right )}{c^2 d^3 (1+m) (3+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right )^{2} \left (B \,x^{2}+A \right )}{\left (d \,x^{2}+c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________