3.1.21 \(\int \frac {(e x)^m (A+B x^2) (c+d x^2)^3}{(a+b x^2)^3} \, dx\) [21]

Optimal. Leaf size=480 \[ -\frac {d \left (A b \left (2 b^2 c^2 \left (3+2 m-m^2\right )+3 a b c d \left (3+4 m+m^2\right )-a^2 d^2 \left (15+8 m+m^2\right )\right )+a B \left (2 b^2 c^2 (1+m)^2-3 a b c d \left (15+8 m+m^2\right )+a^2 d^2 \left (35+12 m+m^2\right )\right )\right ) (e x)^{1+m}}{8 a^2 b^4 e (1+m)}-\frac {d^2 \left (A b (3+m) (b c (3-m)+a d (5+m))+a B \left (b c \left (3+4 m+m^2\right )-a d \left (35+12 m+m^2\right )\right )\right ) (e x)^{3+m}}{8 a^2 b^3 e^3 (3+m)}+\frac {(A b (b c (3-m)+a d (3+m))+a B (b c (1+m)-a d (7+m))) (e x)^{1+m} \left (c+d x^2\right )^2}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^3}{4 a b e \left (a+b x^2\right )^2}+\frac {(b c-a d) \left (A b \left (2 a b c d \left (3-2 m-m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )+a^2 d^2 \left (15+8 m+m^2\right )\right )+a B \left (b^2 c^2 \left (1-m^2\right )+2 a b c d \left (5+6 m+m^2\right )-a^2 d^2 \left (35+12 m+m^2\right )\right )\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{8 a^3 b^4 e (1+m)} \]

[Out]

-1/8*d*(A*b*(2*b^2*c^2*(-m^2+2*m+3)+3*a*b*c*d*(m^2+4*m+3)-a^2*d^2*(m^2+8*m+15))+a*B*(2*b^2*c^2*(1+m)^2-3*a*b*c
*d*(m^2+8*m+15)+a^2*d^2*(m^2+12*m+35)))*(e*x)^(1+m)/a^2/b^4/e/(1+m)-1/8*d^2*(A*b*(3+m)*(b*c*(3-m)+a*d*(5+m))+a
*B*(b*c*(m^2+4*m+3)-a*d*(m^2+12*m+35)))*(e*x)^(3+m)/a^2/b^3/e^3/(3+m)+1/8*(A*b*(b*c*(3-m)+a*d*(3+m))+a*B*(b*c*
(1+m)-a*d*(7+m)))*(e*x)^(1+m)*(d*x^2+c)^2/a^2/b^2/e/(b*x^2+a)+1/4*(A*b-B*a)*(e*x)^(1+m)*(d*x^2+c)^3/a/b/e/(b*x
^2+a)^2+1/8*(-a*d+b*c)*(A*b*(2*a*b*c*d*(-m^2-2*m+3)+b^2*c^2*(m^2-4*m+3)+a^2*d^2*(m^2+8*m+15))+a*B*(b^2*c^2*(-m
^2+1)+2*a*b*c*d*(m^2+6*m+5)-a^2*d^2*(m^2+12*m+35)))*(e*x)^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-b*x^2/a)
/a^3/b^4/e/(1+m)

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Rubi [A]
time = 0.76, antiderivative size = 480, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {d^2 (e x)^{m+3} \left (A b (m+3) (a d (m+5)+b c (3-m))+a B \left (b c \left (m^2+4 m+3\right )-a d \left (m^2+12 m+35\right )\right )\right )}{8 a^2 b^3 e^3 (m+3)}+\frac {\left (c+d x^2\right )^2 (e x)^{m+1} (A b (a d (m+3)+b c (3-m))+a B (b c (m+1)-a d (m+7)))}{8 a^2 b^2 e \left (a+b x^2\right )}-\frac {d (e x)^{m+1} \left (A b \left (-a^2 d^2 \left (m^2+8 m+15\right )+3 a b c d \left (m^2+4 m+3\right )+2 b^2 c^2 \left (-m^2+2 m+3\right )\right )+a B \left (a^2 d^2 \left (m^2+12 m+35\right )-3 a b c d \left (m^2+8 m+15\right )+2 b^2 c^2 (m+1)^2\right )\right )}{8 a^2 b^4 e (m+1)}+\frac {(e x)^{m+1} (b c-a d) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right ) \left (A b \left (a^2 d^2 \left (m^2+8 m+15\right )+2 a b c d \left (-m^2-2 m+3\right )+b^2 c^2 \left (m^2-4 m+3\right )\right )+a B \left (-a^2 d^2 \left (m^2+12 m+35\right )+2 a b c d \left (m^2+6 m+5\right )+b^2 c^2 \left (1-m^2\right )\right )\right )}{8 a^3 b^4 e (m+1)}+\frac {\left (c+d x^2\right )^3 (e x)^{m+1} (A b-a B)}{4 a b e \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(d*(A*b*(2*b^2*c^2*(3 + 2*m - m^2) + 3*a*b*c*d*(3 + 4*m + m^2) - a^2*d^2*(15 + 8*m + m^2)) + a*B*(2*b^2*c
^2*(1 + m)^2 - 3*a*b*c*d*(15 + 8*m + m^2) + a^2*d^2*(35 + 12*m + m^2)))*(e*x)^(1 + m))/(a^2*b^4*e*(1 + m)) - (
d^2*(A*b*(3 + m)*(b*c*(3 - m) + a*d*(5 + m)) + a*B*(b*c*(3 + 4*m + m^2) - a*d*(35 + 12*m + m^2)))*(e*x)^(3 + m
))/(8*a^2*b^3*e^3*(3 + m)) + ((A*b*(b*c*(3 - m) + a*d*(3 + m)) + a*B*(b*c*(1 + m) - a*d*(7 + m)))*(e*x)^(1 + m
)*(c + d*x^2)^2)/(8*a^2*b^2*e*(a + b*x^2)) + ((A*b - a*B)*(e*x)^(1 + m)*(c + d*x^2)^3)/(4*a*b*e*(a + b*x^2)^2)
+ ((b*c - a*d)*(A*b*(2*a*b*c*d*(3 - 2*m - m^2) + b^2*c^2*(3 - 4*m + m^2) + a^2*d^2*(15 + 8*m + m^2)) + a*B*(b
^2*c^2*(1 - m^2) + 2*a*b*c*d*(5 + 6*m + m^2) - a^2*d^2*(35 + 12*m + m^2)))*(e*x)^(1 + m)*Hypergeometric2F1[1,
(1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(8*a^3*b^4*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 ) \left (c+d x^2\right )^3}{\left (a+b x^2\right )^3} \, dx &=\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^3}{4 a b e \left (a+b x^2\right )^2}-\frac {\int \frac {(e x)^m \left (c+d x^2\right )^2 \left (-c (A b (3-m)+a B (1+m))+d (A b (3+m)-a B (7+m)) x^2\right )}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac {(A b (b c (3-m)+a d (3+m))+a B (b c (1+m)-a d (7+m))) (e x)^{1+m} \left (c+d x^2\right )^2}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^3}{4 a b e \left (a+b x^2\right )^2}+\frac {\int \frac {(e x)^m \left (c+d x^2\right ) \left (c \left (a B (1+m) (a d (7+m)+b (c-c m))+A b \left (b c \left (3-4 m+m^2\right )-a d \left (3+4 m+m^2\right )\right )\right )-d \left (A b (3+m) (b c (3-m)+a d (5+m))+a B \left (b c \left (3+4 m+m^2\right )-a d \left (35+12 m+m^2\right )\right )\right ) x^2\right )}{a+b x^2} \, dx}{8 a^2 b^2}\\ &=\frac {(A b (b c (3-m)+a d (3+m))+a B (b c (1+m)-a d (7+m))) (e x)^{1+m} \left (c+d x^2\right )^2}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^3}{4 a b e \left (a+b x^2\right )^2}+\frac {\int \left (-\frac {d \left (A b \left (2 b^2 c^2 \left (3+2 m-m^2\right )+3 a b c d \left (3+4 m+m^2\right )-a^2 d^2 \left (15+8 m+m^2\right )\right )+a B \left (2 b^2 c^2 (1+m)^2-3 a b c d \left (15+8 m+m^2\right )+a^2 d^2 \left (35+12 m+m^2\right )\right )\right ) (e x)^m}{b^2}-\frac {d^2 \left (A b (3+m) (b c (3-m)+a d (5+m))+a B \left (b c \left (3+4 m+m^2\right )-a d \left (35+12 m+m^2\right )\right )\right ) (e x)^{2+m}}{b e^2}+\frac {\left (3 A b^4 c^3+a b^3 B c^3+3 a A b^3 c^2 d+9 a^2 b^2 B c^2 d+9 a^2 A b^2 c d^2-45 a^3 b B c d^2-15 a^3 A b d^3+35 a^4 B d^3-4 A b^4 c^3 m+12 a^2 b^2 B c^2 d m+12 a^2 A b^2 c d^2 m-24 a^3 b B c d^2 m-8 a^3 A b d^3 m+12 a^4 B d^3 m+A b^4 c^3 m^2-a b^3 B c^3 m^2-3 a A b^3 c^2 d m^2+3 a^2 b^2 B c^2 d m^2+3 a^2 A b^2 c d^2 m^2-3 a^3 b B c d^2 m^2-a^3 A b d^3 m^2+a^4 B d^3 m^2\right ) (e x)^m}{b^2 \left (a+b x^2\right )}\right ) \, dx}{8 a^2 b^2}\\ &=-\frac {d \left (A b \left (2 b^2 c^2 \left (3+2 m-m^2\right )+3 a b c d \left (3+4 m+m^2\right )-a^2 d^2 \left (15+8 m+m^2\right )\right )+a B \left (2 b^2 c^2 (1+m)^2-3 a b c d \left (15+8 m+m^2\right )+a^2 d^2 \left (35+12 m+m^2\right )\right )\right ) (e x)^{1+m}}{8 a^2 b^4 e (1+m)}-\frac {d^2 \left (A b (3+m) (b c (3-m)+a d (5+m))+a B \left (b c \left (3+4 m+m^2\right )-a d \left (35+12 m+m^2\right )\right )\right ) (e x)^{3+m}}{8 a^2 b^3 e^3 (3+m)}+\frac {(A b (b c (3-m)+a d (3+m))+a B (b c (1+m)-a d (7+m))) (e x)^{1+m} \left (c+d x^2\right )^2}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^3}{4 a b e \left (a+b x^2\right )^2}+\frac {\left ((b c-a d) \left (A b \left (2 a b c d \left (3-2 m-m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )+a^2 d^2 \left (15+8 m+m^2\right )\right )+a B \left (b^2 c^2 \left (1-m^2\right )+2 a b c d \left (5+6 m+m^2\right )-a^2 d^2 \left (35+12 m+m^2\right )\right )\right )\right ) \int \frac {(e x)^m}{a+b x^2} \, dx}{8 a^2 b^4}\\ &=-\frac {d \left (A b \left (2 b^2 c^2 \left (3+2 m-m^2\right )+3 a b c d \left (3+4 m+m^2\right )-a^2 d^2 \left (15+8 m+m^2\right )\right )+a B \left (2 b^2 c^2 (1+m)^2-3 a b c d \left (15+8 m+m^2\right )+a^2 d^2 \left (35+12 m+m^2\right )\right )\right ) (e x)^{1+m}}{8 a^2 b^4 e (1+m)}-\frac {d^2 \left (A b (3+m) (b c (3-m)+a d (5+m))+a B \left (b c \left (3+4 m+m^2\right )-a d \left (35+12 m+m^2\right )\right )\right ) (e x)^{3+m}}{8 a^2 b^3 e^3 (3+m)}+\frac {(A b (b c (3-m)+a d (3+m))+a B (b c (1+m)-a d (7+m))) (e x)^{1+m} \left (c+d x^2\right )^2}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )^3}{4 a b e \left (a+b x^2\right )^2}+\frac {(b c-a d) \left (A b \left (2 a b c d \left (3-2 m-m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )+a^2 d^2 \left (15+8 m+m^2\right )\right )+a B \left (b^2 c^2 \left (1-m^2\right )+2 a b c d \left (5+6 m+m^2\right )-a^2 d^2 \left (35+12 m+m^2\right )\right )\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{8 a^3 b^4 e (1+m)}\\ \end {align*}

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Mathematica [A]
time = 1.03, size = 579, normalized size = 1.21 \begin {gather*} \frac {x (e x)^m \left (a^3 d^2 (3 b B c+A b d-3 a B d) (3+m)+a^3 b B d^3 (1+m) x^2+3 a^2 b^2 B c^2 d (3+m) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+3 a^2 A b^2 c d^2 (3+m) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )-9 a^3 b B c d^2 (3+m) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )-3 a^3 A b d^3 (3+m) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+6 a^4 B d^3 (3+m) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+a b^3 B c^3 (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+3 a A b^3 c^2 d (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )-6 a^2 b^2 B c^2 d (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )-6 a^2 A b^2 c d^2 (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+9 a^3 b B c d^2 (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+3 a^3 A b d^3 (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )-4 a^4 B d^3 (3+m) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+(A b-a B) (b c-a d)^3 (3+m) \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )\right )}{a^3 b^4 (1+m) (3+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*(a^3*d^2*(3*b*B*c + A*b*d - 3*a*B*d)*(3 + m) + a^3*b*B*d^3*(1 + m)*x^2 + 3*a^2*b^2*B*c^2*d*(3 + m)*
Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] + 3*a^2*A*b^2*c*d^2*(3 + m)*Hypergeometric2F1[1, (1 +
m)/2, (3 + m)/2, -((b*x^2)/a)] - 9*a^3*b*B*c*d^2*(3 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)
/a)] - 3*a^3*A*b*d^3*(3 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] + 6*a^4*B*d^3*(3 + m)*Hy
pergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] + a*b^3*B*c^3*(3 + m)*Hypergeometric2F1[2, (1 + m)/2, (
3 + m)/2, -((b*x^2)/a)] + 3*a*A*b^3*c^2*d*(3 + m)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] - 6
*a^2*b^2*B*c^2*d*(3 + m)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] - 6*a^2*A*b^2*c*d^2*(3 + m)*
Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] + 9*a^3*b*B*c*d^2*(3 + m)*Hypergeometric2F1[2, (1 + m
)/2, (3 + m)/2, -((b*x^2)/a)] + 3*a^3*A*b*d^3*(3 + m)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)]
- 4*a^4*B*d^3*(3 + m)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)] + (A*b - a*B)*(b*c - a*d)^3*(3
+ m)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)]))/(a^3*b^4*(1 + m)*(3 + m))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (B \,x^{2}+A \right ) \left (d \,x^{2}+c \right )^{3}}{\left (b \,x^{2}+a \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(B*x^2+A)*(d*x^2+c)^3/(b*x^2+a)^3,x)

[Out]

int((e*x)^m*(B*x^2+A)*(d*x^2+c)^3/(b*x^2+a)^3,x)

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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)*(d*x^2+c)^3/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(d*x^2 + c)^3*(x*e)^m/(b*x^2 + a)^3, x)

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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)*(d*x^2+c)^3/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

integral((B*d^3*x^8 + (3*B*c*d^2 + A*d^3)*x^6 + 3*(B*c^2*d + A*c*d^2)*x^4 + A*c^3 + (B*c^3 + 3*A*c^2*d)*x^2)*(
x*e)^m/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3), x)

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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 (c + d x^{2}\right )^{3}}{\left (a + b x^{2}\right )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(B*x**2+A)*(d*x**2+c)**3/(b*x**2+a)**3,x)

[Out]

Integral((e*x)**m*(A + B*x**2)*(c + d*x**2)**3/(a + b*x**2)**3, x)

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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)*(d*x^2+c)^3/(b*x^2+a)^3,x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*(d*x^2 + c)^3*(x*e)^m/(b*x^2 + a)^3, x)

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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 (d\,x^2+c\right )}^3}{{\left (b\,x^2+a\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(e*x)^m*(c + d*x^2)^3)/(a + b*x^2)^3,x)

[Out]

int(((A + B*x^2)*(e*x)^m*(c + d*x^2)^3)/(a + b*x^2)^3, x)

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