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

Optimal. Leaf size=304 \[ \frac {b (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right ) (A b (b c (1-m)-a d (5-m))+a B (a d (3-m)+b c (m+1)))}{2 a^2 e (m+1) (b c-a d)^3}-\frac {d (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right ) (a d (A d (1-m)+B c (m+1))+b c (B c (3-m)-A d (5-m)))}{2 c^2 e (m+1) (b c-a d)^3}+\frac {d (e x)^{m+1} (a A d-2 a B c+A b c)}{2 a c e \left (c+d x^2\right ) (b c-a d)^2}+\frac {(e x)^{m+1} (A b-a B)}{2 a e \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.80, antiderivative size = 304, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {579, 584, 364} \[ \frac {b (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right ) (A b (b c (1-m)-a d (5-m))+a B (a d (3-m)+b c (m+1)))}{2 a^2 e (m+1) (b c-a d)^3}-\frac {d (e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right ) (a d (A d (1-m)+B c (m+1))+b c (B c (3-m)-A d (5-m)))}{2 c^2 e (m+1) (b c-a d)^3}+\frac {d (e x)^{m+1} (a A d-2 a B c+A b c)}{2 a c e \left (c+d x^2\right ) (b c-a d)^2}+\frac {(e x)^{m+1} (A b-a B)}{2 a e \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(A*b*c - 2*a*B*c + a*A*d)*(e*x)^(1 + m))/(2*a*c*(b*c - a*d)^2*e*(c + d*x^2)) + ((A*b - a*B)*(e*x)^(1 + m))/
(2*a*(b*c - a*d)*e*(a + b*x^2)*(c + d*x^2)) + (b*(A*b*(b*c*(1 - m) - a*d*(5 - m)) + a*B*(a*d*(3 - m) + b*c*(1
+ m)))*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(2*a^2*(b*c - a*d)^3*e*(1 + m))
- (d*(b*c*(B*c*(3 - m) - A*d*(5 - m)) + a*d*(A*d*(1 - m) + B*c*(1 + m)))*(e*x)^(1 + m)*Hypergeometric2F1[1, (
1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(2*c^2*(b*c - a*d)^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 579

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 + 1))/(a*g*n*(b*c - a*d)*(p +
1)), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)*(
m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 584

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

Rubi steps

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

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Mathematica [A]  time = 0.27, size = 207, normalized size = 0.68 \[ \frac {x (e x)^m \left (-(b c-a d) \left (a^2 d (A d-B c) \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )+b c^2 (A b-a B) \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )\right )+a^2 c d \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right ) (a B d-2 A b d+b B c)-a b c^2 \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right ) (a B d-2 A b d+b B c)\right )}{a^2 c^2 (m+1) (a d-b c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{b^{2} d^{2} x^{8} + 2 \, {\left (b^{2} c d + a b d^{2}\right )} x^{6} + {\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} x^{4} + a^{2} c^{2} + 2 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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