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

Optimal. Leaf size=205 \[ \frac {(e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right ) (a d (A d (1-m)+B c (m+1))+b c (B c (1-m)-A d (3-m)))}{2 c^2 e (m+1) (b c-a d)^2}+\frac {b (e x)^{m+1} (A b-a B) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a e (m+1) (b c-a d)^2}+\frac {(e x)^{m+1} (B c-A d)}{2 c e \left (c+d x^2\right ) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 205, 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 = {579, 584, 364} \[ \frac {(e x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right ) (a d (A d (1-m)+B c (m+1))+b c (B c (1-m)-A d (3-m)))}{2 c^2 e (m+1) (b c-a d)^2}+\frac {b (e x)^{m+1} (A b-a B) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a e (m+1) (b c-a d)^2}+\frac {(e x)^{m+1} (B c-A d)}{2 c e \left (c+d x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.17, size = 147, normalized size = 0.72 \[ \frac {x (e x)^m \left (b c^2 (A b-a B) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )+a c d (a B-A b) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )+a (b c-a d) (B c-A d) \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )\right )}{a c^2 (m+1) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{2} + A\right )} \left (e x\right )^{m}}{b d^{2} x^{6} + {\left (2 \, b c d + a d^{2}\right )} x^{4} + a c^{2} + {\left (b c^{2} + 2 \, a c d\right )} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*x^2 + A)*(e*x)^m/(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), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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