3.1.29 \(\int \frac {(e x)^m (A+B x^2)}{(a+b x^2)^3 (c+d x^2)} \, dx\) [29]

Optimal. Leaf size=342 \[ \frac {(A b-a B) (e x)^{1+m}}{4 a (b c-a d) e \left (a+b x^2\right )^2}+\frac {(A b (b c (3-m)-a d (7-m))+a B (a d (3-m)+b c (1+m))) (e x)^{1+m}}{8 a^2 (b c-a d)^2 e \left (a+b x^2\right )}+\frac {\left (A b \left (a^2 d^2 \left (15-8 m+m^2\right )-2 a b c d \left (5-6 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )+a B \left (b^2 c^2 \left (1-m^2\right )-2 a b c d \left (3+2 m-m^2\right )-a^2 d^2 \left (3-4 m+m^2\right )\right )\right ) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{8 a^3 (b c-a d)^3 e (1+m)}+\frac {d^2 (B c-A d) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{c (b c-a d)^3 e (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.54, antiderivative size = 342, 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 = {593, 598, 371} \begin {gather*} \frac {(e x)^{m+1} (A b (b c (3-m)-a d (7-m))+a B (a d (3-m)+b c (m+1)))}{8 a^2 e \left (a+b x^2\right ) (b c-a d)^2}+\frac {(e x)^{m+1} \, _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^2-6 m+5\right )+b^2 c^2 \left (m^2-4 m+3\right )\right )+a B \left (-a^2 d^2 \left (m^2-4 m+3\right )-2 a b c d \left (-m^2+2 m+3\right )+b^2 c^2 \left (1-m^2\right )\right )\right )}{8 a^3 e (m+1) (b c-a d)^3}+\frac {d^2 (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 )}{c e (m+1) (b c-a d)^3}+\frac {(e x)^{m+1} (A b-a B)}{4 a e \left (a+b x^2\right )^2 (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*(e*x)^(1 + m))/(4*a*(b*c - a*d)*e*(a + b*x^2)^2) + ((A*b*(b*c*(3 - m) - a*d*(7 - m)) + a*B*(a*d*(
3 - m) + b*c*(1 + m)))*(e*x)^(1 + m))/(8*a^2*(b*c - a*d)^2*e*(a + b*x^2)) + ((A*b*(a^2*d^2*(15 - 8*m + m^2) -
2*a*b*c*d*(5 - 6*m + m^2) + b^2*c^2*(3 - 4*m + m^2)) + a*B*(b^2*c^2*(1 - m^2) - 2*a*b*c*d*(3 + 2*m - m^2) - a^
2*d^2*(3 - 4*m + m^2)))*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(8*a^3*(b*c -
a*d)^3*e*(1 + m)) + (d^2*(B*c - A*d)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(
c*(b*c - a*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 593

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 598

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

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Mathematica [A]
time = 0.94, size = 195, normalized size = 0.57 \begin {gather*} \frac {x (e x)^m \left (\frac {b d (-B c+A d) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{a}+\frac {d^2 (B c-A d) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{c}+\frac {b (b c-a d) (B c-A d) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{a^2}+\frac {(A b-a B) (b c-a d)^2 \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{a^3}\right )}{(b c-a d)^3 (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (B \,x^{2}+A \right )}{\left (b \,x^{2}+a \right )^{3} \left (d \,x^{2}+c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

[Out]

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

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