3.49 \(\int \frac {(e x)^m (a+b x^2)^p (A+B x^2)}{(c+d x^2)^3} \, dx\)

Optimal. Leaf size=483 \[ \frac {(e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a^2 d^2 (1-m) (A d (3-m)+B c (m+1))-2 a b c d (A d (1-m) (-m-2 p+3)+B c (m+1) (-m-2 p+1))+b^2 c^2 (-m-2 p+1) (A d (-m-2 p+3)+B c (m+2 p+1))\right ) F_1\left (\frac {m+1}{2};-p,1;\frac {m+3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{8 c^3 d e (m+1) (b c-a d)^2}-\frac {b (m+2 p+1) (e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {m+1}{2},-p;\frac {m+3}{2};-\frac {b x^2}{a}\right ) (a d (A d (3-m)+B c (m+1))+b c (B c (-m-2 p+1)-A d (-m-2 p+5)))}{8 c^2 d e (m+1) (b c-a d)^2}+\frac {(e x)^{m+1} \left (a+b x^2\right )^{p+1} (a d (A d (3-m)+B c (m+1))+b c (B c (-m-2 p+1)-A d (-m-2 p+5)))}{8 c^2 e \left (c+d x^2\right ) (b c-a d)^2}+\frac {(e x)^{m+1} \left (a+b x^2\right )^{p+1} (B c-A d)}{4 c e \left (c+d x^2\right )^2 (b c-a d)} \]

[Out]

1/4*(-A*d+B*c)*(e*x)^(1+m)*(b*x^2+a)^(1+p)/c/(-a*d+b*c)/e/(d*x^2+c)^2+1/8*(a*d*(A*d*(3-m)+B*c*(1+m))+b*c*(B*c*
(1-m-2*p)-A*d*(5-m-2*p)))*(e*x)^(1+m)*(b*x^2+a)^(1+p)/c^2/(-a*d+b*c)^2/e/(d*x^2+c)+1/8*(a^2*d^2*(1-m)*(A*d*(3-
m)+B*c*(1+m))-2*a*b*c*d*(B*c*(1+m)*(1-m-2*p)+A*d*(1-m)*(3-m-2*p))+b^2*c^2*(1-m-2*p)*(A*d*(3-m-2*p)+B*c*(1+m+2*
p)))*(e*x)^(1+m)*(b*x^2+a)^p*AppellF1(1/2+1/2*m,-p,1,3/2+1/2*m,-b*x^2/a,-d*x^2/c)/c^3/d/(-a*d+b*c)^2/e/(1+m)/(
(1+b*x^2/a)^p)-1/8*b*(a*d*(A*d*(3-m)+B*c*(1+m))+b*c*(B*c*(1-m-2*p)-A*d*(5-m-2*p)))*(1+m+2*p)*(e*x)^(1+m)*(b*x^
2+a)^p*hypergeom([-p, 1/2+1/2*m],[3/2+1/2*m],-b*x^2/a)/c^2/d/(-a*d+b*c)^2/e/(1+m)/((1+b*x^2/a)^p)

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Rubi [A]  time = 1.04, antiderivative size = 483, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {579, 584, 365, 364, 511, 510} \[ \frac {(e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (a^2 d^2 (1-m) (A d (3-m)+B c (m+1))-2 a b c d (A d (1-m) (-m-2 p+3)+B c (m+1) (-m-2 p+1))+b^2 c^2 (-m-2 p+1) (A d (-m-2 p+3)+B c (m+2 p+1))\right ) F_1\left (\frac {m+1}{2};-p,1;\frac {m+3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{8 c^3 d e (m+1) (b c-a d)^2}-\frac {b (m+2 p+1) (e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {m+1}{2},-p;\frac {m+3}{2};-\frac {b x^2}{a}\right ) (a d (A d (3-m)+B c (m+1))+b c (B c (-m-2 p+1)-A d (-m-2 p+5)))}{8 c^2 d e (m+1) (b c-a d)^2}+\frac {(e x)^{m+1} \left (a+b x^2\right )^{p+1} (a d (A d (3-m)+B c (m+1))+b c (B c (-m-2 p+1)-A d (-m-2 p+5)))}{8 c^2 e \left (c+d x^2\right ) (b c-a d)^2}+\frac {(e x)^{m+1} \left (a+b x^2\right )^{p+1} (B c-A d)}{4 c e \left (c+d x^2\right )^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((B*c - A*d)*(e*x)^(1 + m)*(a + b*x^2)^(1 + p))/(4*c*(b*c - a*d)*e*(c + d*x^2)^2) + ((a*d*(A*d*(3 - m) + B*c*(
1 + m)) + b*c*(B*c*(1 - m - 2*p) - A*d*(5 - m - 2*p)))*(e*x)^(1 + m)*(a + b*x^2)^(1 + p))/(8*c^2*(b*c - a*d)^2
*e*(c + d*x^2)) + ((a^2*d^2*(1 - m)*(A*d*(3 - m) + B*c*(1 + m)) - 2*a*b*c*d*(B*c*(1 + m)*(1 - m - 2*p) + A*d*(
1 - m)*(3 - m - 2*p)) + b^2*c^2*(1 - m - 2*p)*(A*d*(3 - m - 2*p) + B*c*(1 + m + 2*p)))*(e*x)^(1 + m)*(a + b*x^
2)^p*AppellF1[(1 + m)/2, -p, 1, (3 + m)/2, -((b*x^2)/a), -((d*x^2)/c)])/(8*c^3*d*(b*c - a*d)^2*e*(1 + m)*(1 +
(b*x^2)/a)^p) - (b*(a*d*(A*d*(3 - m) + B*c*(1 + m)) + b*c*(B*c*(1 - m - 2*p) - A*d*(5 - m - 2*p)))*(1 + m + 2*
p)*(e*x)^(1 + m)*(a + b*x^2)^p*Hypergeometric2F1[(1 + m)/2, -p, (3 + m)/2, -((b*x^2)/a)])/(8*c^2*d*(b*c - a*d)
^2*e*(1 + m)*(1 + (b*x^2)/a)^p)

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 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(e*x)^m*(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || 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 )^p \left (A+B x^2\right )}{\left (c+d x^2\right )^3} \, dx &=\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {\int \frac {(e x)^m \left (a+b x^2\right )^p \left (4 A b c-a A d (3-m)-a B c (1+m)+b (B c-A d) (1-m-2 p) x^2\right )}{\left (c+d x^2\right )^2} \, dx}{4 c (b c-a d)}\\ &=\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac {\int \frac {(e x)^m \left (a+b x^2\right )^p \left (a B c (1+m) (a d (1-m)-b c (3-m-2 p))+A \left (8 b^2 c^2+a^2 d^2 \left (3-4 m+m^2\right )-a b c d \left (9+m^2-2 m (3-p)+2 p\right )\right )+b (d (4 A b c-a A d (3-m)-a B c (1+m))-b c (B c-A d) (1-m-2 p)) (1+m+2 p) x^2\right )}{c+d x^2} \, dx}{8 c^2 (b c-a d)^2}\\ &=\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac {\int \left (\frac {b (d (4 A b c-a A d (3-m)-a B c (1+m))-b c (B c-A d) (1-m-2 p)) (1+m+2 p) (e x)^m \left (a+b x^2\right )^p}{d}+\frac {\left (-b c (d (4 A b c-a A d (3-m)-a B c (1+m))-b c (B c-A d) (1-m-2 p)) (1+m+2 p)+d \left (a B c (1+m) (a d (1-m)-b c (3-m-2 p))+A \left (8 b^2 c^2+a^2 d^2 \left (3-4 m+m^2\right )-a b c d \left (9+m^2-2 m (3-p)+2 p\right )\right )\right )\right ) (e x)^m \left (a+b x^2\right )^p}{d \left (c+d x^2\right )}\right ) \, dx}{8 c^2 (b c-a d)^2}\\ &=\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}-\frac {(b (a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (1+m+2 p)) \int (e x)^m \left (a+b x^2\right )^p \, dx}{8 c^2 d (b c-a d)^2}+\frac {\left (a^2 d^2 (1-m) (A d (3-m)+B c (1+m))-2 a b c d (B c (1+m) (1-m-2 p)+A d (1-m) (3-m-2 p))+b^2 c^2 (1-m-2 p) (A d (3-m-2 p)+B c (1+m+2 p))\right ) \int \frac {(e x)^m \left (a+b x^2\right )^p}{c+d x^2} \, dx}{8 c^2 d (b c-a d)^2}\\ &=\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}-\frac {\left (b (a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (1+m+2 p) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int (e x)^m \left (1+\frac {b x^2}{a}\right )^p \, dx}{8 c^2 d (b c-a d)^2}+\frac {\left (\left (a^2 d^2 (1-m) (A d (3-m)+B c (1+m))-2 a b c d (B c (1+m) (1-m-2 p)+A d (1-m) (3-m-2 p))+b^2 c^2 (1-m-2 p) (A d (3-m-2 p)+B c (1+m+2 p))\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \frac {(e x)^m \left (1+\frac {b x^2}{a}\right )^p}{c+d x^2} \, dx}{8 c^2 d (b c-a d)^2}\\ &=\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{4 c (b c-a d) e \left (c+d x^2\right )^2}+\frac {(a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{8 c^2 (b c-a d)^2 e \left (c+d x^2\right )}+\frac {\left (a^2 d^2 (1-m) (A d (3-m)+B c (1+m))-2 a b c d (B c (1+m) (1-m-2 p)+A d (1-m) (3-m-2 p))+b^2 c^2 (1-m-2 p) (A d (3-m-2 p)+B c (1+m+2 p))\right ) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1+m}{2};-p,1;\frac {3+m}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{8 c^3 d (b c-a d)^2 e (1+m)}-\frac {b (a d (A d (3-m)+B c (1+m))+b c (B c (1-m-2 p)-A d (5-m-2 p))) (1+m+2 p) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1+m}{2},-p;\frac {3+m}{2};-\frac {b x^2}{a}\right )}{8 c^2 d (b c-a d)^2 e (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.51, size = 128, normalized size = 0.27 \[ \frac {x (e x)^m \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left ((A d-B c) F_1\left (\frac {m+1}{2};-p,3;\frac {m+3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+B c F_1\left (\frac {m+1}{2};-p,2;\frac {m+3}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}{c^3 d (m+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(e*x)^m*(a + b*x^2)^p*(B*c*AppellF1[(1 + m)/2, -p, 2, (3 + m)/2, -((b*x^2)/a), -((d*x^2)/c)] + (-(B*c) + A*
d)*AppellF1[(1 + m)/2, -p, 3, (3 + m)/2, -((b*x^2)/a), -((d*x^2)/c)]))/(c^3*d*(1 + m)*(1 + (b*x^2)/a)^p)

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fricas [F]  time = 1.00, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{m}}{d^{3} x^{6} + 3 \, c d^{2} x^{4} + 3 \, c^{2} d x^{2} + c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\left (B \,x^{2}+A \right ) \left (e x \right )^{m} \left (b \,x^{2}+a \right )^{p}}{\left (d \,x^{2}+c \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,{\left (b\,x^2+a\right )}^p}{{\left (d\,x^2+c\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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