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

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

[Out]

-(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))/(8*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))

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Rubi [A]  time = 1.07282, antiderivative size = 480, normalized size of antiderivative = 1., 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 = {577, 570, 364} \[ \frac{(e x)^{m+1} (b c-a d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) \left (A b \left (a^2 d^2 \left (m^2+8 m+15\right )+2 a b c d \left (-m^2-2 m+3\right )+b^2 c^2 \left (m^2-4 m+3\right )\right )+a B \left (-a^2 d^2 \left (m^2+12 m+35\right )+2 a b c d \left (m^2+6 m+5\right )+b^2 c^2 \left (1-m^2\right )\right )\right )}{8 a^3 b^4 e (m+1)}-\frac{d (e x)^{m+1} \left (A b \left (-a^2 d^2 \left (m^2+8 m+15\right )+3 a b c d \left (m^2+4 m+3\right )+2 b^2 c^2 \left (-m^2+2 m+3\right )\right )+a B \left (a^2 d^2 \left (m^2+12 m+35\right )-3 a b c d \left (m^2+8 m+15\right )+2 b^2 c^2 (m+1)^2\right )\right )}{8 a^2 b^4 e (m+1)}-\frac{d^2 (e x)^{m+3} \left (A b (m+3) (a d (m+5)+b c (3-m))+a B \left (b c \left (m^2+4 m+3\right )-a d \left (m^2+12 m+35\right )\right )\right )}{8 a^2 b^3 e^3 (m+3)}+\frac{\left (c+d x^2\right )^2 (e x)^{m+1} (A b (a d (m+3)+b c (3-m))+a B (b c (m+1)-a d (m+7)))}{8 a^2 b^2 e \left (a+b x^2\right )}+\frac{\left (c+d x^2\right )^3 (e x)^{m+1} (A b-a B)}{4 a b e \left (a+b x^2\right )^2} \]

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]

-(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))/(8*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 577

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] + Dist[
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])

Rule 570

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 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])

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*}

Mathematica [A]  time = 0.301038, size = 218, normalized size = 0.45 \[ \frac{x (e x)^m \left (\frac{(b c-a d)^2 \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) (-4 a B d+3 A b d+b B c)}{a^2 (m+1)}+\frac{(a B-A b) (a d-b c)^3 \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{a^3 (m+1)}+\frac{d^2 (-3 a B d+A b d+3 b B c)}{m+1}+\frac{3 d (b c-a d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right ) (-2 a B d+A b d+b B c)}{a (m+1)}+\frac{b B d^3 x^2}{m+3}\right )}{b^4} \]

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

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex \right ) ^{m} \left ( B{x}^{2}+A \right ) \left ( d{x}^{2}+c \right ) ^{3}}{ \left ( b{x}^{2}+a \right ) ^{3}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )}{\left (d x^{2} + c\right )}^{3} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{3}}\,{d x} \end{align*}

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*(e*x)^m/(b*x^2 + a)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B d^{3} x^{8} +{\left (3 \, B c d^{2} + A d^{3}\right )} x^{6} + 3 \,{\left (B c^{2} d + A c d^{2}\right )} x^{4} + A c^{3} +{\left (B c^{3} + 3 \, A c^{2} d\right )} x^{2}\right )} \left (e x\right )^{m}}{b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}}, x\right ) \end{align*}

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)*(
e*x)^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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B x^{2} + A\right )}{\left (d x^{2} + c\right )}^{3} \left (e x\right )^{m}}{{\left (b x^{2} + a\right )}^{3}}\,{d x} \end{align*}

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*(e*x)^m/(b*x^2 + a)^3, x)