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

Optimal. Leaf size=340 \[ -\frac{b (e x)^{m+1} \left (3 a^2 d^2 (A d (m+1)-B c (m+3))-3 a b c d (A d (m+3)-B c (m+5))+b^2 c^2 (A d (m+5)-B c (m+7))\right )}{2 c d^4 e (m+1)}-\frac{b^2 (e x)^{m+3} (3 a d (A d (m+3)-B c (m+5))-b c (A d (m+5)-B c (m+7)))}{2 c d^3 e^3 (m+3)}+\frac{(e x)^{m+1} (b c-a d)^2 \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) (a d (A d (1-m)+B c (m+1))+b c (A d (m+5)-B c (m+7)))}{2 c^2 d^4 e (m+1)}-\frac{\left (a+b x^2\right )^3 (e x)^{m+1} (B c-A d)}{2 c d e \left (c+d x^2\right )}-\frac{b^3 (e x)^{m+5} (A d (m+5)-B c (m+7))}{2 c d^2 e^5 (m+5)} \]

[Out]

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

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Rubi [A]  time = 0.715592, antiderivative size = 340, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {577, 570, 364} \[ -\frac{b (e x)^{m+1} \left (3 a^2 d^2 (A d (m+1)-B c (m+3))-3 a b c d (A d (m+3)-B c (m+5))+b^2 c^2 (A d (m+5)-B c (m+7))\right )}{2 c d^4 e (m+1)}-\frac{b^2 (e x)^{m+3} (3 a d (A d (m+3)-B c (m+5))-b c (A d (m+5)-B c (m+7)))}{2 c d^3 e^3 (m+3)}+\frac{(e x)^{m+1} (b c-a d)^2 \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) (a d (A d (1-m)+B c (m+1))+b c (A d (m+5)-B c (m+7)))}{2 c^2 d^4 e (m+1)}-\frac{\left (a+b x^2\right )^3 (e x)^{m+1} (B c-A d)}{2 c d e \left (c+d x^2\right )}-\frac{b^3 (e x)^{m+5} (A d (m+5)-B c (m+7))}{2 c d^2 e^5 (m+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(3*a^2*d^2*(A*d*(1 + m) - B*c*(3 + m)) - 3*a*b*c*d*(A*d*(3 + m) - B*c*(5 + m)) + b^2*c^2*(A*d*(5 + m) - B*
c*(7 + m)))*(e*x)^(1 + m))/(2*c*d^4*e*(1 + m)) - (b^2*(3*a*d*(A*d*(3 + m) - B*c*(5 + m)) - b*c*(A*d*(5 + m) -
B*c*(7 + m)))*(e*x)^(3 + m))/(2*c*d^3*e^3*(3 + m)) - (b^3*(A*d*(5 + m) - B*c*(7 + m))*(e*x)^(5 + m))/(2*c*d^2*
e^5*(5 + m)) - ((B*c - A*d)*(e*x)^(1 + m)*(a + b*x^2)^3)/(2*c*d*e*(c + d*x^2)) + ((b*c - a*d)^2*(a*d*(A*d*(1 -
m) + B*c*(1 + m)) + b*c*(A*d*(5 + m) - B*c*(7 + m)))*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2,
-((d*x^2)/c)])/(2*c^2*d^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 )^3 \left (A+B x^2\right )}{\left (c+d x^2\right )^2} \, dx &=-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^3}{2 c d e \left (c+d x^2\right )}-\frac{\int \frac{(e x)^m \left (a+b x^2\right )^2 \left (-a (A d (1-m)+B c (1+m))+b (A d (5+m)-B c (7+m)) x^2\right )}{c+d x^2} \, dx}{2 c d}\\ &=-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^3}{2 c d e \left (c+d x^2\right )}-\frac{\int \left (\frac{b \left (3 a^2 d^2 (A d (1+m)-B c (3+m))-3 a b c d (A d (3+m)-B c (5+m))+b^2 c^2 (A d (5+m)-B c (7+m))\right ) (e x)^m}{d^3}+\frac{b^2 (3 a d (A d (3+m)-B c (5+m))-b c (A d (5+m)-B c (7+m))) (e x)^{2+m}}{d^2 e^2}+\frac{b^3 (A d (5+m)-B c (7+m)) (e x)^{4+m}}{d e^4}+\frac{\left (7 b^3 B c^4-5 A b^3 c^3 d-15 a b^2 B c^3 d+9 a A b^2 c^2 d^2+9 a^2 b B c^2 d^2-3 a^2 A b c d^3-a^3 B c d^3-a^3 A d^4+b^3 B c^4 m-A b^3 c^3 d m-3 a b^2 B c^3 d m+3 a A b^2 c^2 d^2 m+3 a^2 b B c^2 d^2 m-3 a^2 A b c d^3 m-a^3 B c d^3 m+a^3 A d^4 m\right ) (e x)^m}{d^3 \left (c+d x^2\right )}\right ) \, dx}{2 c d}\\ &=-\frac{b \left (3 a^2 d^2 (A d (1+m)-B c (3+m))-3 a b c d (A d (3+m)-B c (5+m))+b^2 c^2 (A d (5+m)-B c (7+m))\right ) (e x)^{1+m}}{2 c d^4 e (1+m)}-\frac{b^2 (3 a d (A d (3+m)-B c (5+m))-b c (A d (5+m)-B c (7+m))) (e x)^{3+m}}{2 c d^3 e^3 (3+m)}-\frac{b^3 (A d (5+m)-B c (7+m)) (e x)^{5+m}}{2 c d^2 e^5 (5+m)}-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^3}{2 c d e \left (c+d x^2\right )}+\frac{\left ((b c-a d)^2 (a d (A d (1-m)+B c (1+m))+b c (A d (5+m)-B c (7+m)))\right ) \int \frac{(e x)^m}{c+d x^2} \, dx}{2 c d^4}\\ &=-\frac{b \left (3 a^2 d^2 (A d (1+m)-B c (3+m))-3 a b c d (A d (3+m)-B c (5+m))+b^2 c^2 (A d (5+m)-B c (7+m))\right ) (e x)^{1+m}}{2 c d^4 e (1+m)}-\frac{b^2 (3 a d (A d (3+m)-B c (5+m))-b c (A d (5+m)-B c (7+m))) (e x)^{3+m}}{2 c d^3 e^3 (3+m)}-\frac{b^3 (A d (5+m)-B c (7+m)) (e x)^{5+m}}{2 c d^2 e^5 (5+m)}-\frac{(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^3}{2 c d e \left (c+d x^2\right )}+\frac{(b c-a d)^2 (a d (A d (1-m)+B c (1+m))+b c (A d (5+m)-B c (7+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 d^4 e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.34182, size = 212, normalized size = 0.62 \[ \frac{x (e x)^m \left (\frac{b \left (3 a^2 B d^2+3 a b d (A d-2 B c)+b^2 c (3 B c-2 A d)\right )}{m+1}+\frac{b^2 d x^2 (3 a B d+A b d-2 b B c)}{m+3}+\frac{(b c-a d)^3 (B c-A d) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c^2 (m+1)}-\frac{(b c-a d)^2 \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right ) (-a B d-3 A b d+4 b B c)}{c (m+1)}+\frac{b^3 B d^2 x^4}{m+5}\right )}{d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*((b*(3*a^2*B*d^2 + b^2*c*(3*B*c - 2*A*d) + 3*a*b*d*(-2*B*c + A*d)))/(1 + m) + (b^2*d*(-2*b*B*c + A*
b*d + 3*a*B*d)*x^2)/(3 + m) + (b^3*B*d^2*x^4)/(5 + m) - ((b*c - a*d)^2*(4*b*B*c - 3*A*b*d - a*B*d)*Hypergeomet
ric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*(1 + m)) + ((b*c - a*d)^3*(B*c - A*d)*Hypergeometric2F1[2, (
1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c^2*(1 + m))))/d^4

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Maple [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( B{x}^{2}+A \right ) \left ( b{x}^{2}+a \right ) ^{3} \left ( ex \right ) ^{m}}{ \left ( d{x}^{2}+c \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*b^3*x^8 + (3*B*a*b^2 + A*b^3)*x^6 + 3*(B*a^2*b + A*a*b^2)*x^4 + A*a^3 + (B*a^3 + 3*A*a^2*b)*x^2)*(
e*x)^m/(d^2*x^4 + 2*c*d*x^2 + c^2), 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)**3*(B*x**2+A)/(d*x**2+c)**2,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 (b x^{2} + a\right )}^{3} \left (e x\right )^{m}}{{\left (d x^{2} + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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