3.22 \(\int \frac{(e x)^m (a+b x^2)^4 (A+B x^2)}{c+d x^2} \, dx\)

Optimal. Leaf size=363 \[ \frac{b (e x)^{m+3} \left (-6 a^2 b d^2 (B c-A d)+4 a^3 B d^3+4 a b^2 c d (B c-A d)+b^3 \left (-c^2\right ) (B c-A d)\right )}{d^4 e^3 (m+3)}+\frac{(e x)^{m+1} \left (6 a^2 b^2 c d^2 (B c-A d)-4 a^3 b d^3 (B c-A d)+a^4 B d^4-4 a b^3 c^2 d (B c-A d)+b^4 c^3 (B c-A d)\right )}{d^5 e (m+1)}+\frac{b^2 (e x)^{m+5} \left (6 a^2 B d^2-4 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 e^5 (m+5)}-\frac{b^3 (e x)^{m+7} (-4 a B d-A b d+b B c)}{d^2 e^7 (m+7)}-\frac{(e x)^{m+1} (b c-a d)^4 (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c d^5 e (m+1)}+\frac{b^4 B (e x)^{m+9}}{d e^9 (m+9)} \]

[Out]

((a^4*B*d^4 + b^4*c^3*(B*c - A*d) - 4*a*b^3*c^2*d*(B*c - A*d) + 6*a^2*b^2*c*d^2*(B*c - A*d) - 4*a^3*b*d^3*(B*c
- A*d))*(e*x)^(1 + m))/(d^5*e*(1 + m)) + (b*(4*a^3*B*d^3 - b^3*c^2*(B*c - A*d) + 4*a*b^2*c*d*(B*c - A*d) - 6*
a^2*b*d^2*(B*c - A*d))*(e*x)^(3 + m))/(d^4*e^3*(3 + m)) + (b^2*(6*a^2*B*d^2 + b^2*c*(B*c - A*d) - 4*a*b*d*(B*c
- A*d))*(e*x)^(5 + m))/(d^3*e^5*(5 + m)) - (b^3*(b*B*c - A*b*d - 4*a*B*d)*(e*x)^(7 + m))/(d^2*e^7*(7 + m)) +
(b^4*B*(e*x)^(9 + m))/(d*e^9*(9 + m)) - ((b*c - a*d)^4*(B*c - A*d)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/
2, (3 + m)/2, -((d*x^2)/c)])/(c*d^5*e*(1 + m))

________________________________________________________________________________________

Rubi [A]  time = 0.370923, antiderivative size = 363, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {570, 364} \[ \frac{b (e x)^{m+3} \left (-6 a^2 b d^2 (B c-A d)+4 a^3 B d^3+4 a b^2 c d (B c-A d)+b^3 \left (-c^2\right ) (B c-A d)\right )}{d^4 e^3 (m+3)}+\frac{(e x)^{m+1} \left (6 a^2 b^2 c d^2 (B c-A d)-4 a^3 b d^3 (B c-A d)+a^4 B d^4-4 a b^3 c^2 d (B c-A d)+b^4 c^3 (B c-A d)\right )}{d^5 e (m+1)}+\frac{b^2 (e x)^{m+5} \left (6 a^2 B d^2-4 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 e^5 (m+5)}-\frac{b^3 (e x)^{m+7} (-4 a B d-A b d+b B c)}{d^2 e^7 (m+7)}-\frac{(e x)^{m+1} (b c-a d)^4 (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c d^5 e (m+1)}+\frac{b^4 B (e x)^{m+9}}{d e^9 (m+9)} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^m*(a + b*x^2)^4*(A + B*x^2))/(c + d*x^2),x]

[Out]

((a^4*B*d^4 + b^4*c^3*(B*c - A*d) - 4*a*b^3*c^2*d*(B*c - A*d) + 6*a^2*b^2*c*d^2*(B*c - A*d) - 4*a^3*b*d^3*(B*c
- A*d))*(e*x)^(1 + m))/(d^5*e*(1 + m)) + (b*(4*a^3*B*d^3 - b^3*c^2*(B*c - A*d) + 4*a*b^2*c*d*(B*c - A*d) - 6*
a^2*b*d^2*(B*c - A*d))*(e*x)^(3 + m))/(d^4*e^3*(3 + m)) + (b^2*(6*a^2*B*d^2 + b^2*c*(B*c - A*d) - 4*a*b*d*(B*c
- A*d))*(e*x)^(5 + m))/(d^3*e^5*(5 + m)) - (b^3*(b*B*c - A*b*d - 4*a*B*d)*(e*x)^(7 + m))/(d^2*e^7*(7 + m)) +
(b^4*B*(e*x)^(9 + m))/(d*e^9*(9 + m)) - ((b*c - a*d)^4*(B*c - A*d)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/
2, (3 + m)/2, -((d*x^2)/c)])/(c*d^5*e*(1 + m))

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 )^4 \left (A+B x^2\right )}{c+d x^2} \, dx &=\int \left (\frac{\left (a^4 B d^4+b^4 c^3 (B c-A d)-4 a b^3 c^2 d (B c-A d)+6 a^2 b^2 c d^2 (B c-A d)-4 a^3 b d^3 (B c-A d)\right ) (e x)^m}{d^5}+\frac{b \left (4 a^3 B d^3-b^3 c^2 (B c-A d)+4 a b^2 c d (B c-A d)-6 a^2 b d^2 (B c-A d)\right ) (e x)^{2+m}}{d^4 e^2}+\frac{b^2 \left (6 a^2 B d^2+b^2 c (B c-A d)-4 a b d (B c-A d)\right ) (e x)^{4+m}}{d^3 e^4}-\frac{b^3 (b B c-A b d-4 a B d) (e x)^{6+m}}{d^2 e^6}+\frac{b^4 B (e x)^{8+m}}{d e^8}+\frac{\left (-b^4 B c^5+A b^4 c^4 d+4 a b^3 B c^4 d-4 a A b^3 c^3 d^2-6 a^2 b^2 B c^3 d^2+6 a^2 A b^2 c^2 d^3+4 a^3 b B c^2 d^3-4 a^3 A b c d^4-a^4 B c d^4+a^4 A d^5\right ) (e x)^m}{d^5 \left (c+d x^2\right )}\right ) \, dx\\ &=\frac{\left (a^4 B d^4+b^4 c^3 (B c-A d)-4 a b^3 c^2 d (B c-A d)+6 a^2 b^2 c d^2 (B c-A d)-4 a^3 b d^3 (B c-A d)\right ) (e x)^{1+m}}{d^5 e (1+m)}+\frac{b \left (4 a^3 B d^3-b^3 c^2 (B c-A d)+4 a b^2 c d (B c-A d)-6 a^2 b d^2 (B c-A d)\right ) (e x)^{3+m}}{d^4 e^3 (3+m)}+\frac{b^2 \left (6 a^2 B d^2+b^2 c (B c-A d)-4 a b d (B c-A d)\right ) (e x)^{5+m}}{d^3 e^5 (5+m)}-\frac{b^3 (b B c-A b d-4 a B d) (e x)^{7+m}}{d^2 e^7 (7+m)}+\frac{b^4 B (e x)^{9+m}}{d e^9 (9+m)}-\frac{\left ((b c-a d)^4 (B c-A d)\right ) \int \frac{(e x)^m}{c+d x^2} \, dx}{d^5}\\ &=\frac{\left (a^4 B d^4+b^4 c^3 (B c-A d)-4 a b^3 c^2 d (B c-A d)+6 a^2 b^2 c d^2 (B c-A d)-4 a^3 b d^3 (B c-A d)\right ) (e x)^{1+m}}{d^5 e (1+m)}+\frac{b \left (4 a^3 B d^3-b^3 c^2 (B c-A d)+4 a b^2 c d (B c-A d)-6 a^2 b d^2 (B c-A d)\right ) (e x)^{3+m}}{d^4 e^3 (3+m)}+\frac{b^2 \left (6 a^2 B d^2+b^2 c (B c-A d)-4 a b d (B c-A d)\right ) (e x)^{5+m}}{d^3 e^5 (5+m)}-\frac{b^3 (b B c-A b d-4 a B d) (e x)^{7+m}}{d^2 e^7 (7+m)}+\frac{b^4 B (e x)^{9+m}}{d e^9 (9+m)}-\frac{(b c-a d)^4 (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 d^5 e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.495889, size = 315, normalized size = 0.87 \[ \frac{x (e x)^m \left (\frac{b d x^2 \left (6 a^2 b d^2 (A d-B c)+4 a^3 B d^3+4 a b^2 c d (B c-A d)+b^3 c^2 (A d-B c)\right )}{m+3}+\frac{6 a^2 b^2 c d^2 (B c-A d)+4 a^3 b d^3 (A d-B c)+a^4 B d^4+4 a b^3 c^2 d (A d-B c)+b^4 c^3 (B c-A d)}{m+1}+\frac{b^2 d^2 x^4 \left (6 a^2 B d^2+4 a b d (A d-B c)+b^2 c (B c-A d)\right )}{m+5}+\frac{b^3 d^3 x^6 (4 a B d+A b d-b B c)}{m+7}-\frac{(b c-a d)^4 (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c (m+1)}+\frac{b^4 B d^4 x^8}{m+9}\right )}{d^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^m*(a + b*x^2)^4*(A + B*x^2))/(c + d*x^2),x]

[Out]

(x*(e*x)^m*((a^4*B*d^4 + b^4*c^3*(B*c - A*d) + 6*a^2*b^2*c*d^2*(B*c - A*d) + 4*a*b^3*c^2*d*(-(B*c) + A*d) + 4*
a^3*b*d^3*(-(B*c) + A*d))/(1 + m) + (b*d*(4*a^3*B*d^3 + 4*a*b^2*c*d*(B*c - A*d) + b^3*c^2*(-(B*c) + A*d) + 6*a
^2*b*d^2*(-(B*c) + A*d))*x^2)/(3 + m) + (b^2*d^2*(6*a^2*B*d^2 + b^2*c*(B*c - A*d) + 4*a*b*d*(-(B*c) + A*d))*x^
4)/(5 + m) + (b^3*d^3*(-(b*B*c) + A*b*d + 4*a*B*d)*x^6)/(7 + m) + (b^4*B*d^4*x^8)/(9 + m) - ((b*c - a*d)^4*(B*
c - A*d)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*(1 + m))))/d^5

________________________________________________________________________________________

Maple [F]  time = 0.042, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( B{x}^{2}+A \right ) \left ( b{x}^{2}+a \right ) ^{4} \left ( ex \right ) ^{m}}{d{x}^{2}+c}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(b*x^2+a)^4*(B*x^2+A)/(d*x^2+c),x)

[Out]

int((e*x)^m*(b*x^2+a)^4*(B*x^2+A)/(d*x^2+c),x)

________________________________________________________________________________________

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 )}^{4} \left (e x\right )^{m}}{d x^{2} + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(b*x^2+a)^4*(B*x^2+A)/(d*x^2+c),x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(b*x^2 + a)^4*(e*x)^m/(d*x^2 + c), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B b^{4} x^{10} +{\left (4 \, B a b^{3} + A b^{4}\right )} x^{8} + 2 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{6} + A a^{4} + 2 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{4} +{\left (B a^{4} + 4 \, A a^{3} b\right )} x^{2}\right )} \left (e x\right )^{m}}{d x^{2} + c}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(b*x^2+a)^4*(B*x^2+A)/(d*x^2+c),x, algorithm="fricas")

[Out]

integral((B*b^4*x^10 + (4*B*a*b^3 + A*b^4)*x^8 + 2*(3*B*a^2*b^2 + 2*A*a*b^3)*x^6 + A*a^4 + 2*(2*B*a^3*b + 3*A*
a^2*b^2)*x^4 + (B*a^4 + 4*A*a^3*b)*x^2)*(e*x)^m/(d*x^2 + c), x)

________________________________________________________________________________________

Sympy [C]  time = 103.325, size = 1132, normalized size = 3.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**m*(b*x**2+a)**4*(B*x**2+A)/(d*x**2+c),x)

[Out]

A*a**4*e**m*m*x*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(4*c*gamma(m/2 + 3/2))
+ A*a**4*e**m*x*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(4*c*gamma(m/2 + 3/2))
+ A*a**3*b*e**m*m*x**3*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(c*gamma(m/2 + 5
/2)) + 3*A*a**3*b*e**m*x**3*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(c*gamma(m/
2 + 5/2)) + 3*A*a**2*b**2*e**m*m*x**5*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(
2*c*gamma(m/2 + 7/2)) + 15*A*a**2*b**2*e**m*x**5*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 5/2)*gamma(m
/2 + 5/2)/(2*c*gamma(m/2 + 7/2)) + A*a*b**3*e**m*m*x**7*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 7/2)*
gamma(m/2 + 7/2)/(c*gamma(m/2 + 9/2)) + 7*A*a*b**3*e**m*x**7*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 +
7/2)*gamma(m/2 + 7/2)/(c*gamma(m/2 + 9/2)) + A*b**4*e**m*m*x**9*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2
+ 9/2)*gamma(m/2 + 9/2)/(4*c*gamma(m/2 + 11/2)) + 9*A*b**4*e**m*x**9*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c,
1, m/2 + 9/2)*gamma(m/2 + 9/2)/(4*c*gamma(m/2 + 11/2)) + B*a**4*e**m*m*x**3*x**m*lerchphi(d*x**2*exp_polar(I*p
i)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*c*gamma(m/2 + 5/2)) + 3*B*a**4*e**m*x**3*x**m*lerchphi(d*x**2*exp_pola
r(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*c*gamma(m/2 + 5/2)) + B*a**3*b*e**m*m*x**5*x**m*lerchphi(d*x**2*e
xp_polar(I*pi)/c, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(c*gamma(m/2 + 7/2)) + 5*B*a**3*b*e**m*x**5*x**m*lerchphi(d*x
**2*exp_polar(I*pi)/c, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(c*gamma(m/2 + 7/2)) + 3*B*a**2*b**2*e**m*m*x**7*x**m*le
rchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 7/2)*gamma(m/2 + 7/2)/(2*c*gamma(m/2 + 9/2)) + 21*B*a**2*b**2*e**m*x
**7*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 7/2)*gamma(m/2 + 7/2)/(2*c*gamma(m/2 + 9/2)) + B*a*b**3*e
**m*m*x**9*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 9/2)*gamma(m/2 + 9/2)/(c*gamma(m/2 + 11/2)) + 9*B*
a*b**3*e**m*x**9*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 9/2)*gamma(m/2 + 9/2)/(c*gamma(m/2 + 11/2))
+ B*b**4*e**m*m*x**11*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 11/2)*gamma(m/2 + 11/2)/(4*c*gamma(m/2
+ 13/2)) + 11*B*b**4*e**m*x**11*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 11/2)*gamma(m/2 + 11/2)/(4*c*
gamma(m/2 + 13/2))

________________________________________________________________________________________

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 )}^{4} \left (e x\right )^{m}}{d x^{2} + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^m*(b*x^2+a)^4*(B*x^2+A)/(d*x^2+c),x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*(b*x^2 + a)^4*(e*x)^m/(d*x^2 + c), x)