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

Optimal. Leaf size=180 \[ \frac{(e x)^{m+1} \left (a^2 B d^2-2 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 e (m+1)}-\frac{b (e x)^{m+3} (-2 a B d-A b d+b B c)}{d^2 e^3 (m+3)}-\frac{(e x)^{m+1} (b c-a d)^2 (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c d^3 e (m+1)}+\frac{b^2 B (e x)^{m+5}}{d e^5 (m+5)} \]

[Out]

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

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Rubi [A]  time = 0.187503, antiderivative size = 180, 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} \[ \frac{(e x)^{m+1} \left (a^2 B d^2-2 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 e (m+1)}-\frac{b (e x)^{m+3} (-2 a B d-A b d+b B c)}{d^2 e^3 (m+3)}-\frac{(e x)^{m+1} (b c-a d)^2 (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c d^3 e (m+1)}+\frac{b^2 B (e x)^{m+5}}{d e^5 (m+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.204872, size = 147, normalized size = 0.82 \[ \frac{x (e x)^m \left (\frac{a^2 B d^2+2 a b d (A d-B c)+b^2 c (B c-A d)}{m+1}-\frac{(b c-a d)^2 (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c (m+1)}+\frac{b d x^2 (2 a B d+A b d-b B c)}{m+3}+\frac{b^2 B d^2 x^4}{m+5}\right )}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.049, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( B{x}^{2}+A \right ) \left ( b{x}^{2}+a \right ) ^{2} \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)^2*(B*x^2+A)/(d*x^2+c),x)

[Out]

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

[Out]

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

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

[Out]

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

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Sympy [C]  time = 35.2755, size = 666, normalized size = 3.7 \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)**2*(B*x**2+A)/(d*x**2+c),x)

[Out]

A*a**2*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**2*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*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)/(2*c*gamma(m/2 + 5/
2)) + 3*A*a*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)/(2*c*gamma(m/2
+ 5/2)) + A*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)/(4*c*gamma
(m/2 + 7/2)) + 5*A*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)/(4*c*
gamma(m/2 + 7/2)) + B*a**2*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)/
(4*c*gamma(m/2 + 5/2)) + 3*B*a**2*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)/(4*c*gamma(m/2 + 5/2)) + B*a*b*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)) + 5*B*a*b*e**m*x**5*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 5/2)*gamm
a(m/2 + 5/2)/(2*c*gamma(m/2 + 7/2)) + B*b**2*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)/(4*c*gamma(m/2 + 9/2)) + 7*B*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)/(4*c*gamma(m/2 + 9/2))

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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 )}^{2} \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)^2*(B*x^2+A)/(d*x^2+c),x, algorithm="giac")

[Out]

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