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

Optimal. Leaf size=260 \[ \frac{(e x)^{m+1} \left (-3 a^2 b d^2 (B c-A d)+a^3 B d^3+3 a b^2 c d (B c-A d)+b^3 \left (-c^2\right ) (B c-A d)\right )}{d^4 e (m+1)}+\frac{b (e x)^{m+3} \left (3 a^2 B d^2-3 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 e^3 (m+3)}-\frac{b^2 (e x)^{m+5} (-3 a B d-A b d+b B c)}{d^2 e^5 (m+5)}+\frac{(e x)^{m+1} (b c-a d)^3 (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c d^4 e (m+1)}+\frac{b^3 B (e x)^{m+7}}{d e^7 (m+7)} \]

[Out]

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

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Rubi [A]  time = 0.295017, antiderivative size = 260, 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} \[ \frac{(e x)^{m+1} \left (-3 a^2 b d^2 (B c-A d)+a^3 B d^3+3 a b^2 c d (B c-A d)+b^3 \left (-c^2\right ) (B c-A d)\right )}{d^4 e (m+1)}+\frac{b (e x)^{m+3} \left (3 a^2 B d^2-3 a b d (B c-A d)+b^2 c (B c-A d)\right )}{d^3 e^3 (m+3)}-\frac{b^2 (e x)^{m+5} (-3 a B d-A b d+b B c)}{d^2 e^5 (m+5)}+\frac{(e x)^{m+1} (b c-a d)^3 (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c d^4 e (m+1)}+\frac{b^3 B (e x)^{m+7}}{d e^7 (m+7)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^3*B*d^3 - b^3*c^2*(B*c - A*d) + 3*a*b^2*c*d*(B*c - A*d) - 3*a^2*b*d^2*(B*c - A*d))*(e*x)^(1 + m))/(d^4*e*(
1 + m)) + (b*(3*a^2*B*d^2 + b^2*c*(B*c - A*d) - 3*a*b*d*(B*c - A*d))*(e*x)^(3 + m))/(d^3*e^3*(3 + m)) - (b^2*(
b*B*c - A*b*d - 3*a*B*d)*(e*x)^(5 + m))/(d^2*e^5*(5 + m)) + (b^3*B*(e*x)^(7 + m))/(d*e^7*(7 + m)) + ((b*c - a*
d)^3*(B*c - A*d)*(e*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*d^4*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 )^3 \left (A+B x^2\right )}{c+d x^2} \, dx &=\int \left (\frac{\left (a^3 B d^3-b^3 c^2 (B c-A d)+3 a b^2 c d (B c-A d)-3 a^2 b d^2 (B c-A d)\right ) (e x)^m}{d^4}+\frac{b \left (3 a^2 B d^2+b^2 c (B c-A d)-3 a b d (B c-A d)\right ) (e x)^{2+m}}{d^3 e^2}-\frac{b^2 (b B c-A b d-3 a B d) (e x)^{4+m}}{d^2 e^4}+\frac{b^3 B (e x)^{6+m}}{d e^6}+\frac{\left (b^3 B c^4-A b^3 c^3 d-3 a b^2 B c^3 d+3 a A b^2 c^2 d^2+3 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\right ) (e x)^m}{d^4 \left (c+d x^2\right )}\right ) \, dx\\ &=\frac{\left (a^3 B d^3-b^3 c^2 (B c-A d)+3 a b^2 c d (B c-A d)-3 a^2 b d^2 (B c-A d)\right ) (e x)^{1+m}}{d^4 e (1+m)}+\frac{b \left (3 a^2 B d^2+b^2 c (B c-A d)-3 a b d (B c-A d)\right ) (e x)^{3+m}}{d^3 e^3 (3+m)}-\frac{b^2 (b B c-A b d-3 a B d) (e x)^{5+m}}{d^2 e^5 (5+m)}+\frac{b^3 B (e x)^{7+m}}{d e^7 (7+m)}+\frac{\left ((b c-a d)^3 (B c-A d)\right ) \int \frac{(e x)^m}{c+d x^2} \, dx}{d^4}\\ &=\frac{\left (a^3 B d^3-b^3 c^2 (B c-A d)+3 a b^2 c d (B c-A d)-3 a^2 b d^2 (B c-A d)\right ) (e x)^{1+m}}{d^4 e (1+m)}+\frac{b \left (3 a^2 B d^2+b^2 c (B c-A d)-3 a b d (B c-A d)\right ) (e x)^{3+m}}{d^3 e^3 (3+m)}-\frac{b^2 (b B c-A b d-3 a B d) (e x)^{5+m}}{d^2 e^5 (5+m)}+\frac{b^3 B (e x)^{7+m}}{d e^7 (7+m)}+\frac{(b c-a d)^3 (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^4 e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.312702, size = 219, normalized size = 0.84 \[ \frac{x (e x)^m \left (\frac{3 a^2 b d^2 (A d-B c)+a^3 B d^3+3 a b^2 c d (B c-A d)+b^3 c^2 (A d-B c)}{m+1}+\frac{b d x^2 \left (3 a^2 B d^2+3 a b d (A d-B c)+b^2 c (B c-A d)\right )}{m+3}+\frac{b^2 d^2 x^4 (3 a B d+A b d-b B c)}{m+5}+\frac{(b c-a d)^3 (B c-A d) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c (m+1)}+\frac{b^3 B d^3 x^6}{m+7}\right )}{d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.037, 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}}{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)^3*(B*x^2+A)/(d*x^2+c),x)

[Out]

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

[Out]

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

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Sympy [C]  time = 53.9713, size = 911, normalized size = 3.5 \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)**3*(B*x**2+A)/(d*x**2+c),x)

[Out]

A*a**3*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**3*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))
+ 3*A*a**2*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)/(4*c*gamma(m/2
+ 5/2)) + 9*A*a**2*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)/(4*c*ga
mma(m/2 + 5/2)) + 3*A*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)) + 15*A*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)) + A*b**3*e**m*m*x**7*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 7/2)*ga
mma(m/2 + 7/2)/(4*c*gamma(m/2 + 9/2)) + 7*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)/(4*c*gamma(m/2 + 9/2)) + B*a**3*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**3*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)) + 3*B*a**2*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)/(4*c*gamma(m/2 + 7/2)) + 15*B*a**2*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)/(4*c*gamma(m/2 + 7/2)) + 3*B*a*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)) + 21*B*a*b**2*e**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)/(4*c*gamma(m/2 + 9/2)) + B*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)/(4*c*gamma(m/2 + 11/2)) + 9*B*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)/(4*c*gamma(m/2 + 11/2))

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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 )}^{3} \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)^3*(B*x^2+A)/(d*x^2+c),x, algorithm="giac")

[Out]

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