3.1.32 \(\int \frac {(e x)^m (a+b x^2) (A+B x^2)}{(c+d x^2)^2} \, dx\) [32]

Optimal. Leaf size=171 \[ -\frac {B (a d (1+m)-b c (3+m)) (e x)^{1+m}}{2 c d^2 e (1+m)}-\frac {(b c-a d) (e x)^{1+m} \left (A+B x^2\right )}{2 c d e \left (c+d x^2\right )}+\frac {(a d (A d (1-m)+B c (1+m))+b c (A d (1+m)-B c (3+m))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{2 c^2 d^2 e (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {591, 470, 371} \begin {gather*} \frac {(e x)^{m+1} \, _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+1)-B c (m+3)))}{2 c^2 d^2 e (m+1)}-\frac {\left (A+B x^2\right ) (e x)^{m+1} (b c-a d)}{2 c d e \left (c+d x^2\right )}-\frac {B (e x)^{m+1} (a d (m+1)-b c (m+3))}{2 c d^2 e (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(B*(a*d*(1 + m) - b*c*(3 + m))*(e*x)^(1 + m))/(c*d^2*e*(1 + m)) - ((b*c - a*d)*(e*x)^(1 + m)*(A + B*x^2))
/(2*c*d*e*(c + d*x^2)) + ((a*d*(A*d*(1 - m) + B*c*(1 + m)) + b*c*(A*d*(1 + m) - B*c*(3 + m)))*(e*x)^(1 + m)*Hy
pergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(2*c^2*d^2*e*(1 + m))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 591

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] + Dis
t[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])

Rubi steps

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

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Mathematica [A]
time = 0.28, size = 113, normalized size = 0.66 \begin {gather*} -\frac {x (e x)^m \left (-b B c^2+c (2 b B c-A b d-a B d) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )-(b c-a d) (B c-A d) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )\right )}{c^2 d^2 (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right ) \left (B \,x^{2}+A \right )}{\left (d \,x^{2}+c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 30.70, size = 2076, normalized size = 12.14 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a*(-c*e**m*m**2*x*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*c**3*gamma(m/2 +
3/2) + 8*c**2*d*x**2*gamma(m/2 + 3/2)) + 2*c*e**m*m*x*x**m*gamma(m/2 + 1/2)/(8*c**3*gamma(m/2 + 3/2) + 8*c**2
*d*x**2*gamma(m/2 + 3/2)) + c*e**m*x*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8
*c**3*gamma(m/2 + 3/2) + 8*c**2*d*x**2*gamma(m/2 + 3/2)) + 2*c*e**m*x*x**m*gamma(m/2 + 1/2)/(8*c**3*gamma(m/2
+ 3/2) + 8*c**2*d*x**2*gamma(m/2 + 3/2)) - d*e**m*m**2*x**3*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 1
/2)*gamma(m/2 + 1/2)/(8*c**3*gamma(m/2 + 3/2) + 8*c**2*d*x**2*gamma(m/2 + 3/2)) + d*e**m*x**3*x**m*lerchphi(d*
x**2*exp_polar(I*pi)/c, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*c**3*gamma(m/2 + 3/2) + 8*c**2*d*x**2*gamma(m/2 + 3/
2))) + A*b*(-c*e**m*m**2*x**3*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*c**3*g
amma(m/2 + 5/2) + 8*c**2*d*x**2*gamma(m/2 + 5/2)) - 4*c*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)/(8*c**3*gamma(m/2 + 5/2) + 8*c**2*d*x**2*gamma(m/2 + 5/2)) + 2*c*e**m*m*x**3*x**m
*gamma(m/2 + 3/2)/(8*c**3*gamma(m/2 + 5/2) + 8*c**2*d*x**2*gamma(m/2 + 5/2)) - 3*c*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)/(8*c**3*gamma(m/2 + 5/2) + 8*c**2*d*x**2*gamma(m/2 + 5/2
)) + 6*c*e**m*x**3*x**m*gamma(m/2 + 3/2)/(8*c**3*gamma(m/2 + 5/2) + 8*c**2*d*x**2*gamma(m/2 + 5/2)) - d*e**m*m
**2*x**5*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*c**3*gamma(m/2 + 5/2) + 8*c
**2*d*x**2*gamma(m/2 + 5/2)) - 4*d*e**m*m*x**5*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2
+ 3/2)/(8*c**3*gamma(m/2 + 5/2) + 8*c**2*d*x**2*gamma(m/2 + 5/2)) - 3*d*e**m*x**5*x**m*lerchphi(d*x**2*exp_po
lar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*c**3*gamma(m/2 + 5/2) + 8*c**2*d*x**2*gamma(m/2 + 5/2))) + B*a*
(-c*e**m*m**2*x**3*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*c**3*gamma(m/2 +
5/2) + 8*c**2*d*x**2*gamma(m/2 + 5/2)) - 4*c*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)/(8*c**3*gamma(m/2 + 5/2) + 8*c**2*d*x**2*gamma(m/2 + 5/2)) + 2*c*e**m*m*x**3*x**m*gamma(m/2
+ 3/2)/(8*c**3*gamma(m/2 + 5/2) + 8*c**2*d*x**2*gamma(m/2 + 5/2)) - 3*c*e**m*x**3*x**m*lerchphi(d*x**2*exp_pol
ar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*c**3*gamma(m/2 + 5/2) + 8*c**2*d*x**2*gamma(m/2 + 5/2)) + 6*c*e*
*m*x**3*x**m*gamma(m/2 + 3/2)/(8*c**3*gamma(m/2 + 5/2) + 8*c**2*d*x**2*gamma(m/2 + 5/2)) - d*e**m*m**2*x**5*x*
*m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*c**3*gamma(m/2 + 5/2) + 8*c**2*d*x**2*
gamma(m/2 + 5/2)) - 4*d*e**m*m*x**5*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*
c**3*gamma(m/2 + 5/2) + 8*c**2*d*x**2*gamma(m/2 + 5/2)) - 3*d*e**m*x**5*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c
, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*c**3*gamma(m/2 + 5/2) + 8*c**2*d*x**2*gamma(m/2 + 5/2))) + B*b*(-c*e**m*m*
*2*x**5*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(8*c**3*gamma(m/2 + 7/2) + 8*c*
*2*d*x**2*gamma(m/2 + 7/2)) - 8*c*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)/(8*c**3*gamma(m/2 + 7/2) + 8*c**2*d*x**2*gamma(m/2 + 7/2)) + 2*c*e**m*m*x**5*x**m*gamma(m/2 + 5/2)/(8*c
**3*gamma(m/2 + 7/2) + 8*c**2*d*x**2*gamma(m/2 + 7/2)) - 15*c*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)/(8*c**3*gamma(m/2 + 7/2) + 8*c**2*d*x**2*gamma(m/2 + 7/2)) + 10*c*e**m*x**5*x
**m*gamma(m/2 + 5/2)/(8*c**3*gamma(m/2 + 7/2) + 8*c**2*d*x**2*gamma(m/2 + 7/2)) - d*e**m*m**2*x**7*x**m*lerchp
hi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(8*c**3*gamma(m/2 + 7/2) + 8*c**2*d*x**2*gamma(m/2
+ 7/2)) - 8*d*e**m*m*x**7*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(8*c**3*gamm
a(m/2 + 7/2) + 8*c**2*d*x**2*gamma(m/2 + 7/2)) - 15*d*e**m*x**7*x**m*lerchphi(d*x**2*exp_polar(I*pi)/c, 1, m/2
+ 5/2)*gamma(m/2 + 5/2)/(8*c**3*gamma(m/2 + 7/2) + 8*c**2*d*x**2*gamma(m/2 + 7/2)))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,\left (b\,x^2+a\right )}{{\left (d\,x^2+c\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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