3.1.6 \(\int \frac {(e x)^m (A+B x^2) (c+d x^2)}{(a+b x^2)^2} \, dx\) [6]
Optimal. Leaf size=171 \[ -\frac {d (A b (1+m)-a B (3+m)) (e x)^{1+m}}{2 a b^2 e (1+m)}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )}{2 a b e \left (a+b x^2\right )}+\frac {(a B (b c (1+m)-a d (3+m))+A b (a d (1+m)+b (c-c m))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{2 a^2 b^2 e (1+m)} \]
[Out]
-1/2*d*(A*b*(1+m)-a*B*(3+m))*(e*x)^(1+m)/a/b^2/e/(1+m)+1/2*(A*b-B*a)*(e*x)^(1+m)*(d*x^2+c)/a/b/e/(b*x^2+a)+1/2
*(a*B*(b*c*(1+m)-a*d*(3+m))+A*b*(a*d*(1+m)+b*(-c*m+c)))*(e*x)^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-b*x^
2/a)/a^2/b^2/e/(1+m)
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Rubi [A]
time = 0.16, 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 {b x^2}{a}\right ) (A b (a d (m+1)+b (c-c m))+a B (b c (m+1)-a d (m+3)))}{2 a^2 b^2 e (m+1)}-\frac {d (e x)^{m+1} (A b (m+1)-a B (m+3))}{2 a b^2 e (m+1)}+\frac {\left (c+d x^2\right ) (e x)^{m+1} (A b-a B)}{2 a b e \left (a+b x^2\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((e*x)^m*(A + B*x^2)*(c + d*x^2))/(a + b*x^2)^2,x]
[Out]
-1/2*(d*(A*b*(1 + m) - a*B*(3 + m))*(e*x)^(1 + m))/(a*b^2*e*(1 + m)) + ((A*b - a*B)*(e*x)^(1 + m)*(c + d*x^2))
/(2*a*b*e*(a + b*x^2)) + ((a*B*(b*c*(1 + m) - a*d*(3 + m)) + A*b*(a*d*(1 + m) + b*(c - c*m)))*(e*x)^(1 + m)*Hy
pergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(2*a^2*b^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 (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx &=\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )}{2 a b e \left (a+b x^2\right )}-\frac {\int \frac {(e x)^m \left (-c (A b (1-m)+a B (1+m))+d (A b (1+m)-a B (3+m)) x^2\right )}{a+b x^2} \, dx}{2 a b}\\ &=-\frac {d (A b (1+m)-a B (3+m)) (e x)^{1+m}}{2 a b^2 e (1+m)}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )}{2 a b e \left (a+b x^2\right )}+\frac {(a B (b c (1+m)-a d (3+m))+A b (a d (1+m)+b (c-c m))) \int \frac {(e x)^m}{a+b x^2} \, dx}{2 a b^2}\\ &=-\frac {d (A b (1+m)-a B (3+m)) (e x)^{1+m}}{2 a b^2 e (1+m)}+\frac {(A b-a B) (e x)^{1+m} \left (c+d x^2\right )}{2 a b e \left (a+b x^2\right )}+\frac {(a B (b c (1+m)-a d (3+m))+A b (a d (1+m)+b (c-c m))) (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{2 a^2 b^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 (-a^2 B d+a (2 a B d-b (B c+A d)) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )-(A b-a B) (b c-a d) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )\right )}{a^2 b^2 (1+m)} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((e*x)^m*(A + B*x^2)*(c + d*x^2))/(a + b*x^2)^2,x]
[Out]
-((x*(e*x)^m*(-(a^2*B*d) + a*(2*a*B*d - b*(B*c + A*d))*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)
] - (A*b - a*B)*(b*c - a*d)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)]))/(a^2*b^2*(1 + m)))
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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (B \,x^{2}+A \right ) \left (d \,x^{2}+c \right )}{\left (b \,x^{2}+a \right )^{2}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*x)^m*(B*x^2+A)*(d*x^2+c)/(b*x^2+a)^2,x)
[Out]
int((e*x)^m*(B*x^2+A)*(d*x^2+c)/(b*x^2+a)^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)*(d*x^2+c)/(b*x^2+a)^2,x, algorithm="maxima")
[Out]
integrate((B*x^2 + A)*(d*x^2 + c)*(x*e)^m/(b*x^2 + a)^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)*(d*x^2+c)/(b*x^2+a)^2,x, algorithm="fricas")
[Out]
integral((B*d*x^4 + (B*c + A*d)*x^2 + A*c)*(x*e)^m/(b^2*x^4 + 2*a*b*x^2 + a^2), x)
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Sympy [C] Result contains complex when optimal does not.
time = 31.66, 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)*(d*x**2+c)/(b*x**2+a)**2,x)
[Out]
A*c*(-a*e**m*m**2*x*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2 +
3/2) + 8*a**2*b*x**2*gamma(m/2 + 3/2)) + 2*a*e**m*m*x*x**m*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2 + 3/2) + 8*a**2
*b*x**2*gamma(m/2 + 3/2)) + a*e**m*x*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8
*a**3*gamma(m/2 + 3/2) + 8*a**2*b*x**2*gamma(m/2 + 3/2)) + 2*a*e**m*x*x**m*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2
+ 3/2) + 8*a**2*b*x**2*gamma(m/2 + 3/2)) - b*e**m*m**2*x**3*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1
/2)*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2 + 3/2) + 8*a**2*b*x**2*gamma(m/2 + 3/2)) + b*e**m*x**3*x**m*lerchphi(b*
x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**3*gamma(m/2 + 3/2) + 8*a**2*b*x**2*gamma(m/2 + 3/
2))) + A*d*(-a*e**m*m**2*x**3*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*a**3*g
amma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) - 4*a*e**m*m*x**3*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1,
m/2 + 3/2)*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) + 2*a*e**m*m*x**3*x**m
*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) - 3*a*e**m*x**3*x**m*lerchphi(b*x
**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2
)) + 6*a*e**m*x**3*x**m*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) - b*e**m*m
**2*x**5*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a
**2*b*x**2*gamma(m/2 + 5/2)) - 4*b*e**m*m*x**5*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2
+ 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) - 3*b*e**m*x**5*x**m*lerchphi(b*x**2*exp_po
lar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2))) + B*c*
(-a*e**m*m**2*x**3*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 +
5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) - 4*a*e**m*m*x**3*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)
*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) + 2*a*e**m*m*x**3*x**m*gamma(m/2
+ 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) - 3*a*e**m*x**3*x**m*lerchphi(b*x**2*exp_pol
ar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) + 6*a*e*
*m*x**3*x**m*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) - b*e**m*m**2*x**5*x*
*m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*
gamma(m/2 + 5/2)) - 4*b*e**m*m*x**5*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*
a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2)) - 3*b*e**m*x**5*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a
, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(8*a**3*gamma(m/2 + 5/2) + 8*a**2*b*x**2*gamma(m/2 + 5/2))) + B*d*(-a*e**m*m*
*2*x**5*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(8*a**3*gamma(m/2 + 7/2) + 8*a*
*2*b*x**2*gamma(m/2 + 7/2)) - 8*a*e**m*m*x**5*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 5/2)*gamma(m/2
+ 5/2)/(8*a**3*gamma(m/2 + 7/2) + 8*a**2*b*x**2*gamma(m/2 + 7/2)) + 2*a*e**m*m*x**5*x**m*gamma(m/2 + 5/2)/(8*a
**3*gamma(m/2 + 7/2) + 8*a**2*b*x**2*gamma(m/2 + 7/2)) - 15*a*e**m*x**5*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a
, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(8*a**3*gamma(m/2 + 7/2) + 8*a**2*b*x**2*gamma(m/2 + 7/2)) + 10*a*e**m*x**5*x
**m*gamma(m/2 + 5/2)/(8*a**3*gamma(m/2 + 7/2) + 8*a**2*b*x**2*gamma(m/2 + 7/2)) - b*e**m*m**2*x**7*x**m*lerchp
hi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(8*a**3*gamma(m/2 + 7/2) + 8*a**2*b*x**2*gamma(m/2
+ 7/2)) - 8*b*e**m*m*x**7*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 5/2)*gamma(m/2 + 5/2)/(8*a**3*gamm
a(m/2 + 7/2) + 8*a**2*b*x**2*gamma(m/2 + 7/2)) - 15*b*e**m*x**7*x**m*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2
+ 5/2)*gamma(m/2 + 5/2)/(8*a**3*gamma(m/2 + 7/2) + 8*a**2*b*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)*(d*x^2+c)/(b*x^2+a)^2,x, algorithm="giac")
[Out]
integrate((B*x^2 + A)*(d*x^2 + c)*(x*e)^m/(b*x^2 + a)^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 (d\,x^2+c\right )}{{\left (b\,x^2+a\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*(c + d*x^2))/(a + b*x^2)^2,x)
[Out]
int(((A + B*x^2)*(e*x)^m*(c + d*x^2))/(a + b*x^2)^2, x)
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