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

Optimal. Leaf size=258 \[ \frac {d (e x)^{m+3} \left (a^2 B d^2-a b d (A d+3 B c)+3 b^2 c (A d+B c)\right )}{b^3 e^3 (m+3)}-\frac {(e x)^{m+1} \left (a^3 B d^3-a^2 b d^2 (A d+3 B c)+3 a b^2 c d (A d+B c)+b^3 \left (-c^2\right ) (3 A d+B c)\right )}{b^4 e (m+1)}+\frac {(e x)^{m+1} (A b-a B) (b c-a d)^3 \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a b^4 e (m+1)}+\frac {d^2 (e x)^{m+5} (-a B d+A b d+3 b B c)}{b^2 e^5 (m+5)}+\frac {B d^3 (e x)^{m+7}}{b e^7 (m+7)} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 258, normalized size of antiderivative = 1.00, 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} \[ -\frac {(e x)^{m+1} \left (-a^2 b d^2 (A d+3 B c)+a^3 B d^3+3 a b^2 c d (A d+B c)+b^3 \left (-c^2\right ) (3 A d+B c)\right )}{b^4 e (m+1)}+\frac {d (e x)^{m+3} \left (a^2 B d^2-a b d (A d+3 B c)+3 b^2 c (A d+B c)\right )}{b^3 e^3 (m+3)}+\frac {d^2 (e x)^{m+5} (-a B d+A b d+3 b B c)}{b^2 e^5 (m+5)}+\frac {(e x)^{m+1} (A b-a B) (b c-a d)^3 \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a b^4 e (m+1)}+\frac {B d^3 (e x)^{m+7}}{b e^7 (m+7)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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]

Rubi steps

\begin {align*} \int \frac {(e x)^m \left (A+B x^2\right ) \left (c+d x^2\right )^3}{a+b x^2} \, dx &=\int \left (-\frac {\left (a^3 B d^3+3 a b^2 c d (B c+A d)-a^2 b d^2 (3 B c+A d)-b^3 c^2 (B c+3 A d)\right ) (e x)^m}{b^4}+\frac {d \left (a^2 B d^2+3 b^2 c (B c+A d)-a b d (3 B c+A d)\right ) (e x)^{2+m}}{b^3 e^2}+\frac {d^2 (3 b B c+A b d-a B d) (e x)^{4+m}}{b^2 e^4}+\frac {B d^3 (e x)^{6+m}}{b e^6}+\frac {\left (A b^4 c^3-a b^3 B c^3-3 a A b^3 c^2 d+3 a^2 b^2 B c^2 d+3 a^2 A b^2 c d^2-3 a^3 b B c d^2-a^3 A b d^3+a^4 B d^3\right ) (e x)^m}{b^4 \left (a+b x^2\right )}\right ) \, dx\\ &=-\frac {\left (a^3 B d^3+3 a b^2 c d (B c+A d)-a^2 b d^2 (3 B c+A d)-b^3 c^2 (B c+3 A d)\right ) (e x)^{1+m}}{b^4 e (1+m)}+\frac {d \left (a^2 B d^2+3 b^2 c (B c+A d)-a b d (3 B c+A d)\right ) (e x)^{3+m}}{b^3 e^3 (3+m)}+\frac {d^2 (3 b B c+A b d-a B d) (e x)^{5+m}}{b^2 e^5 (5+m)}+\frac {B d^3 (e x)^{7+m}}{b e^7 (7+m)}+\frac {\left ((A b-a B) (b c-a d)^3\right ) \int \frac {(e x)^m}{a+b x^2} \, dx}{b^4}\\ &=-\frac {\left (a^3 B d^3+3 a b^2 c d (B c+A d)-a^2 b d^2 (3 B c+A d)-b^3 c^2 (B c+3 A d)\right ) (e x)^{1+m}}{b^4 e (1+m)}+\frac {d \left (a^2 B d^2+3 b^2 c (B c+A d)-a b d (3 B c+A d)\right ) (e x)^{3+m}}{b^3 e^3 (3+m)}+\frac {d^2 (3 b B c+A b d-a B d) (e x)^{5+m}}{b^2 e^5 (5+m)}+\frac {B d^3 (e x)^{7+m}}{b e^7 (7+m)}+\frac {(A b-a B) (b c-a d)^3 (e x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{a b^4 e (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 217, normalized size = 0.84 \[ \frac {x (e x)^m \left (\frac {b d x^2 \left (a^2 B d^2-a b d (A d+3 B c)+3 b^2 c (A d+B c)\right )}{m+3}+\frac {-a^3 B d^3+a^2 b d^2 (A d+3 B c)-3 a b^2 c d (A d+B c)+b^3 c^2 (3 A d+B c)}{m+1}+\frac {b^2 d^2 x^4 (-a B d+A b d+3 b B c)}{m+5}+\frac {(a B-A b) (a d-b c)^3 \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{a (m+1)}+\frac {b^3 B d^3 x^6}{m+7}\right )}{b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B d^{3} x^{8} + {\left (3 \, B c d^{2} + A d^{3}\right )} x^{6} + 3 \, {\left (B c^{2} d + A c d^{2}\right )} x^{4} + A c^{3} + {\left (B c^{3} + 3 \, A c^{2} d\right )} x^{2}\right )} \left (e x\right )^{m}}{b x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} {\left (d x^{2} + c\right )}^{3} \left (e x\right )^{m}}{b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {\left (B \,x^{2}+A \right ) \left (d \,x^{2}+c \right )^{3} \left (e x \right )^{m}}{b \,x^{2}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B x^{2} + A\right )} {\left (d x^{2} + c\right )}^{3} \left (e x\right )^{m}}{b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,{\left (d\,x^2+c\right )}^3}{b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 29.05, size = 911, normalized size = 3.53 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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