3.1.23 \(\int \frac {(e x)^m (a+b x^2)^3 (A+B x^2)}{c+d x^2} \, dx\) [23]

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

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 260, 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 = {584, 371} \begin {gather*} \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 {(e x)^{m+1} \left (a^3 B d^3-3 a^2 b d^2 (B c-A d)+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^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)} \end {gather*}

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 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 584

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 )^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*}

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Mathematica [A]
time = 0.60, size = 219, normalized size = 0.84 \begin {gather*} \frac {x (e x)^m \left (\frac {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}+\frac {b d \left (3 a^2 B d^2+b^2 c (B c-A d)+3 a b d (-B c+A d)\right ) x^2}{3+m}+\frac {b^2 d^2 (-b B c+A b d+3 a B d) x^4}{5+m}+\frac {b^3 B d^3 x^6}{7+m}+\frac {(b c-a d)^3 (B c-A d) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )}{c (1+m)}\right )}{d^4} \end {gather*}

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 = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right )^{3} \left (B \,x^{2}+A \right )}{d \,x^{2}+c}\, dx\]

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 = 0.00, 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)^3*(B*x^2+A)/(d*x^2+c),x, algorithm="maxima")

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 10.32, size = 911, normalized size = 3.50 \begin {gather*} \frac {A a^{3} e^{m} m x x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A a^{3} e^{m} x x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {3 A a^{2} b e^{m} m x^{3} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {9 A a^{2} b e^{m} x^{3} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 A a b^{2} e^{m} m x^{5} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {15 A a b^{2} e^{m} x^{5} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {A b^{3} e^{m} m x^{7} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {7}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {9}{2}\right )} + \frac {7 A b^{3} e^{m} x^{7} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {7}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {9}{2}\right )} + \frac {B a^{3} e^{m} m x^{3} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 B a^{3} e^{m} x^{3} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 B a^{2} b e^{m} m x^{5} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {15 B a^{2} b e^{m} x^{5} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {5}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} + \frac {3 B a b^{2} e^{m} m x^{7} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {7}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {9}{2}\right )} + \frac {21 B a b^{2} e^{m} x^{7} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {7}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {9}{2}\right )} + \frac {B b^{3} e^{m} m x^{9} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {9}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {9}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {11}{2}\right )} + \frac {9 B b^{3} e^{m} x^{9} x^{m} \Phi \left (\frac {d x^{2} e^{i \pi }}{c}, 1, \frac {m}{2} + \frac {9}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {9}{2}\right )}{4 c \Gamma \left (\frac {m}{2} + \frac {11}{2}\right )} \end {gather*}

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 = 0.00, 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)^3*(B*x^2+A)/(d*x^2+c),x, algorithm="giac")

[Out]

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

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

[Out]

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

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