3.1.46 \(\int (e x)^m (a+b x^2)^p (A+B x^2) (c+d x^2) \, dx\) [46]

Optimal. Leaf size=253 \[ -\frac {(a B d (3+m)-b (2 A d+B c (5+m+2 p))) (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{b^2 e (3+m+2 p) (5+m+2 p)}+\frac {d (e x)^{1+m} \left (a+b x^2\right )^{1+p} \left (A+B x^2\right )}{b e (5+m+2 p)}-\frac {(A b (3+m+2 p) (a d (1+m)-b c (5+m+2 p))-a (1+m) (a B d (3+m)-b (2 A d+B c (5+m+2 p)))) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1+m}{2},-p;\frac {3+m}{2};-\frac {b x^2}{a}\right )}{b^2 e (1+m) (3+m+2 p) (5+m+2 p)} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 238, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {595, 470, 372, 371} \begin {gather*} \frac {(e x)^{m+1} \left (a+b x^2\right )^{p+1} (-a B d (m+3)+2 A b d+b B c (m+2 p+5))}{b^2 e (m+2 p+3) (m+2 p+5)}-\frac {(e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {m+1}{2},-p;\frac {m+3}{2};-\frac {b x^2}{a}\right ) \left (\frac {a (-a B d (m+3)+2 A b d+b B c (m+2 p+5))}{b (m+2 p+3)}+a A d-\frac {A b c (m+2 p+5)}{m+1}\right )}{b e (m+2 p+5)}+\frac {d \left (A+B x^2\right ) (e x)^{m+1} \left (a+b x^2\right )^{p+1}}{b e (m+2 p+5)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], 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 595

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[f*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*g*(m + n*(p + q + 1) + 1))), x] + Dis
t[1/(b*(m + n*(p + q + 1) + 1)), Int[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*((b*e - a*f)*(m + 1) + b
*e*n*(p + q + 1)) + (d*(b*e - a*f)*(m + 1) + f*n*q*(b*c - a*d) + b*e*d*n*(p + q + 1))*x^n, x], x], x] /; FreeQ
[{a, b, c, d, e, f, g, m, p}, x] && IGtQ[n, 0] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rubi steps

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

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Mathematica [A]
time = 0.15, size = 147, normalized size = 0.58 \begin {gather*} x (e x)^m \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (\frac {A c \, _2F_1\left (\frac {1+m}{2},-p;\frac {3+m}{2};-\frac {b x^2}{a}\right )}{1+m}+\frac {(B c+A d) x^2 \, _2F_1\left (\frac {3+m}{2},-p;\frac {5+m}{2};-\frac {b x^2}{a}\right )}{3+m}+\frac {B d x^4 \, _2F_1\left (\frac {5+m}{2},-p;\frac {7+m}{2};-\frac {b x^2}{a}\right )}{5+m}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 204.44, size = 241, normalized size = 0.95 \begin {gather*} \frac {A a^{p} c e^{m} x x^{m} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {A a^{p} d e^{m} x^{3} x^{m} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {B a^{p} c e^{m} x^{3} x^{m} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {B a^{p} d e^{m} x^{5} x^{m} \Gamma \left (\frac {m}{2} + \frac {5}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + \frac {5}{2} \\ \frac {m}{2} + \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {7}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**p*c*e**m*x*x**m*gamma(m/2 + 1/2)*hyper((-p, m/2 + 1/2), (m/2 + 3/2,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(
m/2 + 3/2)) + A*a**p*d*e**m*x**3*x**m*gamma(m/2 + 3/2)*hyper((-p, m/2 + 3/2), (m/2 + 5/2,), b*x**2*exp_polar(I
*pi)/a)/(2*gamma(m/2 + 5/2)) + B*a**p*c*e**m*x**3*x**m*gamma(m/2 + 3/2)*hyper((-p, m/2 + 3/2), (m/2 + 5/2,), b
*x**2*exp_polar(I*pi)/a)/(2*gamma(m/2 + 5/2)) + B*a**p*d*e**m*x**5*x**m*gamma(m/2 + 5/2)*hyper((-p, m/2 + 5/2)
, (m/2 + 7/2,), b*x**2*exp_polar(I*pi)/a)/(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)^p*(B*x^2+A)*(d*x^2+c),x, algorithm="giac")

[Out]

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

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

[Out]

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

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