\(\int \frac {(e x)^m (a+b x^2)^p (A+B x^2)}{c+d x^2} \, dx\) [47]

Optimal result

Integrand size = 31, antiderivative size = 162 \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=-\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {AppellF1}\left (\frac {1+m}{2},-p,1,\frac {3+m}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c d e (1+m)}+\frac {B (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-p,\frac {3+m}{2},-\frac {b x^2}{a}\right )}{d e (1+m)} \]

[Out]

-(-A*d+B*c)*(e*x)^(1+m)*(b*x^2+a)^p*AppellF1(1/2+1/2*m,-p,1,3/2+1/2*m,-b*x^2/a,-d*x^2/c)/c/d/e/(1+m)/((1+b*x^2
/a)^p)+B*(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)/d/e/(1+m)/((1+b*x^2/a)^p)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {598, 372, 371, 525, 524} \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {B (e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},-p,\frac {m+3}{2},-\frac {b x^2}{a}\right )}{d e (m+1)}-\frac {(e x)^{m+1} \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} (B c-A d) \operatorname {AppellF1}\left (\frac {m+1}{2},-p,1,\frac {m+3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c d e (m+1)} \]

[In]

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

[Out]

-(((B*c - A*d)*(e*x)^(1 + m)*(a + b*x^2)^p*AppellF1[(1 + m)/2, -p, 1, (3 + m)/2, -((b*x^2)/a), -((d*x^2)/c)])/
(c*d*e*(1 + m)*(1 + (b*x^2)/a)^p)) + (B*(e*x)^(1 + m)*(a + b*x^2)^p*Hypergeometric2F1[(1 + m)/2, -p, (3 + m)/2
, -((b*x^2)/a)])/(d*e*(1 + m)*(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 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

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

Rule 598

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, p}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {B (e x)^m \left (a+b x^2\right )^p}{d}+\frac {(-B c+A d) (e x)^m \left (a+b x^2\right )^p}{d \left (c+d x^2\right )}\right ) \, dx \\ & = \frac {B \int (e x)^m \left (a+b x^2\right )^p \, dx}{d}+\frac {(-B c+A d) \int \frac {(e x)^m \left (a+b x^2\right )^p}{c+d x^2} \, dx}{d} \\ & = \frac {\left (B \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}{d}+\frac {\left ((-B c+A d) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \frac {(e x)^m \left (1+\frac {b x^2}{a}\right )^p}{c+d x^2} \, dx}{d} \\ & = -\frac {(B c-A d) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} F_1\left (\frac {1+m}{2};-p,1;\frac {3+m}{2};-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c d e (1+m)}+\frac {B (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 )}{d e (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.73 \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=\frac {x (e x)^m \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left ((-B c+A d) \operatorname {AppellF1}\left (\frac {1+m}{2},-p,1,\frac {3+m}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+B c \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-p,\frac {3+m}{2},-\frac {b x^2}{a}\right )\right )}{c d (1+m)} \]

[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*((-(B*c) + A*d)*AppellF1[(1 + m)/2, -p, 1, (3 + m)/2, -((b*x^2)/a), -((d*x^2)/c)] + B
*c*Hypergeometric2F1[(1 + m)/2, -p, (3 + m)/2, -((b*x^2)/a)]))/(c*d*(1 + m)*(1 + (b*x^2)/a)^p)

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (b \,x^{2}+a \right )^{p} \left (x^{2} B +A \right )}{d \,x^{2}+c}d x\]

[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)

Fricas [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{m}}{d x^{2} + c} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{m}}{d x^{2} + c} \,d x } \]

[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)*(b*x^2 + a)^p*(e*x)^m/(d*x^2 + c), x)

Giac [F]

\[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=\int { \frac {{\left (B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{m}}{d x^{2} + c} \,d x } \]

[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)*(b*x^2 + a)^p*(e*x)^m/(d*x^2 + c), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (a+b x^2\right )^p \left (A+B x^2\right )}{c+d x^2} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (e\,x\right )}^m\,{\left (b\,x^2+a\right )}^p}{d\,x^2+c} \,d x \]

[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)