\(\int \frac {(a+b x^2)^p (c+d x^2)^q}{x} \, dx\) [1148]

Optimal result

Integrand size = 22, antiderivative size = 97 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x} \, dx=-\frac {\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^q \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \operatorname {AppellF1}\left (1+p,-q,1,2+p,-\frac {d \left (a+b x^2\right )}{b c-a d},\frac {a+b x^2}{a}\right )}{2 a (1+p)} \]

[Out]

-1/2*(b*x^2+a)^(p+1)*(d*x^2+c)^q*AppellF1(p+1,1,-q,2+p,(b*x^2+a)/a,-d*(b*x^2+a)/(-a*d+b*c))/a/(p+1)/((b*(d*x^2
+c)/(-a*d+b*c))^q)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {457, 142, 141} \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x} \, dx=-\frac {\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^q \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \operatorname {AppellF1}\left (p+1,-q,1,p+2,-\frac {d \left (b x^2+a\right )}{b c-a d},\frac {b x^2+a}{a}\right )}{2 a (p+1)} \]

[In]

Int[((a + b*x^2)^p*(c + d*x^2)^q)/x,x]

[Out]

-1/2*((a + b*x^2)^(1 + p)*(c + d*x^2)^q*AppellF1[1 + p, -q, 1, 2 + p, -((d*(a + b*x^2))/(b*c - a*d)), (a + b*x
^2)/a])/(a*(1 + p)*((b*(c + d*x^2))/(b*c - a*d))^q)

Rule 141

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(b*e - a*f
)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 142

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

Rule 457

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x} \, dx=\frac {\left (1+\frac {a}{b x^2}\right )^{-p} \left (1+\frac {c}{d x^2}\right )^{-q} \left (a+b x^2\right )^p \left (c+d x^2\right )^q \operatorname {AppellF1}\left (-p-q,-p,-q,1-p-q,-\frac {a}{b x^2},-\frac {c}{d x^2}\right )}{2 (p+q)} \]

[In]

Integrate[((a + b*x^2)^p*(c + d*x^2)^q)/x,x]

[Out]

((a + b*x^2)^p*(c + d*x^2)^q*AppellF1[-p - q, -p, -q, 1 - p - q, -(a/(b*x^2)), -(c/(d*x^2))])/(2*(p + q)*(1 +
a/(b*x^2))^p*(1 + c/(d*x^2))^q)

Maple [F]

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

[In]

int((b*x^2+a)^p*(d*x^2+c)^q/x,x)

[Out]

int((b*x^2+a)^p*(d*x^2+c)^q/x,x)

Fricas [F]

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

[In]

integrate((b*x^2+a)^p*(d*x^2+c)^q/x,x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^p*(d*x^2 + c)^q/x, x)

Sympy [F]

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

[In]

integrate((b*x**2+a)**p*(d*x**2+c)**q/x,x)

[Out]

Integral((a + b*x**2)**p*(c + d*x**2)**q/x, x)

Maxima [F]

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

[In]

integrate((b*x^2+a)^p*(d*x^2+c)^q/x,x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^p*(d*x^2 + c)^q/x, x)

Giac [F]

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

[In]

integrate((b*x^2+a)^p*(d*x^2+c)^q/x,x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^p*(d*x^2 + c)^q/x, x)

Mupad [F(-1)]

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

[In]

int(((a + b*x^2)^p*(c + d*x^2)^q)/x,x)

[Out]

int(((a + b*x^2)^p*(c + d*x^2)^q)/x, x)