3.1148 \(\int \frac{(a+b x^2)^p (c+d x^2)^q}{x} \, dx\)

Optimal. Leaf size=97 \[ -\frac{\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^q \left (\frac{b \left (c+d x^2\right )}{b c-a d}\right )^{-q} F_1\left (p+1;-q,1;p+2;-\frac{d \left (b x^2+a\right )}{b c-a d},\frac{b x^2+a}{a}\right )}{2 a (p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0934395, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {446, 137, 136} \[ -\frac{\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^q \left (\frac{b \left (c+d x^2\right )}{b c-a d}\right )^{-q} F_1\left (p+1;-q,1;p+2;-\frac{d \left (b x^2+a\right )}{b c-a d},\frac{b x^2+a}{a}\right )}{2 a (p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 446

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]]

Rule 137

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 136

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)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), 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])

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^p \left (c+d x^2\right )^q}{x} \, dx &=\frac{1}{2} \operatorname{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 ) \operatorname{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]  time = 0.0712278, size = 95, normalized size = 0.98 \[ \frac{\left (\frac{a}{b x^2}+1\right )^{-p} \left (a+b x^2\right )^p \left (\frac{c}{d x^2}+1\right )^{-q} \left (c+d x^2\right )^q F_1\left (-p-q;-p,-q;-p-q+1;-\frac{a}{b x^2},-\frac{c}{d x^2}\right )}{2 (p+q)} \]

Warning: Unable to verify antiderivative.

[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)

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( b{x}^{2}+a \right ) ^{p} \left ( d{x}^{2}+c \right ) ^{q}}{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{p}{\left (d x^{2} + c\right )}^{q}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{2} + a\right )}^{p}{\left (d x^{2} + c\right )}^{q}}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{p}{\left (d x^{2} + c\right )}^{q}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)