3.12.47 \(\int x (a+b x^2)^p (c+d x^2)^q \, dx\) [1147]

Optimal. Leaf size=85 \[ \frac {\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^q \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \, _2F_1\left (1+p,-q;2+p;-\frac {d \left (a+b x^2\right )}{b c-a d}\right )}{2 b (1+p)} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {455, 72, 71} \begin {gather*} \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} \, _2F_1\left (p+1,-q;p+2;-\frac {d \left (b x^2+a\right )}{b c-a d}\right )}{2 b (p+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^2)^p*(c + d*x^2)^q,x]

[Out]

((a + b*x^2)^(1 + p)*(c + d*x^2)^q*Hypergeometric2F1[1 + p, -q, 2 + p, -((d*(a + b*x^2))/(b*c - a*d))])/(2*b*(
1 + p)*((b*(c + d*x^2))/(b*c - a*d))^q)

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps

\begin {align*} \int x \left (a+b x^2\right )^p \left (c+d x^2\right )^q \, dx &=\frac {1}{2} \text {Subst}\left (\int (a+b x)^p (c+d x)^q \, 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 (a+b x)^p \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^q \, 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} \, _2F_1\left (1+p,-q;2+p;-\frac {d \left (a+b x^2\right )}{b c-a d}\right )}{2 b (1+p)}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 84, normalized size = 0.99 \begin {gather*} \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} \, _2F_1\left (1+p,-q;2+p;\frac {d \left (a+b x^2\right )}{-b c+a d}\right )}{2 b (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)^(1 + p)*(c + d*x^2)^q*Hypergeometric2F1[1 + p, -q, 2 + p, (d*(a + b*x^2))/(-(b*c) + a*d)])/(2*b*(
1 + p)*((b*(c + d*x^2))/(b*c - a*d))^q)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [F]
time = 0.00, 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(x*(b*x^2+a)^p*(d*x^2+c)^q,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,{\left (b\,x^2+a\right )}^p\,{\left (d\,x^2+c\right )}^q \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*x^2)^p*(c + d*x^2)^q,x)

[Out]

int(x*(a + b*x^2)^p*(c + d*x^2)^q, x)

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