3.1146 \(\int x^3 (a+b x^2)^p (c+d x^2)^q \, dx\)

Optimal. Leaf size=146 \[ \frac{\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^{q+1}}{2 b d (p+q+2)}-\frac{\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^q (a d (q+1)+b c (p+1)) \left (\frac{b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \, _2F_1\left (p+1,-q;p+2;-\frac{d \left (b x^2+a\right )}{b c-a d}\right )}{2 b^2 d (p+1) (p+q+2)} \]

[Out]

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

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Rubi [A]  time = 0.123154, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {446, 80, 70, 69} \[ \frac{\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^{q+1}}{2 b d (p+q+2)}-\frac{\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^q (a d (q+1)+b c (p+1)) \left (\frac{b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \, _2F_1\left (p+1,-q;p+2;-\frac{d \left (b x^2+a\right )}{b c-a d}\right )}{2 b^2 d (p+1) (p+q+2)} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^2)^p*(c + d*x^2)^q,x]

[Out]

((a + b*x^2)^(1 + p)*(c + d*x^2)^(1 + q))/(2*b*d*(2 + p + q)) - ((b*c*(1 + p) + a*d*(1 + q))*(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^2*d*(1 + p)*(2 + p
+ q)*((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 80

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

Rule 70

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 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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]))

Rubi steps

\begin{align*} \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right )^q \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x (a+b x)^p (c+d x)^q \, dx,x,x^2\right )\\ &=\frac{\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (2+p+q)}-\frac{(b c (1+p)+a d (1+q)) \operatorname{Subst}\left (\int (a+b x)^p (c+d x)^q \, dx,x,x^2\right )}{2 b d (2+p+q)}\\ &=\frac{\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (2+p+q)}-\frac{\left ((b c (1+p)+a d (1+q)) \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 (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 )}{2 b d (2+p+q)}\\ &=\frac{\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (2+p+q)}-\frac{(b c (1+p)+a d (1+q)) \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^2 d (1+p) (2+p+q)}\\ \end{align*}

Mathematica [A]  time = 0.0915058, size = 118, normalized size = 0.81 \[ \frac{\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^q \left (b \left (c+d x^2\right )-\frac{(a d (q+1)+b c (p+1)) \left (\frac{b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \, _2F_1\left (p+1,-q;p+2;\frac{d \left (b x^2+a\right )}{a d-b c}\right )}{p+1}\right )}{2 b^2 d (p+q+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*x^2)^p*(c + d*x^2)^q,x]

[Out]

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

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Maple [F]  time = 0.06, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( b{x}^{2}+a \right ) ^{p} \left ( d{x}^{2}+c \right ) ^{q}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*x^2+a)^p*(d*x^2+c)^q,x)

[Out]

int(x^3*(b*x^2+a)^p*(d*x^2+c)^q,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{p}{\left (d x^{2} + c\right )}^{q} x^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{2} + a\right )}^{p}{\left (d x^{2} + c\right )}^{q} x^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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^3, 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(x**3*(b*x**2+a)**p*(d*x**2+c)**q,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x^{2} + a\right )}^{p}{\left (d x^{2} + c\right )}^{q} x^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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^3, x)