3.274 \(\int x^m (A+B x^2) (b x^2+c x^4)^p \, dx\)

Optimal. Leaf size=140 \[ \frac {B x^{m-1} \left (b x^2+c x^4\right )^{p+1}}{c (m+4 p+3)}-\frac {x^{m+1} \left (\frac {c x^2}{b}+1\right )^{-p} \left (b x^2+c x^4\right )^p (b B (m+2 p+1)-A c (m+4 p+3)) \, _2F_1\left (-p,\frac {1}{2} (m+2 p+1);\frac {1}{2} (m+2 p+3);-\frac {c x^2}{b}\right )}{c (m+2 p+1) (m+4 p+3)} \]

[Out]

B*x^(-1+m)*(c*x^4+b*x^2)^(1+p)/c/(3+m+4*p)-(b*B*(1+m+2*p)-A*c*(3+m+4*p))*x^(1+m)*(c*x^4+b*x^2)^p*hypergeom([-p
, 1/2+1/2*m+p],[3/2+1/2*m+p],-c*x^2/b)/c/(1+m+2*p)/(3+m+4*p)/((1+c*x^2/b)^p)

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Rubi [A]  time = 0.14, antiderivative size = 126, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2039, 2032, 365, 364} \[ x^{m+1} \left (\frac {c x^2}{b}+1\right )^{-p} \left (b x^2+c x^4\right )^p \left (\frac {A}{m+2 p+1}-\frac {b B}{c (m+4 p+3)}\right ) \, _2F_1\left (-p,\frac {1}{2} (m+2 p+1);\frac {1}{2} (m+2 p+3);-\frac {c x^2}{b}\right )+\frac {B x^{m-1} \left (b x^2+c x^4\right )^{p+1}}{c (m+4 p+3)} \]

Antiderivative was successfully verified.

[In]

Int[x^m*(A + B*x^2)*(b*x^2 + c*x^4)^p,x]

[Out]

(B*x^(-1 + m)*(b*x^2 + c*x^4)^(1 + p))/(c*(3 + m + 4*p)) + ((A/(1 + m + 2*p) - (b*B)/(c*(3 + m + 4*p)))*x^(1 +
m)*(b*x^2 + c*x^4)^p*Hypergeometric2F1[-p, (1 + m + 2*p)/2, (3 + m + 2*p)/2, -((c*x^2)/b)])/(1 + (c*x^2)/b)^p

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 365

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 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rule 2039

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

Rubi steps

\begin {align*} \int x^m \left (A+B x^2\right ) \left (b x^2+c x^4\right )^p \, dx &=\frac {B x^{-1+m} \left (b x^2+c x^4\right )^{1+p}}{c (3+m+4 p)}-\left (-A+\frac {b B (1+m+2 p)}{c (3+m+4 p)}\right ) \int x^m \left (b x^2+c x^4\right )^p \, dx\\ &=\frac {B x^{-1+m} \left (b x^2+c x^4\right )^{1+p}}{c (3+m+4 p)}-\left (\left (-A+\frac {b B (1+m+2 p)}{c (3+m+4 p)}\right ) x^{-2 p} \left (b+c x^2\right )^{-p} \left (b x^2+c x^4\right )^p\right ) \int x^{m+2 p} \left (b+c x^2\right )^p \, dx\\ &=\frac {B x^{-1+m} \left (b x^2+c x^4\right )^{1+p}}{c (3+m+4 p)}-\left (\left (-A+\frac {b B (1+m+2 p)}{c (3+m+4 p)}\right ) x^{-2 p} \left (1+\frac {c x^2}{b}\right )^{-p} \left (b x^2+c x^4\right )^p\right ) \int x^{m+2 p} \left (1+\frac {c x^2}{b}\right )^p \, dx\\ &=\frac {B x^{-1+m} \left (b x^2+c x^4\right )^{1+p}}{c (3+m+4 p)}+\left (\frac {A}{1+m+2 p}-\frac {b B}{c (3+m+4 p)}\right ) x^{1+m} \left (1+\frac {c x^2}{b}\right )^{-p} \left (b x^2+c x^4\right )^p \, _2F_1\left (-p,\frac {1}{2} (1+m+2 p);\frac {1}{2} (3+m+2 p);-\frac {c x^2}{b}\right )\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 135, normalized size = 0.96 \[ \frac {x^{m+1} \left (x^2 \left (b+c x^2\right )\right )^p \left (\frac {c x^2}{b}+1\right )^{-p} \left (A (m+2 p+3) \, _2F_1\left (-p,\frac {1}{2} (m+2 p+1);\frac {1}{2} (m+2 p+3);-\frac {c x^2}{b}\right )+B x^2 (m+2 p+1) \, _2F_1\left (-p,\frac {1}{2} (m+2 p+3);\frac {1}{2} (m+2 p+5);-\frac {c x^2}{b}\right )\right )}{(m+2 p+1) (m+2 p+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*(A + B*x^2)*(b*x^2 + c*x^4)^p,x]

[Out]

(x^(1 + m)*(x^2*(b + c*x^2))^p*(A*(3 + m + 2*p)*Hypergeometric2F1[-p, (1 + m + 2*p)/2, (3 + m + 2*p)/2, -((c*x
^2)/b)] + B*(1 + m + 2*p)*x^2*Hypergeometric2F1[-p, (3 + m + 2*p)/2, (5 + m + 2*p)/2, -((c*x^2)/b)]))/((1 + m
+ 2*p)*(3 + m + 2*p)*(1 + (c*x^2)/b)^p)

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B x^{2} + A\right )} {\left (c x^{4} + b x^{2}\right )}^{p} x^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(B*x^2+A)*(c*x^4+b*x^2)^p,x, algorithm="fricas")

[Out]

integral((B*x^2 + A)*(c*x^4 + b*x^2)^p*x^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B x^{2} + A\right )} {\left (c x^{4} + b x^{2}\right )}^{p} x^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(B*x^2+A)*(c*x^4+b*x^2)^p,x, algorithm="giac")

[Out]

integrate((B*x^2 + A)*(c*x^4 + b*x^2)^p*x^m, x)

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maple [F]  time = 0.53, size = 0, normalized size = 0.00 \[ \int \left (B \,x^{2}+A \right ) x^{m} \left (c \,x^{4}+b \,x^{2}\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(B*x^2+A)*(c*x^4+b*x^2)^p,x)

[Out]

int(x^m*(B*x^2+A)*(c*x^4+b*x^2)^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B x^{2} + A\right )} {\left (c x^{4} + b x^{2}\right )}^{p} x^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*(B*x^2+A)*(c*x^4+b*x^2)^p,x, algorithm="maxima")

[Out]

integrate((B*x^2 + A)*(c*x^4 + b*x^2)^p*x^m, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^m\,\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(A + B*x^2)*(b*x^2 + c*x^4)^p,x)

[Out]

int(x^m*(A + B*x^2)*(b*x^2 + c*x^4)^p, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \left (x^{2} \left (b + c x^{2}\right )\right )^{p} \left (A + B x^{2}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(B*x**2+A)*(c*x**4+b*x**2)**p,x)

[Out]

Integral(x**m*(x**2*(b + c*x**2))**p*(A + B*x**2), x)

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