\(\int (d x)^m (a^2+2 a b x^2+b^2 x^4)^p \, dx\) [797]

Optimal result

Integrand size = 26, antiderivative size = 74 \[ \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {(d x)^{1+m} \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (3+m+4 p),\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a d (1+m)} \]

[Out]

(d*x)^(1+m)*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^p*hypergeom([1, 3/2+1/2*m+2*p],[3/2+1/2*m],-b*x^2/a)/a/d/(1+m)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.04, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1127, 371} \[ \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {(d x)^{m+1} \left (\frac {b x^2}{a}+1\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \operatorname {Hypergeometric2F1}\left (\frac {m+1}{2},-2 p,\frac {m+3}{2},-\frac {b x^2}{a}\right )}{d (m+1)} \]

[In]

Int[(d*x)^m*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

((d*x)^(1 + m)*(a^2 + 2*a*b*x^2 + b^2*x^4)^p*Hypergeometric2F1[(1 + m)/2, -2*p, (3 + m)/2, -((b*x^2)/a)])/(d*(
1 + m)*(1 + (b*x^2)/a)^(2*p))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1127

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^2 +
c*x^4)^FracPart[p]/(1 + 2*c*(x^2/b))^(2*FracPart[p])), Int[(d*x)^m*(1 + 2*c*(x^2/b))^(2*p), x], x] /; FreeQ[{
a, b, c, d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = \left (\left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \int (d x)^m \left (1+\frac {b x^2}{a}\right )^{2 p} \, dx \\ & = \frac {(d x)^{1+m} \left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p \, _2F_1\left (\frac {1+m}{2},-2 p;\frac {3+m}{2};-\frac {b x^2}{a}\right )}{d (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.89 \[ \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\frac {x (d x)^m \left (\left (a+b x^2\right )^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-2 p} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-2 p,1+\frac {1+m}{2},-\frac {b x^2}{a}\right )}{1+m} \]

[In]

Integrate[(d*x)^m*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

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

Maple [F]

\[\int \left (d x \right )^{m} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}d x\]

[In]

int((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^p,x)

[Out]

int((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^p,x)

Fricas [F]

\[ \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} \left (d x\right )^{m} \,d x } \]

[In]

integrate((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="fricas")

[Out]

integral((b^2*x^4 + 2*a*b*x^2 + a^2)^p*(d*x)^m, x)

Sympy [F]

\[ \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int \left (d x\right )^{m} \left (\left (a + b x^{2}\right )^{2}\right )^{p}\, dx \]

[In]

integrate((d*x)**m*(b**2*x**4+2*a*b*x**2+a**2)**p,x)

[Out]

Integral((d*x)**m*((a + b*x**2)**2)**p, x)

Maxima [F]

\[ \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} \left (d x\right )^{m} \,d x } \]

[In]

integrate((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="maxima")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p*(d*x)^m, x)

Giac [F]

\[ \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int { {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} \left (d x\right )^{m} \,d x } \]

[In]

integrate((d*x)^m*(b^2*x^4+2*a*b*x^2+a^2)^p,x, algorithm="giac")

[Out]

integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p*(d*x)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (d x)^m \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx=\int {\left (d\,x\right )}^m\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p \,d x \]

[In]

int((d*x)^m*(a^2 + b^2*x^4 + 2*a*b*x^2)^p,x)

[Out]

int((d*x)^m*(a^2 + b^2*x^4 + 2*a*b*x^2)^p, x)