\(\int (d x)^m (a+b x^2+c x^4)^p \, dx\) [1114]

Optimal result

Integrand size = 20, antiderivative size = 155 \[ \int (d x)^m \left (a+b x^2+c x^4\right )^p \, dx=\frac {(d x)^{1+m} \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (\frac {1+m}{2},-p,-p,\frac {3+m}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (1+m)} \]

[Out]

(d*x)^(1+m)*(c*x^4+b*x^2+a)^p*AppellF1(1/2+1/2*m,-p,-p,3/2+1/2*m,-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^2/(b+(
-4*a*c+b^2)^(1/2)))/d/(1+m)/((1+2*c*x^2/(b-(-4*a*c+b^2)^(1/2)))^p)/((1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^p)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1155, 524} \[ \int (d x)^m \left (a+b x^2+c x^4\right )^p \, dx=\frac {(d x)^{m+1} \left (\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (\frac {m+1}{2},-p,-p,\frac {m+3}{2},-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1)} \]

[In]

Int[(d*x)^m*(a + b*x^2 + c*x^4)^p,x]

[Out]

((d*x)^(1 + m)*(a + b*x^2 + c*x^4)^p*AppellF1[(1 + m)/2, -p, -p, (3 + m)/2, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])
, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(d*(1 + m)*(1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x^2)/(b
+ Sqrt[b^2 - 4*a*c]))^p)

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 1155

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 + Rt[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^2/(b - Rt[b^2 - 4*a*c, 2
])))^FracPart[p])), Int[(d*x)^m*(1 + 2*c*(x^2/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2*c*(x^2/(b - Sqrt[b^2 - 4*a*c]
)))^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x]

Rubi steps \begin{align*} \text {integral}& = \left (\left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p\right ) \int (d x)^m \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^p \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^p \, dx \\ & = \frac {(d x)^{1+m} \left (1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p F_1\left (\frac {1+m}{2};-p,-p;\frac {3+m}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (1+m)} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 0.25 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.15 \[ \int (d x)^m \left (a+b x^2+c x^4\right )^p \, dx=\frac {x (d x)^m \left (\frac {b-\sqrt {b^2-4 a c}+2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x^2+c x^4\right )^p \operatorname {AppellF1}\left (\frac {1+m}{2},-p,-p,\frac {3+m}{2},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{-b+\sqrt {b^2-4 a c}}\right )}{1+m} \]

[In]

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

[Out]

(x*(d*x)^m*(a + b*x^2 + c*x^4)^p*AppellF1[(1 + m)/2, -p, -p, (3 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2
*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])])/((1 + m)*((b - Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b - Sqrt[b^2 - 4*a*c]))^p*((b
+ Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))^p)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((d*x)^m*(a + b*x^2 + c*x^4)^p,x)

[Out]

int((d*x)^m*(a + b*x^2 + c*x^4)^p, x)