3.1114 \(\int (d x)^m (a+b x^2+c x^4)^p \, dx\)

Optimal. Leaf size=155 \[ \frac {(d x)^{m+1} \left (\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^2+c x^4\right )^p F_1\left (\frac {m+1}{2};-p,-p;\frac {m+3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1)} \]

[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)

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Rubi [A]  time = 0.10, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1141, 510} \[ \frac {(d x)^{m+1} \left (\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x^2}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x^2+c x^4\right )^p F_1\left (\frac {m+1}{2};-p,-p;\frac {m+3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1)} \]

Antiderivative was successfully verified.

[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 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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 1141

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

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Mathematica [A]  time = 0.24, size = 179, normalized size = 1.15 \[ \frac {x (d x)^m \left (\frac {-\sqrt {b^2-4 a c}+b+2 c x^2}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}\right )^{-p} \left (a+b x^2+c x^4\right )^p F_1\left (\frac {m+1}{2};-p,-p;\frac {m+3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}\right )}{m+1} \]

Warning: Unable to verify antiderivative.

[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)

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (c x^{4} + b x^{2} + a\right )}^{p} \left (d x\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \left (d x \right )^{m} \left (c \,x^{4}+b \,x^{2}+a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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