3.12.8 \(\int \frac {(d x)^m}{a+b x^2+c x^4} \, dx\) [1108]

Optimal. Leaf size=173 \[ \frac {2 c (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) d (1+m)}-\frac {2 c (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) d (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1145, 371} \begin {gather*} \frac {2 c (d x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{d (m+1) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c (d x)^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*x)^m/(a + b*x^2 + c*x^4),x]

[Out]

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

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 1145

Int[((d_.)*(x_))^(m_.)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
c/q, Int[(d*x)^m/(b/2 - q/2 + c*x^2), x], x] - Dist[c/q, Int[(d*x)^m/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a,
 b, c, d, m}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
time = 0.08, size = 82, normalized size = 0.47 \begin {gather*} \frac {(d x)^m \text {RootSum}\left [a+b \text {$\#$1}^2+c \text {$\#$1}^4\&,\frac {\, _2F_1\left (-m,-m;1-m;-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m}}{b \text {$\#$1}+2 c \text {$\#$1}^3}\&\right ]}{2 m} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (d x \right )^{m}}{c \,x^{4}+b \,x^{2}+a}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d x\right )^{m}}{a + b x^{2} + c x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d*x)**m/(a + b*x**2 + c*x**4), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d\,x\right )}^m}{c\,x^4+b\,x^2+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^m/(a + b*x^2 + c*x^4),x)

[Out]

int((d*x)^m/(a + b*x^2 + c*x^4), x)

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