3.1108 \(\int \frac{(d x)^m}{a+b x^2+c x^4} \, dx\)

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

[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))

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Rubi [A]  time = 0.250873, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {1131, 364} \[ \frac{2 c (d x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}\right )}{d (m+1) \sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}\right )}-\frac{2 c (d x)^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{d (m+1) \sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}+b\right )} \]

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 1131

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]

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])

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*}

Mathematica [C]  time = 0.0418205, size = 59, normalized size = 0.34 \[ \frac{x (d x)^m \text{RootSum}\left [\text{$\#$1}^2 b+\text{$\#$1}^4 c+a\& ,\frac{\, _2F_1\left (1,m+1;m+2;\frac{x}{\text{$\#$1}}\right )}{\text{$\#$1}^2 b+2 a}\& \right ]}{2 (m+1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.065, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx \right ) ^{m}}{c{x}^{4}+b{x}^{2}+a}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}}{c x^{4} + b x^{2} + a}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (d x\right )^{m}}{c x^{4} + b x^{2} + a}, x\right ) \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}}{a + b x^{2} + c x^{4}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d x\right )^{m}}{c x^{4} + b x^{2} + a}\,{d x} \end{align*}

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)