3.1.40 \(\int \frac {(d x)^m (A+B x+C x^2)}{a+b x^2+c x^4} \, dx\) [40]

Optimal. Leaf size=368 \[ \frac {\left (C+\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) d (1+m)}+\frac {\left (C-\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) (d x)^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) d (1+m)}+\frac {2 B c (d x)^{2+m} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right ) d^2 (2+m)}-\frac {2 B c (d x)^{2+m} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+m}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right ) d^2 (2+m)} \]

[Out]

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

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Rubi [A]
time = 0.43, antiderivative size = 368, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1676, 1299, 371, 12, 1145} \begin {gather*} \frac {(d x)^{m+1} \left (\frac {2 A c-b C}{\sqrt {b^2-4 a c}}+C\right ) \, _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) \left (b-\sqrt {b^2-4 a c}\right )}+\frac {(d x)^{m+1} \left (C-\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) \, _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) \left (\sqrt {b^2-4 a c}+b\right )}+\frac {2 B c (d x)^{m+2} \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{d^2 (m+2) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 B c (d x)^{m+2} \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{d^2 (m+2) \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 + C*x^2))/(a + b*x^2 + c*x^4),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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]

Rule 1299

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

Rule 1676

Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x],
k}, Int[(d*x)^m*Sum[Coeff[Pq, x, 2*k]*x^(2*k), {k, 0, q/2 + 1}]*(a + b*x^2 + c*x^4)^p, x] + Dist[1/d, Int[(d*
x)^(m + 1)*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0, (q - 1)/2 + 1}]*(a + b*x^2 + c*x^4)^p, x], x]] /; FreeQ[{
a, b, c, d, m, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2]

Rubi steps

\begin {align*} \int \frac {(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx &=\frac {\int \frac {B (d x)^{1+m}}{a+b x^2+c x^4} \, dx}{d}+\int \frac {(d x)^m \left (A+C x^2\right )}{a+b x^2+c x^4} \, dx\\ &=\frac {1}{2} \left (C-\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) \int \frac {(d x)^m}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^2} \, dx+\frac {1}{2} \left (C+\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) \int \frac {(d x)^m}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^2} \, dx+\frac {B \int \frac {(d x)^{1+m}}{a+b x^2+c x^4} \, dx}{d}\\ &=\frac {\left (C+\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) (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 )}{\left (b-\sqrt {b^2-4 a c}\right ) d (1+m)}+\frac {\left (C-\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) (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 )}{\left (b+\sqrt {b^2-4 a c}\right ) d (1+m)}+\frac {(B c) \int \frac {(d x)^{1+m}}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^2} \, dx}{\sqrt {b^2-4 a c} d}-\frac {(B c) \int \frac {(d x)^{1+m}}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^2} \, dx}{\sqrt {b^2-4 a c} d}\\ &=\frac {\left (C+\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) (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 )}{\left (b-\sqrt {b^2-4 a c}\right ) d (1+m)}+\frac {\left (C-\frac {2 A c-b C}{\sqrt {b^2-4 a c}}\right ) (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 )}{\left (b+\sqrt {b^2-4 a c}\right ) d (1+m)}+\frac {2 B c (d x)^{2+m} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+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^2 (2+m)}-\frac {2 B c (d x)^{2+m} \, _2F_1\left (1,\frac {2+m}{2};\frac {4+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^2 (2+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 = 1.15, size = 438, normalized size = 1.19 \begin {gather*} \frac {(d x)^m \left (A \left (2+3 m+m^2\right ) \text {RootSum}\left [a+b \text {$\#$1}^2+c \text {$\#$1}^4\&,\frac {\, _2F_1\left (-m,-m;1-m;-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m}}{b \text {$\#$1}+2 c \text {$\#$1}^3}\&\right ]+B (2+m) \text {RootSum}\left [a+b \text {$\#$1}^2+c \text {$\#$1}^4\&,\frac {m x+\, _2F_1\left (-m,-m;1-m;-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}+m \, _2F_1\left (-m,-m;1-m;-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}}{b \text {$\#$1}+2 c \text {$\#$1}^3}\&\right ]+C \text {RootSum}\left [a+b \text {$\#$1}^2+c \text {$\#$1}^4\&,\frac {m x^2+m^2 x^2+2 m x \text {$\#$1}+m^2 x \text {$\#$1}+2 \, _2F_1\left (-m,-m;1-m;-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+3 m \, _2F_1\left (-m,-m;1-m;-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+m^2 \, _2F_1\left (-m,-m;1-m;-\frac {\text {$\#$1}}{x-\text {$\#$1}}\right ) \left (\frac {x}{x-\text {$\#$1}}\right )^{-m} \text {$\#$1}^2+m \left (\frac {x}{\text {$\#$1}}\right )^{-m} \text {$\#$1}^2}{b \text {$\#$1}+2 c \text {$\#$1}^3}\&\right ]\right )}{2 m (1+m) (2+m)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((d*x)^m*(C*x^2+B*x+A)/(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^2+B*x+A)/(c*x^4+b*x^2+a),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.36, size = 32, normalized size = 0.09 \begin {gather*} {\rm integral}\left (\frac {{\left (C x^{2} + B x + A\right )} \left (d x\right )^{m}}{c x^{4} + b x^{2} + a}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*x^2 + B*x + A)*(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} \left (A + B x + C x^{2}\right )}{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**2+B*x+A)/(c*x**4+b*x**2+a),x)

[Out]

Integral((d*x)**m*(A + B*x + C*x**2)/(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^2+B*x+A)/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

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

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

[Out]

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

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