3.40 \(\int \frac{(d x)^m \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx\)

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

[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)*Hypergeometric2F
1[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))

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Rubi [A]  time = 1.48125, antiderivative size = 368, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167 \[ \frac{(d x)^{m+1} \left (\frac{2 A c-b C}{\sqrt{b^2-4 a c}}+C\right ) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}\right )}{d (m+1) \left (b-\sqrt{b^2-4 a c}\right )}+\frac{(d x)^{m+1} \left (C-\frac{2 A c-b C}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{d (m+1) \left (\sqrt{b^2-4 a c}+b\right )}+\frac{2 B c (d x)^{m+2} \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}\right )}{d^2 (m+2) \sqrt{b^2-4 a c} \left (b-\sqrt{b^2-4 a c}\right )}-\frac{2 B c (d x)^{m+2} \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )}{d^2 (m+2) \sqrt{b^2-4 a c} \left (\sqrt{b^2-4 a c}+b\right )} \]

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)*Hypergeometric2F
1[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))

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Rubi in Sympy [A]  time = 117.687, size = 335, normalized size = 0.91 \[ - \frac{2 B c \left (d x\right )^{m + 2}{{}_{2}F_{1}\left (\begin{matrix} 1, \frac{m}{2} + 1 \\ \frac{m}{2} + 2 \end{matrix}\middle |{- \frac{2 c x^{2}}{b + \sqrt{- 4 a c + b^{2}}}} \right )}}{d^{2} \left (b + \sqrt{- 4 a c + b^{2}}\right ) \left (m + 2\right ) \sqrt{- 4 a c + b^{2}}} + \frac{2 B c \left (d x\right )^{m + 2}{{}_{2}F_{1}\left (\begin{matrix} 1, \frac{m}{2} + 1 \\ \frac{m}{2} + 2 \end{matrix}\middle |{- \frac{2 c x^{2}}{b - \sqrt{- 4 a c + b^{2}}}} \right )}}{d^{2} \left (b - \sqrt{- 4 a c + b^{2}}\right ) \left (m + 2\right ) \sqrt{- 4 a c + b^{2}}} - \frac{\left (d x\right )^{m + 1} \left (2 A c - C b - C \sqrt{- 4 a c + b^{2}}\right ){{}_{2}F_{1}\left (\begin{matrix} 1, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{- \frac{2 c x^{2}}{b + \sqrt{- 4 a c + b^{2}}}} \right )}}{d \left (b + \sqrt{- 4 a c + b^{2}}\right ) \left (m + 1\right ) \sqrt{- 4 a c + b^{2}}} + \frac{\left (d x\right )^{m + 1} \left (2 A c - C b + C \sqrt{- 4 a c + b^{2}}\right ){{}_{2}F_{1}\left (\begin{matrix} 1, \frac{m}{2} + \frac{1}{2} \\ \frac{m}{2} + \frac{3}{2} \end{matrix}\middle |{- \frac{2 c x^{2}}{b - \sqrt{- 4 a c + b^{2}}}} \right )}}{d \left (b - \sqrt{- 4 a c + b^{2}}\right ) \left (m + 1\right ) \sqrt{- 4 a c + b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-2*B*c*(d*x)**(m + 2)*hyper((1, m/2 + 1), (m/2 + 2,), -2*c*x**2/(b + sqrt(-4*a*c
+ b**2)))/(d**2*(b + sqrt(-4*a*c + b**2))*(m + 2)*sqrt(-4*a*c + b**2)) + 2*B*c*
(d*x)**(m + 2)*hyper((1, m/2 + 1), (m/2 + 2,), -2*c*x**2/(b - sqrt(-4*a*c + b**2
)))/(d**2*(b - sqrt(-4*a*c + b**2))*(m + 2)*sqrt(-4*a*c + b**2)) - (d*x)**(m + 1
)*(2*A*c - C*b - C*sqrt(-4*a*c + b**2))*hyper((1, m/2 + 1/2), (m/2 + 3/2,), -2*c
*x**2/(b + sqrt(-4*a*c + b**2)))/(d*(b + sqrt(-4*a*c + b**2))*(m + 1)*sqrt(-4*a*
c + b**2)) + (d*x)**(m + 1)*(2*A*c - C*b + C*sqrt(-4*a*c + b**2))*hyper((1, m/2
+ 1/2), (m/2 + 3/2,), -2*c*x**2/(b - sqrt(-4*a*c + b**2)))/(d*(b - sqrt(-4*a*c +
b**2))*(m + 1)*sqrt(-4*a*c + b**2))

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Mathematica [C]  time = 0.428581, size = 438, normalized size = 1.19 \[ \frac{(d x)^m \left (A \left (m^2+3 m+2\right ) \text{RootSum}\left [\text{$\#$1}^4 c+\text{$\#$1}^2 b+a\&,\frac{\left (\frac{x}{x-\text{$\#$1}}\right )^{-m} \, _2F_1\left (-m,-m;1-m;-\frac{\text{$\#$1}}{x-\text{$\#$1}}\right )}{2 \text{$\#$1}^3 c+\text{$\#$1} b}\&\right ]+B (m+2) \text{RootSum}\left [\text{$\#$1}^4 c+\text{$\#$1}^2 b+a\&,\frac{\text{$\#$1} m \left (\frac{x}{x-\text{$\#$1}}\right )^{-m} \, _2F_1\left (-m,-m;1-m;-\frac{\text{$\#$1}}{x-\text{$\#$1}}\right )+\text{$\#$1} \left (\frac{x}{x-\text{$\#$1}}\right )^{-m} \, _2F_1\left (-m,-m;1-m;-\frac{\text{$\#$1}}{x-\text{$\#$1}}\right )+m x}{2 \text{$\#$1}^3 c+\text{$\#$1} b}\&\right ]+C \text{RootSum}\left [\text{$\#$1}^4 c+\text{$\#$1}^2 b+a\&,\frac{\text{$\#$1}^2 m^2 \left (\frac{x}{x-\text{$\#$1}}\right )^{-m} \, _2F_1\left (-m,-m;1-m;-\frac{\text{$\#$1}}{x-\text{$\#$1}}\right )+3 \text{$\#$1}^2 m \left (\frac{x}{x-\text{$\#$1}}\right )^{-m} \, _2F_1\left (-m,-m;1-m;-\frac{\text{$\#$1}}{x-\text{$\#$1}}\right )+2 \text{$\#$1}^2 \left (\frac{x}{x-\text{$\#$1}}\right )^{-m} \, _2F_1\left (-m,-m;1-m;-\frac{\text{$\#$1}}{x-\text{$\#$1}}\right )+\text{$\#$1}^2 m \left (\frac{x}{\text{$\#$1}}\right )^{-m}+\text{$\#$1} m^2 x+2 \text{$\#$1} m x+m^2 x^2+m x^2}{2 \text{$\#$1}^3 c+\text{$\#$1} b}\&\right ]\right )}{2 m (m+1) (m+2)} \]

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 = 0.041, size = 0, normalized size = 0. \[ \int{\frac{ \left ( dx \right ) ^{m} \left ( C{x}^{2}+Bx+A \right ) }{c{x}^{4}+b{x}^{2}+a}}\, dx \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((C*x^2 + B*x + A)*(d*x)^m/(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., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (C x^{2} + B x + A\right )} \left (d x\right )^{m}}{c x^{4} + b x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((C*x^2 + B*x + A)*(d*x)^m/(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., size = 0, normalized size = 0. \[ \int \frac{\left (d x\right )^{m} \left (A + B x + C x^{2}\right )}{a + b x^{2} + c x^{4}}\, dx \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((C*x^2 + B*x + A)*(d*x)^m/(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)