3.142 \(\int \frac{(f x)^m \left (d+e x^n\right )}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx\)

Optimal. Leaf size=374 \[ -\frac{c (f x)^{m+1} \left ((m-n+1) (b d-2 a e)-\frac{2 a b e n+4 a c d (m-2 n+1)+b^2 (-d) (m-n+1)}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{a f (m+1) n \left (b^2-4 a c\right ) \left (b-\sqrt{b^2-4 a c}\right )}-\frac{c (f x)^{m+1} \left (\frac{2 a b e n+4 a c d (m-2 n+1)+b^2 (-d) (m-n+1)}{\sqrt{b^2-4 a c}}+(m-n+1) (b d-2 a e)\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{a f (m+1) n \left (b^2-4 a c\right ) \left (\sqrt{b^2-4 a c}+b\right )}+\frac{(f x)^{m+1} \left (c x^n (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{a f n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )} \]

[Out]

((f*x)^(1 + m)*(b^2*d - 2*a*c*d - a*b*e + c*(b*d - 2*a*e)*x^n))/(a*(b^2 - 4*a*c)
*f*n*(a + b*x^n + c*x^(2*n))) - (c*((b*d - 2*a*e)*(1 + m - n) - (4*a*c*d*(1 + m
- 2*n) - b^2*d*(1 + m - n) + 2*a*b*e*n)/Sqrt[b^2 - 4*a*c])*(f*x)^(1 + m)*Hyperge
ometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*
(b^2 - 4*a*c)*(b - Sqrt[b^2 - 4*a*c])*f*(1 + m)*n) - (c*((b*d - 2*a*e)*(1 + m -
n) + (4*a*c*d*(1 + m - 2*n) - b^2*d*(1 + m - n) + 2*a*b*e*n)/Sqrt[b^2 - 4*a*c])*
(f*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b + Sqr
t[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b + Sqrt[b^2 - 4*a*c])*f*(1 + m)*n)

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Rubi [A]  time = 2.72955, antiderivative size = 374, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103 \[ -\frac{c (f x)^{m+1} \left ((m-n+1) (b d-2 a e)-\frac{2 a b e n+4 a c d (m-2 n+1)+b^2 (-d) (m-n+1)}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{a f (m+1) n \left (b^2-4 a c\right ) \left (b-\sqrt{b^2-4 a c}\right )}-\frac{c (f x)^{m+1} \left (\frac{2 a b e n+4 a c d (m-2 n+1)+b^2 (-d) (m-n+1)}{\sqrt{b^2-4 a c}}+(m-n+1) (b d-2 a e)\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{a f (m+1) n \left (b^2-4 a c\right ) \left (\sqrt{b^2-4 a c}+b\right )}+\frac{(f x)^{m+1} \left (c x^n (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{a f n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )} \]

Antiderivative was successfully verified.

[In]  Int[((f*x)^m*(d + e*x^n))/(a + b*x^n + c*x^(2*n))^2,x]

[Out]

((f*x)^(1 + m)*(b^2*d - 2*a*c*d - a*b*e + c*(b*d - 2*a*e)*x^n))/(a*(b^2 - 4*a*c)
*f*n*(a + b*x^n + c*x^(2*n))) - (c*((b*d - 2*a*e)*(1 + m - n) - (4*a*c*d*(1 + m
- 2*n) - b^2*d*(1 + m - n) + 2*a*b*e*n)/Sqrt[b^2 - 4*a*c])*(f*x)^(1 + m)*Hyperge
ometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*
(b^2 - 4*a*c)*(b - Sqrt[b^2 - 4*a*c])*f*(1 + m)*n) - (c*((b*d - 2*a*e)*(1 + m -
n) + (4*a*c*d*(1 + m - 2*n) - b^2*d*(1 + m - n) + 2*a*b*e*n)/Sqrt[b^2 - 4*a*c])*
(f*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b + Sqr
t[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b + Sqrt[b^2 - 4*a*c])*f*(1 + m)*n)

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Rubi in Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \[ \text{Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((f*x)**m*(d+e*x**n)/(a+b*x**n+c*x**(2*n))**2,x)

[Out]

Timed out

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Mathematica [B]  time = 6.59179, size = 5363, normalized size = 14.34 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]  Integrate[((f*x)^m*(d + e*x^n))/(a + b*x^n + c*x^(2*n))^2,x]

[Out]

Result too large to show

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Maple [F]  time = 0.06, size = 0, normalized size = 0. \[ \int{\frac{ \left ( fx \right ) ^{m} \left ( d+e{x}^{n} \right ) }{ \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) ^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n))^2,x)

[Out]

int((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n))^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \frac{{\left (b^{2} d f^{m} -{\left (2 \, c d f^{m} + b e f^{m}\right )} a\right )} x x^{m} +{\left (b c d f^{m} - 2 \, a c e f^{m}\right )} x e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{a^{2} b^{2} n - 4 \, a^{3} c n +{\left (a b^{2} c n - 4 \, a^{2} c^{2} n\right )} x^{2 \, n} +{\left (a b^{3} n - 4 \, a^{2} b c n\right )} x^{n}} - \int \frac{{\left (b^{2} d f^{m}{\left (m - n + 1\right )} -{\left (2 \, c d f^{m}{\left (m - 2 \, n + 1\right )} + b e f^{m}{\left (m + 1\right )}\right )} a\right )} x^{m} +{\left (b c d f^{m}{\left (m - n + 1\right )} - 2 \, a c e f^{m}{\left (m - n + 1\right )}\right )} e^{\left (m \log \left (x\right ) + n \log \left (x\right )\right )}}{a^{2} b^{2} n - 4 \, a^{3} c n +{\left (a b^{2} c n - 4 \, a^{2} c^{2} n\right )} x^{2 \, n} +{\left (a b^{3} n - 4 \, a^{2} b c n\right )} x^{n}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x^n + d)*(f*x)^m/(c*x^(2*n) + b*x^n + a)^2,x, algorithm="maxima")

[Out]

((b^2*d*f^m - (2*c*d*f^m + b*e*f^m)*a)*x*x^m + (b*c*d*f^m - 2*a*c*e*f^m)*x*e^(m*
log(x) + n*log(x)))/(a^2*b^2*n - 4*a^3*c*n + (a*b^2*c*n - 4*a^2*c^2*n)*x^(2*n) +
(a*b^3*n - 4*a^2*b*c*n)*x^n) - integrate(((b^2*d*f^m*(m - n + 1) - (2*c*d*f^m*(
m - 2*n + 1) + b*e*f^m*(m + 1))*a)*x^m + (b*c*d*f^m*(m - n + 1) - 2*a*c*e*f^m*(m
- n + 1))*e^(m*log(x) + n*log(x)))/(a^2*b^2*n - 4*a^3*c*n + (a*b^2*c*n - 4*a^2*
c^2*n)*x^(2*n) + (a*b^3*n - 4*a^2*b*c*n)*x^n), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left (\frac{{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{c^{2} x^{4 \, n} + 2 \, a b x^{n} + a^{2} +{\left (2 \, b c x^{n} + b^{2} + 2 \, a c\right )} x^{2 \, n}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x^n + d)*(f*x)^m/(c*x^(2*n) + b*x^n + a)^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((f*x)**m*(d+e*x**n)/(a+b*x**n+c*x**(2*n))**2,x)

[Out]

Timed out

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((e*x^n + d)*(f*x)^m/(c*x^(2*n) + b*x^n + a)^2,x, algorithm="giac")

[Out]

integrate((e*x^n + d)*(f*x)^m/(c*x^(2*n) + b*x^n + a)^2, x)