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

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*(-2*a*e+b*d)*x^n)/a/(-4*a*c+b^2)/f/n/(a+b*x^n+c*x^(2*n))-c*(f*x)^(1+m)*hype
rgeom([1, (1+m)/n],[(1+m+n)/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))*((-2*a*e+b*d)*(1+m-n)+(-4*a*c*d*(1+m-2*n)+b^2*
d*(1+m-n)-2*a*b*e*n)/(-4*a*c+b^2)^(1/2))/a/(-4*a*c+b^2)/f/(1+m)/n/(b-(-4*a*c+b^2)^(1/2))-c*(f*x)^(1+m)*hyperge
om([1, (1+m)/n],[(1+m+n)/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))*((-2*a*e+b*d)*(1+m-n)+(4*a*c*d*(1+m-2*n)-b^2*d*(1
+m-n)+2*a*b*e*n)/(-4*a*c+b^2)^(1/2))/a/(-4*a*c+b^2)/f/(1+m)/n/(b+(-4*a*c+b^2)^(1/2))

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Rubi [A]  time = 1.38, antiderivative size = 374, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1558, 1560, 364} \[ -\frac {c (f x)^{m+1} \left ((m-n+1) (b d-2 a e)-\frac {2 a b e n+4 a c d (m-2 n+1)+b^2 (-d) (m-n+1)}{\sqrt {b^2-4 a c}}\right ) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a f (m+1) n \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right )}-\frac {c (f x)^{m+1} \left (\frac {2 a b e n+4 a c d (m-2 n+1)+b^2 (-d) (m-n+1)}{\sqrt {b^2-4 a c}}+(m-n+1) (b d-2 a e)\right ) \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a f (m+1) n \left (b^2-4 a c\right ) \left (\sqrt {b^2-4 a c}+b\right )}+\frac {(f x)^{m+1} \left (c x^n (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{a f n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )} \]

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)*Hypergeometric2F1[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 + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b + Sqrt[b^2 - 4*a*c])*f*(1 + m)*n)

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])

Rule 1558

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :>
-Simp[((f*x)^(m + 1)*(a + b*x^n + c*x^(2*n))^(p + 1)*(d*(b^2 - 2*a*c) - a*b*e + (b*d - 2*a*e)*c*x^n))/(a*f*n*
(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(a*n*(p + 1)*(b^2 - 4*a*c)), Int[(f*x)^m*(a + b*x^n + c*x^(2*n))^(p + 1)*S
imp[d*(b^2*(m + n*(p + 1) + 1) - 2*a*c*(m + 2*n*(p + 1) + 1)) - a*b*e*(m + 1) + (m + n*(2*p + 3) + 1)*(b*d - 2
*a*e)*c*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && ILtQ[p
+ 1, 0]

Rule 1560

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Sy
mbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^n)^q*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e,
f, m, n, p, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0])

Rubi steps

\begin {align*} \int \frac {(f x)^m \left (d+e x^n\right )}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx &=\frac {(f x)^{1+m} \left (b^2 d-2 a c d-a b e+c (b d-2 a e) x^n\right )}{a \left (b^2-4 a c\right ) f n \left (a+b x^n+c x^{2 n}\right )}-\frac {\int \frac {(f x)^m \left (-a b e (1+m)-2 a c d (1+m-2 n)+b^2 d (1+m-n)+c (b d-2 a e) (1+m-n) x^n\right )}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}\\ &=\frac {(f x)^{1+m} \left (b^2 d-2 a c d-a b e+c (b d-2 a e) x^n\right )}{a \left (b^2-4 a c\right ) f n \left (a+b x^n+c x^{2 n}\right )}-\frac {\int \left (\frac {\left (c (b d-2 a e) (1+m-n)+\frac {c \left (b^2 d-4 a c d+b^2 d m-4 a c d m-b^2 d n+8 a c d n-2 a b e n\right )}{\sqrt {b^2-4 a c}}\right ) (f x)^m}{b-\sqrt {b^2-4 a c}+2 c x^n}+\frac {\left (c (b d-2 a e) (1+m-n)-\frac {c \left (b^2 d-4 a c d+b^2 d m-4 a c d m-b^2 d n+8 a c d n-2 a b e n\right )}{\sqrt {b^2-4 a c}}\right ) (f x)^m}{b+\sqrt {b^2-4 a c}+2 c x^n}\right ) \, dx}{a \left (b^2-4 a c\right ) n}\\ &=\frac {(f x)^{1+m} \left (b^2 d-2 a c d-a b e+c (b d-2 a e) x^n\right )}{a \left (b^2-4 a c\right ) f n \left (a+b x^n+c x^{2 n}\right )}-\frac {\left (c \left ((b d-2 a e) (1+m-n)-\frac {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}}\right )\right ) \int \frac {(f x)^m}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right ) n}-\frac {\left (c \left ((b d-2 a e) (1+m-n)+\frac {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}}\right )\right ) \int \frac {(f x)^m}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right ) n}\\ &=\frac {(f x)^{1+m} \left (b^2 d-2 a c d-a b e+c (b d-2 a e) x^n\right )}{a \left (b^2-4 a c\right ) f n \left (a+b x^n+c x^{2 n}\right )}-\frac {c \left ((b d-2 a e) (1+m-n)-\frac {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}}\right ) (f x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right ) f (1+m) n}-\frac {c \left ((b d-2 a e) (1+m-n)+\frac {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}}\right ) (f x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b+\sqrt {b^2-4 a c}\right ) f (1+m) n}\\ \end {align*}

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Mathematica [B]  time = 6.52, 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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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{c^{2} x^{4 \, n} + b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2} + 2 \, {\left (b c x^{n} + a c\right )} x^{2 \, n}}, x\right ) \]

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, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \,x^{n}+d \right ) \left (f x \right )^{m}}{\left (b \,x^{n}+c \,x^{2 n}+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (b^{2} d f^{m} - {\left (2 \, c d f^{m} + b e f^{m}\right )} a\right )} x x^{m} + {\left (b c d f^{m} - 2 \, a c e f^{m}\right )} x e^{\left (m \log \relax (x) + n \log \relax (x)\right )}}{a^{2} b^{2} n - 4 \, a^{3} c n + {\left (a b^{2} c n - 4 \, a^{2} c^{2} n\right )} x^{2 \, n} + {\left (a b^{3} n - 4 \, a^{2} b c n\right )} x^{n}} - \int \frac {{\left (b^{2} d f^{m} {\left (m - n + 1\right )} - {\left (2 \, c d f^{m} {\left (m - 2 \, n + 1\right )} + b e f^{m} {\left (m + 1\right )}\right )} a\right )} x^{m} + {\left (b c d f^{m} {\left (m - n + 1\right )} - 2 \, a c e f^{m} {\left (m - n + 1\right )}\right )} e^{\left (m \log \relax (x) + n \log \relax (x)\right )}}{a^{2} b^{2} n - 4 \, a^{3} c n + {\left (a b^{2} c n - 4 \, a^{2} c^{2} n\right )} x^{2 \, n} + {\left (a b^{3} n - 4 \, a^{2} b c n\right )} x^{n}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n))^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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (f\,x\right )}^m\,\left (d+e\,x^n\right )}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^2} \,d x \]

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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