3.2.41 \(\int \frac {(f x)^m (d+e x^n)}{a+b x^n+c x^{2 n}} \, dx\) [141]

Optimal. Leaf size=196 \[ \frac {\left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b-\sqrt {b^2-4 a c}\right ) f (1+m)}+\frac {\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) (f x)^{1+m} \, _2F_1\left (1,\frac {1+m}{n};\frac {1+m+n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b+\sqrt {b^2-4 a c}\right ) f (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1574, 371} \begin {gather*} \frac {(f x)^{m+1} \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+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 )}{f (m+1) \left (b-\sqrt {b^2-4 a c}\right )}+\frac {(f x)^{m+1} \left (e-\frac {2 c d-b e}{\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 )}{f (m+1) \left (\sqrt {b^2-4 a c}+b\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1574

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

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Mathematica [A]
time = 0.64, size = 318, normalized size = 1.62 \begin {gather*} \frac {2^{-\frac {1+m+n}{n}} x (f x)^m \left (2^{\frac {1+m+n}{n}} \sqrt {b^2-4 a c} d-\left (b d+\sqrt {b^2-4 a c} d-2 a e\right ) \left (\frac {c x^n}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )^{-\frac {1+m}{n}} \, _2F_1\left (-\frac {1+m}{n},-\frac {1+m}{n};1-\frac {1+m}{n};\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )-\left (-b d+\sqrt {b^2-4 a c} d+2 a e\right ) \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-\frac {1+m}{n}} \, _2F_1\left (-\frac {1+m}{n},-\frac {1+m}{n};1-\frac {1+m}{n};\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )\right )}{a \sqrt {b^2-4 a c} (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (f x \right )^{m} \left (d +e \,x^{n}\right )}{a +b \,x^{n}+c \,x^{2 n}}\, dx\]

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)),x)

[Out]

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

[Out]

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

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

[Out]

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

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

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)),x)

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (f\,x\right )}^m\,\left (d+e\,x^n\right )}{a+b\,x^n+c\,x^{2\,n}} \,d x \end {gather*}

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)),x)

[Out]

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

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