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

Optimal. Leaf size=175 \[ \frac {2 c (d x)^{m+1} \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{d (m+1) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c (d x)^{m+1} \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1383, 364} \[ \frac {2 c (d x)^{m+1} \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{d (m+1) \sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {2 c (d x)^{m+1} \, _2F_1\left (1,\frac {m+1}{n};\frac {m+n+1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1) \sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1383

Int[((d_.)*(x_))^(m_.)/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]
}, Dist[(2*c)/q, Int[(d*x)^m/(b - q + 2*c*x^n), x], x] - Dist[(2*c)/q, Int[(d*x)^m/(b + q + 2*c*x^n), x], x]]
/; FreeQ[{a, b, c, d, m, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

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

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Mathematica [A]  time = 0.85, size = 307, normalized size = 1.75 \[ -\frac {x (d x)^m \left (\frac {2 c \left (1-2^{-\frac {m+1}{n}} \left (\frac {c x^n}{-\sqrt {b^2-4 a c}+b+2 c x^n}\right )^{-\frac {m+1}{n}} \, _2F_1\left (-\frac {m+1}{n},-\frac {m+1}{n};1-\frac {m+1}{n};\frac {b-\sqrt {b^2-4 a c}}{2 c x^n+b-\sqrt {b^2-4 a c}}\right )\right )}{-b \sqrt {b^2-4 a c}-4 a c+b^2}+\frac {2 c \left (1-2^{-\frac {m+1}{n}} \left (\frac {c x^n}{\sqrt {b^2-4 a c}+b+2 c x^n}\right )^{-\frac {m+1}{n}} \, _2F_1\left (-\frac {m+1}{n},-\frac {m+1}{n};\frac {-m+n-1}{n};\frac {b+\sqrt {b^2-4 a c}}{2 c x^n+b+\sqrt {b^2-4 a c}}\right )\right )}{\sqrt {b^2-4 a c} \left (\sqrt {b^2-4 a c}+b\right )}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (d x\right )^{m}}{c x^{2 \, n} + b x^{n} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {\left (d x \right )^{m}}{b \,x^{n}+c \,x^{2 n}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*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 \[ \int \frac {{\left (d\,x\right )}^m}{a+b\,x^n+c\,x^{2\,n}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d*x)**m/(a + b*x**n + c*x**(2*n)), x)

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