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

Optimal. Leaf size=224 \[ \frac {x \left (\frac {-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2}{\sqrt {b^2-4 a c}}-b e^2+2 c d e\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{c \left (b-\sqrt {b^2-4 a c}\right )}+\frac {x \left (-\frac {-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2}{\sqrt {b^2-4 a c}}-b e^2+2 c d e\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{c \left (\sqrt {b^2-4 a c}+b\right )}+\frac {e^2 x}{c} \]

[Out]

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

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Rubi [A]  time = 0.48, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1424, 1422, 245} \[ \frac {x \left (\frac {-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2}{\sqrt {b^2-4 a c}}-b e^2+2 c d e\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{c \left (b-\sqrt {b^2-4 a c}\right )}+\frac {x \left (-\frac {-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2}{\sqrt {b^2-4 a c}}-b e^2+2 c d e\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{c \left (\sqrt {b^2-4 a c}+b\right )}+\frac {e^2 x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*x)/c + ((2*c*d*e - b*e^2 + (2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2
F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(c*(b - Sqrt[b^2 - 4*a*c])) + ((2*c*d*e - b*e^2
- (2*c^2*d^2 + b^2*e^2 - 2*c*e*(b*d + a*e))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2
*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(c*(b + Sqrt[b^2 - 4*a*c]))

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 1422

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*
c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^n), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), In
t[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 - 4*a*c] ||  !IGtQ[n/2, 0])

Rule 1424

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Int[ExpandIntegran
d[(d + e*x^n)^q/(a + b*x^n + c*x^(2*n)), x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4
*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[q]

Rubi steps

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

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Mathematica [A]  time = 0.52, size = 216, normalized size = 0.96 \[ \frac {x \left (\frac {\left (\frac {-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2}{\sqrt {b^2-4 a c}}-b e^2+2 c d e\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};\frac {2 c x^n}{\sqrt {b^2-4 a c}-b}\right )}{b-\sqrt {b^2-4 a c}}+\frac {\left (-\frac {-2 c e (a e+b d)+b^2 e^2+2 c^2 d^2}{\sqrt {b^2-4 a c}}-b e^2+2 c d e\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}+b}+e^2\right )}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e^{2} x^{2 \, n} + 2 \, d e x^{n} + d^{2}}{c x^{2 \, n} + b x^{n} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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