3.542 \(\int \frac {x^3 (e+f x)^n}{a+b x+c x^2} \, dx\)

Optimal. Leaf size=290 \[ \frac {\left (\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}+a-\frac {b^2}{c}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (b-\sqrt {b^2-4 a c}\right )\right )}+\frac {\left (-\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}+a-\frac {b^2}{c}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (\sqrt {b^2-4 a c}+b\right )\right )}-\frac {(b f+c e) (e+f x)^{n+1}}{c^2 f^2 (n+1)}+\frac {(e+f x)^{n+2}}{c f^2 (n+2)} \]

[Out]

-(b*f+c*e)*(f*x+e)^(1+n)/c^2/f^2/(1+n)+(f*x+e)^(2+n)/c/f^2/(2+n)+(f*x+e)^(1+n)*hypergeom([1, 1+n],[2+n],2*c*(f
*x+e)/(2*c*e-f*(b-(-4*a*c+b^2)^(1/2))))*(a-1/c*b^2+b*(-3*a*c+b^2)/c/(-4*a*c+b^2)^(1/2))/c/(1+n)/(2*c*e-f*(b-(-
4*a*c+b^2)^(1/2)))+(f*x+e)^(1+n)*hypergeom([1, 1+n],[2+n],2*c*(f*x+e)/(2*c*e-f*(b+(-4*a*c+b^2)^(1/2))))*(a-1/c
*b^2-b*(-3*a*c+b^2)/c/(-4*a*c+b^2)^(1/2))/c/(1+n)/(2*c*e-f*(b+(-4*a*c+b^2)^(1/2)))

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Rubi [A]  time = 0.77, antiderivative size = 290, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1628, 68} \[ \frac {\left (\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}+a-\frac {b^2}{c}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (b-\sqrt {b^2-4 a c}\right )\right )}+\frac {\left (-\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}+a-\frac {b^2}{c}\right ) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (\sqrt {b^2-4 a c}+b\right )\right )}-\frac {(b f+c e) (e+f x)^{n+1}}{c^2 f^2 (n+1)}+\frac {(e+f x)^{n+2}}{c f^2 (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(e + f*x)^n)/(a + b*x + c*x^2),x]

[Out]

-(((c*e + b*f)*(e + f*x)^(1 + n))/(c^2*f^2*(1 + n))) + (e + f*x)^(2 + n)/(c*f^2*(2 + n)) + ((a - b^2/c + (b*(b
^2 - 3*a*c))/(c*Sqrt[b^2 - 4*a*c]))*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*
e - (b - Sqrt[b^2 - 4*a*c])*f)])/(c*(2*c*e - (b - Sqrt[b^2 - 4*a*c])*f)*(1 + n)) + ((a - b^2/c - (b*(b^2 - 3*a
*c))/(c*Sqrt[b^2 - 4*a*c]))*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*e - (b +
Sqrt[b^2 - 4*a*c])*f)])/(c*(2*c*e - (b + Sqrt[b^2 - 4*a*c])*f)*(1 + n))

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 1628

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {align*} \int \frac {x^3 (e+f x)^n}{a+b x+c x^2} \, dx &=\int \left (\frac {(-c e-b f) (e+f x)^n}{c^2 f}+\frac {\left (\frac {b^2}{c^2}-\frac {a}{c}-\frac {b \left (b^2-3 a c\right )}{c^2 \sqrt {b^2-4 a c}}\right ) (e+f x)^n}{b-\sqrt {b^2-4 a c}+2 c x}+\frac {\left (\frac {b^2}{c^2}-\frac {a}{c}+\frac {b \left (b^2-3 a c\right )}{c^2 \sqrt {b^2-4 a c}}\right ) (e+f x)^n}{b+\sqrt {b^2-4 a c}+2 c x}+\frac {(e+f x)^{1+n}}{c f}\right ) \, dx\\ &=-\frac {(c e+b f) (e+f x)^{1+n}}{c^2 f^2 (1+n)}+\frac {(e+f x)^{2+n}}{c f^2 (2+n)}+\left (\frac {b^2}{c^2}-\frac {a}{c}+\frac {b \left (b^2-3 a c\right )}{c^2 \sqrt {b^2-4 a c}}\right ) \int \frac {(e+f x)^n}{b+\sqrt {b^2-4 a c}+2 c x} \, dx-\frac {\left (a-\frac {b^2}{c}+\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\right ) \int \frac {(e+f x)^n}{b-\sqrt {b^2-4 a c}+2 c x} \, dx}{c}\\ &=-\frac {(c e+b f) (e+f x)^{1+n}}{c^2 f^2 (1+n)}+\frac {(e+f x)^{2+n}}{c f^2 (2+n)}+\frac {\left (a-\frac {b^2}{c}+\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{c \left (2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}+\frac {\left (a-\frac {b^2}{c}-\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{c \left (2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.82, size = 261, normalized size = 0.90 \[ \frac {(e+f x)^{n+1} \left (\frac {c \left (\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}+a-\frac {b^2}{c}\right ) \, _2F_1\left (1,n+1;n+2;\frac {2 c (e+f x)}{2 c e+\left (\sqrt {b^2-4 a c}-b\right ) f}\right )}{(n+1) \left (f \left (\sqrt {b^2-4 a c}-b\right )+2 c e\right )}+\frac {c \left (-\frac {b \left (b^2-3 a c\right )}{c \sqrt {b^2-4 a c}}+a-\frac {b^2}{c}\right ) \, _2F_1\left (1,n+1;n+2;\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{(n+1) \left (2 c e-f \left (\sqrt {b^2-4 a c}+b\right )\right )}-\frac {b f+c e}{f^2 (n+1)}+\frac {c (e+f x)}{f^2 (n+2)}\right )}{c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(e + f*x)^n)/(a + b*x + c*x^2),x]

[Out]

((e + f*x)^(1 + n)*(-((c*e + b*f)/(f^2*(1 + n))) + (c*(e + f*x))/(f^2*(2 + n)) + (c*(a - b^2/c + (b*(b^2 - 3*a
*c))/(c*Sqrt[b^2 - 4*a*c]))*Hypergeometric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*e + (-b + Sqrt[b^2 - 4*a*c
])*f)])/((2*c*e + (-b + Sqrt[b^2 - 4*a*c])*f)*(1 + n)) + (c*(a - b^2/c - (b*(b^2 - 3*a*c))/(c*Sqrt[b^2 - 4*a*c
]))*Hypergeometric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*e - (b + Sqrt[b^2 - 4*a*c])*f)])/((2*c*e - (b + Sq
rt[b^2 - 4*a*c])*f)*(1 + n))))/c^2

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fricas [F]  time = 1.07, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (f x + e\right )}^{n} x^{3}}{c x^{2} + b x + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(f*x+e)^n/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

integral((f*x + e)^n*x^3/(c*x^2 + b*x + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{n} x^{3}}{c x^{2} + b x + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(f*x+e)^n/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

integrate((f*x + e)^n*x^3/(c*x^2 + b*x + a), x)

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \left (f x +e \right )^{n}}{c \,x^{2}+b x +a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(f*x+e)^n/(c*x^2+b*x+a),x)

[Out]

int(x^3*(f*x+e)^n/(c*x^2+b*x+a),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(f*x+e)^n/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

integrate((f*x + e)^n*x^3/(c*x^2 + b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(e + f*x)^n)/(a + b*x + c*x^2),x)

[Out]

int((x^3*(e + f*x)^n)/(a + b*x + c*x^2), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(f*x+e)**n/(c*x**2+b*x+a),x)

[Out]

Exception raised: HeuristicGCDFailed

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