3.6.42 \(\int \frac {x^3 (e+f x)^n}{a+b x+c x^2} \, dx\) [542]

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

[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-b^2/c+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-b^2/c
-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 = 0.53, 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 = {1642, 70} \begin {gather*} \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)} \end {gather*}

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 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 1642

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 = 0.60, size = 353, normalized size = 1.22 \begin {gather*} \frac {2^{-1-n} (e+f x)^n \left (\left (-b^3 f+3 a b c f+b^2 \sqrt {\left (b^2-4 a c\right ) f^2}-a c \sqrt {\left (b^2-4 a c\right ) f^2}\right ) \left (\frac {c (e+f x)}{b f-\sqrt {\left (b^2-4 a c\right ) f^2}+2 c f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;\frac {2 c e-b f+\sqrt {\left (b^2-4 a c\right ) f^2}}{-b f+\sqrt {\left (b^2-4 a c\right ) f^2}-2 c f x}\right )+\left (b^3 f-3 a b c f+b^2 \sqrt {\left (b^2-4 a c\right ) f^2}-a c \sqrt {\left (b^2-4 a c\right ) f^2}\right ) \left (\frac {c (e+f x)}{b f+\sqrt {\left (b^2-4 a c\right ) f^2}+2 c f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;\frac {-2 c e+b f+\sqrt {\left (b^2-4 a c\right ) f^2}}{b f+\sqrt {\left (b^2-4 a c\right ) f^2}+2 c f x}\right )\right )}{c^3 \sqrt {\left (b^2-4 a c\right ) f^2} n} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2^(-1 - n)*(e + f*x)^n*(((-(b^3*f) + 3*a*b*c*f + b^2*Sqrt[(b^2 - 4*a*c)*f^2] - a*c*Sqrt[(b^2 - 4*a*c)*f^2])*H
ypergeometric2F1[-n, -n, 1 - n, (2*c*e - b*f + Sqrt[(b^2 - 4*a*c)*f^2])/(-(b*f) + Sqrt[(b^2 - 4*a*c)*f^2] - 2*
c*f*x)])/((c*(e + f*x))/(b*f - Sqrt[(b^2 - 4*a*c)*f^2] + 2*c*f*x))^n + ((b^3*f - 3*a*b*c*f + b^2*Sqrt[(b^2 - 4
*a*c)*f^2] - a*c*Sqrt[(b^2 - 4*a*c)*f^2])*Hypergeometric2F1[-n, -n, 1 - n, (-2*c*e + b*f + Sqrt[(b^2 - 4*a*c)*
f^2])/(b*f + Sqrt[(b^2 - 4*a*c)*f^2] + 2*c*f*x)])/((c*(e + f*x))/(b*f + Sqrt[(b^2 - 4*a*c)*f^2] + 2*c*f*x))^n)
)/(c^3*Sqrt[(b^2 - 4*a*c)*f^2]*n)

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Maple [F]
time = 0.07, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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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(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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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

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 >> no luck

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

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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