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

Optimal. Leaf size=222 \[ -\frac {b^3 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b (e+f x)}{b e-a f}\right )}{a^2 (n+1) (b c-a d) (b e-a f)}+\frac {(a d+b c) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {f x}{e}+1\right )}{a^2 c^2 e (n+1)}+\frac {d^3 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {d (e+f x)}{d e-c f}\right )}{c^2 (n+1) (b c-a d) (d e-c f)}+\frac {f (e+f x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac {f x}{e}+1\right )}{a c e^2 (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {180, 65, 68} \[ -\frac {b^3 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b (e+f x)}{b e-a f}\right )}{a^2 (n+1) (b c-a d) (b e-a f)}+\frac {(a d+b c) (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {f x}{e}+1\right )}{a^2 c^2 e (n+1)}+\frac {d^3 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {d (e+f x)}{d e-c f}\right )}{c^2 (n+1) (b c-a d) (d e-c f)}+\frac {f (e+f x)^{n+1} \, _2F_1\left (2,n+1;n+2;\frac {f x}{e}+1\right )}{a c e^2 (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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

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 180

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 177, normalized size = 0.80 \[ \frac {(e+f x)^{n+1} \left (\frac {\frac {e (a d+b c) \, _2F_1\left (1,n+1;n+2;\frac {f x}{e}+1\right )+a c f \, _2F_1\left (2,n+1;n+2;\frac {f x}{e}+1\right )}{a^2 e^2}-\frac {d^3 \, _2F_1\left (1,n+1;n+2;\frac {d (e+f x)}{d e-c f}\right )}{(b c-a d) (c f-d e)}}{c^2}-\frac {b^3 \, _2F_1\left (1,n+1;n+2;\frac {b (e+f x)}{b e-a f}\right )}{a^2 (b c-a d) (b e-a f)}\right )}{n+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (f x + e\right )}^{n}}{b d x^{4} + a c x^{2} + {\left (b c + a d\right )} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((f*x + e)^n/(b*d*x^4 + a*c*x^2 + (b*c + a*d)*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.25, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{n}}{\left (b x +a \right ) \left (d x +c \right ) x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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