3.443 \(\int \frac {(d+e x)^m}{b x+c x^2} \, dx\)

Optimal. Leaf size=93 \[ \frac {c (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {c (d+e x)}{c d-b e}\right )}{b (m+1) (c d-b e)}-\frac {(d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {e x}{d}+1\right )}{b d (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {711, 65, 68} \[ \frac {c (d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {c (d+e x)}{c d-b e}\right )}{b (m+1) (c d-b e)}-\frac {(d+e x)^{m+1} \, _2F_1\left (1,m+1;m+2;\frac {e x}{d}+1\right )}{b d (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^m/(b*x + c*x^2),x]

[Out]

(c*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (c*(d + e*x))/(c*d - b*e)])/(b*(c*d - b*e)*(1 + m)) -
((d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, 1 + (e*x)/d])/(b*d*(1 + m))

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 711

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^
m, 1/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a
*e^2, 0] && NeQ[2*c*d - b*e, 0] &&  !IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 86, normalized size = 0.92 \[ -\frac {(d+e x)^{m+1} \left (c d \, _2F_1\left (1,m+1;m+2;\frac {c (d+e x)}{c d-b e}\right )+(b e-c d) \, _2F_1\left (1,m+1;m+2;\frac {e x}{d}+1\right )\right )}{b d (m+1) (b e-c d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((d + e*x)^(1 + m)*(c*d*Hypergeometric2F1[1, 1 + m, 2 + m, (c*(d + e*x))/(c*d - b*e)] + (-(c*d) + b*e)*Hyper
geometric2F1[1, 1 + m, 2 + m, 1 + (e*x)/d]))/(b*d*(-(c*d) + b*e)*(1 + m)))

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fricas [F]  time = 1.19, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x + d\right )}^{m}}{c x^{2} + b x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.73, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right )^{m}}{c \,x^{2}+b x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^m}{c\,x^2+b\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^m/(b*x + c*x^2),x)

[Out]

int((d + e*x)^m/(b*x + c*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x)**m/(x*(b + c*x)), x)

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