3.2.29 \(\int \frac {(b x+c x^2)^p}{x^2} \, dx\) [129]

Optimal. Leaf size=50 \[ -\frac {\left (1+\frac {c x}{b}\right )^{-p} \left (b x+c x^2\right )^p \, _2F_1\left (-1+p,-p;p;-\frac {c x}{b}\right )}{(1-p) x} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {688, 68, 66} \begin {gather*} -\frac {\left (\frac {c x}{b}+1\right )^{-p} \left (b x+c x^2\right )^p \, _2F_1\left (p-1,-p;p;-\frac {c x}{b}\right )}{(1-p) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^p/x^2,x]

[Out]

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

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))

Rule 68

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(
x/c))^FracPart[n]), Int[(b*x)^m*(1 + d*(x/c))^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0] && ((RationalQ[m] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0])) |
|  !RationalQ[n])

Rule 688

Int[((e_.)*(x_))^(m_)*((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(e*x)^m*((b*x + c*x^2)^p/(x^(m + p)*
(b + c*x)^p)), Int[x^(m + p)*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, m}, x] &&  !IntegerQ[p]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 45, normalized size = 0.90 \begin {gather*} \frac {(x (b+c x))^p \left (1+\frac {c x}{b}\right )^{-p} \, _2F_1\left (-1+p,-p;p;-\frac {c x}{b}\right )}{(-1+p) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^p/x^2,x]

[Out]

((x*(b + c*x))^p*Hypergeometric2F1[-1 + p, -p, p, -((c*x)/b)])/((-1 + p)*x*(1 + (c*x)/b)^p)

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (c \,x^{2}+b x \right )^{p}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^p/x^2,x)

[Out]

int((c*x^2+b*x)^p/x^2,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((c*x^2+b*x)^p/x^2,x, algorithm="maxima")

[Out]

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

[Out]

integral((c*x^2 + b*x)^p/x^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x \left (b + c x\right )\right )^{p}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**p/x**2,x)

[Out]

Integral((x*(b + c*x))**p/x**2, x)

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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((c*x^2+b*x)^p/x^2,x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x)^p/x^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x\right )}^p}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^p/x^2,x)

[Out]

int((b*x + c*x^2)^p/x^2, x)

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