3.134 \(\int \frac {(b x+c x^2)^p}{\sqrt {d x}} \, dx\)

Optimal. Leaf size=61 \[ \frac {2 x \left (\frac {c x}{b}+1\right )^{-p} \left (b x+c x^2\right )^p \, _2F_1\left (-p,p+\frac {1}{2};p+\frac {3}{2};-\frac {c x}{b}\right )}{(2 p+1) \sqrt {d x}} \]

[Out]

2*x*(c*x^2+b*x)^p*hypergeom([-p, 1/2+p],[3/2+p],-c*x/b)/(1+2*p)/((c*x/b+1)^p)/(d*x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {674, 66, 64} \[ \frac {2 x \left (\frac {c x}{b}+1\right )^{-p} \left (b x+c x^2\right )^p \, _2F_1\left (-p,p+\frac {1}{2};p+\frac {3}{2};-\frac {c x}{b}\right )}{(2 p+1) \sqrt {d x}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^p/Sqrt[d*x],x]

[Out]

(2*x*(b*x + c*x^2)^p*Hypergeometric2F1[-p, 1/2 + p, 3/2 + p, -((c*x)/b)])/((1 + 2*p)*Sqrt[d*x]*(1 + (c*x)/b)^p
)

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), 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 66

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 674

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

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Mathematica [A]  time = 0.01, size = 58, normalized size = 0.95 \[ \frac {x (x (b+c x))^p \left (\frac {c x}{b}+1\right )^{-p} \, _2F_1\left (-p,p+\frac {1}{2};p+\frac {3}{2};-\frac {c x}{b}\right )}{\left (p+\frac {1}{2}\right ) \sqrt {d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^p/Sqrt[d*x],x]

[Out]

(x*(x*(b + c*x))^p*Hypergeometric2F1[-p, 1/2 + p, 3/2 + p, -((c*x)/b)])/((1/2 + p)*Sqrt[d*x]*(1 + (c*x)/b)^p)

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fricas [F]  time = 0.94, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d x} {\left (c x^{2} + b x\right )}^{p}}{d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^p/(d*x)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*x)*(c*x^2 + b*x)^p/(d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^p/(d*x)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.52, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \,x^{2}+b x \right )^{p}}{\sqrt {d x}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^p/(d*x)^(1/2),x)

[Out]

int((c*x^2+b*x)^p/(d*x)^(1/2),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^p/(d*x)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x + c*x^2)^p/(d*x)^(1/2),x)

[Out]

int((b*x + c*x^2)^p/(d*x)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**p/(d*x)**(1/2),x)

[Out]

Integral((x*(b + c*x))**p/sqrt(d*x), x)

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