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*(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
)

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Rubi [A]  time = 0.0234973, antiderivative size = 61, normalized size of antiderivative = 1., 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 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]

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 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])))

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*}

Mathematica [A]  time = 0.0120283, 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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Maple [F]  time = 0.361, size = 0, normalized size = 0. \begin{align*} \int{ \left ( c{x}^{2}+bx \right ) ^{p}{\frac{1}{\sqrt{dx}}}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x\right )}^{p}}{\sqrt{d x}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d x}{\left (c x^{2} + b x\right )}^{p}}{d x}, x\right ) \end{align*}

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

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c x^{2} + b x\right )}^{p}}{\sqrt{d x}}\,{d x} \end{align*}

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)