3.136 \(\int \frac{(b x+c x^2)^p}{(d x)^{5/2}} \, dx\)

Optimal. Leaf size=61 \[ -\frac{2 x \left (\frac{c x}{b}+1\right )^{-p} \left (b x+c x^2\right )^p \, _2F_1\left (p-\frac{3}{2},-p;p-\frac{1}{2};-\frac{c x}{b}\right )}{(3-2 p) (d x)^{5/2}} \]

[Out]

(-2*x*(b*x + c*x^2)^p*Hypergeometric2F1[-3/2 + p, -p, -1/2 + p, -((c*x)/b)])/((3 - 2*p)*(d*x)^(5/2)*(1 + (c*x)
/b)^p)

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Rubi [A]  time = 0.0258183, 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} \[ -\frac{2 x \left (\frac{c x}{b}+1\right )^{-p} \left (b x+c x^2\right )^p \, _2F_1\left (p-\frac{3}{2},-p;p-\frac{1}{2};-\frac{c x}{b}\right )}{(3-2 p) (d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^p/(d*x)^(5/2),x]

[Out]

(-2*x*(b*x + c*x^2)^p*Hypergeometric2F1[-3/2 + p, -p, -1/2 + p, -((c*x)/b)])/((3 - 2*p)*(d*x)^(5/2)*(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}{(d x)^{5/2}} \, dx &=\frac{\left (x^{\frac{5}{2}-p} (b+c x)^{-p} \left (b x+c x^2\right )^p\right ) \int x^{-\frac{5}{2}+p} (b+c x)^p \, dx}{(d x)^{5/2}}\\ &=\frac{\left (x^{\frac{5}{2}-p} \left (1+\frac{c x}{b}\right )^{-p} \left (b x+c x^2\right )^p\right ) \int x^{-\frac{5}{2}+p} \left (1+\frac{c x}{b}\right )^p \, dx}{(d x)^{5/2}}\\ &=-\frac{2 x \left (1+\frac{c x}{b}\right )^{-p} \left (b x+c x^2\right )^p \, _2F_1\left (-\frac{3}{2}+p,-p;-\frac{1}{2}+p;-\frac{c x}{b}\right )}{(3-2 p) (d x)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0145333, size = 58, normalized size = 0.95 \[ \frac{x (x (b+c x))^p \left (\frac{c x}{b}+1\right )^{-p} \, _2F_1\left (p-\frac{3}{2},-p;p-\frac{1}{2};-\frac{c x}{b}\right )}{\left (p-\frac{3}{2}\right ) (d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^p/(d*x)^(5/2),x]

[Out]

(x*(x*(b + c*x))^p*Hypergeometric2F1[-3/2 + p, -p, -1/2 + p, -((c*x)/b)])/((-3/2 + p)*(d*x)^(5/2)*(1 + (c*x)/b
)^p)

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Maple [F]  time = 0.335, size = 0, normalized size = 0. \begin{align*} \int{ \left ( c{x}^{2}+bx \right ) ^{p} \left ( dx \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^p/(d*x)^(5/2),x)

[Out]

int((c*x^2+b*x)^p/(d*x)^(5/2),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}}{\left (d x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^p/(d*x)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^p/(d*x)^(5/2), 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^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^p/(d*x)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*x)*(c*x^2 + b*x)^p/(d^3*x^3), 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}}{\left (d x\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**p/(d*x)**(5/2),x)

[Out]

Integral((x*(b + c*x))**p/(d*x)**(5/2), 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}}{\left (d x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^p/(d*x)^(5/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x)^p/(d*x)^(5/2), x)