\(\int \frac {(b x+c x^2)^p}{\sqrt {d x}} \, dx\) [134]

Optimal result

Integrand size = 19, antiderivative size = 61 \[ \int \frac {\left (b x+c x^2\right )^p}{\sqrt {d x}} \, dx=\frac {2 x \left (1+\frac {c x}{b}\right )^{-p} \left (b x+c x^2\right )^p \operatorname {Hypergeometric2F1}\left (-p,\frac {1}{2}+p,\frac {3}{2}+p,-\frac {c x}{b}\right )}{(1+2 p) \sqrt {d x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {688, 68, 66} \[ \int \frac {\left (b x+c x^2\right )^p}{\sqrt {d x}} \, dx=\frac {2 x \left (\frac {c x}{b}+1\right )^{-p} \left (b x+c x^2\right )^p \operatorname {Hypergeometric2F1}\left (-p,p+\frac {1}{2},p+\frac {3}{2},-\frac {c x}{b}\right )}{(2 p+1) \sqrt {d x}} \]

[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 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*} \text {integral}& = \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] (verified)

Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.95 \[ \int \frac {\left (b x+c x^2\right )^p}{\sqrt {d x}} \, dx=\frac {x (x (b+c x))^p \left (1+\frac {c x}{b}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-p,\frac {1}{2}+p,\frac {3}{2}+p,-\frac {c x}{b}\right )}{\left (\frac {1}{2}+p\right ) \sqrt {d x}} \]

[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)

Maple [F]

\[\int \frac {\left (c \,x^{2}+b x \right )^{p}}{\sqrt {d x}}d x\]

[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)

Fricas [F]

\[ \int \frac {\left (b x+c x^2\right )^p}{\sqrt {d x}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{p}}{\sqrt {d x}} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {\left (b x+c x^2\right )^p}{\sqrt {d x}} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{p}}{\sqrt {d x}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\left (b x+c x^2\right )^p}{\sqrt {d x}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{p}}{\sqrt {d x}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {\left (b x+c x^2\right )^p}{\sqrt {d x}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{p}}{\sqrt {d x}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x+c x^2\right )^p}{\sqrt {d x}} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^p}{\sqrt {d\,x}} \,d x \]

[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)