\(\int \frac {(b x+c x^2)^p}{(d x)^{3/2}} \, dx\) [135]

Optimal result

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

[Out]

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

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} \[ \int \frac {\left (b x+c x^2\right )^p}{(d x)^{3/2}} \, dx=-\frac {2 x \left (\frac {c x}{b}+1\right )^{-p} \left (b x+c x^2\right )^p \operatorname {Hypergeometric2F1}\left (p-\frac {1}{2},-p,p+\frac {1}{2},-\frac {c x}{b}\right )}{(1-2 p) (d x)^{3/2}} \]

[In]

Int[(b*x + c*x^2)^p/(d*x)^(3/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.95 \[ \int \frac {\left (b x+c x^2\right )^p}{(d x)^{3/2}} \, dx=\frac {x (x (b+c x))^p \left (1+\frac {c x}{b}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2}+p,-p,\frac {1}{2}+p,-\frac {c x}{b}\right )}{\left (-\frac {1}{2}+p\right ) (d x)^{3/2}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (c \,x^{2}+b x \right )^{p}}{\left (d x \right )^{\frac {3}{2}}}d x\]

[In]

int((c*x^2+b*x)^p/(d*x)^(3/2),x)

[Out]

int((c*x^2+b*x)^p/(d*x)^(3/2),x)

Fricas [F]

\[ \int \frac {\left (b x+c x^2\right )^p}{(d x)^{3/2}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{p}}{\left (d x\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((c*x^2+b*x)^p/(d*x)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\left (b x+c x^2\right )^p}{(d x)^{3/2}} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{p}}{\left (d x\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((c*x**2+b*x)**p/(d*x)**(3/2),x)

[Out]

Integral((x*(b + c*x))**p/(d*x)**(3/2), x)

Maxima [F]

\[ \int \frac {\left (b x+c x^2\right )^p}{(d x)^{3/2}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{p}}{\left (d x\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((c*x^2+b*x)^p/(d*x)^(3/2),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^p/(d*x)^(3/2), x)

Giac [F]

\[ \int \frac {\left (b x+c x^2\right )^p}{(d x)^{3/2}} \, dx=\int { \frac {{\left (c x^{2} + b x\right )}^{p}}{\left (d x\right )^{\frac {3}{2}}} \,d x } \]

[In]

integrate((c*x^2+b*x)^p/(d*x)^(3/2),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x)^p/(d*x)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (b x+c x^2\right )^p}{(d x)^{3/2}} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^p}{{\left (d\,x\right )}^{3/2}} \,d x \]

[In]

int((b*x + c*x^2)^p/(d*x)^(3/2),x)

[Out]

int((b*x + c*x^2)^p/(d*x)^(3/2), x)