\(\int (d x)^m \sqrt {b x+c x^2} \, dx\) [120]

Optimal result

Integrand size = 19, antiderivative size = 67 \[ \int (d x)^m \sqrt {b x+c x^2} \, dx=\frac {2 \left (-\frac {c x}{b}\right )^{-\frac {1}{2}-m} (d x)^m (b+c x) \sqrt {b x+c x^2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {1}{2}-m,\frac {5}{2},1+\frac {c x}{b}\right )}{3 c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 67, 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, 69, 67} \[ \int (d x)^m \sqrt {b x+c x^2} \, dx=\frac {2 (b+c x) \sqrt {b x+c x^2} (d x)^m \left (-\frac {c x}{b}\right )^{-m-\frac {1}{2}} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m-\frac {1}{2},\frac {5}{2},\frac {c x}{b}+1\right )}{3 c} \]

[In]

Int[(d*x)^m*Sqrt[b*x + c*x^2],x]

[Out]

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

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 69

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[((-b)*(c/d))^IntPart[m]*((b*x)^FracPart[m]/(
(-d)*(x/c))^FracPart[m]), Int[((-d)*(x/c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0]

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.90 \[ \int (d x)^m \sqrt {b x+c x^2} \, dx=-\frac {2 \left (-\frac {c x}{b}\right )^{-\frac {3}{2}-m} (d x)^m (x (b+c x))^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-\frac {1}{2}-m,\frac {5}{2},1+\frac {c x}{b}\right )}{3 b} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral((d*x)**m*sqrt(x*(b + c*x)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((b*x + c*x^2)^(1/2)*(d*x)^m,x)

[Out]

int((b*x + c*x^2)^(1/2)*(d*x)^m, x)