\(\int \frac {(c+d x)^n (e+f x)^p}{\sqrt {b x}} \, dx\) [965]

Optimal result

Integrand size = 22, antiderivative size = 77 \[ \int \frac {(c+d x)^n (e+f x)^p}{\sqrt {b x}} \, dx=\frac {2 \sqrt {b x} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} (e+f x)^p \left (1+\frac {f x}{e}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-n,-p,\frac {3}{2},-\frac {d x}{c},-\frac {f x}{e}\right )}{b} \]

[Out]

2*(d*x+c)^n*(f*x+e)^p*AppellF1(1/2,-n,-p,3/2,-d*x/c,-f*x/e)*(b*x)^(1/2)/b/((1+d*x/c)^n)/((1+f*x/e)^p)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {140, 138} \[ \int \frac {(c+d x)^n (e+f x)^p}{\sqrt {b x}} \, dx=\frac {2 \sqrt {b x} (c+d x)^n \left (\frac {d x}{c}+1\right )^{-n} (e+f x)^p \left (\frac {f x}{e}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-n,-p,\frac {3}{2},-\frac {d x}{c},-\frac {f x}{e}\right )}{b} \]

[In]

Int[((c + d*x)^n*(e + f*x)^p)/Sqrt[b*x],x]

[Out]

(2*Sqrt[b*x]*(c + d*x)^n*(e + f*x)^p*AppellF1[1/2, -n, -p, 3/2, -((d*x)/c), -((f*x)/e)])/(b*(1 + (d*x)/c)^n*(1
+ (f*x)/e)^p)

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 140

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), 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*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rubi steps \begin{align*} \text {integral}& = \left ((c+d x)^n \left (1+\frac {d x}{c}\right )^{-n}\right ) \int \frac {\left (1+\frac {d x}{c}\right )^n (e+f x)^p}{\sqrt {b x}} \, dx \\ & = \left ((c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} (e+f x)^p \left (1+\frac {f x}{e}\right )^{-p}\right ) \int \frac {\left (1+\frac {d x}{c}\right )^n \left (1+\frac {f x}{e}\right )^p}{\sqrt {b x}} \, dx \\ & = \frac {2 \sqrt {b x} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} (e+f x)^p \left (1+\frac {f x}{e}\right )^{-p} F_1\left (\frac {1}{2};-n,-p;\frac {3}{2};-\frac {d x}{c},-\frac {f x}{e}\right )}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int \frac {(c+d x)^n (e+f x)^p}{\sqrt {b x}} \, dx=\frac {2 x (c+d x)^n \left (\frac {c+d x}{c}\right )^{-n} (e+f x)^p \left (\frac {e+f x}{e}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-n,-p,\frac {3}{2},-\frac {d x}{c},-\frac {f x}{e}\right )}{\sqrt {b x}} \]

[In]

Integrate[((c + d*x)^n*(e + f*x)^p)/Sqrt[b*x],x]

[Out]

(2*x*(c + d*x)^n*(e + f*x)^p*AppellF1[1/2, -n, -p, 3/2, -((d*x)/c), -((f*x)/e)])/(Sqrt[b*x]*((c + d*x)/c)^n*((
e + f*x)/e)^p)

Maple [F]

\[\int \frac {\left (d x +c \right )^{n} \left (f x +e \right )^{p}}{\sqrt {b x}}d x\]

[In]

int((d*x+c)^n*(f*x+e)^p/(b*x)^(1/2),x)

[Out]

int((d*x+c)^n*(f*x+e)^p/(b*x)^(1/2),x)

Fricas [F]

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

[In]

integrate((d*x+c)^n*(f*x+e)^p/(b*x)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x)*(d*x + c)^n*(f*x + e)^p/(b*x), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate((d*x+c)^n*(f*x+e)^p/(b*x)^(1/2),x, algorithm="maxima")

[Out]

integrate((d*x + c)^n*(f*x + e)^p/sqrt(b*x), x)

Giac [F]

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

[In]

integrate((d*x+c)^n*(f*x+e)^p/(b*x)^(1/2),x, algorithm="giac")

[Out]

integrate((d*x + c)^n*(f*x + e)^p/sqrt(b*x), x)

Mupad [F(-1)]

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

[In]

int(((e + f*x)^p*(c + d*x)^n)/(b*x)^(1/2),x)

[Out]

int(((e + f*x)^p*(c + d*x)^n)/(b*x)^(1/2), x)