\(\int \frac {(A+B x) (c+d x)^n (e+f x)^p}{\sqrt {a+b x}} \, dx\) [138]

Optimal result

Integrand size = 29, antiderivative size = 250 \[ \int \frac {(A+B x) (c+d x)^n (e+f x)^p}{\sqrt {a+b x}} \, dx=\frac {2 (A b-a B) \sqrt {a+b x} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-n,-p,\frac {3}{2},-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^2}+\frac {2 B (a+b x)^{3/2} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-n,-p,\frac {5}{2},-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{3 b^2} \]

[Out]

2/3*B*(b*x+a)^(3/2)*(d*x+c)^n*(f*x+e)^p*AppellF1(3/2,-n,-p,5/2,-d*(b*x+a)/(-a*d+b*c),-f*(b*x+a)/(-a*f+b*e))/b^
2/((b*(d*x+c)/(-a*d+b*c))^n)/((b*(f*x+e)/(-a*f+b*e))^p)+2*(A*b-B*a)*(d*x+c)^n*(f*x+e)^p*AppellF1(1/2,-n,-p,3/2
,-d*(b*x+a)/(-a*d+b*c),-f*(b*x+a)/(-a*f+b*e))*(b*x+a)^(1/2)/b^2/((b*(d*x+c)/(-a*d+b*c))^n)/((b*(f*x+e)/(-a*f+b
*e))^p)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {165, 145, 144, 143} \[ \int \frac {(A+B x) (c+d x)^n (e+f x)^p}{\sqrt {a+b x}} \, dx=\frac {2 \sqrt {a+b x} (A b-a B) (c+d x)^n (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-n,-p,\frac {3}{2},-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^2}+\frac {2 B (a+b x)^{3/2} (c+d x)^n (e+f x)^p \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-n,-p,\frac {5}{2},-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{3 b^2} \]

[In]

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

[Out]

(2*(A*b - a*B)*Sqrt[a + b*x]*(c + d*x)^n*(e + f*x)^p*AppellF1[1/2, -n, -p, 3/2, -((d*(a + b*x))/(b*c - a*d)),
-((f*(a + b*x))/(b*e - a*f))])/(b^2*((b*(c + d*x))/(b*c - a*d))^n*((b*(e + f*x))/(b*e - a*f))^p) + (2*B*(a + b
*x)^(3/2)*(c + d*x)^n*(e + f*x)^p*AppellF1[3/2, -n, -p, 5/2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*
e - a*f))])/(3*b^2*((b*(c + d*x))/(b*c - a*d))^n*((b*(e + f*x))/(b*e - a*f))^p)

Rule 143

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

Rule 144

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

Rule 145

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

Rule 165

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Dist[h/b, Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(a + b*x)^m*(
c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n, p}, x] && (SumSimplerQ[m, 1] || ( !SumS
implerQ[n, 1] &&  !SumSimplerQ[p, 1]))

Rubi steps \begin{align*} \text {integral}& = \frac {B \int \sqrt {a+b x} (c+d x)^n (e+f x)^p \, dx}{b}+\frac {(A b-a B) \int \frac {(c+d x)^n (e+f x)^p}{\sqrt {a+b x}} \, dx}{b} \\ & = \frac {\left (B (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n}\right ) \int \sqrt {a+b x} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n (e+f x)^p \, dx}{b}+\frac {\left ((A b-a B) (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n}\right ) \int \frac {\left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n (e+f x)^p}{\sqrt {a+b x}} \, dx}{b} \\ & = \frac {\left (B (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p}\right ) \int \sqrt {a+b x} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n \left (\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}\right )^p \, dx}{b}+\frac {\left ((A b-a B) (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p}\right ) \int \frac {\left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^n \left (\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}\right )^p}{\sqrt {a+b x}} \, dx}{b} \\ & = \frac {2 (A b-a B) \sqrt {a+b x} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (\frac {1}{2};-n,-p;\frac {3}{2};-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b^2}+\frac {2 B (a+b x)^{3/2} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} F_1\left (\frac {3}{2};-n,-p;\frac {5}{2};-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{3 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.74 \[ \int \frac {(A+B x) (c+d x)^n (e+f x)^p}{\sqrt {a+b x}} \, dx=\frac {2 \sqrt {a+b x} (c+d x)^n \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac {b (e+f x)}{b e-a f}\right )^{-p} \left (3 (A b-a B) \operatorname {AppellF1}\left (\frac {1}{2},-n,-p,\frac {3}{2},\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )+B (a+b x) \operatorname {AppellF1}\left (\frac {3}{2},-n,-p,\frac {5}{2},\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )\right )}{3 b^2} \]

[In]

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

[Out]

(2*Sqrt[a + b*x]*(c + d*x)^n*(e + f*x)^p*(3*(A*b - a*B)*AppellF1[1/2, -n, -p, 3/2, (d*(a + b*x))/(-(b*c) + a*d
), (f*(a + b*x))/(-(b*e) + a*f)] + B*(a + b*x)*AppellF1[3/2, -n, -p, 5/2, (d*(a + b*x))/(-(b*c) + a*d), (f*(a
+ b*x))/(-(b*e) + a*f)]))/(3*b^2*((b*(c + d*x))/(b*c - a*d))^n*((b*(e + f*x))/(b*e - a*f))^p)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {(A+B x) (c+d x)^n (e+f x)^p}{\sqrt {a+b x}} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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