\(\int \frac {(e+f x)^n}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx\) [3174]

Optimal result

Integrand size = 26, antiderivative size = 128 \[ \int \frac {(e+f x)^n}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\frac {2 \sqrt {a+b x} \sqrt {\frac {b (c+d x)}{b c-a d}} (e+f x)^n \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{2},-n,\frac {3}{2},-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{(b c-a d) \sqrt {c+d x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {145, 144, 143} \[ \int \frac {(e+f x)^n}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\frac {2 \sqrt {a+b x} (e+f x)^n \sqrt {\frac {b (c+d x)}{b c-a d}} \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{2},-n,\frac {3}{2},-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{\sqrt {c+d x} (b c-a d)} \]

[In]

Int[(e + f*x)^n/(Sqrt[a + b*x]*(c + d*x)^(3/2)),x]

[Out]

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

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]

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

Mathematica [A] (verified)

Time = 2.98 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.88 \[ \int \frac {(e+f x)^n}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=-\frac {2 \sqrt {a+b x} \sqrt {\frac {b (c+d x)}{b c-a d}} (e+f x)^n \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} \left ((-3 b c+3 a d) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-n,\frac {3}{2},\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )+d (a+b x) \left (\operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2},-n,\frac {5}{2},\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )+\operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{2},-n,\frac {5}{2},\frac {d (a+b x)}{-b c+a d},\frac {f (a+b x)}{-b e+a f}\right )\right )\right )}{3 (b c-a d)^2 \sqrt {c+d x}} \]

[In]

Integrate[(e + f*x)^n/(Sqrt[a + b*x]*(c + d*x)^(3/2)),x]

[Out]

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

Maple [F]

\[\int \frac {\left (f x +e \right )^{n}}{\left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}}d x\]

[In]

int((f*x+e)^n/(d*x+c)^(3/2)/(b*x+a)^(1/2),x)

[Out]

int((f*x+e)^n/(d*x+c)^(3/2)/(b*x+a)^(1/2),x)

Fricas [F]

\[ \int \frac {(e+f x)^n}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{\sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((f*x+e)^n/(d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x + a)*sqrt(d*x + c)*(f*x + e)^n/(b*d^2*x^3 + a*c^2 + (2*b*c*d + a*d^2)*x^2 + (b*c^2 + 2*a*c*d
)*x), x)

Sympy [F]

\[ \int \frac {(e+f x)^n}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\int \frac {\left (e + f x\right )^{n}}{\sqrt {a + b x} \left (c + d x\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((f*x+e)**n/(d*x+c)**(3/2)/(b*x+a)**(1/2),x)

[Out]

Integral((e + f*x)**n/(sqrt(a + b*x)*(c + d*x)**(3/2)), x)

Maxima [F]

\[ \int \frac {(e+f x)^n}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{\sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((f*x+e)^n/(d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((f*x + e)^n/(sqrt(b*x + a)*(d*x + c)^(3/2)), x)

Giac [F]

\[ \int \frac {(e+f x)^n}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{\sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate((f*x+e)^n/(d*x+c)^(3/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

integrate((f*x + e)^n/(sqrt(b*x + a)*(d*x + c)^(3/2)), x)

Mupad [F(-1)]

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

[In]

int((e + f*x)^n/((a + b*x)^(1/2)*(c + d*x)^(3/2)),x)

[Out]

int((e + f*x)^n/((a + b*x)^(1/2)*(c + d*x)^(3/2)), x)