3.32.73 \(\int \frac {\sqrt {c+d x} (e+f x)^n}{\sqrt {a+b x}} \, dx\) [3173]

Optimal. Leaf size=121 \[ \frac {2 \sqrt {a+b x} \sqrt {c+d x} (e+f x)^n \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} F_1\left (\frac {1}{2};-\frac {1}{2},-n;\frac {3}{2};-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{b \sqrt {\frac {b (c+d x)}{b c-a d}}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {145, 144, 143} \begin {gather*} \frac {2 \sqrt {a+b x} \sqrt {c+d x} (e+f x)^n \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} F_1\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 )}{b \sqrt {\frac {b (c+d x)}{b c-a d}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + b*x]*Sqrt[c + d*x]*(e + f*x)^n*AppellF1[1/2, -1/2, -n, 3/2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a
+ b*x))/(b*e - a*f))])/(b*Sqrt[(b*(c + d*x))/(b*c - a*d)]*((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*} \int \frac {\sqrt {c+d x} (e+f x)^n}{\sqrt {a+b x}} \, dx &=\frac {\sqrt {c+d x} \int \frac {\sqrt {\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}} (e+f x)^n}{\sqrt {a+b x}} \, dx}{\sqrt {\frac {b (c+d x)}{b c-a d}}}\\ &=\frac {\left (\sqrt {c+d x} (e+f x)^n \left (\frac {b (e+f x)}{b e-a f}\right )^{-n}\right ) \int \frac {\sqrt {\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}} \left (\frac {b e}{b e-a f}+\frac {b f x}{b e-a f}\right )^n}{\sqrt {a+b x}} \, dx}{\sqrt {\frac {b (c+d x)}{b c-a d}}}\\ &=\frac {2 \sqrt {a+b x} \sqrt {c+d x} (e+f x)^n \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} F_1\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 )}{b \sqrt {\frac {b (c+d x)}{b c-a d}}}\\ \end {align*}

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Mathematica [A]
time = 1.75, size = 119, normalized size = 0.98 \begin {gather*} \frac {2 \sqrt {a+b x} \sqrt {c+d x} (e+f x)^n \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} F_1\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 )}{b \sqrt {\frac {b (c+d x)}{b c-a d}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (f x +e \right )^{n} \sqrt {d x +c}}{\sqrt {b x +a}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e+f\,x\right )}^n\,\sqrt {c+d\,x}}{\sqrt {a+b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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