3.3172 \(\int \frac{\sqrt{a+b x} (e+f x)^n}{\sqrt{c+d x}} \, dx\)

Optimal. Leaf size=123 \[ \frac{2 (a+b x)^{3/2} (e+f x)^n \sqrt{\frac{b (c+d x)}{b c-a d}} \left (\frac{b (e+f x)}{b e-a f}\right )^{-n} F_1\left (\frac{3}{2};\frac{1}{2},-n;\frac{5}{2};-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{3 b \sqrt{c+d x}} \]

[Out]

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

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Rubi [A]  time = 0.0783641, antiderivative size = 123, normalized size of antiderivative = 1., 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 = {140, 139, 138} \[ \frac{2 (a+b x)^{3/2} (e+f x)^n \sqrt{\frac{b (c+d x)}{b c-a d}} \left (\frac{b (e+f x)}{b e-a f}\right )^{-n} F_1\left (\frac{3}{2};\frac{1}{2},-n;\frac{5}{2};-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{3 b \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 140

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 139

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 138

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f))])/(b*(m + 1
)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p), 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])

Rubi steps

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

Mathematica [A]  time = 0.092202, size = 121, normalized size = 0.98 \[ \frac{2 (a+b x)^{3/2} (e+f x)^n \sqrt{\frac{b (c+d x)}{b c-a d}} \left (\frac{b (e+f x)}{b e-a f}\right )^{-n} F_1\left (\frac{3}{2};\frac{1}{2},-n;\frac{5}{2};\frac{d (a+b x)}{a d-b c},\frac{f (a+b x)}{a f-b e}\right )}{3 b \sqrt{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int{ \left ( fx+e \right ) ^{n}\sqrt{bx+a}{\frac{1}{\sqrt{dx+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b x + a}{\left (f x + e\right )}^{n}}{\sqrt{d x + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x + a}{\left (f x + e\right )}^{n}}{\sqrt{d x + c}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b x + a}{\left (f x + e\right )}^{n}}{\sqrt{d x + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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