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

Optimal. Leaf size=250 \[ \frac{2 \sqrt{a+b x} (A b-a B) (c+d x)^n (e+f x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (\frac{1}{2};-n,-p;\frac{3}{2};-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^2}+\frac{2 B (a+b x)^{3/2} (c+d x)^n (e+f x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (\frac{3}{2};-n,-p;\frac{5}{2};-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{3 b^2} \]

[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)

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Rubi [A]  time = 0.217165, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {159, 140, 139, 138} \[ \frac{2 \sqrt{a+b x} (A b-a B) (c+d x)^n (e+f x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (\frac{1}{2};-n,-p;\frac{3}{2};-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^2}+\frac{2 B (a+b x)^{3/2} (c+d x)^n (e+f x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (\frac{3}{2};-n,-p;\frac{5}{2};-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{3 b^2} \]

Antiderivative was successfully verified.

[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 159

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]))

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{(A+B x) (c+d x)^n (e+f x)^p}{\sqrt{a+b x}} \, dx &=\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]  time = 0.20808, size = 184, normalized size = 0.74 \[ \frac{2 \sqrt{a+b x} (c+d x)^n (e+f x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} \left (3 (A b-a B) F_1\left (\frac{1}{2};-n,-p;\frac{3}{2};\frac{d (a+b x)}{a d-b c},\frac{f (a+b x)}{a f-b e}\right )+B (a+b x) F_1\left (\frac{3}{2};-n,-p;\frac{5}{2};\frac{d (a+b x)}{a d-b c},\frac{f (a+b x)}{a f-b e}\right )\right )}{3 b^2} \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int{ \left ( Bx+A \right ) \left ( dx+c \right ) ^{n} \left ( fx+e \right ) ^{p}{\frac{1}{\sqrt{bx+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)