3.3174 \(\int \frac {(e+f x)^n}{\sqrt {a+b x} (c+d x)^{3/2}} \, dx\)

Optimal. Leaf size=128 \[ \frac {2 \sqrt {a+b x} (e+f x)^n \sqrt {\frac {b (c+d x)}{b c-a d}} \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} F_1\left (\frac {1}{2};\frac {3}{2},-n;\frac {3}{2};-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{\sqrt {c+d x} (b c-a d)} \]

[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)

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Rubi [A]  time = 0.08, antiderivative size = 128, 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 = {140, 139, 138} \[ \frac {2 \sqrt {a+b x} (e+f x)^n \sqrt {\frac {b (c+d x)}{b c-a d}} \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} F_1\left (\frac {1}{2};\frac {3}{2},-n;\frac {3}{2};-\frac {d (a+b x)}{b c-a d},-\frac {f (a+b x)}{b e-a f}\right )}{\sqrt {c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[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 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])

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 241, normalized size = 1.88 \[ -\frac {2 \sqrt {a+b x} (e+f x)^n \sqrt {\frac {b (c+d x)}{b c-a d}} \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} \left ((3 a d-3 b c) F_1\left (\frac {1}{2};-\frac {1}{2},-n;\frac {3}{2};\frac {d (a+b x)}{a d-b c},\frac {f (a+b x)}{a f-b e}\right )+d (a+b x) \left (F_1\left (\frac {3}{2};\frac {1}{2},-n;\frac {5}{2};\frac {d (a+b x)}{a d-b c},\frac {f (a+b x)}{a f-b e}\right )+F_1\left (\frac {3}{2};\frac {3}{2},-n;\frac {5}{2};\frac {d (a+b x)}{a d-b c},\frac {f (a+b x)}{a f-b e}\right )\right )\right )}{3 \sqrt {c+d x} (b c-a d)^2} \]

Warning: Unable to verify antiderivative.

[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)

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fricas [F]  time = 1.10, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x + a} \sqrt {d x + c} {\left (f x + e\right )}^{n}}{b d^{2} x^{3} + a c^{2} + {\left (2 \, b c d + a d^{2}\right )} x^{2} + {\left (b c^{2} + 2 \, a c d\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{n}}{\sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{n}}{\left (d x +c \right )^{\frac {3}{2}} \sqrt {b x +a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{n}}{\sqrt {b x + a} {\left (d x + c\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (e+f\,x\right )}^n}{\sqrt {a+b\,x}\,{\left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e + f x\right )^{n}}{\sqrt {a + b x} \left (c + d x\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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