3.3175 \(\int \frac {(e+f x)^n}{(a+b x)^{3/2} \sqrt {c+d x}} \, dx\)

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

[Out]

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

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [B]  time = 0.23, size = 262, normalized size = 2.17 \[ \frac {2 \sqrt {c+d x} (e+f x)^n \left (\frac {b (e+f x)}{b e-a f}\right )^{-n} \left (d^2 (a+b x)^2 F_1\left (\frac {3}{2};\frac {1}{2},-n;\frac {5}{2};\frac {d (a+b x)}{a d-b c},\frac {f (a+b x)}{a f-b e}\right )-3 (b c-a d)^2 F_1\left (-\frac {1}{2};-\frac {1}{2},-n;\frac {1}{2};\frac {d (a+b x)}{a d-b c},\frac {f (a+b x)}{a f-b e}\right )+3 d (a+b x) (a d-b c) F_1\left (\frac {1}{2};-\frac {1}{2},-n;\frac {3}{2};\frac {d (a+b x)}{a d-b c},\frac {f (a+b x)}{a f-b e}\right )\right )}{3 \sqrt {a+b x} (b c-a d)^3 \sqrt {\frac {b (c+d x)}{b c-a d}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x + a} \sqrt {d x + c} {\left (f x + e\right )}^{n}}{b^{2} d x^{3} + a^{2} c + {\left (b^{2} c + 2 \, a b d\right )} x^{2} + {\left (2 \, a b c + a^{2} d\right )} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.25, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{n}}{\left (b x +a \right )^{\frac {3}{2}} \sqrt {d x +c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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