3.8.64 \(\int \frac {(f+g x)^n \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}}{\sqrt {d+e x}} \, dx\) [764]

Optimal. Leaf size=104 \[ -\frac {(a e+c d x) (f+g x)^{1+n} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, _2F_1\left (1,\frac {5}{2}+n;2+n;\frac {c d (f+g x)}{c d f-a e g}\right )}{(c d f-a e g) (1+n) \sqrt {d+e x}} \]

[Out]

-(c*d*x+a*e)*(g*x+f)^(1+n)*hypergeom([1, 5/2+n],[2+n],c*d*(g*x+f)/(-a*e*g+c*d*f))*(a*d*e+(a*e^2+c*d^2)*x+c*d*e
*x^2)^(1/2)/(-a*e*g+c*d*f)/(1+n)/(e*x+d)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 120, normalized size of antiderivative = 1.15, number of steps used = 3, number of rules used = 3, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {905, 72, 71} \begin {gather*} \frac {2 (f+g x)^n (a e+c d x) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \, _2F_1\left (\frac {3}{2},-n;\frac {5}{2};-\frac {g (a e+c d x)}{c d f-a e g}\right )}{3 c d \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a*e + c*d*x)*(f + g*x)^n*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]*Hypergeometric2F1[3/2, -n, 5/2, -((g*
(a*e + c*d*x))/(c*d*f - a*e*g))])/(3*c*d*Sqrt[d + e*x]*((c*d*(f + g*x))/(c*d*f - a*e*g))^n)

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 905

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Dist[(a + b*x + c*x^2)^FracPart[p]/((d + e*x)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p]), Int[(d + e*x)^(m + p)*
(f + g*x)^n*(a/d + (c/e)*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2
- 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] &&  !IGtQ[m, 0] &&  !IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(f+g x)^n \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt {d+e x}} \, dx &=\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \int \sqrt {a e+c d x} (f+g x)^n \, dx}{\sqrt {a e+c d x} \sqrt {d+e x}}\\ &=\frac {\left ((f+g x)^n \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}\right ) \int \sqrt {a e+c d x} \left (\frac {c d f}{c d f-a e g}+\frac {c d g x}{c d f-a e g}\right )^n \, dx}{\sqrt {a e+c d x} \sqrt {d+e x}}\\ &=\frac {2 (a e+c d x) (f+g x)^n \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, _2F_1\left (\frac {3}{2},-n;\frac {5}{2};-\frac {g (a e+c d x)}{c d f-a e g}\right )}{3 c d \sqrt {d+e x}}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 100, normalized size = 0.96 \begin {gather*} \frac {2 ((a e+c d x) (d+e x))^{3/2} (f+g x)^n \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \, _2F_1\left (\frac {3}{2},-n;\frac {5}{2};\frac {g (a e+c d x)}{-c d f+a e g}\right )}{3 c d (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((a*e + c*d*x)*(d + e*x))^(3/2)*(f + g*x)^n*Hypergeometric2F1[3/2, -n, 5/2, (g*(a*e + c*d*x))/(-(c*d*f) + a
*e*g)])/(3*c*d*(d + e*x)^(3/2)*((c*d*(f + g*x))/(c*d*f - a*e*g))^n)

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (g x +f \right )^{n} \sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d e \,x^{2}}}{\sqrt {e x +d}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (f+g\,x\right )}^n\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{\sqrt {d+e\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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