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

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

[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 = 0.10, 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 = {891, 70, 69} \[ \frac {2 (f+g x)^n (a e+c d x) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \, _2F_1\left (\frac {3}{2},-n;\frac {5}{2};-\frac {g (a e+c d x)}{c d f-a e g}\right )}{3 c d \sqrt {d+e x}} \]

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 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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 70

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 891

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*x)/e)^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 = 0.07, size = 100, normalized size = 0.96 \[ \frac {2 (f+g x)^n ((d+e x) (a e+c d x))^{3/2} \left (\frac {c d (f+g x)}{c d f-a e g}\right )^{-n} \, _2F_1\left (\frac {3}{2},-n;\frac {5}{2};\frac {g (a e+c d x)}{a e g-c d f}\right )}{3 c d (d+e x)^{3/2}} \]

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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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (g x + f\right )}^{n}}{\sqrt {e x + d}}, x\right ) \]

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

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

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

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, \left (g x +f \right )^{n}}{\sqrt {e x +d}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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

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

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]

Integral(sqrt((d + e*x)*(a*e + c*d*x))*(f + g*x)**n/sqrt(d + e*x), x)

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