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

Optimal. Leaf size=166 \[ \frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+1} \, _2F_1\left (2,n+1;n+2;\frac {2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )}{2 d e-b f^2}\right )}{2 e (n+1) \left (2 d e-b f^2\right )^2}+\frac {\left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+1}}{2 e (n+1)} \]

[Out]

1/2*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1+n)/e/(1+n)+1/2*f^2*(-b^2*f^2+4*a*e^2)*hypergeom([2, 1+n],[2+n],2*e*
(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))/(-b*f^2+2*d*e))*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1+n)/e/(-b*f^2+2*d*e)
^2/(1+n)

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Rubi [A]  time = 0.18, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {2116, 947, 64} \[ \frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+1} \, _2F_1\left (2,n+1;n+2;\frac {2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )}{2 d e-b f^2}\right )}{2 e (n+1) \left (2 d e-b f^2\right )^2}+\frac {\left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+1}}{2 e (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(1 + n)/(2*e*(1 + n)) + (f^2*(4*a*e^2 - b^2*f^2)*(d + e*x + f*Sqrt
[a + b*x + (e^2*x^2)/f^2])^(1 + n)*Hypergeometric2F1[2, 1 + n, 2 + n, (2*e*(d + e*x + f*Sqrt[a + b*x + (e^2*x^
2)/f^2]))/(2*d*e - b*f^2)])/(2*e*(2*d*e - b*f^2)^2*(1 + n))

Rule 64

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

Rule 947

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && (IGtQ[m, 0] || (E
qQ[m, -2] && EqQ[p, 1] && EqQ[2*c*d - b*e, 0]))

Rule 2116

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

Rubi steps

\begin {align*} \int \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^n \, dx &=2 \operatorname {Subst}\left (\int \frac {x^n \left (d^2 e-(b d-a e) f^2-\left (2 d e-b f^2\right ) x+e x^2\right )}{\left (-2 d e+b f^2+2 e x\right )^2} \, dx,x,d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )\\ &=2 \operatorname {Subst}\left (\int \left (\frac {x^n}{4 e}+\frac {\left (4 a e^2 f^2-b^2 f^4\right ) x^n}{4 e \left (2 d e-b f^2-2 e x\right )^2}\right ) \, dx,x,d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )\\ &=\frac {\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{1+n}}{2 e (1+n)}+\frac {\left (4 a e^2 f^2-b^2 f^4\right ) \operatorname {Subst}\left (\int \frac {x^n}{\left (2 d e-b f^2-2 e x\right )^2} \, dx,x,d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )}{2 e}\\ &=\frac {\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{1+n}}{2 e (1+n)}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{1+n} \, _2F_1\left (2,1+n;2+n;\frac {2 e \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )}{2 d e-b f^2}\right )}{2 e \left (2 d e-b f^2\right )^2 (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 134, normalized size = 0.81 \[ \frac {\left (f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}+d+e x\right )^{n+1} \left (\left (4 a e^2 f^2-b^2 f^4\right ) \, _2F_1\left (2,n+1;n+2;\frac {2 e \left (d+e x+f \sqrt {a+x \left (\frac {x e^2}{f^2}+b\right )}\right )}{2 d e-b f^2}\right )+\left (b f^2-2 d e\right )^2\right )}{2 e (n+1) \left (b f^2-2 d e\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])^(1 + n)*((-2*d*e + b*f^2)^2 + (4*a*e^2*f^2 - b^2*f^4)*Hypergeomet
ric2F1[2, 1 + n, 2 + n, (2*e*(d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]))/(2*d*e - b*f^2)]))/(2*e*(-2*d*e + b*
f^2)^2*(1 + n))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \left (e x +d +\sqrt {b x +\frac {e^{2} x^{2}}{f^{2}}+a}\, f \right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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