3.5.72 \(\int (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}})^n \, dx\) [472]

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

[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 = 0.14, 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 = {2141, 961, 66} \begin {gather*} \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)} \end {gather*}

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 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], 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 961

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 2141

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 \text {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 \text {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 ) \text {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 = 1.12, size = 134, normalized size = 0.81 \begin {gather*} \frac {\left (d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )^{1+n} \left (\left (-2 d e+b f^2\right )^2+\left (4 a e^2 f^2-b^2 f^4\right ) \, _2F_1\left (2,1+n;2+n;\frac {2 e \left (d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )}{2 d e-b f^2}\right )\right )}{2 e \left (-2 d e+b f^2\right )^2 (1+n)} \end {gather*}

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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Maple [F]
time = 0.01, 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]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

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]

Exception raised: HeuristicGCDFailed >> no luck

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

[Out]

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

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

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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