3.6.14 \(\int (a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2})^{3/2} (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}})^n \, dx\) [514]

Optimal. Leaf size=297 \[ -\frac {\left (d^2-a f^2\right )^4 \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{-4+n}}{16 e f^3 (4-n)}+\frac {\left (d^2-a f^2\right )^3 \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{-2+n}}{4 e f^3 (2-n)}+\frac {3 \left (d^2-a f^2\right )^2 \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n}{8 e f^3 n}-\frac {\left (d^2-a f^2\right ) \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{2+n}}{4 e f^3 (2+n)}+\frac {\left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{4+n}}{16 e f^3 (4+n)} \]

[Out]

-1/16*(-a*f^2+d^2)^4*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^(-4+n)/e/f^3/(4-n)+1/4*(-a*f^2+d^2)^3*(d+e*x+
f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^(-2+n)/e/f^3/(2-n)+3/8*(-a*f^2+d^2)^2*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2
)^(1/2))^n/e/f^3/n-1/4*(-a*f^2+d^2)*(d+e*x+f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^(2+n)/e/f^3/(2+n)+1/16*(d+e*x+
f*(a+2*d*e*x/f^2+e^2*x^2/f^2)^(1/2))^(4+n)/e/f^3/(4+n)

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Rubi [A]
time = 0.28, antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {2146, 12, 276} \begin {gather*} -\frac {\left (d^2-a f^2\right )^4 \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n-4}}{16 e f^3 (4-n)}+\frac {\left (d^2-a f^2\right )^3 \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n-2}}{4 e f^3 (2-n)}+\frac {3 \left (d^2-a f^2\right )^2 \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^n}{8 e f^3 n}-\frac {\left (d^2-a f^2\right ) \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+2}}{4 e f^3 (n+2)}+\frac {\left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^{n+4}}{16 e f^3 (n+4)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*((d^2 - a*f^2)^4*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(-4 + n))/(e*f^3*(4 - n)) + ((d^2
- a*f^2)^3*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(-2 + n))/(4*e*f^3*(2 - n)) + (3*(d^2 - a*f^
2)^2*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^n)/(8*e*f^3*n) - ((d^2 - a*f^2)*(d + e*x + f*Sqrt[a
+ (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^(2 + n))/(4*e*f^3*(2 + n)) + (d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2
)/f^2])^(4 + n)/(16*e*f^3*(4 + n))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2146

Int[((g_.) + (h_.)*(x_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)
^2])^(n_.), x_Symbol] :> Dist[(2/f^(2*m))*(i/c)^m, Subst[Int[x^n*((d^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x
+ e*x^2)^(2*m + 1)/(-2*d*e + b*f^2 + 2*e*x)^(2*(m + 1))), x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; Fr
eeQ[{a, b, c, d, e, f, g, h, i, n}, x] && EqQ[e^2 - c*f^2, 0] && EqQ[c*g - a*i, 0] && EqQ[c*h - b*i, 0] && Int
egerQ[2*m] && (IntegerQ[m] || GtQ[i/c, 0])

Rubi steps

\begin {align*} \int \left (a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}\right )^{3/2} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n \, dx &=\frac {2 \text {Subst}\left (\int \frac {x^{-5+n} \left (d^2 e-\left (-a e+\frac {2 d^2 e}{f^2}\right ) f^2+e x^2\right )^4}{32 e^5} \, dx,x,d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )}{f^3}\\ &=\frac {\text {Subst}\left (\int x^{-5+n} \left (d^2 e-\left (-a e+\frac {2 d^2 e}{f^2}\right ) f^2+e x^2\right )^4 \, dx,x,d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )}{16 e^5 f^3}\\ &=\frac {\text {Subst}\left (\int \left (e^4 \left (d^2-a f^2\right )^4 x^{-5+n}-4 e^4 \left (d^2-a f^2\right )^3 x^{-3+n}+6 e^4 \left (d^2-a f^2\right )^2 x^{-1+n}-4 e^4 \left (d^2-a f^2\right ) x^{1+n}+e^4 x^{3+n}\right ) \, dx,x,d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )}{16 e^5 f^3}\\ &=-\frac {\left (d^2-a f^2\right )^4 \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{-4+n}}{16 e f^3 (4-n)}+\frac {\left (d^2-a f^2\right )^3 \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{-2+n}}{4 e f^3 (2-n)}+\frac {3 \left (d^2-a f^2\right )^2 \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n}{8 e f^3 n}-\frac {\left (d^2-a f^2\right ) \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{2+n}}{4 e f^3 (2+n)}+\frac {\left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^{4+n}}{16 e f^3 (4+n)}\\ \end {align*}

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Mathematica [A]
time = 6.54, size = 228, normalized size = 0.77 \begin {gather*} \frac {\left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^n \left (\frac {6 \left (d^2-a f^2\right )^2}{n}+\frac {\left (d^2-a f^2\right )^4}{(-4+n) \left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^4}-\frac {4 \left (d^2-a f^2\right )^3}{(-2+n) \left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^2}-\frac {4 \left (d^2-a f^2\right ) \left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^2}{2+n}+\frac {\left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^4}{4+n}\right )}{16 e f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^n*((6*(d^2 - a*f^2)^2)/n + (d^2 - a*f^2)^4/((-4 + n)*(d + e*x +
f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^4) - (4*(d^2 - a*f^2)^3)/((-2 + n)*(d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))
/f^2])^2) - (4*(d^2 - a*f^2)*(d + e*x + f*Sqrt[a + (e*x*(2*d + e*x))/f^2])^2)/(2 + n) + (d + e*x + f*Sqrt[a +
(e*x*(2*d + e*x))/f^2])^4/(4 + n)))/(16*e*f^3)

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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \left (a +\frac {2 d e x}{f^{2}}+\frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}} \left (d +e x +f \sqrt {a +\frac {2 d e x}{f^{2}}+\frac {e^{2} x^{2}}{f^{2}}}\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.39, size = 346, normalized size = 1.16 \begin {gather*} \frac {{\left (a^{2} f^{4} n^{4} + 24 \, a^{2} f^{4} + {\left (n^{4} - 4 \, n^{2}\right )} x^{4} e^{4} - 48 \, a d^{2} f^{2} + 4 \, {\left (d n^{4} - 4 \, d n^{2}\right )} x^{3} e^{3} + 24 \, d^{4} + 2 \, {\left ({\left (a f^{2} + 2 \, d^{2}\right )} n^{4} - 2 \, {\left (5 \, a f^{2} + d^{2}\right )} n^{2}\right )} x^{2} e^{2} - 4 \, {\left (4 \, a^{2} f^{4} - 3 \, a d^{2} f^{2}\right )} n^{2} + 4 \, {\left (a d f^{2} n^{4} - 2 \, {\left (5 \, a d f^{2} - 3 \, d^{3}\right )} n^{2}\right )} x e - 4 \, {\left (a d f^{3} n^{3} + {\left (f n^{3} - 4 \, f n\right )} x^{3} e^{3} + 3 \, {\left (d f n^{3} - 4 \, d f n\right )} x^{2} e^{2} + {\left ({\left (a f^{3} + 2 \, d^{2} f\right )} n^{3} - 2 \, {\left (5 \, a f^{3} + d^{2} f\right )} n\right )} x e - 2 \, {\left (5 \, a d f^{3} - 3 \, d^{3} f\right )} n\right )} \sqrt {\frac {a f^{2} + x^{2} e^{2} + 2 \, d x e}{f^{2}}}\right )} {\left (x e + f \sqrt {\frac {a f^{2} + x^{2} e^{2} + 2 \, d x e}{f^{2}}} + d\right )}^{n} e^{\left (-1\right )}}{f^{3} n^{5} - 20 \, f^{3} n^{3} + 64 \, f^{3} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*f^4*n^4 + 24*a^2*f^4 + (n^4 - 4*n^2)*x^4*e^4 - 48*a*d^2*f^2 + 4*(d*n^4 - 4*d*n^2)*x^3*e^3 + 24*d^4 + 2*((
a*f^2 + 2*d^2)*n^4 - 2*(5*a*f^2 + d^2)*n^2)*x^2*e^2 - 4*(4*a^2*f^4 - 3*a*d^2*f^2)*n^2 + 4*(a*d*f^2*n^4 - 2*(5*
a*d*f^2 - 3*d^3)*n^2)*x*e - 4*(a*d*f^3*n^3 + (f*n^3 - 4*f*n)*x^3*e^3 + 3*(d*f*n^3 - 4*d*f*n)*x^2*e^2 + ((a*f^3
+ 2*d^2*f)*n^3 - 2*(5*a*f^3 + d^2*f)*n)*x*e - 2*(5*a*d*f^3 - 3*d^3*f)*n)*sqrt((a*f^2 + x^2*e^2 + 2*d*x*e)/f^2
))*(x*e + f*sqrt((a*f^2 + x^2*e^2 + 2*d*x*e)/f^2) + d)^n*e^(-1)/(f^3*n^5 - 20*f^3*n^3 + 64*f^3*n)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + \frac {2 d e x}{f^{2}} + \frac {e^{2} x^{2}}{f^{2}}\right )^{\frac {3}{2}} \left (d + e x + f \sqrt {a + \frac {2 d e x}{f^{2}} + \frac {e^{2} x^{2}}{f^{2}}}\right )^{n}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (d+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}+\frac {2\,d\,e\,x}{f^2}}+e\,x\right )}^n\,{\left (a+\frac {e^2\,x^2}{f^2}+\frac {2\,d\,e\,x}{f^2}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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