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

Optimal. Leaf size=41 \[ \frac {f \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^n}{e n} \]

[Out]

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

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Rubi [A]  time = 0.44, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2127, 2121, 12, 30} \[ \frac {f \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^n}{e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*(d + e*x + f*Sqrt[a + (2*d*e*x)/f^2 + (e^2*x^2)/f^2])^n)/(e*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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2121

Int[((g_.) + (h_.)*(x_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)
^2])^(n_.), x_Symbol] :> Dist[(2*(i/c)^m)/f^(2*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])

Rule 2127

Int[((u_) + (f_.)*((j_.) + (k_.)*Sqrt[v_]))^(n_.)*(w_)^(m_.), x_Symbol] :> Int[ExpandToSum[w, x]^m*(ExpandToSu
m[u + f*j, x] + f*k*Sqrt[ExpandToSum[v, x]])^n, x] /; FreeQ[{f, j, k, m, n}, x] && LinearQ[u, x] && QuadraticQ
[{v, w}, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[{v, w}, x] && (EqQ[j, 0] || EqQ[f, 1])) && EqQ[Coeffic
ient[u, x, 1]^2 - Coefficient[v, x, 2]*f^2*k^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 36, normalized size = 0.88 \[ \frac {f \left (f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}+d+e x\right )^n}{e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 41, normalized size = 1.00 \[ \frac {{\left (e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}} + d\right )}^{n} f}{e n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.08, size = 39, normalized size = 0.95 \[ \frac {f\,{\left (d+e\,x+f\,\sqrt {\frac {a\,f^2+e\,x\,\left (2\,d+e\,x\right )}{f^2}}\right )}^n}{e\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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*f**2+e*x*(e*x+2*d))/f**2)**(1/2))**n/((a*f**2+e*x*(e*x+2*d))/f**2)**(1/2),x)

[Out]

Timed out

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