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

Optimal. Leaf size=93 \[ \frac{f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}} \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^n}{e n \sqrt{a g+\frac{2 d e g x}{f^2}+\frac{e^2 g x^2}{f^2}}} \]

[Out]

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

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Rubi [A]  time = 0.735419, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 60, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {2127, 2125, 2121, 12, 30} \[ \frac{f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}} \left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}+d+e x\right )^n}{e n \sqrt{a g+\frac{2 d e g x}{f^2}+\frac{e^2 g x^2}{f^2}}} \]

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*g + e*g*x*(2*d + e*x))/f^2],x]

[Out]

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

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]

Rule 2125

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

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 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]

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 g+e g 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 g+\frac{2 d e g x}{f^2}+\frac{e^2 g x^2}{f^2}}} \, dx\\ &=\frac{\sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}} \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}{\sqrt{a g+\frac{2 d e g x}{f^2}+\frac{e^2 g x^2}{f^2}}}\\ &=\frac{\left (2 f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right ) \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 )}{\sqrt{a g+\frac{2 d e g x}{f^2}+\frac{e^2 g x^2}{f^2}}}\\ &=\frac{\left (f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}}\right ) \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 \sqrt{a g+\frac{2 d e g x}{f^2}+\frac{e^2 g x^2}{f^2}}}\\ &=\frac{f \sqrt{a+\frac{2 d e x}{f^2}+\frac{e^2 x^2}{f^2}} \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 \sqrt{a g+\frac{2 d e g x}{f^2}+\frac{e^2 g x^2}{f^2}}}\\ \end{align*}

Mathematica [A]  time = 0.0394468, size = 76, normalized size = 0.82 \[ \frac{f \sqrt{a+\frac{e x (2 d+e x)}{f^2}} \left (f \sqrt{a+\frac{e x (2 d+e x)}{f^2}}+d+e x\right )^n}{e n \sqrt{g \left (a+\frac{e x (2 d+e x)}{f^2}\right )}} \]

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*g + e*g*x*(2*d + e*x))/f^2],x]

[Out]

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

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Maple [F]  time = 0.076, size = 0, normalized size = 0. \begin{align*} \int{ \left ( d+ex+f\sqrt{{\frac{a{f}^{2}+ex \left ( ex+2\,d \right ) }{{f}^{2}}}} \right ) ^{n}{\frac{1}{\sqrt{{\frac{a{f}^{2}g+egx \left ( ex+2\,d \right ) }{{f}^{2}}}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} g +{\left (e x + 2 \, d\right )} e g x}{f^{2}}}}\,{d x} \end{align*}

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*g+e*g*x*(e*x+2*d))/f^2)^(1/2),x, algorithm="ma
xima")

[Out]

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

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Fricas [A]  time = 1.06035, size = 250, normalized size = 2.69 \begin{align*} \frac{{\left (e x + f \sqrt{\frac{e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}} + d\right )}^{n} f^{3} \sqrt{\frac{e^{2} g x^{2} + a f^{2} g + 2 \, d e g x}{f^{2}}} \sqrt{\frac{e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}}}{e^{3} g n x^{2} + a e f^{2} g n + 2 \, d e^{2} g n x} \end{align*}

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*g+e*g*x*(e*x+2*d))/f^2)^(1/2),x, algorithm="fr
icas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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*g+e*g*x*(e*x+2*d))/f**2)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} g +{\left (e x + 2 \, d\right )} e g x}{f^{2}}}}\,{d x} \end{align*}

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*g+e*g*x*(e*x+2*d))/f^2)^(1/2),x, algorithm="gi
ac")

[Out]

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