3.6.22 \(\int \frac {(d+e x+f \sqrt {\frac {a f^2+e x (2 d+e x)}{f^2}})^n}{\sqrt {\frac {a f^2 g+e g x (2 d+e x)}{f^2}}} \, dx\) [522]

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

[Out]

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

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Rubi [A]
time = 0.48, antiderivative size = 93, normalized size of antiderivative = 1.00, 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 = {2152, 2150, 2146, 12, 30} \begin {gather*} \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}}} \end {gather*}

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 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 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])

Rule 2150

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 2152

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 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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.02, size = 76, normalized size = 0.82 \begin {gather*} \frac {f \sqrt {a+\frac {e x (2 d+e x)}{f^2}} \left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^n}{e n \sqrt {g \left (a+\frac {e x (2 d+e x)}{f^2}\right )}} \end {gather*}

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 = 0.09, size = 0, normalized size = 0.00 \[\int \frac {\left (d +e x +f \sqrt {\frac {a \,f^{2}+e x \left (e x +2 d \right )}{f^{2}}}\right )^{n}}{\sqrt {\frac {a \,f^{2} g +e g x \left (e x +2 d \right )}{f^{2}}}}\, dx\]

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 = 0.00, 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*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]

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

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Fricas [A]
time = 0.36, size = 117, normalized size = 1.26 \begin {gather*} \frac {{\left (x e + f \sqrt {\frac {a f^{2} + x^{2} e^{2} + 2 \, d x e}{f^{2}}} + d\right )}^{n} f^{3} \sqrt {\frac {a f^{2} g + g x^{2} e^{2} + 2 \, d g x e}{f^{2}}} \sqrt {\frac {a f^{2} + x^{2} e^{2} + 2 \, d x e}{f^{2}}}}{a f^{2} g n e + g n x^{2} e^{3} + 2 \, d g n x e^{2}} \end {gather*}

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]

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

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

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]

Exception raised: SystemError >> excessive stack use: stack is 3436 deep

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

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

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*g + e*g*x*(2*d + e*x))/f^2)^(1/2),x)

[Out]

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

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