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]

3.6.22.1 Optimal result

Integrand size = 60, antiderivative size = 93 \[ \int \frac {\left (d+e x+f \sqrt {\frac {a f^2+e x (2 d+e x)}{f^2}}\right )^n}{\sqrt {\frac {a f^2 g+e g x (2 d+e x)}{f^2}}} \, dx=\frac {f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}} \left (d+e x+f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}\right )^n}{e n \sqrt {a g+\frac {2 d e g x}{f^2}+\frac {e^2 g x^2}{f^2}}} \]

output

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)
 

3.6.22.2 Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.82 \[ \int \frac {\left (d+e x+f \sqrt {\frac {a f^2+e x (2 d+e x)}{f^2}}\right )^n}{\sqrt {\frac {a f^2 g+e g x (2 d+e x)}{f^2}}} \, dx=\frac {f \sqrt {a+\frac {e x (2 d+e x)}{f^2}} \left (d+e x+f \sqrt {a+\frac {e x (2 d+e x)}{f^2}}\right )^n}{e n \sqrt {g \left (a+\frac {e x (2 d+e x)}{f^2}\right )}} \]

input

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]
 

output

(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)])
 

3.6.22.3 Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2552, 2550, 2546, 15}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2552

\(\displaystyle \int \frac {\left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^n}{\sqrt {a g+\frac {2 d e g x}{f^2}+\frac {e^2 g x^2}{f^2}}}dx\)

\(\Big \downarrow \) 2550

\(\displaystyle \frac {\sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}} \int \frac {\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^n}{\sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}}dx}{\sqrt {a g+\frac {2 d e g x}{f^2}+\frac {e^2 g x^2}{f^2}}}\)

\(\Big \downarrow \) 2546

\(\displaystyle \frac {2 f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}} \int \frac {\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )^{n-1}}{2 e}d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+\frac {2 d e x}{f^2}+a}\right )}{\sqrt {a g+\frac {2 d e g x}{f^2}+\frac {e^2 g x^2}{f^2}}}\)

\(\Big \downarrow \) 15

\(\displaystyle \frac {f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}} \left (f \sqrt {a+\frac {2 d e x}{f^2}+\frac {e^2 x^2}{f^2}}+d+e x\right )^n}{e n \sqrt {a g+\frac {2 d e g x}{f^2}+\frac {e^2 g x^2}{f^2}}}\)

input

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]
 

output

(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])
 

3.6.22.3.1 Defintions of rubi rules used

rule 15

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

rule 2546

Int[((g_.) + (h_.)*(x_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*S 
qrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Simp[(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*S 
qrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n}, x] && Eq 
Q[e^2 - c*f^2, 0] && EqQ[c*g - a*i, 0] && EqQ[c*h - b*i, 0] && IntegerQ[2*m 
] && (IntegerQ[m] || GtQ[i/c, 0])
 

rule 2550

Int[((g_.) + (h_.)*(x_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*S 
qrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Simp[(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 2552

Int[((u_) + (f_.)*((j_.) + (k_.)*Sqrt[v_]))^(n_.)*(w_)^(m_.), x_Symbol] :> 
Int[ExpandToSum[w, x]^m*(ExpandToSum[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[Coefficient[u, x, 1]^2 - Coefficient[v, x, 2]*f^2*k^ 
2, 0]
 

3.6.22.4 Maple [F]

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

input

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)
 

output

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)
 

3.6.22.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.26 \[ \int \frac {\left (d+e x+f \sqrt {\frac {a f^2+e x (2 d+e x)}{f^2}}\right )^n}{\sqrt {\frac {a f^2 g+e g x (2 d+e x)}{f^2}}} \, dx=\frac {{\left (e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}} + d\right )}^{n} f^{3} \sqrt {\frac {e^{2} g x^{2} + a f^{2} g + 2 \, d e g x}{f^{2}}} \sqrt {\frac {e^{2} x^{2} + a f^{2} + 2 \, d e x}{f^{2}}}}{e^{3} g n x^{2} + a e f^{2} g n + 2 \, d e^{2} g n x} \]

input

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="fricas")
 

output

(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)
 

3.6.22.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\left (d+e x+f \sqrt {\frac {a f^2+e x (2 d+e x)}{f^2}}\right )^n}{\sqrt {\frac {a f^2 g+e g x (2 d+e x)}{f^2}}} \, dx=\text {Timed out} \]

input

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)
 

output

Timed out
 

3.6.22.7 Maxima [F]

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

input

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="maxima")
 

output

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

3.6.22.8 Giac [F]

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

input

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="giac")
 

output

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)
 

3.6.22.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x+f \sqrt {\frac {a f^2+e x (2 d+e x)}{f^2}}\right )^n}{\sqrt {\frac {a f^2 g+e g x (2 d+e x)}{f^2}}} \, dx=\int \frac {{\left (d+e\,x+f\,\sqrt {\frac {a\,f^2+e\,x\,\left (2\,d+e\,x\right )}{f^2}}\right )}^n}{\sqrt {\frac {a\,g\,f^2+e\,g\,x\,\left (2\,d+e\,x\right )}{f^2}}} \,d x \]

input

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)
 

output

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)