3.5.62 \(\int \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}} \, dx\) [462]

3.5.62.1 Optimal result

Integrand size = 27, antiderivative size = 147 \[ \int \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}} \, dx=-\frac {a f^2 \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{2 e \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}+\frac {\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2}}{3 e}-\frac {a f^2 \text {arctanh}\left (\frac {\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{\sqrt {d}}\right )}{2 \sqrt {d} e} \]

output

-1/2*a*f^2*arctanh((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2)/d^(1/2))/e/d^(1/2 
)+1/3*(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(3/2)/e-1/2*a*f^2*(d+e*x+f*(a+e^2*x^ 
2/f^2)^(1/2))^(1/2)/e/(e*x+f*(a+e^2*x^2/f^2)^(1/2))
 

3.5.62.2 Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.97 \[ \int \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}} \, dx=\frac {\frac {\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}} \left (-a f^2+2 (d+2 e x) \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )\right )}{e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}-\frac {3 a f^2 \text {arctanh}\left (\frac {\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{\sqrt {d}}\right )}{\sqrt {d}}}{6 e} \]

input

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

output

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

3.5.62.3 Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2542, 1192, 1580, 25, 1467, 2009}

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.

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

\(\Big \downarrow \) 2542

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

\(\Big \downarrow \) 1192

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

\(\Big \downarrow \) 1580

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1467

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2}{3} \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^{3/2}-\frac {a f^2 \text {arctanh}\left (\frac {\sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}\right )+\frac {a f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{2 \left (f \left (-\sqrt {a+\frac {e^2 x^2}{f^2}}\right )-e x\right )}}{e}\)

input

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

output

((a*f^2*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/(2*(-(e*x) - f*Sqrt[a + 
 (e^2*x^2)/f^2])) + ((2*(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(3/2))/3 - ( 
a*f^2*ArcTanh[Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]]/Sqrt[d]])/Sqrt[d]) 
/2)/e
 

3.5.62.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 1192

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1)   Subst[Int[x^( 
2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + 
 c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && 
IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
 

rule 1467

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
 x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], 
x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e 
 + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

rule 1580

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_) 
^4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d 
 + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[1/(2*e^(2*p + m/2)* 
(q + 1))   Int[(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2* 
e^(2*p + m/2)*(q + 1)*x^m*(a + b*x^2 + c*x^4)^p - (-d)^(m/2 - 1)*(c*d^2 - b 
*d*e + a*e^2)^p*(d + e*(2*q + 3)*x^2))], x], x], x] /; FreeQ[{a, b, c, d, e 
}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && IGtQ[m/2, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2542

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^( 
n_))^(p_.), x_Symbol] :> Simp[1/(2*e)   Subst[Int[(g + h*x^n)^p*((d^2 + a*f 
^2 - 2*d*x + x^2)/(d - x)^2), x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /; Fr 
eeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]
 

3.5.62.4 Maple [F]

\[\int \sqrt {d +e x +f \sqrt {a +\frac {e^{2} x^{2}}{f^{2}}}}d x\]

input

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

output

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

3.5.62.5 Fricas [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.05 \[ \int \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}} \, dx=\left [\frac {3 \, a \sqrt {d} f^{2} \log \left (a f^{2} - 2 \, d e x + 2 \, d f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + 2 \, {\left (\sqrt {d} e x - \sqrt {d} f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}\right ) + 2 \, {\left (5 \, d e x - d f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + 2 \, d^{2}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{12 \, d e}, \frac {3 \, a \sqrt {-d} f^{2} \arctan \left (\frac {\sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d} \sqrt {-d}}{d}\right ) + {\left (5 \, d e x - d f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + 2 \, d^{2}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{6 \, d e}\right ] \]

input

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

output

[1/12*(3*a*sqrt(d)*f^2*log(a*f^2 - 2*d*e*x + 2*d*f*sqrt((e^2*x^2 + a*f^2)/ 
f^2) + 2*(sqrt(d)*e*x - sqrt(d)*f*sqrt((e^2*x^2 + a*f^2)/f^2))*sqrt(e*x + 
f*sqrt((e^2*x^2 + a*f^2)/f^2) + d)) + 2*(5*d*e*x - d*f*sqrt((e^2*x^2 + a*f 
^2)/f^2) + 2*d^2)*sqrt(e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d))/(d*e), 1/ 
6*(3*a*sqrt(-d)*f^2*arctan(sqrt(e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d)*s 
qrt(-d)/d) + (5*d*e*x - d*f*sqrt((e^2*x^2 + a*f^2)/f^2) + 2*d^2)*sqrt(e*x 
+ f*sqrt((e^2*x^2 + a*f^2)/f^2) + d))/(d*e)]
 

3.5.62.6 Sympy [F]

\[ \int \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}} \, dx=\int \sqrt {d + e x + f \sqrt {a + \frac {e^{2} x^{2}}{f^{2}}}}\, dx \]

input

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

output

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

3.5.62.7 Maxima [F]

\[ \int \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}} \, dx=\int { \sqrt {e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a} f + d} \,d x } \]

input

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

output

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

3.5.62.8 Giac [F]

\[ \int \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}} \, dx=\int { \sqrt {e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a} f + d} \,d x } \]

input

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

output

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

3.5.62.9 Mupad [F(-1)]

Timed out. \[ \int \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}} \, dx=\int \sqrt {d+e\,x+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}}} \,d x \]

input

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

output

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