3.5.61 \(\int (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}})^{3/2} \, dx\) [461]

3.5.61.1 Optimal result

Integrand size = 27, antiderivative size = 183 \[ \int \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx=\frac {a f^2 \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{e}-\frac {a d f^2 \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{2 e \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}+\frac {\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{5/2}}{5 e}-\frac {3 a \sqrt {d} f^2 \text {arctanh}\left (\frac {\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{\sqrt {d}}\right )}{2 e} \]

output

-3/2*a*f^2*arctanh((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2)/d^(1/2))*d^(1/2)/ 
e+1/5*(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(5/2)/e+a*f^2*(d+e*x+f*(a+e^2*x^2/f^ 
2)^(1/2))^(1/2)/e-1/2*a*d*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))
 

3.5.61.2 Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.93 \[ \int \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx=\frac {\frac {\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}} \left (2 (d+2 e x)^2 \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )+a f^2 \left (-d+16 e x+12 f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )\right )}{e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}-15 a \sqrt {d} f^2 \text {arctanh}\left (\frac {\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{\sqrt {d}}\right )}{10 e} \]

input

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

output

((Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]]*(2*(d + 2*e*x)^2*(e*x + f*Sqrt 
[a + (e^2*x^2)/f^2]) + a*f^2*(-d + 16*e*x + 12*f*Sqrt[a + (e^2*x^2)/f^2])) 
)/(e*x + f*Sqrt[a + (e^2*x^2)/f^2]) - 15*a*Sqrt[d]*f^2*ArcTanh[Sqrt[d + e* 
x + f*Sqrt[a + (e^2*x^2)/f^2]]/Sqrt[d]])/(10*e)
 

3.5.61.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2542, 1192, 1580, 2341, 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.

\(\displaystyle \int \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^{3/2} \, dx\)

\(\Big \downarrow \) 2542

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

\(\Big \downarrow \) 1192

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

\(\Big \downarrow \) 1580

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

\(\Big \downarrow \) 2341

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{2} \left (-3 a \sqrt {d} f^2 \text {arctanh}\left (\frac {\sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {d}}\right )+\frac {2}{5} \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^{5/2}+2 a f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}\right )+\frac {a d f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{2 \left (f \left (-\sqrt {a+\frac {e^2 x^2}{f^2}}\right )-e x\right )}}{e}\)

input

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

output

((a*d*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*a*f^2*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]] + 
 (2*(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(5/2))/5 - 3*a*Sqrt[d]*f^2*ArcTa 
nh[Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]]/Sqrt[d]])/2)/e
 

3.5.61.3.1 Defintions of rubi rules used

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 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 2341

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 

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]
 

3.5.61.4 Maple [F]

\[\int {\left (d +e x +f \sqrt {a +\frac {e^{2} x^{2}}{f^{2}}}\right )}^{\frac {3}{2}}d x\]

input

int((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(3/2),x)
 

output

int((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(3/2),x)
 

3.5.61.5 Fricas [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.84 \[ \int \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx=\left [\frac {15 \, a \sqrt {d} f^{2} \log \left (a f^{2} - 2 \, d e x + 2 \, d f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + 2 \, {\left (\sqrt {d} e x - \sqrt {d} f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}\right ) + 2 \, {\left (4 \, e^{2} x^{2} + 12 \, a f^{2} + 9 \, d e x + 2 \, d^{2} + {\left (4 \, e f x - d f\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{20 \, e}, \frac {15 \, a \sqrt {-d} f^{2} \arctan \left (\frac {\sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d} \sqrt {-d}}{d}\right ) + {\left (4 \, e^{2} x^{2} + 12 \, a f^{2} + 9 \, d e x + 2 \, d^{2} + {\left (4 \, e f x - d f\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{10 \, e}\right ] \]

input

integrate((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(3/2),x, algorithm="fricas")
 

output

[1/20*(15*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*(4*e^2*x^2 + 12*a*f^2 + 9*d*e*x + 
 2*d^2 + (4*e*f*x - 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))/e, 1/10*(15*a*sqrt(-d)*f^2*arctan(sqrt(e*x + f* 
sqrt((e^2*x^2 + a*f^2)/f^2) + d)*sqrt(-d)/d) + (4*e^2*x^2 + 12*a*f^2 + 9*d 
*e*x + 2*d^2 + (4*e*f*x - d*f)*sqrt((e^2*x^2 + a*f^2)/f^2))*sqrt(e*x + f*s 
qrt((e^2*x^2 + a*f^2)/f^2) + d))/e]
 

3.5.61.6 Sympy [F]

\[ \int \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx=\int \left (d + e x + f \sqrt {a + \frac {e^{2} x^{2}}{f^{2}}}\right )^{\frac {3}{2}}\, dx \]

input

integrate((d+e*x+f*(a+e**2*x**2/f**2)**(1/2))**(3/2),x)
 

output

Integral((d + e*x + f*sqrt(a + e**2*x**2/f**2))**(3/2), x)
 

3.5.61.7 Maxima [F]

\[ \int \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx=\int { {\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {3}{2}} \,d x } \]

input

integrate((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(3/2),x, algorithm="maxima")
 

output

integrate((e*x + sqrt(e^2*x^2/f^2 + a)*f + d)^(3/2), x)
 

3.5.61.8 Giac [F]

\[ \int \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx=\int { {\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {3}{2}} \,d x } \]

input

integrate((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(3/2),x, algorithm="giac")
 

output

integrate((e*x + sqrt(e^2*x^2/f^2 + a)*f + d)^(3/2), x)
 

3.5.61.9 Mupad [F(-1)]

Timed out. \[ \int \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2} \, dx=\int {\left (d+e\,x+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}}\right )}^{3/2} \,d x \]

input

int((d + e*x + f*(a + (e^2*x^2)/f^2)^(1/2))^(3/2),x)
 

output

int((d + e*x + f*(a + (e^2*x^2)/f^2)^(1/2))^(3/2), x)