Integrand size = 27, antiderivative size = 158 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx=-\frac {1+\frac {a f^2}{d^2}}{e \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}-\frac {a f^2 \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{2 d^2 e \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}+\frac {3 a f^2 \text {arctanh}\left (\frac {\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{\sqrt {d}}\right )}{2 d^{5/2} e} \]
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^(5/2)/e +(-1-a*f^2/d^2)/e/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2)-1/2*a*f^2*(d+e*x+f *(a+e^2*x^2/f^2)^(1/2))^(1/2)/d^2/e/(e*x+f*(a+e^2*x^2/f^2)^(1/2))
Time = 1.31 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx=\frac {-\frac {\sqrt {d} \left (2 d^2 \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )+a f^2 \left (d+3 e x+3 f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )\right )}{\left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right ) \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}+3 a f^2 \text {arctanh}\left (\frac {\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{\sqrt {d}}\right )}{2 d^{5/2} e} \]
(-((Sqrt[d]*(2*d^2*(e*x + f*Sqrt[a + (e^2*x^2)/f^2]) + a*f^2*(d + 3*e*x + 3*f*Sqrt[a + (e^2*x^2)/f^2])))/((e*x + f*Sqrt[a + (e^2*x^2)/f^2])*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])) + 3*a*f^2*ArcTanh[Sqrt[d + e*x + f*Sqr t[a + (e^2*x^2)/f^2]]/Sqrt[d]])/(2*d^(5/2)*e)
Time = 0.31 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2542, 1192, 1582, 25, 359, 219}
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.
| \(\displaystyle \int \frac {1}{\left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2542 |
| \(\displaystyle \frac {\int \frac {d^2-2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right ) d+a f^2+\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )^2}{\left (-\sqrt {\frac {e^2 x^2}{f^2}+a} f-e x\right )^2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )^{3/2}}d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )}{2 e}\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle \frac {\int \frac {d^2-2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right ) d+a f^2+\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )^2}{\left (-\sqrt {\frac {e^2 x^2}{f^2}+a} f-e x\right )^2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}}}{e}\) |
| \(\Big \downarrow \) 1582 |
| \(\displaystyle \frac {\frac {a f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{2 d^2 \left (f \left (-\sqrt {a+\frac {e^2 x^2}{f^2}}\right )-e x\right )}-\frac {\int -\frac {2 d \left (d^2+a f^2\right )-\left (2 d^2-a f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )}{\left (-\sqrt {\frac {e^2 x^2}{f^2}+a} f-e x\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}}}{2 d^2}}{e}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {2 d \left (d^2+a f^2\right )-\left (2 d^2-a f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )}{\left (-\sqrt {\frac {e^2 x^2}{f^2}+a} f-e x\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}}}{2 d^2}+\frac {a f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{2 d^2 \left (f \left (-\sqrt {a+\frac {e^2 x^2}{f^2}}\right )-e x\right )}}{e}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {3 a f^2 \int \frac {1}{-\sqrt {\frac {e^2 x^2}{f^2}+a} f-e x}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}}-\frac {2 \left (a f^2+d^2\right )}{\sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}}{2 d^2}+\frac {a f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{2 d^2 \left (f \left (-\sqrt {a+\frac {e^2 x^2}{f^2}}\right )-e x\right )}}{e}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {3 a f^2 \text {arctanh}\left (\frac {\sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}-\frac {2 \left (a f^2+d^2\right )}{\sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}}{2 d^2}+\frac {a f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{2 d^2 \left (f \left (-\sqrt {a+\frac {e^2 x^2}{f^2}}\right )-e x\right )}}{e}\) |
((a*f^2*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/(2*d^2*(-(e*x) - f*Sqrt [a + (e^2*x^2)/f^2])) + ((-2*(d^2 + a*f^2))/Sqrt[d + 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]]/S qrt[d]])/Sqrt[d])/(2*d^2))/e
3.5.64.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
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]
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[(-d)^(m/2 - 1)/(2*e^ (2*p)*(q + 1)) Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e *x^2))*(2*(-d)^(-m/2 + 1)*e^(2*p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]
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]
\[\int \frac {1}{{\left (d +e x +f \sqrt {a +\frac {e^{2} x^{2}}{f^{2}}}\right )}^{\frac {3}{2}}}d x\]
Time = 0.47 (sec) , antiderivative size = 487, normalized size of antiderivative = 3.08 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (a^{2} f^{4} - 2 \, a d e f^{2} x - a d^{2} f^{2}\right )} \sqrt {d} \log \left (a f^{2} - 2 \, d e x + 2 \, d f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} - 2 \, {\left (\sqrt {d} e x - \sqrt {d} f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}\right ) - 2 \, {\left (2 \, d^{2} e^{2} x^{2} - 2 \, a d^{2} f^{2} - 2 \, d^{4} - {\left (3 \, a d e f^{2} + d^{3} e\right )} x + {\left (3 \, a d f^{3} - 2 \, d^{2} e f x + d^{3} f\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{4 \, {\left (a d^{3} e f^{2} - 2 \, d^{4} e^{2} x - d^{5} e\right )}}, -\frac {3 \, {\left (a^{2} f^{4} - 2 \, a d e f^{2} x - a d^{2} f^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d} \sqrt {-d}}{d}\right ) + {\left (2 \, d^{2} e^{2} x^{2} - 2 \, a d^{2} f^{2} - 2 \, d^{4} - {\left (3 \, a d e f^{2} + d^{3} e\right )} x + {\left (3 \, a d f^{3} - 2 \, d^{2} e f x + d^{3} f\right )} \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}}\right )} \sqrt {e x + f \sqrt {\frac {e^{2} x^{2} + a f^{2}}{f^{2}}} + d}}{2 \, {\left (a d^{3} e f^{2} - 2 \, d^{4} e^{2} x - d^{5} e\right )}}\right ] \]
[1/4*(3*(a^2*f^4 - 2*a*d*e*f^2*x - a*d^2*f^2)*sqrt(d)*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*(2* d^2*e^2*x^2 - 2*a*d^2*f^2 - 2*d^4 - (3*a*d*e*f^2 + d^3*e)*x + (3*a*d*f^3 - 2*d^2*e*f*x + d^3*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))/(a*d^3*e*f^2 - 2*d^4*e^2*x - d^5*e), -1/2*(3*(a^2* f^4 - 2*a*d*e*f^2*x - a*d^2*f^2)*sqrt(-d)*arctan(sqrt(e*x + f*sqrt((e^2*x^ 2 + a*f^2)/f^2) + d)*sqrt(-d)/d) + (2*d^2*e^2*x^2 - 2*a*d^2*f^2 - 2*d^4 - (3*a*d*e*f^2 + d^3*e)*x + (3*a*d*f^3 - 2*d^2*e*f*x + d^3*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))/(a*d^3*e*f^2 - 2*d^4*e^2*x - d^5*e)]
\[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx=\int \frac {1}{\left (d + e x + f \sqrt {a + \frac {e^{2} x^{2}}{f^{2}}}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx=\int { \frac {1}{{\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx=\int { \frac {1}{{\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx=\int \frac {1}{{\left (d+e\,x+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}}\right )}^{3/2}} \,d x \]