Integrand size = 27, antiderivative size = 199 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=-\frac {1+\frac {a f^2}{d^2}}{3 e \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2}}-\frac {2 a f^2}{d^3 e \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}-\frac {a f^2 \sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{2 d^3 e \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )}+\frac {5 a f^2 \text {arctanh}\left (\frac {\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{\sqrt {d}}\right )}{2 d^{7/2} e} \]
5/2*a*f^2*arctanh((d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(1/2)/d^(1/2))/d^(7/2)/e +1/3*(-1-a*f^2/d^2)/e/(d+e*x+f*(a+e^2*x^2/f^2)^(1/2))^(3/2)-2*a*f^2/d^3/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^3/e/(e*x+f*(a+e^2*x^2/f^2)^(1/2))
Time = 1.37 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=\frac {-\frac {\sqrt {d} \left (15 a^2 f^4+2 d^3 \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )+a f^2 \left (3 d^2+20 d \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )+30 e x \left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )\right )\right )}{\left (e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right ) \left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{3/2}}+15 a f^2 \text {arctanh}\left (\frac {\sqrt {d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}}}{\sqrt {d}}\right )}{6 d^{7/2} e} \]
(-((Sqrt[d]*(15*a^2*f^4 + 2*d^3*(e*x + f*Sqrt[a + (e^2*x^2)/f^2]) + a*f^2* (3*d^2 + 20*d*(e*x + f*Sqrt[a + (e^2*x^2)/f^2]) + 30*e*x*(e*x + f*Sqrt[a + (e^2*x^2)/f^2]))))/((e*x + f*Sqrt[a + (e^2*x^2)/f^2])*(d + e*x + f*Sqrt[a + (e^2*x^2)/f^2])^(3/2))) + 15*a*f^2*ArcTanh[Sqrt[d + e*x + f*Sqrt[a + (e ^2*x^2)/f^2]]/Sqrt[d]])/(6*d^(7/2)*e)
Time = 0.37 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2542, 1192, 1582, 1584, 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.
| \(\displaystyle \int \frac {1}{\left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2542 |
| \(\displaystyle \frac {\int \frac {d^2-2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right ) d+a f^2+\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )^2}{\left (-\sqrt {\frac {e^2 x^2}{f^2}+a} f-e x\right )^2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )^{5/2}}d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )}{2 e}\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle \frac {\int \frac {d^2-2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right ) d+a f^2+\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )^2}{\left (-\sqrt {\frac {e^2 x^2}{f^2}+a} f-e x\right )^2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )^2}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}}}{e}\) |
| \(\Big \downarrow \) 1582 |
| \(\displaystyle \frac {\frac {\int \frac {2 \left (d^2+a f^2\right ) d^2-2 \left (d^2-a f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right ) d+a f^2 \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )^2}{\left (-\sqrt {\frac {e^2 x^2}{f^2}+a} f-e x\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )^2}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}}}{2 d^3}+\frac {a f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{2 d^3 \left (f \left (-\sqrt {a+\frac {e^2 x^2}{f^2}}\right )-e x\right )}}{e}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {\frac {\int \left (\frac {5 a f^2}{-\sqrt {\frac {e^2 x^2}{f^2}+a} f-e x}+\frac {4 a f^2}{d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}}+\frac {2 \left (d^3+a f^2 d\right )}{\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}\right )^2}\right )d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+a}}}{2 d^3}+\frac {a f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{2 d^3 \left (f \left (-\sqrt {a+\frac {e^2 x^2}{f^2}}\right )-e x\right )}}{e}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\frac {5 a f^2 \text {arctanh}\left (\frac {\sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}-\frac {2 d \left (a f^2+d^2\right )}{3 \left (f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x\right )^{3/2}}-\frac {4 a f^2}{\sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}}{2 d^3}+\frac {a f^2 \sqrt {f \sqrt {a+\frac {e^2 x^2}{f^2}}+d+e x}}{2 d^3 \left (f \left (-\sqrt {a+\frac {e^2 x^2}{f^2}}\right )-e x\right )}}{e}\) |
((a*f^2*Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]])/(2*d^3*(-(e*x) - f*Sqrt [a + (e^2*x^2)/f^2])) + ((-2*d*(d^2 + a*f^2))/(3*(d + e*x + f*Sqrt[a + (e^ 2*x^2)/f^2])^(3/2)) - (4*a*f^2)/Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]] + (5*a*f^2*ArcTanh[Sqrt[d + e*x + f*Sqrt[a + (e^2*x^2)/f^2]]/Sqrt[d]])/Sqr t[d])/(2*d^3))/e
3.5.65.3.1 Defintions of rubi rules used
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[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
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]
\[\int \frac {1}{{\left (d +e x +f \sqrt {a +\frac {e^{2} x^{2}}{f^{2}}}\right )}^{\frac {5}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 377 vs. \(2 (171) = 342\).
Time = 0.66 (sec) , antiderivative size = 812, normalized size of antiderivative = 4.08 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=\left [\frac {15 \, {\left (a^{3} f^{6} + 4 \, a d^{2} e^{2} f^{2} x^{2} - 2 \, a^{2} d^{2} f^{4} + a d^{4} f^{2} - 4 \, {\left (a^{2} d e f^{4} - a d^{3} e f^{2}\right )} x\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 (12 \, d^{3} e^{3} x^{3} + 10 \, a^{2} d^{2} f^{4} - 16 \, a d^{4} f^{2} - 2 \, d^{6} - 8 \, {\left (5 \, a d^{2} e^{2} f^{2} - d^{4} e^{2}\right )} x^{2} + {\left (15 \, a^{2} d e f^{4} - 46 \, a d^{3} e f^{2} - d^{5} e\right )} x - {\left (15 \, a^{2} d f^{5} + 12 \, d^{3} e^{2} f x^{2} - 22 \, a d^{3} f^{3} - d^{5} f - 8 \, {\left (5 \, a d^{2} e f^{3} - d^{4} e f\right )} x\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}}{12 \, {\left (a^{2} d^{4} e f^{4} + 4 \, d^{6} e^{3} x^{2} - 2 \, a d^{6} e f^{2} + d^{8} e - 4 \, {\left (a d^{5} e^{2} f^{2} - d^{7} e^{2}\right )} x\right )}}, -\frac {15 \, {\left (a^{3} f^{6} + 4 \, a d^{2} e^{2} f^{2} x^{2} - 2 \, a^{2} d^{2} f^{4} + a d^{4} f^{2} - 4 \, {\left (a^{2} d e f^{4} - a d^{3} e f^{2}\right )} x\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 (12 \, d^{3} e^{3} x^{3} + 10 \, a^{2} d^{2} f^{4} - 16 \, a d^{4} f^{2} - 2 \, d^{6} - 8 \, {\left (5 \, a d^{2} e^{2} f^{2} - d^{4} e^{2}\right )} x^{2} + {\left (15 \, a^{2} d e f^{4} - 46 \, a d^{3} e f^{2} - d^{5} e\right )} x - {\left (15 \, a^{2} d f^{5} + 12 \, d^{3} e^{2} f x^{2} - 22 \, a d^{3} f^{3} - d^{5} f - 8 \, {\left (5 \, a d^{2} e f^{3} - d^{4} e f\right )} x\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}}{6 \, {\left (a^{2} d^{4} e f^{4} + 4 \, d^{6} e^{3} x^{2} - 2 \, a d^{6} e f^{2} + d^{8} e - 4 \, {\left (a d^{5} e^{2} f^{2} - d^{7} e^{2}\right )} x\right )}}\right ] \]
[1/12*(15*(a^3*f^6 + 4*a*d^2*e^2*f^2*x^2 - 2*a^2*d^2*f^4 + a*d^4*f^2 - 4*( a^2*d*e*f^4 - a*d^3*e*f^2)*x)*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*(12*d^3*e^3*x^3 + 1 0*a^2*d^2*f^4 - 16*a*d^4*f^2 - 2*d^6 - 8*(5*a*d^2*e^2*f^2 - d^4*e^2)*x^2 + (15*a^2*d*e*f^4 - 46*a*d^3*e*f^2 - d^5*e)*x - (15*a^2*d*f^5 + 12*d^3*e^2* f*x^2 - 22*a*d^3*f^3 - d^5*f - 8*(5*a*d^2*e*f^3 - d^4*e*f)*x)*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^2*d^4*e *f^4 + 4*d^6*e^3*x^2 - 2*a*d^6*e*f^2 + d^8*e - 4*(a*d^5*e^2*f^2 - d^7*e^2) *x), -1/6*(15*(a^3*f^6 + 4*a*d^2*e^2*f^2*x^2 - 2*a^2*d^2*f^4 + a*d^4*f^2 - 4*(a^2*d*e*f^4 - a*d^3*e*f^2)*x)*sqrt(-d)*arctan(sqrt(e*x + f*sqrt((e^2*x ^2 + a*f^2)/f^2) + d)*sqrt(-d)/d) - (12*d^3*e^3*x^3 + 10*a^2*d^2*f^4 - 16* a*d^4*f^2 - 2*d^6 - 8*(5*a*d^2*e^2*f^2 - d^4*e^2)*x^2 + (15*a^2*d*e*f^4 - 46*a*d^3*e*f^2 - d^5*e)*x - (15*a^2*d*f^5 + 12*d^3*e^2*f*x^2 - 22*a*d^3*f^ 3 - d^5*f - 8*(5*a*d^2*e*f^3 - d^4*e*f)*x)*sqrt((e^2*x^2 + a*f^2)/f^2))*sq rt(e*x + f*sqrt((e^2*x^2 + a*f^2)/f^2) + d))/(a^2*d^4*e*f^4 + 4*d^6*e^3*x^ 2 - 2*a*d^6*e*f^2 + d^8*e - 4*(a*d^5*e^2*f^2 - d^7*e^2)*x)]
\[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=\int \frac {1}{\left (d + e x + f \sqrt {a + \frac {e^{2} x^{2}}{f^{2}}}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=\int { \frac {1}{{\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=\int { \frac {1}{{\left (e x + \sqrt {\frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (d+e x+f \sqrt {a+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=\int \frac {1}{{\left (d+e\,x+f\,\sqrt {a+\frac {e^2\,x^2}{f^2}}\right )}^{5/2}} \,d x \]