Integrand size = 30, antiderivative size = 219 \[ \int \frac {1}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x)}{5 (b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 b^2 (a+b x)}{(b d-a e)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 b^{5/2} (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
2/5*(b*x+a)/(-a*e+b*d)/(e*x+d)^(5/2)/((b*x+a)^2)^(1/2)+2/3*b*(b*x+a)/(-a*e +b*d)^2/(e*x+d)^(3/2)/((b*x+a)^2)^(1/2)-2*b^(5/2)*(b*x+a)*arctanh(b^(1/2)* (e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/(-a*e+b*d)^(7/2)/((b*x+a)^2)^(1/2)+2*b^2*( b*x+a)/(-a*e+b*d)^3/(e*x+d)^(1/2)/((b*x+a)^2)^(1/2)
Time = 0.25 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2 (a+b x) \left (\frac {3 a^2 e^2-a b e (11 d+5 e x)+b^2 \left (23 d^2+35 d e x+15 e^2 x^2\right )}{(b d-a e)^3 (d+e x)^{5/2}}-\frac {15 b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{7/2}}\right )}{15 \sqrt {(a+b x)^2}} \]
(2*(a + b*x)*((3*a^2*e^2 - a*b*e*(11*d + 5*e*x) + b^2*(23*d^2 + 35*d*e*x + 15*e^2*x^2))/((b*d - a*e)^3*(d + e*x)^(5/2)) - (15*b^(5/2)*ArcTan[(Sqrt[b ]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(-(b*d) + a*e)^(7/2)))/(15*Sqrt[(a + b*x)^2])
Time = 0.27 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.76, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {1102, 27, 61, 61, 61, 73, 221}
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.
| \(\displaystyle \int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1102 |
| \(\displaystyle \frac {b (a+b x) \int \frac {1}{b (a+b x) (d+e x)^{7/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {1}{(a+b x) (d+e x)^{7/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a+b x) \left (\frac {b \int \frac {1}{(a+b x) (d+e x)^{5/2}}dx}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a+b x) \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) (d+e x)^{3/2}}dx}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a+b x) \left (\frac {b \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {b \left (\frac {b \left (\frac {2 b \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (\frac {b \left (\frac {b \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(2/(5*(b*d - a*e)*(d + e*x)^(5/2)) + (b*(2/(3*(b*d - a*e)*(d + e*x)^(3/2)) + (b*(2/((b*d - a*e)*Sqrt[d + e*x]) - (2*Sqrt[b]*ArcTanh[(Sqrt [b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^(3/2)))/(b*d - a*e)))/(b* d - a*e)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.18.10.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F racPart[p])) Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
Time = 2.16 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.92
| method | result | size |
| default | \(-\frac {2 \left (b x +a \right ) \left (15 b^{3} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) \left (e x +d \right )^{\frac {5}{2}}+15 \sqrt {\left (a e -b d \right ) b}\, b^{2} e^{2} x^{2}-5 \sqrt {\left (a e -b d \right ) b}\, a b \,e^{2} x +35 \sqrt {\left (a e -b d \right ) b}\, b^{2} d e x +3 \sqrt {\left (a e -b d \right ) b}\, a^{2} e^{2}-11 \sqrt {\left (a e -b d \right ) b}\, a b d e +23 \sqrt {\left (a e -b d \right ) b}\, b^{2} d^{2}\right )}{15 \sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}}\) | \(202\) |
-2/15*(b*x+a)*(15*b^3*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*(e*x+d)^ (5/2)+15*((a*e-b*d)*b)^(1/2)*b^2*e^2*x^2-5*((a*e-b*d)*b)^(1/2)*a*b*e^2*x+3 5*((a*e-b*d)*b)^(1/2)*b^2*d*e*x+3*((a*e-b*d)*b)^(1/2)*a^2*e^2-11*((a*e-b*d )*b)^(1/2)*a*b*d*e+23*((a*e-b*d)*b)^(1/2)*b^2*d^2)/((b*x+a)^2)^(1/2)/(a*e- b*d)^3/(e*x+d)^(5/2)/((a*e-b*d)*b)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (155) = 310\).
Time = 0.31 (sec) , antiderivative size = 706, normalized size of antiderivative = 3.22 \[ \int \frac {1}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\left [-\frac {15 \, {\left (b^{2} e^{3} x^{3} + 3 \, b^{2} d e^{2} x^{2} + 3 \, b^{2} d^{2} e x + b^{2} d^{3}\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, {\left (15 \, b^{2} e^{2} x^{2} + 23 \, b^{2} d^{2} - 11 \, a b d e + 3 \, a^{2} e^{2} + 5 \, {\left (7 \, b^{2} d e - a b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (b^{3} d^{6} - 3 \, a b^{2} d^{5} e + 3 \, a^{2} b d^{4} e^{2} - a^{3} d^{3} e^{3} + {\left (b^{3} d^{3} e^{3} - 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} b d e^{5} - a^{3} e^{6}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{2} - 3 \, a b^{2} d^{3} e^{3} + 3 \, a^{2} b d^{2} e^{4} - a^{3} d e^{5}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e - 3 \, a b^{2} d^{4} e^{2} + 3 \, a^{2} b d^{3} e^{3} - a^{3} d^{2} e^{4}\right )} x\right )}}, -\frac {2 \, {\left (15 \, {\left (b^{2} e^{3} x^{3} + 3 \, b^{2} d e^{2} x^{2} + 3 \, b^{2} d^{2} e x + b^{2} d^{3}\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (15 \, b^{2} e^{2} x^{2} + 23 \, b^{2} d^{2} - 11 \, a b d e + 3 \, a^{2} e^{2} + 5 \, {\left (7 \, b^{2} d e - a b e^{2}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, {\left (b^{3} d^{6} - 3 \, a b^{2} d^{5} e + 3 \, a^{2} b d^{4} e^{2} - a^{3} d^{3} e^{3} + {\left (b^{3} d^{3} e^{3} - 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} b d e^{5} - a^{3} e^{6}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{2} - 3 \, a b^{2} d^{3} e^{3} + 3 \, a^{2} b d^{2} e^{4} - a^{3} d e^{5}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e - 3 \, a b^{2} d^{4} e^{2} + 3 \, a^{2} b d^{3} e^{3} - a^{3} d^{2} e^{4}\right )} x\right )}}\right ] \]
[-1/15*(15*(b^2*e^3*x^3 + 3*b^2*d*e^2*x^2 + 3*b^2*d^2*e*x + b^2*d^3)*sqrt( b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*sqrt(e*x + d)*sqrt (b/(b*d - a*e)))/(b*x + a)) - 2*(15*b^2*e^2*x^2 + 23*b^2*d^2 - 11*a*b*d*e + 3*a^2*e^2 + 5*(7*b^2*d*e - a*b*e^2)*x)*sqrt(e*x + d))/(b^3*d^6 - 3*a*b^2 *d^5*e + 3*a^2*b*d^4*e^2 - a^3*d^3*e^3 + (b^3*d^3*e^3 - 3*a*b^2*d^2*e^4 + 3*a^2*b*d*e^5 - a^3*e^6)*x^3 + 3*(b^3*d^4*e^2 - 3*a*b^2*d^3*e^3 + 3*a^2*b* d^2*e^4 - a^3*d*e^5)*x^2 + 3*(b^3*d^5*e - 3*a*b^2*d^4*e^2 + 3*a^2*b*d^3*e^ 3 - a^3*d^2*e^4)*x), -2/15*(15*(b^2*e^3*x^3 + 3*b^2*d*e^2*x^2 + 3*b^2*d^2* e*x + b^2*d^3)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt (-b/(b*d - a*e))/(b*e*x + b*d)) - (15*b^2*e^2*x^2 + 23*b^2*d^2 - 11*a*b*d* e + 3*a^2*e^2 + 5*(7*b^2*d*e - a*b*e^2)*x)*sqrt(e*x + d))/(b^3*d^6 - 3*a*b ^2*d^5*e + 3*a^2*b*d^4*e^2 - a^3*d^3*e^3 + (b^3*d^3*e^3 - 3*a*b^2*d^2*e^4 + 3*a^2*b*d*e^5 - a^3*e^6)*x^3 + 3*(b^3*d^4*e^2 - 3*a*b^2*d^3*e^3 + 3*a^2* b*d^2*e^4 - a^3*d*e^5)*x^2 + 3*(b^3*d^5*e - 3*a*b^2*d^4*e^2 + 3*a^2*b*d^3* e^3 - a^3*d^2*e^4)*x)]
Timed out. \[ \int \frac {1}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int { \frac {1}{\sqrt {{\left (b x + a\right )}^{2}} {\left (e x + d\right )}^{\frac {7}{2}}} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {2}{15} \, {\left (\frac {15 \, b^{3} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e}} + \frac {15 \, {\left (e x + d\right )}^{2} b^{2} + 5 \, {\left (e x + d\right )} b^{2} d + 3 \, b^{2} d^{2} - 5 \, {\left (e x + d\right )} a b e - 6 \, a b d e + 3 \, a^{2} e^{2}}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} {\left (e x + d\right )}^{\frac {5}{2}}}\right )} \mathrm {sgn}\left (b x + a\right ) \]
2/15*(15*b^3*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d^3 - 3*a* b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*sqrt(-b^2*d + a*b*e)) + (15*(e*x + d) ^2*b^2 + 5*(e*x + d)*b^2*d + 3*b^2*d^2 - 5*(e*x + d)*a*b*e - 6*a*b*d*e + 3 *a^2*e^2)/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*(e*x + d)^( 5/2)))*sgn(b*x + a)
Timed out. \[ \int \frac {1}{(d+e x)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\int \frac {1}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^{7/2}} \,d x \]