Integrand size = 30, antiderivative size = 202 \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {15 e^2 (a+b x) \sqrt {d+e x}}{4 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {5 e (d+e x)^{3/2}}{4 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{5/2}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {15 e^2 \sqrt {b d-a e} (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
-5/4*e*(e*x+d)^(3/2)/b^2/((b*x+a)^2)^(1/2)-1/2*(e*x+d)^(5/2)/b/(b*x+a)/((b *x+a)^2)^(1/2)-15/4*e^2*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^( 1/2))*(-a*e+b*d)^(1/2)/b^(7/2)/((b*x+a)^2)^(1/2)+15/4*e^2*(b*x+a)*(e*x+d)^ (1/2)/b^3/((b*x+a)^2)^(1/2)
Time = 0.40 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.68 \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {b} \sqrt {d+e x} \left (15 a^2 e^2-5 a b e (d-5 e x)+b^2 \left (-2 d^2-9 d e x+8 e^2 x^2\right )\right )-15 e^2 \sqrt {-b d+a e} (a+b x)^2 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{4 b^{7/2} (a+b x) \sqrt {(a+b x)^2}} \]
(Sqrt[b]*Sqrt[d + e*x]*(15*a^2*e^2 - 5*a*b*e*(d - 5*e*x) + b^2*(-2*d^2 - 9 *d*e*x + 8*e^2*x^2)) - 15*e^2*Sqrt[-(b*d) + a*e]*(a + b*x)^2*ArcTan[(Sqrt[ b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(4*b^(7/2)*(a + b*x)*Sqrt[(a + b*x) ^2])
Time = 0.26 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.74, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {1102, 27, 51, 51, 60, 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.
| \(\displaystyle \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1102 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {(d+e x)^{5/2}}{b^3 (a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {(d+e x)^{5/2}}{(a+b x)^3}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {5 e \int \frac {(d+e x)^{3/2}}{(a+b x)^2}dx}{4 b}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {5 e \left (\frac {3 e \int \frac {\sqrt {d+e x}}{a+b x}dx}{2 b}-\frac {(d+e x)^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {5 e \left (\frac {3 e \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{2 b}-\frac {(d+e x)^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {5 e \left (\frac {3 e \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{2 b}-\frac {(d+e x)^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (\frac {5 e \left (\frac {3 e \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{2 b}-\frac {(d+e x)^{3/2}}{b (a+b x)}\right )}{4 b}-\frac {(d+e x)^{5/2}}{2 b (a+b x)^2}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/2*(d + e*x)^(5/2)/(b*(a + b*x)^2) + (5*e*(-((d + e*x)^(3/2) /(b*(a + b*x))) + (3*e*((2*Sqrt[d + e*x])/b - (2*Sqrt[b*d - a*e]*ArcTanh[( Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(3/2)))/(2*b)))/(4*b)))/Sqrt[a^ 2 + 2*a*b*x + b^2*x^2]
3.18.13.3.1 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/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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 = 2.31 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.74
| method | result | size |
| risch | \(\frac {2 e^{2} \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{b^{3} \left (b x +a \right )}-\frac {\left (2 a e -2 b d \right ) e^{2} \left (\frac {-\frac {9 \left (e x +d \right )^{\frac {3}{2}} b}{8}+\left (-\frac {7 a e}{8}+\frac {7 b d}{8}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{3} \left (b x +a \right )}\) | \(149\) |
| default | \(\frac {\left (-15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{2} e^{3} x^{2}+15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{3} d \,e^{2} x^{2}+8 b^{2} e^{2} x^{2} \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}-30 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b \,e^{3} x +30 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{2} d \,e^{2} x +9 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a b e -9 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{2} d +16 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a b \,e^{2} x -15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{3} e^{3}+15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b d \,e^{2}+15 \sqrt {e x +d}\, a^{2} e^{2} \sqrt {\left (a e -b d \right ) b}-14 \sqrt {e x +d}\, a d e b \sqrt {\left (a e -b d \right ) b}+7 \sqrt {e x +d}\, d^{2} b^{2} \sqrt {\left (a e -b d \right ) b}\right ) \left (b x +a \right )}{4 \sqrt {\left (a e -b d \right ) b}\, b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(413\) |
2/b^3*e^2*(e*x+d)^(1/2)*((b*x+a)^2)^(1/2)/(b*x+a)-1/b^3*(2*a*e-2*b*d)*e^2* ((-9/8*(e*x+d)^(3/2)*b+(-7/8*a*e+7/8*b*d)*(e*x+d)^(1/2))/(b*(e*x+d)+a*e-b* d)^2+15/8/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))) *((b*x+a)^2)^(1/2)/(b*x+a)
Time = 0.28 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.70 \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [\frac {15 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (8 \, b^{2} e^{2} x^{2} - 2 \, b^{2} d^{2} - 5 \, a b d e + 15 \, a^{2} e^{2} - {\left (9 \, b^{2} d e - 25 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}, -\frac {15 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (8 \, b^{2} e^{2} x^{2} - 2 \, b^{2} d^{2} - 5 \, a b d e + 15 \, a^{2} e^{2} - {\left (9 \, b^{2} d e - 25 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}}\right ] \]
[1/8*(15*(b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt((b*d - a*e)/b)*log((b* e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b*x + a)) + 2* (8*b^2*e^2*x^2 - 2*b^2*d^2 - 5*a*b*d*e + 15*a^2*e^2 - (9*b^2*d*e - 25*a*b* e^2)*x)*sqrt(e*x + d))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), -1/4*(15*(b^2*e^2* x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b* sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (8*b^2*e^2*x^2 - 2*b^2*d^2 - 5*a*b*d*e + 15*a^2*e^2 - (9*b^2*d*e - 25*a*b*e^2)*x)*sqrt(e*x + d))/(b^5*x^2 + 2*a* b^4*x + a^2*b^3)]
\[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{\frac {5}{2}}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {5}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.97 \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {2 \, \sqrt {e x + d} e^{2}}{b^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {15 \, {\left (b d e^{2} - a e^{3}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, \sqrt {-b^{2} d + a b e} b^{3} \mathrm {sgn}\left (b x + a\right )} - \frac {9 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{2} d e^{2} - 7 \, \sqrt {e x + d} b^{2} d^{2} e^{2} - 9 \, {\left (e x + d\right )}^{\frac {3}{2}} a b e^{3} + 14 \, \sqrt {e x + d} a b d e^{3} - 7 \, \sqrt {e x + d} a^{2} e^{4}}{4 \, {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2} b^{3} \mathrm {sgn}\left (b x + a\right )} \]
2*sqrt(e*x + d)*e^2/(b^3*sgn(b*x + a)) + 15/4*(b*d*e^2 - a*e^3)*arctan(sqr t(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^3*sgn(b*x + a)) - 1/4*(9*(e*x + d)^(3/2)*b^2*d*e^2 - 7*sqrt(e*x + d)*b^2*d^2*e^2 - 9*(e*x + d)^(3/2)*a*b*e^3 + 14*sqrt(e*x + d)*a*b*d*e^3 - 7*sqrt(e*x + d)*a^2*e^4 )/(((e*x + d)*b - b*d + a*e)^2*b^3*sgn(b*x + a))
Timed out. \[ \int \frac {(d+e x)^{5/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^{5/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]