Integrand size = 35, antiderivative size = 264 \[ \int \frac {a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {1}{(b d-a e) (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (a+b x)}{5 (b d-a e)^2 (d+e x)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b e (a+b x)}{3 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 b^2 e (a+b x)}{(b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 b^{5/2} e (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
-1/(-a*e+b*d)/(e*x+d)^(5/2)/((b*x+a)^2)^(1/2)-7/5*e*(b*x+a)/(-a*e+b*d)^2/( e*x+d)^(5/2)/((b*x+a)^2)^(1/2)-7/3*b*e*(b*x+a)/(-a*e+b*d)^3/(e*x+d)^(3/2)/ ((b*x+a)^2)^(1/2)+7*b^(5/2)*e*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+ b*d)^(1/2))/(-a*e+b*d)^(9/2)/((b*x+a)^2)^(1/2)-7*b^2*e*(b*x+a)/(-a*e+b*d)^ 4/(e*x+d)^(1/2)/((b*x+a)^2)^(1/2)
Time = 0.18 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.70 \[ \int \frac {a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e (a+b x) \left (\frac {-6 a^3 e^3+2 a^2 b e^2 (16 d+7 e x)-2 a b^2 e \left (58 d^2+84 d e x+35 e^2 x^2\right )-b^3 \left (15 d^3+161 d^2 e x+245 d e^2 x^2+105 e^3 x^3\right )}{e (b d-a e)^4 (a+b x) (d+e x)^{5/2}}-\frac {105 b^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{9/2}}\right )}{15 \sqrt {(a+b x)^2}} \]
(e*(a + b*x)*((-6*a^3*e^3 + 2*a^2*b*e^2*(16*d + 7*e*x) - 2*a*b^2*e*(58*d^2 + 84*d*e*x + 35*e^2*x^2) - b^3*(15*d^3 + 161*d^2*e*x + 245*d*e^2*x^2 + 10 5*e^3*x^3))/(e*(b*d - a*e)^4*(a + b*x)*(d + e*x)^(5/2)) - (105*b^(5/2)*Arc Tan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(-(b*d) + a*e)^(9/2)))/(1 5*Sqrt[(a + b*x)^2])
Time = 0.32 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.80, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {1187, 27, 52, 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.
| \(\displaystyle \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2} (d+e x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {1}{b^3 (a+b x)^2 (d+e x)^{7/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {1}{(a+b x)^2 (d+e x)^{7/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {7 e \int \frac {1}{(a+b x) (d+e x)^{7/2}}dx}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {7 e \left (\frac {b \int \frac {1}{(a+b x) (d+e x)^{5/2}}dx}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {7 e \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) (d+e x)^{3/2}}dx}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {7 e \left (\frac {b \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {7 e \left (\frac {b \left (\frac {b \left (\frac {2 b \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {7 e \left (\frac {b \left (\frac {b \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{5/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-(1/((b*d - a*e)*(a + b*x)*(d + e*x)^(5/2))) - (7*e*(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])/Sq rt[b*d - a*e]])/(b*d - a*e)^(3/2)))/(b*d - a*e)))/(b*d - a*e)))/(2*(b*d - a*e))))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.22.37.3.1 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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ IntPart[p]*(b/2 + c*x)^(2*FracPart[p])) Int[(d + e*x)^m*(f + g*x)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Time = 0.32 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.30
| method | result | size |
| default | \(-\frac {\left (105 \left (e x +d \right )^{\frac {5}{2}} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{4} e x +105 \left (e x +d \right )^{\frac {5}{2}} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{3} e +105 \sqrt {\left (a e -b d \right ) b}\, b^{3} e^{3} x^{3}+70 \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} e^{3} x^{2}+245 \sqrt {\left (a e -b d \right ) b}\, b^{3} d \,e^{2} x^{2}-14 \sqrt {\left (a e -b d \right ) b}\, a^{2} b \,e^{3} x +168 \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d \,e^{2} x +161 \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{2} e x +6 \sqrt {\left (a e -b d \right ) b}\, a^{3} e^{3}-32 \sqrt {\left (a e -b d \right ) b}\, a^{2} b d \,e^{2}+116 \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d^{2} e +15 \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{3}\right ) \left (b x +a \right )^{2}}{15 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, \left (a e -b d \right )^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(343\) |
-1/15*(105*(e*x+d)^(5/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*b^4*e *x+105*(e*x+d)^(5/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a*b^3*e+1 05*((a*e-b*d)*b)^(1/2)*b^3*e^3*x^3+70*((a*e-b*d)*b)^(1/2)*a*b^2*e^3*x^2+24 5*((a*e-b*d)*b)^(1/2)*b^3*d*e^2*x^2-14*((a*e-b*d)*b)^(1/2)*a^2*b*e^3*x+168 *((a*e-b*d)*b)^(1/2)*a*b^2*d*e^2*x+161*((a*e-b*d)*b)^(1/2)*b^3*d^2*e*x+6*( (a*e-b*d)*b)^(1/2)*a^3*e^3-32*((a*e-b*d)*b)^(1/2)*a^2*b*d*e^2+116*((a*e-b* d)*b)^(1/2)*a*b^2*d^2*e+15*((a*e-b*d)*b)^(1/2)*b^3*d^3)*(b*x+a)^2/(e*x+d)^ (5/2)/((a*e-b*d)*b)^(1/2)/(a*e-b*d)^4/((b*x+a)^2)^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 604 vs. \(2 (187) = 374\).
Time = 0.37 (sec) , antiderivative size = 1218, normalized size of antiderivative = 4.61 \[ \int \frac {a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [\frac {105 \, {\left (b^{3} e^{4} x^{4} + a b^{2} d^{3} e + {\left (3 \, b^{3} d e^{3} + a b^{2} e^{4}\right )} x^{3} + 3 \, {\left (b^{3} d^{2} e^{2} + a b^{2} d e^{3}\right )} x^{2} + {\left (b^{3} d^{3} e + 3 \, a b^{2} d^{2} e^{2}\right )} x\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 (105 \, b^{3} e^{3} x^{3} + 15 \, b^{3} d^{3} + 116 \, a b^{2} d^{2} e - 32 \, a^{2} b d e^{2} + 6 \, a^{3} e^{3} + 35 \, {\left (7 \, b^{3} d e^{2} + 2 \, a b^{2} e^{3}\right )} x^{2} + 7 \, {\left (23 \, b^{3} d^{2} e + 24 \, a b^{2} d e^{2} - 2 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{30 \, {\left (a b^{4} d^{7} - 4 \, a^{2} b^{3} d^{6} e + 6 \, a^{3} b^{2} d^{5} e^{2} - 4 \, a^{4} b d^{4} e^{3} + a^{5} d^{3} e^{4} + {\left (b^{5} d^{4} e^{3} - 4 \, a b^{4} d^{3} e^{4} + 6 \, a^{2} b^{3} d^{2} e^{5} - 4 \, a^{3} b^{2} d e^{6} + a^{4} b e^{7}\right )} x^{4} + {\left (3 \, b^{5} d^{5} e^{2} - 11 \, a b^{4} d^{4} e^{3} + 14 \, a^{2} b^{3} d^{3} e^{4} - 6 \, a^{3} b^{2} d^{2} e^{5} - a^{4} b d e^{6} + a^{5} e^{7}\right )} x^{3} + 3 \, {\left (b^{5} d^{6} e - 3 \, a b^{4} d^{5} e^{2} + 2 \, a^{2} b^{3} d^{4} e^{3} + 2 \, a^{3} b^{2} d^{3} e^{4} - 3 \, a^{4} b d^{2} e^{5} + a^{5} d e^{6}\right )} x^{2} + {\left (b^{5} d^{7} - a b^{4} d^{6} e - 6 \, a^{2} b^{3} d^{5} e^{2} + 14 \, a^{3} b^{2} d^{4} e^{3} - 11 \, a^{4} b d^{3} e^{4} + 3 \, a^{5} d^{2} e^{5}\right )} x\right )}}, \frac {105 \, {\left (b^{3} e^{4} x^{4} + a b^{2} d^{3} e + {\left (3 \, b^{3} d e^{3} + a b^{2} e^{4}\right )} x^{3} + 3 \, {\left (b^{3} d^{2} e^{2} + a b^{2} d e^{3}\right )} x^{2} + {\left (b^{3} d^{3} e + 3 \, a b^{2} d^{2} e^{2}\right )} x\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 (105 \, b^{3} e^{3} x^{3} + 15 \, b^{3} d^{3} + 116 \, a b^{2} d^{2} e - 32 \, a^{2} b d e^{2} + 6 \, a^{3} e^{3} + 35 \, {\left (7 \, b^{3} d e^{2} + 2 \, a b^{2} e^{3}\right )} x^{2} + 7 \, {\left (23 \, b^{3} d^{2} e + 24 \, a b^{2} d e^{2} - 2 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (a b^{4} d^{7} - 4 \, a^{2} b^{3} d^{6} e + 6 \, a^{3} b^{2} d^{5} e^{2} - 4 \, a^{4} b d^{4} e^{3} + a^{5} d^{3} e^{4} + {\left (b^{5} d^{4} e^{3} - 4 \, a b^{4} d^{3} e^{4} + 6 \, a^{2} b^{3} d^{2} e^{5} - 4 \, a^{3} b^{2} d e^{6} + a^{4} b e^{7}\right )} x^{4} + {\left (3 \, b^{5} d^{5} e^{2} - 11 \, a b^{4} d^{4} e^{3} + 14 \, a^{2} b^{3} d^{3} e^{4} - 6 \, a^{3} b^{2} d^{2} e^{5} - a^{4} b d e^{6} + a^{5} e^{7}\right )} x^{3} + 3 \, {\left (b^{5} d^{6} e - 3 \, a b^{4} d^{5} e^{2} + 2 \, a^{2} b^{3} d^{4} e^{3} + 2 \, a^{3} b^{2} d^{3} e^{4} - 3 \, a^{4} b d^{2} e^{5} + a^{5} d e^{6}\right )} x^{2} + {\left (b^{5} d^{7} - a b^{4} d^{6} e - 6 \, a^{2} b^{3} d^{5} e^{2} + 14 \, a^{3} b^{2} d^{4} e^{3} - 11 \, a^{4} b d^{3} e^{4} + 3 \, a^{5} d^{2} e^{5}\right )} x\right )}}\right ] \]
[1/30*(105*(b^3*e^4*x^4 + a*b^2*d^3*e + (3*b^3*d*e^3 + a*b^2*e^4)*x^3 + 3* (b^3*d^2*e^2 + a*b^2*d*e^3)*x^2 + (b^3*d^3*e + 3*a*b^2*d^2*e^2)*x)*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*(105*b^3*e^3*x^3 + 15*b^3*d^3 + 116*a*b^2*d^ 2*e - 32*a^2*b*d*e^2 + 6*a^3*e^3 + 35*(7*b^3*d*e^2 + 2*a*b^2*e^3)*x^2 + 7* (23*b^3*d^2*e + 24*a*b^2*d*e^2 - 2*a^2*b*e^3)*x)*sqrt(e*x + d))/(a*b^4*d^7 - 4*a^2*b^3*d^6*e + 6*a^3*b^2*d^5*e^2 - 4*a^4*b*d^4*e^3 + a^5*d^3*e^4 + ( b^5*d^4*e^3 - 4*a*b^4*d^3*e^4 + 6*a^2*b^3*d^2*e^5 - 4*a^3*b^2*d*e^6 + a^4* b*e^7)*x^4 + (3*b^5*d^5*e^2 - 11*a*b^4*d^4*e^3 + 14*a^2*b^3*d^3*e^4 - 6*a^ 3*b^2*d^2*e^5 - a^4*b*d*e^6 + a^5*e^7)*x^3 + 3*(b^5*d^6*e - 3*a*b^4*d^5*e^ 2 + 2*a^2*b^3*d^4*e^3 + 2*a^3*b^2*d^3*e^4 - 3*a^4*b*d^2*e^5 + a^5*d*e^6)*x ^2 + (b^5*d^7 - a*b^4*d^6*e - 6*a^2*b^3*d^5*e^2 + 14*a^3*b^2*d^4*e^3 - 11* a^4*b*d^3*e^4 + 3*a^5*d^2*e^5)*x), 1/15*(105*(b^3*e^4*x^4 + a*b^2*d^3*e + (3*b^3*d*e^3 + a*b^2*e^4)*x^3 + 3*(b^3*d^2*e^2 + a*b^2*d*e^3)*x^2 + (b^3*d ^3*e + 3*a*b^2*d^2*e^2)*x)*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)) - (105*b^3*e^3*x^3 + 15*b^3*d^ 3 + 116*a*b^2*d^2*e - 32*a^2*b*d*e^2 + 6*a^3*e^3 + 35*(7*b^3*d*e^2 + 2*a*b ^2*e^3)*x^2 + 7*(23*b^3*d^2*e + 24*a*b^2*d*e^2 - 2*a^2*b*e^3)*x)*sqrt(e*x + d))/(a*b^4*d^7 - 4*a^2*b^3*d^6*e + 6*a^3*b^2*d^5*e^2 - 4*a^4*b*d^4*e^3 + a^5*d^3*e^4 + (b^5*d^4*e^3 - 4*a*b^4*d^3*e^4 + 6*a^2*b^3*d^2*e^5 - 4*a...
\[ \int \frac {a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {a + b x}{\left (d + e x\right )^{\frac {7}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {b x + a}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{\frac {7}{2}}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (187) = 374\).
Time = 0.29 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.47 \[ \int \frac {a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {7 \, b^{3} e \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {-b^{2} d + a b e}} - \frac {\sqrt {e x + d} b^{3} e}{{\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}} - \frac {2 \, {\left (45 \, {\left (e x + d\right )}^{2} b^{2} e + 10 \, {\left (e x + d\right )} b^{2} d e + 3 \, b^{2} d^{2} e - 10 \, {\left (e x + d\right )} a b e^{2} - 6 \, a b d e^{2} + 3 \, a^{2} e^{3}\right )}}{15 \, {\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} {\left (e x + d\right )}^{\frac {5}{2}}} \]
-7*b^3*e*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^4*d^4*sgn(b*x + a) - 4*a*b^3*d^3*e*sgn(b*x + a) + 6*a^2*b^2*d^2*e^2*sgn(b*x + a) - 4*a^3*b *d*e^3*sgn(b*x + a) + a^4*e^4*sgn(b*x + a))*sqrt(-b^2*d + a*b*e)) - sqrt(e *x + d)*b^3*e/((b^4*d^4*sgn(b*x + a) - 4*a*b^3*d^3*e*sgn(b*x + a) + 6*a^2* b^2*d^2*e^2*sgn(b*x + a) - 4*a^3*b*d*e^3*sgn(b*x + a) + a^4*e^4*sgn(b*x + a))*((e*x + d)*b - b*d + a*e)) - 2/15*(45*(e*x + d)^2*b^2*e + 10*(e*x + d) *b^2*d*e + 3*b^2*d^2*e - 10*(e*x + d)*a*b*e^2 - 6*a*b*d*e^2 + 3*a^2*e^3)/( (b^4*d^4*sgn(b*x + a) - 4*a*b^3*d^3*e*sgn(b*x + a) + 6*a^2*b^2*d^2*e^2*sgn (b*x + a) - 4*a^3*b*d*e^3*sgn(b*x + a) + a^4*e^4*sgn(b*x + a))*(e*x + d)^( 5/2))
Timed out. \[ \int \frac {a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {a+b\,x}{{\left (d+e\,x\right )}^{7/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]