Integrand size = 35, antiderivative size = 250 \[ \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {7 e (b d-a e)^2 (a+b x) \sqrt {d+e x}}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 e (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 e (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^{7/2}}{b \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {7 e (b d-a e)^{5/2} (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
7/3*e*(-a*e+b*d)*(b*x+a)*(e*x+d)^(3/2)/b^3/((b*x+a)^2)^(1/2)+7/5*e*(b*x+a) *(e*x+d)^(5/2)/b^2/((b*x+a)^2)^(1/2)-(e*x+d)^(7/2)/b/((b*x+a)^2)^(1/2)-7*e *(-a*e+b*d)^(5/2)*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/ b^(9/2)/((b*x+a)^2)^(1/2)+7*e*(-a*e+b*d)^2*(b*x+a)*(e*x+d)^(1/2)/b^4/((b*x +a)^2)^(1/2)
Time = 0.10 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.72 \[ \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e (a+b x) \left (\frac {\sqrt {b} \sqrt {d+e x} \left (105 a^3 e^3+35 a^2 b e^2 (-7 d+2 e x)-7 a b^2 e \left (-23 d^2+24 d e x+2 e^2 x^2\right )+b^3 \left (-15 d^3+116 d^2 e x+32 d e^2 x^2+6 e^3 x^3\right )\right )}{e (a+b x)}-105 (-b d+a e)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )\right )}{15 b^{9/2} \sqrt {(a+b x)^2}} \]
(e*(a + b*x)*((Sqrt[b]*Sqrt[d + e*x]*(105*a^3*e^3 + 35*a^2*b*e^2*(-7*d + 2 *e*x) - 7*a*b^2*e*(-23*d^2 + 24*d*e*x + 2*e^2*x^2) + b^3*(-15*d^3 + 116*d^ 2*e*x + 32*d*e^2*x^2 + 6*e^3*x^3)))/(e*(a + b*x)) - 105*(-(b*d) + a*e)^(5/ 2)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]]))/(15*b^(9/2)*Sqrt[( a + b*x)^2])
Time = 0.30 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.70, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {1187, 27, 51, 60, 60, 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.
| \(\displaystyle \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^3 (a+b x) \int \frac {(d+e x)^{7/2}}{b^3 (a+b x)^2}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {(d+e x)^{7/2}}{(a+b x)^2}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {7 e \int \frac {(d+e x)^{5/2}}{a+b x}dx}{2 b}-\frac {(d+e x)^{7/2}}{b (a+b x)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {(b d-a e) \int \frac {(d+e x)^{3/2}}{a+b x}dx}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b}-\frac {(d+e x)^{7/2}}{b (a+b x)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {\sqrt {d+e x}}{a+b x}dx}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b}-\frac {(d+e x)^{7/2}}{b (a+b x)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b}-\frac {(d+e x)^{7/2}}{b (a+b x)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b}-\frac {(d+e x)^{7/2}}{b (a+b x)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (\frac {7 e \left (\frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\right )}{2 b}-\frac {(d+e x)^{7/2}}{b (a+b x)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-((d + e*x)^(7/2)/(b*(a + b*x))) + (7*e*((2*(d + e*x)^(5/2))/( 5*b) + ((b*d - a*e)*((2*(d + e*x)^(3/2))/(3*b) + ((b*d - a*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)))/b))/b))/(2*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.22.30.3.1 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_.)*((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 = 0.33 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.81
| method | result | size |
| risch | \(\frac {2 e \left (3 b^{2} e^{2} x^{2}-10 a b \,e^{2} x +16 b^{2} d e x +45 e^{2} a^{2}-100 a b d e +58 b^{2} d^{2}\right ) \sqrt {e x +d}\, \sqrt {\left (b x +a \right )^{2}}}{15 b^{4} \left (b x +a \right )}-\frac {\left (2 a^{3} e^{3}-6 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -2 b^{3} d^{3}\right ) e \left (-\frac {\sqrt {e x +d}}{2 \left (b \left (e x +d \right )+a e -b d \right )}+\frac {7 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right ) \sqrt {\left (b x +a \right )^{2}}}{b^{4} \left (b x +a \right )}\) | \(203\) |
| default | \(\frac {\left (6 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{3} e x +6 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} e -20 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} e^{2} x +20 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{3} d e x -105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{3} b \,e^{4} x +315 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b^{2} d \,e^{3} x -315 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{3} d^{2} e^{2} x +105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{4} d^{3} e x -20 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a^{2} b \,e^{2}+20 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d e +90 a^{2} b \,e^{3} x \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}-180 a \,b^{2} d \,e^{2} x \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}+90 b^{3} d^{2} e x \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}-105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{4} e^{4}+315 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{3} b d \,e^{3}-315 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b^{2} d^{2} e^{2}+105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{3} d^{3} e +105 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{3} e^{3}-225 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a^{2} b d \,e^{2}+135 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d^{2} e -15 \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{3}\right ) \left (b x +a \right )^{2}}{15 \sqrt {\left (a e -b d \right ) b}\, b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) | \(662\) |
2/15*e*(3*b^2*e^2*x^2-10*a*b*e^2*x+16*b^2*d*e*x+45*a^2*e^2-100*a*b*d*e+58* b^2*d^2)*(e*x+d)^(1/2)/b^4*((b*x+a)^2)^(1/2)/(b*x+a)-1/b^4*(2*a^3*e^3-6*a^ 2*b*d*e^2+6*a*b^2*d^2*e-2*b^3*d^3)*e*(-1/2*(e*x+d)^(1/2)/(b*(e*x+d)+a*e-b* d)+7/2/((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 = 0.31 (sec) , antiderivative size = 486, normalized size of antiderivative = 1.94 \[ \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [\frac {105 \, {\left (a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} + a^{3} e^{3} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (6 \, b^{3} e^{3} x^{3} - 15 \, b^{3} d^{3} + 161 \, a b^{2} d^{2} e - 245 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} + 2 \, {\left (16 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (58 \, b^{3} d^{2} e - 84 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{30 \, {\left (b^{5} x + a b^{4}\right )}}, -\frac {105 \, {\left (a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} + a^{3} e^{3} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (6 \, b^{3} e^{3} x^{3} - 15 \, b^{3} d^{3} + 161 \, a b^{2} d^{2} e - 245 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} + 2 \, {\left (16 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (58 \, b^{3} d^{2} e - 84 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (b^{5} x + a b^{4}\right )}}\right ] \]
[1/30*(105*(a*b^2*d^2*e - 2*a^2*b*d*e^2 + a^3*e^3 + (b^3*d^2*e - 2*a*b^2*d *e^2 + a^2*b*e^3)*x)*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*(6*b^3*e^3*x^3 - 15*b^3*d^ 3 + 161*a*b^2*d^2*e - 245*a^2*b*d*e^2 + 105*a^3*e^3 + 2*(16*b^3*d*e^2 - 7* a*b^2*e^3)*x^2 + 2*(58*b^3*d^2*e - 84*a*b^2*d*e^2 + 35*a^2*b*e^3)*x)*sqrt( e*x + d))/(b^5*x + a*b^4), -1/15*(105*(a*b^2*d^2*e - 2*a^2*b*d*e^2 + a^3*e ^3 + (b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*x)*sqrt(-(b*d - a*e)/b)*arcta n(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (6*b^3*e^3*x^3 - 15 *b^3*d^3 + 161*a*b^2*d^2*e - 245*a^2*b*d*e^2 + 105*a^3*e^3 + 2*(16*b^3*d*e ^2 - 7*a*b^2*e^3)*x^2 + 2*(58*b^3*d^2*e - 84*a*b^2*d*e^2 + 35*a^2*b*e^3)*x )*sqrt(e*x + d))/(b^5*x + a*b^4)]
Timed out. \[ \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b x + a\right )} {\left (e x + d\right )}^{\frac {7}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.27 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.18 \[ \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {7 \, {\left (b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {\sqrt {e x + d} b^{3} d^{3} e - 3 \, \sqrt {e x + d} a b^{2} d^{2} e^{2} + 3 \, \sqrt {e x + d} a^{2} b d e^{3} - \sqrt {e x + d} a^{3} e^{4}}{{\left ({\left (e x + d\right )} b - b d + a e\right )} b^{4} \mathrm {sgn}\left (b x + a\right )} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{8} e + 10 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{8} d e + 45 \, \sqrt {e x + d} b^{8} d^{2} e - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{7} e^{2} - 90 \, \sqrt {e x + d} a b^{7} d e^{2} + 45 \, \sqrt {e x + d} a^{2} b^{6} e^{3}\right )}}{15 \, b^{10} \mathrm {sgn}\left (b x + a\right )} \]
7*(b^3*d^3*e - 3*a*b^2*d^2*e^2 + 3*a^2*b*d*e^3 - a^3*e^4)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^4*sgn(b*x + a)) - (sq rt(e*x + d)*b^3*d^3*e - 3*sqrt(e*x + d)*a*b^2*d^2*e^2 + 3*sqrt(e*x + d)*a^ 2*b*d*e^3 - sqrt(e*x + d)*a^3*e^4)/(((e*x + d)*b - b*d + a*e)*b^4*sgn(b*x + a)) + 2/15*(3*(e*x + d)^(5/2)*b^8*e + 10*(e*x + d)^(3/2)*b^8*d*e + 45*sq rt(e*x + d)*b^8*d^2*e - 10*(e*x + d)^(3/2)*a*b^7*e^2 - 90*sqrt(e*x + d)*a* b^7*d*e^2 + 45*sqrt(e*x + d)*a^2*b^6*e^3)/(b^10*sgn(b*x + a))
Timed out. \[ \int \frac {(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^{7/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]