Integrand size = 35, antiderivative size = 207 \[ \int \frac {(a+b x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {(d+e x)^{3/2}}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {e^2 \sqrt {d+e x}}{8 b^2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e \sqrt {d+e x}}{4 b^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^3 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{5/2} (b d-a e)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
-1/3*(e*x+d)^(3/2)/b/(b^2*x^2+2*a*b*x+a^2)^(3/2)+1/8*e^3*(b*x+a)*arctanh(b ^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(5/2)/(-a*e+b*d)^(3/2)/((b*x+a)^2 )^(1/2)-1/8*e^2*(e*x+d)^(1/2)/b^2/(-a*e+b*d)/((b*x+a)^2)^(1/2)-1/4*e*(e*x+ d)^(1/2)/b^2/(b*x+a)/((b*x+a)^2)^(1/2)
Time = 0.10 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.72 \[ \int \frac {(a+b x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^3 (a+b x) \left (\frac {\sqrt {b} \sqrt {d+e x} \left (-3 a^2 e^2-2 a b e (d+4 e x)+b^2 \left (8 d^2+14 d e x+3 e^2 x^2\right )\right )}{e^3 (-b d+a e) (a+b x)^3}+\frac {3 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{3/2}}\right )}{24 b^{5/2} \sqrt {(a+b x)^2}} \]
(e^3*(a + b*x)*((Sqrt[b]*Sqrt[d + e*x]*(-3*a^2*e^2 - 2*a*b*e*(d + 4*e*x) + b^2*(8*d^2 + 14*d*e*x + 3*e^2*x^2)))/(e^3*(-(b*d) + a*e)*(a + b*x)^3) + ( 3*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(-(b*d) + a*e)^(3/2) ))/(24*b^(5/2)*Sqrt[(a + b*x)^2])
Time = 0.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.80, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1187, 27, 51, 51, 52, 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.
| \(\displaystyle \int \frac {(a+b x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^5 (a+b x) \int \frac {(d+e x)^{3/2}}{b^5 (a+b x)^4}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {(d+e x)^{3/2}}{(a+b x)^4}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {e \int \frac {\sqrt {d+e x}}{(a+b x)^3}dx}{2 b}-\frac {(d+e x)^{3/2}}{3 b (a+b x)^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {(a+b x) \left (\frac {e \left (\frac {e \int \frac {1}{(a+b x)^2 \sqrt {d+e x}}dx}{4 b}-\frac {\sqrt {d+e x}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(d+e x)^{3/2}}{3 b (a+b x)^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(a+b x) \left (\frac {e \left (\frac {e \left (-\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{2 (b d-a e)}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 b}-\frac {\sqrt {d+e x}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(d+e x)^{3/2}}{3 b (a+b x)^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (\frac {e \left (\frac {e \left (-\frac {\int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b d-a e}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 b}-\frac {\sqrt {d+e x}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(d+e x)^{3/2}}{3 b (a+b x)^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (\frac {e \left (\frac {e \left (\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2}}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 b}-\frac {\sqrt {d+e x}}{2 b (a+b x)^2}\right )}{2 b}-\frac {(d+e x)^{3/2}}{3 b (a+b x)^3}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/3*(d + e*x)^(3/2)/(b*(a + b*x)^3) + (e*(-1/2*Sqrt[d + e*x]/ (b*(a + b*x)^2) + (e*(-(Sqrt[d + e*x]/((b*d - a*e)*(a + b*x))) + (e*ArcTan h[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(Sqrt[b]*(b*d - a*e)^(3/2))))/ (4*b)))/(2*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
3.22.40.3.1 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 + 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] :> 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]
Leaf count of result is larger than twice the leaf count of optimal. \(325\) vs. \(2(148)=296\).
Time = 0.31 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.57
| method | result | size |
| default | \(\frac {\left (3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{3} e^{3} x^{3}+9 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{2} e^{3} x^{2}+3 \left (e x +d \right )^{\frac {5}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{2}+9 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b \,e^{3} x -8 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, a b e +8 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, b^{2} d +3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{3} e^{3}-3 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a^{2} e^{2}+6 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a b d e -3 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, b^{2} d^{2}\right ) \left (b x +a \right )^{2}}{24 \sqrt {\left (a e -b d \right ) b}\, b^{2} \left (a e -b d \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(326\) |
1/24*(3*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*b^3*e^3*x^3+9*arctan(b *(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a*b^2*e^3*x^2+3*(e*x+d)^(5/2)*((a*e-b* d)*b)^(1/2)*b^2+9*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^2*b*e^3*x- 8*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1/2)*a*b*e+8*(e*x+d)^(3/2)*((a*e-b*d)*b)^(1 /2)*b^2*d+3*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^3*e^3-3*((a*e-b* d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*e^2+6*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*b* d*e-3*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b^2*d^2)*(b*x+a)^2/((a*e-b*d)*b)^( 1/2)/b^2/(a*e-b*d)/((b*x+a)^2)^(5/2)
Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (148) = 296\).
Time = 0.38 (sec) , antiderivative size = 666, normalized size of antiderivative = 3.22 \[ \int \frac {(a+b x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\left [-\frac {3 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) + 2 \, {\left (8 \, b^{4} d^{3} - 10 \, a b^{3} d^{2} e - a^{2} b^{2} d e^{2} + 3 \, a^{3} b e^{3} + 3 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \, {\left (7 \, b^{4} d^{2} e - 11 \, a b^{3} d e^{2} + 4 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (a^{3} b^{5} d^{2} - 2 \, a^{4} b^{4} d e + a^{5} b^{3} e^{2} + {\left (b^{8} d^{2} - 2 \, a b^{7} d e + a^{2} b^{6} e^{2}\right )} x^{3} + 3 \, {\left (a b^{7} d^{2} - 2 \, a^{2} b^{6} d e + a^{3} b^{5} e^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{2} - 2 \, a^{3} b^{5} d e + a^{4} b^{4} e^{2}\right )} x\right )}}, -\frac {3 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) + {\left (8 \, b^{4} d^{3} - 10 \, a b^{3} d^{2} e - a^{2} b^{2} d e^{2} + 3 \, a^{3} b e^{3} + 3 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \, {\left (7 \, b^{4} d^{2} e - 11 \, a b^{3} d e^{2} + 4 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (a^{3} b^{5} d^{2} - 2 \, a^{4} b^{4} d e + a^{5} b^{3} e^{2} + {\left (b^{8} d^{2} - 2 \, a b^{7} d e + a^{2} b^{6} e^{2}\right )} x^{3} + 3 \, {\left (a b^{7} d^{2} - 2 \, a^{2} b^{6} d e + a^{3} b^{5} e^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{2} - 2 \, a^{3} b^{5} d e + a^{4} b^{4} e^{2}\right )} x\right )}}\right ] \]
[-1/48*(3*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(b ^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) + 2*(8*b^4*d^3 - 10*a*b^3*d^2*e - a^2*b^2*d*e^2 + 3*a^3*b*e ^3 + 3*(b^4*d*e^2 - a*b^3*e^3)*x^2 + 2*(7*b^4*d^2*e - 11*a*b^3*d*e^2 + 4*a ^2*b^2*e^3)*x)*sqrt(e*x + d))/(a^3*b^5*d^2 - 2*a^4*b^4*d*e + a^5*b^3*e^2 + (b^8*d^2 - 2*a*b^7*d*e + a^2*b^6*e^2)*x^3 + 3*(a*b^7*d^2 - 2*a^2*b^6*d*e + a^3*b^5*e^2)*x^2 + 3*(a^2*b^6*d^2 - 2*a^3*b^5*d*e + a^4*b^4*e^2)*x), -1/ 24*(3*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(-b^2* d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) + (8*b ^4*d^3 - 10*a*b^3*d^2*e - a^2*b^2*d*e^2 + 3*a^3*b*e^3 + 3*(b^4*d*e^2 - a*b ^3*e^3)*x^2 + 2*(7*b^4*d^2*e - 11*a*b^3*d*e^2 + 4*a^2*b^2*e^3)*x)*sqrt(e*x + d))/(a^3*b^5*d^2 - 2*a^4*b^4*d*e + a^5*b^3*e^2 + (b^8*d^2 - 2*a*b^7*d*e + a^2*b^6*e^2)*x^3 + 3*(a*b^7*d^2 - 2*a^2*b^6*d*e + a^3*b^5*e^2)*x^2 + 3* (a^2*b^6*d^2 - 2*a^3*b^5*d*e + a^4*b^4*e^2)*x)]
Timed out. \[ \int \frac {(a+b x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b x + a\right )} {\left (e x + d\right )}^{\frac {3}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {e^{3} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, {\left (b^{3} d \mathrm {sgn}\left (b x + a\right ) - a b^{2} e \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {-b^{2} d + a b e}} - \frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{2} e^{3} + 8 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{2} d e^{3} - 3 \, \sqrt {e x + d} b^{2} d^{2} e^{3} - 8 \, {\left (e x + d\right )}^{\frac {3}{2}} a b e^{4} + 6 \, \sqrt {e x + d} a b d e^{4} - 3 \, \sqrt {e x + d} a^{2} e^{5}}{24 \, {\left (b^{3} d \mathrm {sgn}\left (b x + a\right ) - a b^{2} e \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{3}} \]
-1/8*e^3*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d*sgn(b*x + a) - a*b^2*e*sgn(b*x + a))*sqrt(-b^2*d + a*b*e)) - 1/24*(3*(e*x + d)^(5/2)*b ^2*e^3 + 8*(e*x + d)^(3/2)*b^2*d*e^3 - 3*sqrt(e*x + d)*b^2*d^2*e^3 - 8*(e* x + d)^(3/2)*a*b*e^4 + 6*sqrt(e*x + d)*a*b*d*e^4 - 3*sqrt(e*x + d)*a^2*e^5 )/((b^3*d*sgn(b*x + a) - a*b^2*e*sgn(b*x + a))*((e*x + d)*b - b*d + a*e)^3 )
Timed out. \[ \int \frac {(a+b x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^{3/2}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]