Integrand size = 35, antiderivative size = 225 \[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {5 e^2 \sqrt {d+e x}}{8 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {d+e x}}{3 (b d-a e) (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e \sqrt {d+e x}}{12 (b d-a e)^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 e^3 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 \sqrt {b} (b d-a e)^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]
5/8*e^3*(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^(1/2)/((b*x+a)^2)^(1/2)-5/8*e^2*(e*x+d)^(1/2)/(-a*e+b*d)^3/((b*x+ a)^2)^(1/2)-1/3*(e*x+d)^(1/2)/(-a*e+b*d)/(b*x+a)^2/((b*x+a)^2)^(1/2)+5/12* e*(e*x+d)^(1/2)/(-a*e+b*d)^2/(b*x+a)/((b*x+a)^2)^(1/2)
Time = 0.11 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.65 \[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^3 (a+b x) \left (\frac {\sqrt {d+e x} \left (33 a^2 e^2+2 a b e (-13 d+20 e x)+b^2 \left (8 d^2-10 d e x+15 e^2 x^2\right )\right )}{e^3 (-b d+a e)^3 (a+b x)^3}+\frac {15 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{\sqrt {b} (-b d+a e)^{7/2}}\right )}{24 \sqrt {(a+b x)^2}} \]
(e^3*(a + b*x)*((Sqrt[d + e*x]*(33*a^2*e^2 + 2*a*b*e*(-13*d + 20*e*x) + b^ 2*(8*d^2 - 10*d*e*x + 15*e^2*x^2)))/(e^3*(-(b*d) + a*e)^3*(a + b*x)^3) + ( 15*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(Sqrt[b]*(-(b*d) + a*e)^(7/2))))/(24*Sqrt[(a + b*x)^2])
Time = 0.30 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.86, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1187, 27, 52, 52, 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.
| \(\displaystyle \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2} \sqrt {d+e x}} \, dx\) |
| \(\Big \downarrow \) 1187 |
| \(\displaystyle \frac {b^5 (a+b x) \int \frac {1}{b^5 (a+b x)^4 \sqrt {d+e x}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(a+b x) \int \frac {1}{(a+b x)^4 \sqrt {d+e x}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {5 e \int \frac {1}{(a+b x)^3 \sqrt {d+e x}}dx}{6 (b d-a e)}-\frac {\sqrt {d+e x}}{3 (a+b x)^3 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {5 e \left (-\frac {3 e \int \frac {1}{(a+b x)^2 \sqrt {d+e x}}dx}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 (b d-a e)}-\frac {\sqrt {d+e x}}{3 (a+b x)^3 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {5 e \left (-\frac {3 e \left (-\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{2 (b d-a e)}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 (b d-a e)}-\frac {\sqrt {d+e x}}{3 (a+b x)^3 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {5 e \left (-\frac {3 e \left (-\frac {\int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b d-a e}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 (b d-a e)}-\frac {\sqrt {d+e x}}{3 (a+b x)^3 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(a+b x) \left (-\frac {5 e \left (-\frac {3 e \left (\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2}}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 (b d-a e)}-\frac {\sqrt {d+e x}}{3 (a+b x)^3 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
((a + b*x)*(-1/3*Sqrt[d + e*x]/((b*d - a*e)*(a + b*x)^3) - (5*e*(-1/2*Sqrt [d + e*x]/((b*d - a*e)*(a + b*x)^2) - (3*e*(-(Sqrt[d + e*x]/((b*d - a*e)*( a + b*x))) + (e*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(Sqrt[b] *(b*d - a*e)^(3/2))))/(4*(b*d - a*e))))/(6*(b*d - a*e))))/Sqrt[a^2 + 2*a*b *x + b^2*x^2]
3.22.42.3.1 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] :> 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. \(333\) vs. \(2(157)=314\).
Time = 0.30 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.48
| method | result | size |
| default | \(\frac {\left (15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{3} e^{3} x^{3}+45 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{2} e^{3} x^{2}+15 b^{2} e^{2} x^{2} \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}+45 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b \,e^{3} x +40 a b \,e^{2} x \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}-10 b^{2} d e x \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}+15 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{3} e^{3}+33 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a^{2} e^{2}-26 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, a b d e +8 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, b^{2} d^{2}\right ) \left (b x +a \right )^{2}}{24 \sqrt {\left (a e -b d \right ) b}\, \left (a e -b d \right )^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) | \(334\) |
1/24*(15*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*b^3*e^3*x^3+45*arctan (b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a*b^2*e^3*x^2+15*b^2*e^2*x^2*((a*e-b *d)*b)^(1/2)*(e*x+d)^(1/2)+45*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))* a^2*b*e^3*x+40*a*b*e^2*x*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)-10*b^2*d*e*x*(( a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)+15*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1 /2))*a^3*e^3+33*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a^2*e^2-26*((a*e-b*d)*b) ^(1/2)*(e*x+d)^(1/2)*a*b*d*e+8*((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)/(a*e-b*d)^3/((b*x+a)^2)^(5/2)
Leaf count of result is larger than twice the leaf count of optimal. 435 vs. \(2 (157) = 314\).
Time = 0.75 (sec) , antiderivative size = 884, normalized size of antiderivative = 3.93 \[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\left [-\frac {15 \, {\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} - 34 \, a b^{3} d^{2} e + 59 \, a^{2} b^{2} d e^{2} - 33 \, a^{3} b e^{3} + 15 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} - 10 \, {\left (b^{4} d^{2} e - 5 \, 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^{4} - 4 \, a^{4} b^{4} d^{3} e + 6 \, a^{5} b^{3} d^{2} e^{2} - 4 \, a^{6} b^{2} d e^{3} + a^{7} b e^{4} + {\left (b^{8} d^{4} - 4 \, a b^{7} d^{3} e + 6 \, a^{2} b^{6} d^{2} e^{2} - 4 \, a^{3} b^{5} d e^{3} + a^{4} b^{4} e^{4}\right )} x^{3} + 3 \, {\left (a b^{7} d^{4} - 4 \, a^{2} b^{6} d^{3} e + 6 \, a^{3} b^{5} d^{2} e^{2} - 4 \, a^{4} b^{4} d e^{3} + a^{5} b^{3} e^{4}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{4} - 4 \, a^{3} b^{5} d^{3} e + 6 \, a^{4} b^{4} d^{2} e^{2} - 4 \, a^{5} b^{3} d e^{3} + a^{6} b^{2} e^{4}\right )} x\right )}}, -\frac {15 \, {\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} - 34 \, a b^{3} d^{2} e + 59 \, a^{2} b^{2} d e^{2} - 33 \, a^{3} b e^{3} + 15 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} - 10 \, {\left (b^{4} d^{2} e - 5 \, 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^{4} - 4 \, a^{4} b^{4} d^{3} e + 6 \, a^{5} b^{3} d^{2} e^{2} - 4 \, a^{6} b^{2} d e^{3} + a^{7} b e^{4} + {\left (b^{8} d^{4} - 4 \, a b^{7} d^{3} e + 6 \, a^{2} b^{6} d^{2} e^{2} - 4 \, a^{3} b^{5} d e^{3} + a^{4} b^{4} e^{4}\right )} x^{3} + 3 \, {\left (a b^{7} d^{4} - 4 \, a^{2} b^{6} d^{3} e + 6 \, a^{3} b^{5} d^{2} e^{2} - 4 \, a^{4} b^{4} d e^{3} + a^{5} b^{3} e^{4}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{4} - 4 \, a^{3} b^{5} d^{3} e + 6 \, a^{4} b^{4} d^{2} e^{2} - 4 \, a^{5} b^{3} d e^{3} + a^{6} b^{2} e^{4}\right )} x\right )}}\right ] \]
[-1/48*(15*(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 - 34*a*b^3*d^2*e + 59*a^2*b^2*d*e^2 - 33*a^ 3*b*e^3 + 15*(b^4*d*e^2 - a*b^3*e^3)*x^2 - 10*(b^4*d^2*e - 5*a*b^3*d*e^2 + 4*a^2*b^2*e^3)*x)*sqrt(e*x + d))/(a^3*b^5*d^4 - 4*a^4*b^4*d^3*e + 6*a^5*b ^3*d^2*e^2 - 4*a^6*b^2*d*e^3 + a^7*b*e^4 + (b^8*d^4 - 4*a*b^7*d^3*e + 6*a^ 2*b^6*d^2*e^2 - 4*a^3*b^5*d*e^3 + a^4*b^4*e^4)*x^3 + 3*(a*b^7*d^4 - 4*a^2* b^6*d^3*e + 6*a^3*b^5*d^2*e^2 - 4*a^4*b^4*d*e^3 + a^5*b^3*e^4)*x^2 + 3*(a^ 2*b^6*d^4 - 4*a^3*b^5*d^3*e + 6*a^4*b^4*d^2*e^2 - 4*a^5*b^3*d*e^3 + a^6*b^ 2*e^4)*x), -1/24*(15*(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 - 34*a*b^3*d^2*e + 59*a^2*b^2*d*e^2 - 33*a^3*b*e^3 + 15*(b^4*d*e^2 - a*b^3*e^3)*x^2 - 10*(b^4*d^2*e - 5*a*b^3*d*e^2 + 4*a^2*b^ 2*e^3)*x)*sqrt(e*x + d))/(a^3*b^5*d^4 - 4*a^4*b^4*d^3*e + 6*a^5*b^3*d^2*e^ 2 - 4*a^6*b^2*d*e^3 + a^7*b*e^4 + (b^8*d^4 - 4*a*b^7*d^3*e + 6*a^2*b^6*d^2 *e^2 - 4*a^3*b^5*d*e^3 + a^4*b^4*e^4)*x^3 + 3*(a*b^7*d^4 - 4*a^2*b^6*d^3*e + 6*a^3*b^5*d^2*e^2 - 4*a^4*b^4*d*e^3 + a^5*b^3*e^4)*x^2 + 3*(a^2*b^6*d^4 - 4*a^3*b^5*d^3*e + 6*a^4*b^4*d^2*e^2 - 4*a^5*b^3*d*e^3 + a^6*b^2*e^4)*x) ]
Timed out. \[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int { \frac {b x + a}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} \sqrt {e x + d}} \,d x } \]
Time = 0.27 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.24 \[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {5 \, e^{3} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, {\left (b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {-b^{2} d + a b e}} - \frac {15 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{2} e^{3} - 40 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{2} d e^{3} + 33 \, \sqrt {e x + d} b^{2} d^{2} e^{3} + 40 \, {\left (e x + d\right )}^{\frac {3}{2}} a b e^{4} - 66 \, \sqrt {e x + d} a b d e^{4} + 33 \, \sqrt {e x + d} a^{2} e^{5}}{24 \, {\left (b^{3} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm {sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{3}} \]
-5/8*e^3*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d^3*sgn(b*x + a) - 3*a*b^2*d^2*e*sgn(b*x + a) + 3*a^2*b*d*e^2*sgn(b*x + a) - a^3*e^3*sgn (b*x + a))*sqrt(-b^2*d + a*b*e)) - 1/24*(15*(e*x + d)^(5/2)*b^2*e^3 - 40*( e*x + d)^(3/2)*b^2*d*e^3 + 33*sqrt(e*x + d)*b^2*d^2*e^3 + 40*(e*x + d)^(3/ 2)*a*b*e^4 - 66*sqrt(e*x + d)*a*b*d*e^4 + 33*sqrt(e*x + d)*a^2*e^5)/((b^3* d^3*sgn(b*x + a) - 3*a*b^2*d^2*e*sgn(b*x + a) + 3*a^2*b*d*e^2*sgn(b*x + a) - a^3*e^3*sgn(b*x + a))*((e*x + d)*b - b*d + a*e)^3)
Timed out. \[ \int \frac {a+b x}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {a+b\,x}{\sqrt {d+e\,x}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]