Integrand size = 33, antiderivative size = 112 \[ \int \frac {(a+b x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {2 (b d-a e)^2 \sqrt {d+e x}}{b^3}+\frac {2 (b d-a e) (d+e x)^{3/2}}{3 b^2}+\frac {2 (d+e x)^{5/2}}{5 b}-\frac {2 (b d-a e)^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{7/2}} \]
2/3*(-a*e+b*d)*(e*x+d)^(3/2)/b^2+2/5*(e*x+d)^(5/2)/b-2*(-a*e+b*d)^(5/2)*ar ctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(7/2)+2*(-a*e+b*d)^2*(e*x+ d)^(1/2)/b^3
Time = 0.04 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \sqrt {d+e x} \left (15 a^2 e^2-5 a b e (7 d+e x)+b^2 \left (23 d^2+11 d e x+3 e^2 x^2\right )\right )}{15 b^3}-\frac {2 (-b d+a e)^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{b^{7/2}} \]
(2*Sqrt[d + e*x]*(15*a^2*e^2 - 5*a*b*e*(7*d + e*x) + b^2*(23*d^2 + 11*d*e* x + 3*e^2*x^2)))/(15*b^3) - (2*(-(b*d) + a*e)^(5/2)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/b^(7/2)
Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {1184, 27, 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.
\(\displaystyle \int \frac {(a+b x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^2 \int \frac {(d+e x)^{5/2}}{b^2 (a+b x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(d+e x)^{5/2}}{a+b x}dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b d-a e) \int \frac {(d+e x)^{3/2}}{a+b x}dx}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {\sqrt {d+e x}}{a+b x}dx}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {(b d-a e) \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {2 (b d-a e) \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b e}+\frac {2 \sqrt {d+e x}}{b}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(b d-a e) \left (\frac {(b d-a e) \left (\frac {2 \sqrt {d+e x}}{b}-\frac {2 \sqrt {b d-a e} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2}}\right )}{b}+\frac {2 (d+e x)^{3/2}}{3 b}\right )}{b}+\frac {2 (d+e x)^{5/2}}{5 b}\) |
(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
3.21.67.3.1 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 + 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[1/c^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}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.36 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.03
method | result | size |
pseudoelliptic | \(-\frac {2 \left (\left (a e -b d \right )^{3} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )-\left (\frac {\left (e^{2} x^{2}+\frac {11}{3} d e x +\frac {23}{3} d^{2}\right ) b^{2}}{5}-\frac {7 e \left (\frac {e x}{7}+d \right ) a b}{3}+e^{2} a^{2}\right ) \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{3}}\) | \(115\) |
risch | \(\frac {2 \left (3 b^{2} e^{2} x^{2}-5 a b \,e^{2} x +11 b^{2} d e x +15 e^{2} a^{2}-35 a b d e +23 b^{2} d^{2}\right ) \sqrt {e x +d}}{15 b^{3}}-\frac {2 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{b^{3} \sqrt {\left (a e -b d \right ) b}}\) | \(139\) |
derivativedivides | \(\frac {\frac {2 \left (e x +d \right )^{\frac {5}{2}} b^{2}}{5}-\frac {2 a b e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 b^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+2 a^{2} e^{2} \sqrt {e x +d}-4 a b d e \sqrt {e x +d}+2 b^{2} d^{2} \sqrt {e x +d}}{b^{3}}+\frac {2 \left (-a^{3} e^{3}+3 a^{2} b d \,e^{2}-3 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{b^{3} \sqrt {\left (a e -b d \right ) b}}\) | \(161\) |
default | \(\frac {\frac {2 \left (e x +d \right )^{\frac {5}{2}} b^{2}}{5}-\frac {2 a b e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 b^{2} d \left (e x +d \right )^{\frac {3}{2}}}{3}+2 a^{2} e^{2} \sqrt {e x +d}-4 a b d e \sqrt {e x +d}+2 b^{2} d^{2} \sqrt {e x +d}}{b^{3}}+\frac {2 \left (-a^{3} e^{3}+3 a^{2} b d \,e^{2}-3 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{b^{3} \sqrt {\left (a e -b d \right ) b}}\) | \(161\) |
-2/((a*e-b*d)*b)^(1/2)*((a*e-b*d)^3*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^( 1/2))-(1/5*(e^2*x^2+11/3*d*e*x+23/3*d^2)*b^2-7/3*e*(1/7*e*x+d)*a*b+e^2*a^2 )*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2))/b^3
Time = 0.36 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.59 \[ \int \frac {(a+b x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\left [\frac {15 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) + 2 \, {\left (3 \, b^{2} e^{2} x^{2} + 23 \, b^{2} d^{2} - 35 \, a b d e + 15 \, a^{2} e^{2} + {\left (11 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{15 \, b^{3}}, -\frac {2 \, {\left (15 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (3 \, b^{2} e^{2} x^{2} + 23 \, b^{2} d^{2} - 35 \, a b d e + 15 \, a^{2} e^{2} + {\left (11 \, b^{2} d e - 5 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, b^{3}}\right ] \]
[1/15*(15*(b^2*d^2 - 2*a*b*d*e + 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*(3*b^ 2*e^2*x^2 + 23*b^2*d^2 - 35*a*b*d*e + 15*a^2*e^2 + (11*b^2*d*e - 5*a*b*e^2 )*x)*sqrt(e*x + d))/b^3, -2/15*(15*(b^2*d^2 - 2*a*b*d*e + 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)) - (3*b^2*e^2*x^2 + 23*b^2*d^2 - 35*a*b*d*e + 15*a^2*e^2 + (11*b^2*d*e - 5*a* b*e^2)*x)*sqrt(e*x + d))/b^3]
Time = 17.74 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.39 \[ \int \frac {(a+b x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\begin {cases} \frac {2 \left (\frac {e \left (d + e x\right )^{\frac {5}{2}}}{5 b} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (- a e^{2} + b d e\right )}{3 b^{2}} + \frac {\sqrt {d + e x} \left (a^{2} e^{3} - 2 a b d e^{2} + b^{2} d^{2} e\right )}{b^{3}} - \frac {e \left (a e - b d\right )^{3} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e - b d}{b}}} \right )}}{b^{4} \sqrt {\frac {a e - b d}{b}}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {d^{\frac {5}{2}} \log {\left (a^{2} + 2 a b x + b^{2} x^{2} \right )}}{2 b} & \text {otherwise} \end {cases} \]
Piecewise((2*(e*(d + e*x)**(5/2)/(5*b) + (d + e*x)**(3/2)*(-a*e**2 + b*d*e )/(3*b**2) + sqrt(d + e*x)*(a**2*e**3 - 2*a*b*d*e**2 + b**2*d**2*e)/b**3 - e*(a*e - b*d)**3*atan(sqrt(d + e*x)/sqrt((a*e - b*d)/b))/(b**4*sqrt((a*e - b*d)/b)))/e, Ne(e, 0)), (d**(5/2)*log(a**2 + 2*a*b*x + b**2*x**2)/(2*b), True))
Exception generated. \[ \int \frac {(a+b x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for m ore detail
Time = 0.27 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.53 \[ \int \frac {(a+b x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{3}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{4} + 5 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{4} d + 15 \, \sqrt {e x + d} b^{4} d^{2} - 5 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{3} e - 30 \, \sqrt {e x + d} a b^{3} d e + 15 \, \sqrt {e x + d} a^{2} b^{2} e^{2}\right )}}{15 \, b^{5}} \]
2*(b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*arctan(sqrt(e*x + d) *b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^3) + 2/15*(3*(e*x + d)^(5 /2)*b^4 + 5*(e*x + d)^(3/2)*b^4*d + 15*sqrt(e*x + d)*b^4*d^2 - 5*(e*x + d) ^(3/2)*a*b^3*e - 30*sqrt(e*x + d)*a*b^3*d*e + 15*sqrt(e*x + d)*a^2*b^2*e^2 )/b^5
Time = 0.07 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.16 \[ \int \frac {(a+b x) (d+e x)^{5/2}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {2\,{\left (d+e\,x\right )}^{5/2}}{5\,b}-\frac {2\,\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{3/2}}{3\,b^2}+\frac {2\,{\left (a\,e-b\,d\right )}^2\,\sqrt {d+e\,x}}{b^3}-\frac {2\,\mathrm {atan}\left (\frac {\sqrt {b}\,{\left (a\,e-b\,d\right )}^{5/2}\,\sqrt {d+e\,x}}{a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3}\right )\,{\left (a\,e-b\,d\right )}^{5/2}}{b^{7/2}} \]