Integrand size = 33, antiderivative size = 181 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {2 b B d-5 A b e+3 a B e}{3 b (b d-a e)^2 (d+e x)^{3/2}}-\frac {A b-a B}{b (b d-a e) (a+b x) (d+e x)^{3/2}}+\frac {2 b B d-5 A b e+3 a B e}{(b d-a e)^3 \sqrt {d+e x}}-\frac {\sqrt {b} (2 b B d-5 A b e+3 a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2}} \]
1/3*(-5*A*b*e+3*B*a*e+2*B*b*d)/b/(-a*e+b*d)^2/(e*x+d)^(3/2)+(-A*b+B*a)/b/( -a*e+b*d)/(b*x+a)/(e*x+d)^(3/2)-(-5*A*b*e+3*B*a*e+2*B*b*d)*arctanh(b^(1/2) *(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))*b^(1/2)/(-a*e+b*d)^(7/2)+(-5*A*b*e+3*B*a* e+2*B*b*d)/(-a*e+b*d)^3/(e*x+d)^(1/2)
Time = 0.15 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {B \left (2 a^2 e (2 d+3 e x)+2 b^2 d x (4 d+3 e x)+a b \left (11 d^2+16 d e x+9 e^2 x^2\right )\right )-A \left (-2 a^2 e^2+2 a b e (7 d+5 e x)+b^2 \left (3 d^2+20 d e x+15 e^2 x^2\right )\right )}{3 (b d-a e)^3 (a+b x) (d+e x)^{3/2}}-\frac {\sqrt {b} (2 b B d-5 A b e+3 a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{7/2}} \]
(B*(2*a^2*e*(2*d + 3*e*x) + 2*b^2*d*x*(4*d + 3*e*x) + a*b*(11*d^2 + 16*d*e *x + 9*e^2*x^2)) - A*(-2*a^2*e^2 + 2*a*b*e*(7*d + 5*e*x) + b^2*(3*d^2 + 20 *d*e*x + 15*e^2*x^2)))/(3*(b*d - a*e)^3*(a + b*x)*(d + e*x)^(3/2)) - (Sqrt [b]*(2*b*B*d - 5*A*b*e + 3*a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b* d) + a*e]])/(-(b*d) + a*e)^(7/2)
Time = 0.32 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {1184, 27, 87, 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.
| \(\displaystyle \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right ) (d+e x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1184 |
| \(\displaystyle b^2 \int \frac {A+B x}{b^2 (a+b x)^2 (d+e x)^{5/2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {A+B x}{(a+b x)^2 (d+e x)^{5/2}}dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(3 a B e-5 A b e+2 b B d) \int \frac {1}{(a+b x) (d+e x)^{5/2}}dx}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) (d+e x)^{3/2} (b d-a e)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(3 a B e-5 A b e+2 b B d) \left (\frac {b \int \frac {1}{(a+b x) (d+e x)^{3/2}}dx}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) (d+e x)^{3/2} (b d-a e)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(3 a B e-5 A b e+2 b B d) \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) (d+e x)^{3/2} (b d-a e)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(3 a B e-5 A b e+2 b B d) \left (\frac {b \left (\frac {2 b \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) (d+e x)^{3/2} (b d-a e)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(3 a B e-5 A b e+2 b B d) \left (\frac {b \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) (d+e x)^{3/2} (b d-a e)}\) |
-((A*b - a*B)/(b*(b*d - a*e)*(a + b*x)*(d + e*x)^(3/2))) + ((2*b*B*d - 5*A *b*e + 3*a*B*e)*(2/(3*(b*d - a*e)*(d + e*x)^(3/2)) + (b*(2/((b*d - a*e)*Sq rt[d + e*x]) - (2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]] )/(b*d - a*e)^(3/2)))/(b*d - a*e)))/(2*b*(b*d - a*e))
3.19.14.3.1 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] && 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 0.38 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.91
| method | result | size |
| derivativedivides | \(-\frac {2 \left (A e -B d \right )}{3 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 \left (-2 A b e +B a e +B b d \right )}{\left (a e -b d \right )^{3} \sqrt {e x +d}}+\frac {2 b \left (\frac {\left (\frac {1}{2} A b e -\frac {1}{2} B a e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (5 A b e -3 B a e -2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{3}}\) | \(164\) |
| default | \(-\frac {2 \left (A e -B d \right )}{3 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 \left (-2 A b e +B a e +B b d \right )}{\left (a e -b d \right )^{3} \sqrt {e x +d}}+\frac {2 b \left (\frac {\left (\frac {1}{2} A b e -\frac {1}{2} B a e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (5 A b e -3 B a e -2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{3}}\) | \(164\) |
| pseudoelliptic | \(\frac {5 b \left (\left (A e -\frac {2 B d}{5}\right ) b -\frac {3 B a e}{5}\right ) \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )-\frac {2 \sqrt {\left (a e -b d \right ) b}\, \left (\left (-\frac {15 A \,e^{2} x^{2}}{2}-10 x d \left (-\frac {3 B x}{10}+A \right ) e -\frac {3 d^{2} \left (-\frac {8 B x}{3}+A \right )}{2}\right ) b^{2}-7 \left (\frac {5 x \left (-\frac {9 B x}{10}+A \right ) e^{2}}{7}+d \left (-\frac {8 B x}{7}+A \right ) e -\frac {11 B \,d^{2}}{14}\right ) a b +\left (\left (3 B x +A \right ) e +2 B d \right ) e \,a^{2}\right )}{3}}{\left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right ) \sqrt {\left (a e -b d \right ) b}}\) | \(193\) |
-2/3*(A*e-B*d)/(a*e-b*d)^2/(e*x+d)^(3/2)-2/(a*e-b*d)^3*(-2*A*b*e+B*a*e+B*b *d)/(e*x+d)^(1/2)+2/(a*e-b*d)^3*b*((1/2*A*b*e-1/2*B*a*e)*(e*x+d)^(1/2)/(b* (e*x+d)+a*e-b*d)+1/2*(5*A*b*e-3*B*a*e-2*B*b*d)/((a*e-b*d)*b)^(1/2)*arctan( b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 548 vs. \(2 (162) = 324\).
Time = 0.35 (sec) , antiderivative size = 1106, normalized size of antiderivative = 6.11 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\text {Too large to display} \]
[1/6*(3*(2*B*a*b*d^3 + (3*B*a^2 - 5*A*a*b)*d^2*e + (2*B*b^2*d*e^2 + (3*B*a *b - 5*A*b^2)*e^3)*x^3 + (4*B*b^2*d^2*e + 2*(4*B*a*b - 5*A*b^2)*d*e^2 + (3 *B*a^2 - 5*A*a*b)*e^3)*x^2 + (2*B*b^2*d^3 + (7*B*a*b - 5*A*b^2)*d^2*e + 2* (3*B*a^2 - 5*A*a*b)*d*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*(2*A*a^ 2*e^2 + (11*B*a*b - 3*A*b^2)*d^2 + 2*(2*B*a^2 - 7*A*a*b)*d*e + 3*(2*B*b^2* d*e + (3*B*a*b - 5*A*b^2)*e^2)*x^2 + 2*(4*B*b^2*d^2 + 2*(4*B*a*b - 5*A*b^2 )*d*e + (3*B*a^2 - 5*A*a*b)*e^2)*x)*sqrt(e*x + d))/(a*b^3*d^5 - 3*a^2*b^2* d^4*e + 3*a^3*b*d^3*e^2 - a^4*d^2*e^3 + (b^4*d^3*e^2 - 3*a*b^3*d^2*e^3 + 3 *a^2*b^2*d*e^4 - a^3*b*e^5)*x^3 + (2*b^4*d^4*e - 5*a*b^3*d^3*e^2 + 3*a^2*b ^2*d^2*e^3 + a^3*b*d*e^4 - a^4*e^5)*x^2 + (b^4*d^5 - a*b^3*d^4*e - 3*a^2*b ^2*d^3*e^2 + 5*a^3*b*d^2*e^3 - 2*a^4*d*e^4)*x), -1/3*(3*(2*B*a*b*d^3 + (3* B*a^2 - 5*A*a*b)*d^2*e + (2*B*b^2*d*e^2 + (3*B*a*b - 5*A*b^2)*e^3)*x^3 + ( 4*B*b^2*d^2*e + 2*(4*B*a*b - 5*A*b^2)*d*e^2 + (3*B*a^2 - 5*A*a*b)*e^3)*x^2 + (2*B*b^2*d^3 + (7*B*a*b - 5*A*b^2)*d^2*e + 2*(3*B*a^2 - 5*A*a*b)*d*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)) - (2*A*a^2*e^2 + (11*B*a*b - 3*A*b^2)*d^2 + 2*(2*B*a^ 2 - 7*A*a*b)*d*e + 3*(2*B*b^2*d*e + (3*B*a*b - 5*A*b^2)*e^2)*x^2 + 2*(4*B* b^2*d^2 + 2*(4*B*a*b - 5*A*b^2)*d*e + (3*B*a^2 - 5*A*a*b)*e^2)*x)*sqrt(e*x + d))/(a*b^3*d^5 - 3*a^2*b^2*d^4*e + 3*a^3*b*d^3*e^2 - a^4*d^2*e^3 + (...
\[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\int \frac {A + B x}{\left (a + b x\right )^{2} \left (d + e x\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.56 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {{\left (2 \, B b^{2} d + 3 \, B a b e - 5 \, A b^{2} e\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e}} + \frac {\sqrt {e x + d} B a b e - \sqrt {e x + d} A b^{2} e}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}} + \frac {2 \, {\left (3 \, {\left (e x + d\right )} B b d + B b d^{2} + 3 \, {\left (e x + d\right )} B a e - 6 \, {\left (e x + d\right )} A b e - B a d e - A b d e + A a e^{2}\right )}}{3 \, {\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} {\left (e x + d\right )}^{\frac {3}{2}}} \]
(2*B*b^2*d + 3*B*a*b*e - 5*A*b^2*e)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a *b*e))/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*sqrt(-b^2*d + a*b*e)) + (sqrt(e*x + d)*B*a*b*e - sqrt(e*x + d)*A*b^2*e)/((b^3*d^3 - 3*a* b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*((e*x + d)*b - b*d + a*e)) + 2/3*(3*( e*x + d)*B*b*d + B*b*d^2 + 3*(e*x + d)*B*a*e - 6*(e*x + d)*A*b*e - B*a*d*e - A*b*d*e + A*a*e^2)/((b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3) *(e*x + d)^(3/2))
Time = 10.86 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {\frac {2\,\left (A\,e-B\,d\right )}{3\,\left (a\,e-b\,d\right )}+\frac {2\,\left (d+e\,x\right )\,\left (3\,B\,a\,e-5\,A\,b\,e+2\,B\,b\,d\right )}{3\,{\left (a\,e-b\,d\right )}^2}+\frac {b\,{\left (d+e\,x\right )}^2\,\left (3\,B\,a\,e-5\,A\,b\,e+2\,B\,b\,d\right )}{{\left (a\,e-b\,d\right )}^3}}{b\,{\left (d+e\,x\right )}^{5/2}+\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{3/2}}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}{{\left (a\,e-b\,d\right )}^{7/2}}\right )\,\left (3\,B\,a\,e-5\,A\,b\,e+2\,B\,b\,d\right )}{{\left (a\,e-b\,d\right )}^{7/2}} \]
- ((2*(A*e - B*d))/(3*(a*e - b*d)) + (2*(d + e*x)*(3*B*a*e - 5*A*b*e + 2*B *b*d))/(3*(a*e - b*d)^2) + (b*(d + e*x)^2*(3*B*a*e - 5*A*b*e + 2*B*b*d))/( a*e - b*d)^3)/(b*(d + e*x)^(5/2) + (a*e - b*d)*(d + e*x)^(3/2)) - (b^(1/2) *atan((b^(1/2)*(d + e*x)^(1/2)*(a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2* b*d*e^2))/(a*e - b*d)^(7/2))*(3*B*a*e - 5*A*b*e + 2*B*b*d))/(a*e - b*d)^(7 /2)