Integrand size = 35, antiderivative size = 474 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {2 e^3 (B d-A e) (a+b x)}{3 (b d-a e)^5 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 e^3 (4 b B d-5 A b e+a B e) (a+b x)}{(b d-a e)^6 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b e^2 (328 b B d-515 A b e+187 a B e) \sqrt {d+e x}}{64 (b d-a e)^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (A b-a B) \sqrt {d+e x}}{4 (b d-a e)^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (8 b B d-23 A b e+15 a B e) \sqrt {d+e x}}{24 (b d-a e)^4 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b e (136 b B d-259 A b e+123 a B e) \sqrt {d+e x}}{96 (b d-a e)^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {105 \sqrt {b} e^3 (8 b B d-11 A b e+3 a B e) (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 (b d-a e)^{13/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:
-2/3*e^3*(-A*e+B*d)*(b*x+a)/(-a*e+b*d)^5/(e*x+d)^(3/2)/((b*x+a)^2)^(1/2)-2 *e^3*(-5*A*b*e+B*a*e+4*B*b*d)*(b*x+a)/(-a*e+b*d)^6/(e*x+d)^(1/2)/((b*x+a)^ 2)^(1/2)-1/64*b*e^2*(-515*A*b*e+187*B*a*e+328*B*b*d)*(e*x+d)^(1/2)/(-a*e+b *d)^6/((b*x+a)^2)^(1/2)-1/4*b*(A*b-B*a)*(e*x+d)^(1/2)/(-a*e+b*d)^3/(b*x+a) ^3/((b*x+a)^2)^(1/2)-1/24*b*(-23*A*b*e+15*B*a*e+8*B*b*d)*(e*x+d)^(1/2)/(-a *e+b*d)^4/(b*x+a)^2/((b*x+a)^2)^(1/2)+1/96*b*e*(-259*A*b*e+123*B*a*e+136*B *b*d)*(e*x+d)^(1/2)/(-a*e+b*d)^5/(b*x+a)/((b*x+a)^2)^(1/2)+105/64*b^(1/2)* e^3*(-11*A*b*e+3*B*a*e+8*B*b*d)*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a* e+b*d)^(1/2))/(-a*e+b*d)^(13/2)/((b*x+a)^2)^(1/2)
Time = 4.84 (sec) , antiderivative size = 559, normalized size of antiderivative = 1.18 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^3 (a+b x) \left (\frac {-B \left (128 a^5 e^4 (2 d+3 e x)+a^4 b e^3 \left (2639 d^2+4510 d e x+2511 e^2 x^2\right )+a^3 b^2 e^2 \left (690 d^3+10331 d^2 e x+12960 d e^2 x^2+4599 e^3 x^3\right )+8 b^5 d x \left (8 d^4-18 d^3 e x+63 d^2 e^2 x^2+420 d e^3 x^3+315 e^4 x^4\right )+a^2 b^3 e \left (-136 d^4+2556 d^3 e x+17433 d^2 e^2 x^2+16926 d e^3 x^3+3465 e^4 x^4\right )+a b^4 \left (16 d^5-520 d^4 e x+1890 d^3 e^2 x^2+12621 d^2 e^3 x^3+10500 d e^4 x^4+945 e^5 x^5\right )\right )+A \left (-128 a^5 e^5+128 a^4 b e^4 (16 d+11 e x)+a^3 b^2 e^3 \left (2295 d^2+12782 d e x+9207 e^2 x^2\right )+a^2 b^3 e^2 \left (-1030 d^3+3795 d^2 e x+22968 d e^2 x^2+16863 e^3 x^3\right )+a b^4 e \left (328 d^4-748 d^3 e x+2673 d^2 e^2 x^2+17094 d e^3 x^3+12705 e^4 x^4\right )+b^5 \left (-48 d^5+88 d^4 e x-198 d^3 e^2 x^2+693 d^2 e^3 x^3+4620 d e^4 x^4+3465 e^5 x^5\right )\right )}{e^3 (b d-a e)^6 (a+b x)^4 (d+e x)^{3/2}}-\frac {315 \sqrt {b} (8 b B d-11 A b e+3 a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{13/2}}\right )}{192 \sqrt {(a+b x)^2}} \] Input:
Integrate[(A + B*x)/((d + e*x)^(5/2)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]
Output:
(e^3*(a + b*x)*((-(B*(128*a^5*e^4*(2*d + 3*e*x) + a^4*b*e^3*(2639*d^2 + 45 10*d*e*x + 2511*e^2*x^2) + a^3*b^2*e^2*(690*d^3 + 10331*d^2*e*x + 12960*d* e^2*x^2 + 4599*e^3*x^3) + 8*b^5*d*x*(8*d^4 - 18*d^3*e*x + 63*d^2*e^2*x^2 + 420*d*e^3*x^3 + 315*e^4*x^4) + a^2*b^3*e*(-136*d^4 + 2556*d^3*e*x + 17433 *d^2*e^2*x^2 + 16926*d*e^3*x^3 + 3465*e^4*x^4) + a*b^4*(16*d^5 - 520*d^4*e *x + 1890*d^3*e^2*x^2 + 12621*d^2*e^3*x^3 + 10500*d*e^4*x^4 + 945*e^5*x^5) )) + A*(-128*a^5*e^5 + 128*a^4*b*e^4*(16*d + 11*e*x) + a^3*b^2*e^3*(2295*d ^2 + 12782*d*e*x + 9207*e^2*x^2) + a^2*b^3*e^2*(-1030*d^3 + 3795*d^2*e*x + 22968*d*e^2*x^2 + 16863*e^3*x^3) + a*b^4*e*(328*d^4 - 748*d^3*e*x + 2673* d^2*e^2*x^2 + 17094*d*e^3*x^3 + 12705*e^4*x^4) + b^5*(-48*d^5 + 88*d^4*e*x - 198*d^3*e^2*x^2 + 693*d^2*e^3*x^3 + 4620*d*e^4*x^4 + 3465*e^5*x^5)))/(e ^3*(b*d - a*e)^6*(a + b*x)^4*(d + e*x)^(3/2)) - (315*Sqrt[b]*(8*b*B*d - 11 *A*b*e + 3*a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(-(b *d) + a*e)^(13/2)))/(192*Sqrt[(a + b*x)^2])
Time = 0.74 (sec) , antiderivative size = 342, normalized size of antiderivative = 0.72, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1187, 27, 87, 52, 52, 52, 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.
\(\displaystyle \int \frac {A+B x}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2} (d+e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 1187 |
\(\displaystyle \frac {b^5 (a+b x) \int \frac {A+B x}{b^5 (a+b x)^5 (d+e x)^{5/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a+b x) \int \frac {A+B x}{(a+b x)^5 (d+e x)^{5/2}}dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {(a+b x) \left (\frac {(3 a B e-11 A b e+8 b B d) \int \frac {1}{(a+b x)^4 (d+e x)^{5/2}}dx}{8 b (b d-a e)}-\frac {A b-a B}{4 b (a+b x)^4 (d+e x)^{3/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {(a+b x) \left (\frac {(3 a B e-11 A b e+8 b B d) \left (-\frac {3 e \int \frac {1}{(a+b x)^3 (d+e x)^{5/2}}dx}{2 (b d-a e)}-\frac {1}{3 (a+b x)^3 (d+e x)^{3/2} (b d-a e)}\right )}{8 b (b d-a e)}-\frac {A b-a B}{4 b (a+b x)^4 (d+e x)^{3/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {(a+b x) \left (\frac {(3 a B e-11 A b e+8 b B d) \left (-\frac {3 e \left (-\frac {7 e \int \frac {1}{(a+b x)^2 (d+e x)^{5/2}}dx}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{3 (a+b x)^3 (d+e x)^{3/2} (b d-a e)}\right )}{8 b (b d-a e)}-\frac {A b-a B}{4 b (a+b x)^4 (d+e x)^{3/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {(a+b x) \left (\frac {(3 a B e-11 A b e+8 b B d) \left (-\frac {3 e \left (-\frac {7 e \left (-\frac {5 e \int \frac {1}{(a+b x) (d+e x)^{5/2}}dx}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{3 (a+b x)^3 (d+e x)^{3/2} (b d-a e)}\right )}{8 b (b d-a e)}-\frac {A b-a B}{4 b (a+b x)^4 (d+e x)^{3/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {(a+b x) \left (\frac {(3 a B e-11 A b e+8 b B d) \left (-\frac {3 e \left (-\frac {7 e \left (-\frac {5 e \left (\frac {b \int \frac {1}{(a+b x) (d+e x)^{3/2}}dx}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{3 (a+b x)^3 (d+e x)^{3/2} (b d-a e)}\right )}{8 b (b d-a e)}-\frac {A b-a B}{4 b (a+b x)^4 (d+e x)^{3/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {(a+b x) \left (\frac {(3 a B e-11 A b e+8 b B d) \left (-\frac {3 e \left (-\frac {7 e \left (-\frac {5 e \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{3 (a+b x)^3 (d+e x)^{3/2} (b d-a e)}\right )}{8 b (b d-a e)}-\frac {A b-a B}{4 b (a+b x)^4 (d+e x)^{3/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(a+b x) \left (\frac {(3 a B e-11 A b e+8 b B d) \left (-\frac {3 e \left (-\frac {7 e \left (-\frac {5 e \left (\frac {b \left (\frac {2 b \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{3 (a+b x)^3 (d+e x)^{3/2} (b d-a e)}\right )}{8 b (b d-a e)}-\frac {A b-a B}{4 b (a+b x)^4 (d+e x)^{3/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(a+b x) \left (\frac {(3 a B e-11 A b e+8 b B d) \left (-\frac {3 e \left (-\frac {7 e \left (-\frac {5 e \left (\frac {b \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{3/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{3/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{3 (a+b x)^3 (d+e x)^{3/2} (b d-a e)}\right )}{8 b (b d-a e)}-\frac {A b-a B}{4 b (a+b x)^4 (d+e x)^{3/2} (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\) |
Input:
Int[(A + B*x)/((d + e*x)^(5/2)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]
Output:
((a + b*x)*(-1/4*(A*b - a*B)/(b*(b*d - a*e)*(a + b*x)^4*(d + e*x)^(3/2)) + ((8*b*B*d - 11*A*b*e + 3*a*B*e)*(-1/3*1/((b*d - a*e)*(a + b*x)^3*(d + e*x )^(3/2)) - (3*e*(-1/2*1/((b*d - a*e)*(a + b*x)^2*(d + e*x)^(3/2)) - (7*e*( -(1/((b*d - a*e)*(a + b*x)*(d + e*x)^(3/2))) - (5*e*(2/(3*(b*d - a*e)*(d + e*x)^(3/2)) + (b*(2/((b*d - a*e)*Sqrt[d + e*x]) - (2*Sqrt[b]*ArcTanh[(Sqr t[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^(3/2)))/(b*d - a*e)))/(2 *(b*d - a*e))))/(4*(b*d - a*e))))/(2*(b*d - a*e))))/(8*b*(b*d - a*e))))/Sq rt[a^2 + 2*a*b*x + b^2*x^2]
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] :> 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[(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. \(1859\) vs. \(2(363)=726\).
Time = 1.52 (sec) , antiderivative size = 1860, normalized size of antiderivative = 3.92
Input:
int((B*x+A)/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERB OSE)
Output:
1/192*(-2520*B*(e*x+d)^(3/2)*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a ^4*b^2*d*e^3-16926*B*(b*(a*e-b*d))^(1/2)*a^2*b^3*d*e^4*x^3+22968*A*(b*(a*e -b*d))^(1/2)*a^2*b^3*d*e^4*x^2+2673*A*(b*(a*e-b*d))^(1/2)*a*b^4*d^2*e^3*x^ 2-1890*B*(b*(a*e-b*d))^(1/2)*a*b^4*d^3*e^2*x^2+12782*A*(b*(a*e-b*d))^(1/2) *a^3*b^2*d*e^4*x+3795*A*(b*(a*e-b*d))^(1/2)*a^2*b^3*d^2*e^3*x-10080*B*(e*x +d)^(3/2)*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a^3*b^3*d*e^3*x-4599 *B*(b*(a*e-b*d))^(1/2)*a^3*b^2*e^5*x^3-198*A*(b*(a*e-b*d))^(1/2)*b^5*d^3*e ^2*x^2-10331*B*(b*(a*e-b*d))^(1/2)*a^3*b^2*d^2*e^3*x-2556*B*(b*(a*e-b*d))^ (1/2)*a^2*b^3*d^3*e^2*x+520*B*(b*(a*e-b*d))^(1/2)*a*b^4*d^4*e*x-12960*B*(b *(a*e-b*d))^(1/2)*a^3*b^2*d*e^4*x^2-17433*B*(b*(a*e-b*d))^(1/2)*a^2*b^3*d^ 2*e^3*x^2-748*A*(b*(a*e-b*d))^(1/2)*a*b^4*d^3*e^2*x-10080*B*(e*x+d)^(3/2)* arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a*b^5*d*e^3*x^3-12621*B*(b*(a* e-b*d))^(1/2)*a*b^4*d^2*e^3*x^3-128*A*(b*(a*e-b*d))^(1/2)*a^5*e^5-48*A*(b* (a*e-b*d))^(1/2)*b^5*d^5-4510*B*(b*(a*e-b*d))^(1/2)*a^4*b*d*e^4*x-15120*B* (e*x+d)^(3/2)*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a^2*b^4*d*e^3*x^ 2-945*B*(e*x+d)^(3/2)*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a*b^5*e^ 4*x^4-2520*B*(e*x+d)^(3/2)*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*b^6 *d*e^3*x^4+13860*A*(e*x+d)^(3/2)*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2 ))*a*b^5*e^4*x^3-3780*B*(e*x+d)^(3/2)*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d)) ^(1/2))*a^2*b^4*e^4*x^3+20790*A*(e*x+d)^(3/2)*arctan(b*(e*x+d)^(1/2)/(b...
Leaf count of result is larger than twice the leaf count of optimal. 1773 vs. \(2 (363) = 726\).
Time = 0.79 (sec) , antiderivative size = 3576, normalized size of antiderivative = 7.54 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate((B*x+A)/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm=" fricas")
Output:
Too large to include
Timed out. \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((B*x+A)/(e*x+d)**(5/2)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
Output:
Timed out
\[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int { \frac {B x + A}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} {\left (e x + d\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((B*x+A)/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm=" maxima")
Output:
integrate((B*x + A)/((b^2*x^2 + 2*a*b*x + a^2)^(5/2)*(e*x + d)^(5/2)), x)
Leaf count of result is larger than twice the leaf count of optimal. 978 vs. \(2 (363) = 726\).
Time = 0.23 (sec) , antiderivative size = 978, normalized size of antiderivative = 2.06 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate((B*x+A)/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm=" giac")
Output:
-105/64*(8*B*b^2*d*e^3 + 3*B*a*b*e^4 - 11*A*b^2*e^4)*arctan(sqrt(e*x + d)* b/sqrt(-b^2*d + a*b*e))/((b^6*d^6*sgn(b*x + a) - 6*a*b^5*d^5*e*sgn(b*x + a ) + 15*a^2*b^4*d^4*e^2*sgn(b*x + a) - 20*a^3*b^3*d^3*e^3*sgn(b*x + a) + 15 *a^4*b^2*d^2*e^4*sgn(b*x + a) - 6*a^5*b*d*e^5*sgn(b*x + a) + a^6*e^6*sgn(b *x + a))*sqrt(-b^2*d + a*b*e)) - 2/3*(12*(e*x + d)*B*b*d*e^3 + B*b*d^2*e^3 + 3*(e*x + d)*B*a*e^4 - 15*(e*x + d)*A*b*e^4 - B*a*d*e^4 - A*b*d*e^4 + A* a*e^5)/((b^6*d^6*sgn(b*x + a) - 6*a*b^5*d^5*e*sgn(b*x + a) + 15*a^2*b^4*d^ 4*e^2*sgn(b*x + a) - 20*a^3*b^3*d^3*e^3*sgn(b*x + a) + 15*a^4*b^2*d^2*e^4* sgn(b*x + a) - 6*a^5*b*d*e^5*sgn(b*x + a) + a^6*e^6*sgn(b*x + a))*(e*x + d )^(3/2)) - 1/192*(984*(e*x + d)^(7/2)*B*b^5*d*e^3 - 3224*(e*x + d)^(5/2)*B *b^5*d^2*e^3 + 3560*(e*x + d)^(3/2)*B*b^5*d^3*e^3 - 1320*sqrt(e*x + d)*B*b ^5*d^4*e^3 + 561*(e*x + d)^(7/2)*B*a*b^4*e^4 - 1545*(e*x + d)^(7/2)*A*b^5* e^4 + 1295*(e*x + d)^(5/2)*B*a*b^4*d*e^4 + 5153*(e*x + d)^(5/2)*A*b^5*d*e^ 4 - 4825*(e*x + d)^(3/2)*B*a*b^4*d^2*e^4 - 5855*(e*x + d)^(3/2)*A*b^5*d^2* e^4 + 2985*sqrt(e*x + d)*B*a*b^4*d^3*e^4 + 2295*sqrt(e*x + d)*A*b^5*d^3*e^ 4 + 1929*(e*x + d)^(5/2)*B*a^2*b^3*e^5 - 5153*(e*x + d)^(5/2)*A*a*b^4*e^5 - 1030*(e*x + d)^(3/2)*B*a^2*b^3*d*e^5 + 11710*(e*x + d)^(3/2)*A*a*b^4*d*e ^5 - 1035*sqrt(e*x + d)*B*a^2*b^3*d^2*e^5 - 6885*sqrt(e*x + d)*A*a*b^4*d^2 *e^5 + 2295*(e*x + d)^(3/2)*B*a^3*b^2*e^6 - 5855*(e*x + d)^(3/2)*A*a^2*b^3 *e^6 - 1605*sqrt(e*x + d)*B*a^3*b^2*d*e^6 + 6885*sqrt(e*x + d)*A*a^2*b^...
Timed out. \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {A+B\,x}{{\left (d+e\,x\right )}^{5/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \] Input:
int((A + B*x)/((d + e*x)^(5/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)),x)
Output:
int((A + B*x)/((d + e*x)^(5/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)), x)
Time = 0.24 (sec) , antiderivative size = 1220, normalized size of antiderivative = 2.57 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:
int((B*x+A)/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
Output:
(315*sqrt(b)*sqrt(d + e*x)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b) *sqrt(a*e - b*d)))*a**3*b*d*e**3 + 315*sqrt(b)*sqrt(d + e*x)*sqrt(a*e - b* d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*a**3*b*e**4*x + 945*s qrt(b)*sqrt(d + e*x)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt( a*e - b*d)))*a**2*b**2*d*e**3*x + 945*sqrt(b)*sqrt(d + e*x)*sqrt(a*e - b*d )*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*a**2*b**2*e**4*x**2 + 945*sqrt(b)*sqrt(d + e*x)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)* sqrt(a*e - b*d)))*a*b**3*d*e**3*x**2 + 945*sqrt(b)*sqrt(d + e*x)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*a*b**3*e**4*x**3 + 315*sqrt(b)*sqrt(d + e*x)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b )*sqrt(a*e - b*d)))*b**4*d*e**3*x**3 + 315*sqrt(b)*sqrt(d + e*x)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*b**4*e**4*x**4 - 16*a**5*e**5 + 224*a**4*b*d*e**4 + 144*a**4*b*e**5*x - 43*a**3*b**2*d**2*e **3 + 810*a**3*b**2*d*e**4*x + 693*a**3*b**2*e**5*x**2 - 215*a**2*b**3*d** 3*e**2 - 774*a**2*b**3*d**2*e**3*x + 441*a**2*b**3*d*e**4*x**2 + 840*a**2* b**3*e**5*x**3 + 58*a*b**4*d**4*e - 198*a*b**4*d**3*e**2*x - 1071*a*b**4*d **2*e**3*x**2 - 420*a*b**4*d*e**4*x**3 + 315*a*b**4*e**5*x**4 - 8*b**5*d** 5 + 18*b**5*d**4*e*x - 63*b**5*d**3*e**2*x**2 - 420*b**5*d**2*e**3*x**3 - 315*b**5*d*e**4*x**4)/(24*sqrt(d + e*x)*(a**9*d*e**6 + a**9*e**7*x - 6*a** 8*b*d**2*e**5 - 3*a**8*b*d*e**6*x + 3*a**8*b*e**7*x**2 + 15*a**7*b**2*d...