Integrand size = 32, antiderivative size = 253 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{3/2}} \, dx=\frac {a b^2 B d^3-a^2 b C d^3+a^3 d^3 D-b^3 \left (2 c^2 C d-2 B c d^2+3 A d^3-2 c^3 D\right )}{b^3 d^2 (b c-a d)^2 \sqrt {c+d x}}-\frac {A-\frac {a \left (b^2 B-a b C+a^2 D\right )}{b^3}}{(b c-a d) (a+b x) \sqrt {c+d x}}+\frac {2 D \sqrt {c+d x}}{b^2 d^2}-\frac {\left (b^3 (2 B c-3 A d)-a b^2 (4 c C-B d)-3 a^3 d D+a^2 b (C d+6 c D)\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{b^{5/2} (b c-a d)^{5/2}} \]
-(b^3*(-3*A*d+2*B*c)-a*b^2*(-B*d+4*C*c)-3*a^3*d*D+a^2*b*(C*d+6*D*c))*arcta nh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(5/2)/(-a*d+b*c)^(5/2)+(a*b^2 *B*d^3-a^2*b*C*d^3+a^3*d^3*D-b^3*(3*A*d^3-2*B*c*d^2+2*C*c^2*d-2*D*c^3))/b^ 3/d^2/(-a*d+b*c)^2/(d*x+c)^(1/2)+(-A+a*(B*b^2-C*a*b+D*a^2)/b^3)/(-a*d+b*c) /(b*x+a)/(d*x+c)^(1/2)+2*D*(d*x+c)^(1/2)/b^2/d^2
Time = 0.65 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.02 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{3/2}} \, dx=\frac {3 a^3 d^2 D (c+d x)+a^2 b d (c+d x) (-C d-4 c D+2 d D x)+b^3 \left (-A d^2 (c+3 d x)+2 c x \left (-c C d+B d^2+2 c^2 D+c d D x\right )\right )+a b^2 \left (4 c^3 D+d^3 (-2 A+B x)-2 c^2 d (C+D x)+c d^2 \left (3 B-4 D x^2\right )\right )}{b^2 d^2 (b c-a d)^2 (a+b x) \sqrt {c+d x}}+\frac {\left (b^3 (2 B c-3 A d)+a b^2 (-4 c C+B d)-3 a^3 d D+a^2 b (C d+6 c D)\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{b^{5/2} (-b c+a d)^{5/2}} \]
(3*a^3*d^2*D*(c + d*x) + a^2*b*d*(c + d*x)*(-(C*d) - 4*c*D + 2*d*D*x) + b^ 3*(-(A*d^2*(c + 3*d*x)) + 2*c*x*(-(c*C*d) + B*d^2 + 2*c^2*D + c*d*D*x)) + a*b^2*(4*c^3*D + d^3*(-2*A + B*x) - 2*c^2*d*(C + D*x) + c*d^2*(3*B - 4*D*x ^2)))/(b^2*d^2*(b*c - a*d)^2*(a + b*x)*Sqrt[c + d*x]) + ((b^3*(2*B*c - 3*A *d) + a*b^2*(-4*c*C + B*d) - 3*a^3*d*D + a^2*b*(C*d + 6*c*D))*ArcTan[(Sqrt [b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(b^(5/2)*(-(b*c) + a*d)^(5/2))
Time = 0.78 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {2124, 27, 1192, 1584, 2009}
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.
| \(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2124 |
| \(\displaystyle -\frac {\int -\frac {2 \left (c-\frac {a d}{b}\right ) D x^2+\frac {2 (b c-a d) (b C-a D) x}{b^2}+\frac {d D a^3-b (C d-2 c D) a^2-b^2 (2 c C-B d) a+b^3 (2 B c-3 A d)}{b^3}}{2 (a+b x) (c+d x)^{3/2}}dx}{b c-a d}-\frac {A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}}{(a+b x) \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\frac {d D a^3}{b^3}-\frac {(C d-2 c D) a^2}{b^2}-\frac {(2 c C-B d) a}{b}+2 \left (c-\frac {a d}{b}\right ) D x^2+2 B c-3 A d+\frac {2 (b c-a d) (b C-a D) x}{b^2}}{(a+b x) (c+d x)^{3/2}}dx}{2 (b c-a d)}-\frac {A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}}{(a+b x) \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle \frac {\int \frac {-2 D c^3+2 C d c^2-2 B d^2 c-2 \left (c-\frac {a d}{b}\right ) D (c+d x)^2+d^3 \left (3 A-\frac {a \left (D a^2-b C a+b^2 B\right )}{b^3}\right )-\frac {2 (b c-a d) (b C d-a D d-2 b c D) (c+d x)}{b^2}}{(c+d x) (b c-a d-b (c+d x))}d\sqrt {c+d x}}{d^2 (b c-a d)}-\frac {A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}}{(a+b x) \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {\int \left (\frac {\left (3 d D a^3-b (C d+6 c D) a^2+b^2 (4 c C-B d) a-b^3 (2 B c-3 A d)\right ) d^2}{b^2 (b c-a d) (b c-a d-b (c+d x))}+\frac {2 (b c-a d) D}{b^2}+\frac {\left (-2 D c^3+2 C d c^2-2 B d^2 c+3 A d^3\right ) b^3-a B d^3 b^2+a^2 C d^3 b-a^3 d^3 D}{b^3 (b c-a d) (c+d x)}\right )d\sqrt {c+d x}}{d^2 (b c-a d)}-\frac {A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}}{(a+b x) \sqrt {c+d x} (b c-a d)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {d^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right ) \left (-3 a^3 d D+a^2 b (6 c D+C d)-a b^2 (4 c C-B d)+b^3 (2 B c-3 A d)\right )}{b^{5/2} (b c-a d)^{3/2}}+\frac {a^3 d^3 D-a^2 b C d^3+a b^2 B d^3-\left (b^3 \left (3 A d^3-2 B c d^2-2 c^3 D+2 c^2 C d\right )\right )}{b^3 \sqrt {c+d x} (b c-a d)}+\frac {2 D \sqrt {c+d x} (b c-a d)}{b^2}}{d^2 (b c-a d)}-\frac {A-\frac {a \left (a^2 D-a b C+b^2 B\right )}{b^3}}{(a+b x) \sqrt {c+d x} (b c-a d)}\) |
-((A - (a*(b^2*B - a*b*C + a^2*D))/b^3)/((b*c - a*d)*(a + b*x)*Sqrt[c + d* x])) + ((a*b^2*B*d^3 - a^2*b*C*d^3 + a^3*d^3*D - b^3*(2*c^2*C*d - 2*B*c*d^ 2 + 3*A*d^3 - 2*c^3*D))/(b^3*(b*c - a*d)*Sqrt[c + d*x]) + (2*(b*c - a*d)*D *Sqrt[c + d*x])/b^2 - (d^2*(b^3*(2*B*c - 3*A*d) - a*b^2*(4*c*C - B*d) - 3* a^3*d*D + a^2*b*(C*d + 6*c*D))*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(b^(5/2)*(b*c - a*d)^(3/2)))/(d^2*(b*c - a*d))
3.1.15.3.1 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[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^( 2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] : > With[{Qx = PolynomialQuotient[Px, a + b*x, x], R = PolynomialRemainder[Px , a + b*x, x]}, Simp[R*(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((m + 1)*(b*c - a*d))), x] + Simp[1/((m + 1)*(b*c - a*d)) Int[(a + b*x)^(m + 1)*(c + d*x )^n*ExpandToSum[(m + 1)*(b*c - a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; Fr eeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && LtQ[m, -1] && (IntegerQ[m] || ! ILtQ[n, -1])
Time = 1.83 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.90
| method | result | size |
| derivativedivides | \(\frac {\frac {2 D \sqrt {d x +c}}{b^{2}}-\frac {2 d^{2} \left (\frac {\left (\frac {1}{2} A \,b^{3} d -\frac {1}{2} B a \,b^{2} d +\frac {1}{2} C \,a^{2} b d -\frac {1}{2} a^{3} d D\right ) \sqrt {d x +c}}{\left (d x +c \right ) b +a d -b c}+\frac {\left (3 A \,b^{3} d -B a \,b^{2} d -2 B \,b^{3} c -C \,a^{2} b d +4 C a \,b^{2} c +3 a^{3} d D-6 D a^{2} b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{2} b^{2}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\left (a d -b c \right )^{2} \sqrt {d x +c}}}{d^{2}}\) | \(228\) |
| default | \(\frac {\frac {2 D \sqrt {d x +c}}{b^{2}}-\frac {2 d^{2} \left (\frac {\left (\frac {1}{2} A \,b^{3} d -\frac {1}{2} B a \,b^{2} d +\frac {1}{2} C \,a^{2} b d -\frac {1}{2} a^{3} d D\right ) \sqrt {d x +c}}{\left (d x +c \right ) b +a d -b c}+\frac {\left (3 A \,b^{3} d -B a \,b^{2} d -2 B \,b^{3} c -C \,a^{2} b d +4 C a \,b^{2} c +3 a^{3} d D-6 D a^{2} b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{2} b^{2}}-\frac {2 \left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right )}{\left (a d -b c \right )^{2} \sqrt {d x +c}}}{d^{2}}\) | \(228\) |
| pseudoelliptic | \(-\frac {3 \left (\left (\left (b^{3} A -\frac {1}{3} a \,b^{2} B -\frac {1}{3} C \,a^{2} b +D a^{3}\right ) d -\frac {2 b c \left (B \,b^{2}-2 C a b +3 D a^{2}\right )}{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right ) \left (b x +a \right ) d^{2} \sqrt {d x +c}+\frac {2 \sqrt {\left (a d -b c \right ) b}\, \left (\left (\frac {3 A \,b^{3} x}{2}+\left (-\frac {B x}{2}+A \right ) a \,b^{2}+\frac {a^{2} x \left (-2 D x +C \right ) b}{2}-\frac {3 D a^{3} x}{2}\right ) d^{3}+\frac {c \left (\left (-2 B x +A \right ) b^{3}-3 a \left (-\frac {4 D x^{2}}{3}+B \right ) b^{2}+a^{2} \left (2 D x +C \right ) b -3 D a^{3}\right ) d^{2}}{2}+\left (\left (-D x +C \right ) b +2 D a \right ) b \,c^{2} \left (b x +a \right ) d -2 D b^{2} c^{3} \left (b x +a \right )\right )}{3}\right )}{\sqrt {d x +c}\, \sqrt {\left (a d -b c \right ) b}\, d^{2} b^{2} \left (b x +a \right ) \left (a d -b c \right )^{2}}\) | \(277\) |
2/d^2*(D/b^2*(d*x+c)^(1/2)-(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/(a*d-b*c)^2/(d*x+ c)^(1/2)-d^2/(a*d-b*c)^2/b^2*((1/2*A*b^3*d-1/2*B*a*b^2*d+1/2*C*a^2*b*d-1/2 *a^3*d*D)*(d*x+c)^(1/2)/((d*x+c)*b+a*d-b*c)+1/2*(3*A*b^3*d-B*a*b^2*d-2*B*b ^3*c-C*a^2*b*d+4*C*a*b^2*c+3*D*a^3*d-6*D*a^2*b*c)/((a*d-b*c)*b)^(1/2)*arct an(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 785 vs. \(2 (236) = 472\).
Time = 0.33 (sec) , antiderivative size = 1583, normalized size of antiderivative = 6.26 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{3/2}} \, dx=\text {Too large to display} \]
[1/2*(((3*D*a^4*c - (C*a^3*b + B*a^2*b^2 - 3*A*a*b^3)*c)*d^3 - 2*(3*D*a^3* b*c^2 - (2*C*a^2*b^2 - B*a*b^3)*c^2)*d^2 + ((3*D*a^3*b - C*a^2*b^2 - B*a*b ^3 + 3*A*b^4)*d^4 - 2*(3*D*a^2*b^2*c - (2*C*a*b^3 - B*b^4)*c)*d^3)*x^2 + ( (3*D*a^4 - C*a^3*b - B*a^2*b^2 + 3*A*a*b^3)*d^4 - 3*(D*a^3*b*c - (C*a^2*b^ 2 - B*a*b^3 + A*b^4)*c)*d^3 - 2*(3*D*a^2*b^2*c^2 - (2*C*a*b^3 - B*b^4)*c^2 )*d^2)*x)*sqrt(b^2*c - a*b*d)*log((b*d*x + 2*b*c - a*d + 2*sqrt(b^2*c - a* b*d)*sqrt(d*x + c))/(b*x + a)) + 2*(4*D*a*b^4*c^4 + 2*A*a^2*b^3*d^4 - (3*D *a^4*b*c - (C*a^3*b^2 - 3*B*a^2*b^3 - A*a*b^4)*c)*d^3 + (7*D*a^3*b^2*c^2 + (C*a^2*b^3 + 3*B*a*b^4 - A*b^5)*c^2)*d^2 + 2*(D*b^5*c^3*d - 3*D*a*b^4*c^2 *d^2 + 3*D*a^2*b^3*c*d^3 - D*a^3*b^2*d^4)*x^2 - 2*(4*D*a^2*b^3*c^3 + C*a*b ^4*c^3)*d + (4*D*b^5*c^4 + 2*(C*a*b^4 + B*b^5)*c^2*d^2 - (3*D*a^4*b - C*a^ 3*b^2 + B*a^2*b^3 - 3*A*a*b^4)*d^4 + (5*D*a^3*b^2*c - (C*a^2*b^3 + B*a*b^4 + 3*A*b^5)*c)*d^3 - 2*(3*D*a*b^4*c^3 + C*b^5*c^3)*d)*x)*sqrt(d*x + c))/(a *b^6*c^4*d^2 - 3*a^2*b^5*c^3*d^3 + 3*a^3*b^4*c^2*d^4 - a^4*b^3*c*d^5 + (b^ 7*c^3*d^3 - 3*a*b^6*c^2*d^4 + 3*a^2*b^5*c*d^5 - a^3*b^4*d^6)*x^2 + (b^7*c^ 4*d^2 - 2*a*b^6*c^3*d^3 + 2*a^3*b^4*c*d^5 - a^4*b^3*d^6)*x), -(((3*D*a^4*c - (C*a^3*b + B*a^2*b^2 - 3*A*a*b^3)*c)*d^3 - 2*(3*D*a^3*b*c^2 - (2*C*a^2* b^2 - B*a*b^3)*c^2)*d^2 + ((3*D*a^3*b - C*a^2*b^2 - B*a*b^3 + 3*A*b^4)*d^4 - 2*(3*D*a^2*b^2*c - (2*C*a*b^3 - B*b^4)*c)*d^3)*x^2 + ((3*D*a^4 - C*a^3* b - B*a^2*b^2 + 3*A*a*b^3)*d^4 - 3*(D*a^3*b*c - (C*a^2*b^2 - B*a*b^3 + ...
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{3/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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*d-b*c>0)', see `assume?` for m ore detail
Time = 0.29 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.53 \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{3/2}} \, dx=\frac {{\left (6 \, D a^{2} b c - 4 \, C a b^{2} c + 2 \, B b^{3} c - 3 \, D a^{3} d + C a^{2} b d + B a b^{2} d - 3 \, A b^{3} d\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} \sqrt {-b^{2} c + a b d}} + \frac {2 \, {\left (d x + c\right )} D b^{3} c^{3} - 2 \, D b^{3} c^{4} - 2 \, {\left (d x + c\right )} C b^{3} c^{2} d + 2 \, D a b^{2} c^{3} d + 2 \, C b^{3} c^{3} d + 2 \, {\left (d x + c\right )} B b^{3} c d^{2} - 2 \, C a b^{2} c^{2} d^{2} - 2 \, B b^{3} c^{2} d^{2} + {\left (d x + c\right )} D a^{3} d^{3} - {\left (d x + c\right )} C a^{2} b d^{3} + {\left (d x + c\right )} B a b^{2} d^{3} - 3 \, {\left (d x + c\right )} A b^{3} d^{3} + 2 \, B a b^{2} c d^{3} + 2 \, A b^{3} c d^{3} - 2 \, A a b^{2} d^{4}}{{\left (b^{4} c^{2} d^{2} - 2 \, a b^{3} c d^{3} + a^{2} b^{2} d^{4}\right )} {\left ({\left (d x + c\right )}^{\frac {3}{2}} b - \sqrt {d x + c} b c + \sqrt {d x + c} a d\right )}} + \frac {2 \, \sqrt {d x + c} D}{b^{2} d^{2}} \]
(6*D*a^2*b*c - 4*C*a*b^2*c + 2*B*b^3*c - 3*D*a^3*d + C*a^2*b*d + B*a*b^2*d - 3*A*b^3*d)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c^2 - 2*a *b^3*c*d + a^2*b^2*d^2)*sqrt(-b^2*c + a*b*d)) + (2*(d*x + c)*D*b^3*c^3 - 2 *D*b^3*c^4 - 2*(d*x + c)*C*b^3*c^2*d + 2*D*a*b^2*c^3*d + 2*C*b^3*c^3*d + 2 *(d*x + c)*B*b^3*c*d^2 - 2*C*a*b^2*c^2*d^2 - 2*B*b^3*c^2*d^2 + (d*x + c)*D *a^3*d^3 - (d*x + c)*C*a^2*b*d^3 + (d*x + c)*B*a*b^2*d^3 - 3*(d*x + c)*A*b ^3*d^3 + 2*B*a*b^2*c*d^3 + 2*A*b^3*c*d^3 - 2*A*a*b^2*d^4)/((b^4*c^2*d^2 - 2*a*b^3*c*d^3 + a^2*b^2*d^4)*((d*x + c)^(3/2)*b - sqrt(d*x + c)*b*c + sqrt (d*x + c)*a*d)) + 2*sqrt(d*x + c)*D/(b^2*d^2)
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(a+b x)^2 (c+d x)^{3/2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^{3/2}} \,d x \]