Integrand size = 26, antiderivative size = 228 \[ \int \frac {(A+B x) (d+e x)^{7/2}}{b x+c x^2} \, dx=\frac {2 \left (B (c d-b e)^3+A c e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right )\right ) \sqrt {d+e x}}{c^4}+\frac {2 \left (B (c d-b e)^2+A c e (2 c d-b e)\right ) (d+e x)^{3/2}}{3 c^3}+\frac {2 (B c d-b B e+A c e) (d+e x)^{5/2}}{5 c^2}+\frac {2 B (d+e x)^{7/2}}{7 c}-\frac {2 A d^{7/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}-\frac {2 (b B-A c) (c d-b e)^{7/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b c^{9/2}} \]
2/3*(B*(-b*e+c*d)^2+A*c*e*(-b*e+2*c*d))*(e*x+d)^(3/2)/c^3+2/5*(A*c*e-B*b*e +B*c*d)*(e*x+d)^(5/2)/c^2+2/7*B*(e*x+d)^(7/2)/c-2*A*d^(7/2)*arctanh((e*x+d )^(1/2)/d^(1/2))/b-2*(-A*c+B*b)*(-b*e+c*d)^(7/2)*arctanh(c^(1/2)*(e*x+d)^( 1/2)/(-b*e+c*d)^(1/2))/b/c^(9/2)+2*(B*(-b*e+c*d)^3+A*c*e*(b^2*e^2-3*b*c*d* e+3*c^2*d^2))*(e*x+d)^(1/2)/c^4
Time = 0.32 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.05 \[ \int \frac {(A+B x) (d+e x)^{7/2}}{b x+c x^2} \, dx=\frac {2 \sqrt {d+e x} \left (7 A c e \left (15 b^2 e^2-5 b c e (10 d+e x)+c^2 \left (58 d^2+16 d e x+3 e^2 x^2\right )\right )+B \left (-105 b^3 e^3+35 b^2 c e^2 (10 d+e x)-7 b c^2 e \left (58 d^2+16 d e x+3 e^2 x^2\right )+c^3 \left (176 d^3+122 d^2 e x+66 d e^2 x^2+15 e^3 x^3\right )\right )\right )}{105 c^4}-\frac {2 (-b B+A c) (-c d+b e)^{7/2} \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )}{b c^{9/2}}-\frac {2 A d^{7/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b} \]
(2*Sqrt[d + e*x]*(7*A*c*e*(15*b^2*e^2 - 5*b*c*e*(10*d + e*x) + c^2*(58*d^2 + 16*d*e*x + 3*e^2*x^2)) + B*(-105*b^3*e^3 + 35*b^2*c*e^2*(10*d + e*x) - 7*b*c^2*e*(58*d^2 + 16*d*e*x + 3*e^2*x^2) + c^3*(176*d^3 + 122*d^2*e*x + 6 6*d*e^2*x^2 + 15*e^3*x^3))))/(105*c^4) - (2*(-(b*B) + A*c)*(-(c*d) + b*e)^ (7/2)*ArcTan[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[-(c*d) + b*e]])/(b*c^(9/2)) - (2 *A*d^(7/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b
Time = 0.72 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1196, 1196, 1196, 1196, 1197, 25, 1480, 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.
| \(\displaystyle \int \frac {(A+B x) (d+e x)^{7/2}}{b x+c x^2} \, dx\) |
| \(\Big \downarrow \) 1196 |
| \(\displaystyle \frac {\int \frac {(d+e x)^{5/2} (A c d+(B c d-b B e+A c e) x)}{c x^2+b x}dx}{c}+\frac {2 B (d+e x)^{7/2}}{7 c}\) |
| \(\Big \downarrow \) 1196 |
| \(\displaystyle \frac {\frac {\int \frac {(d+e x)^{3/2} \left (A c^2 d^2+\left (B (c d-b e)^2+A c e (2 c d-b e)\right ) x\right )}{c x^2+b x}dx}{c}+\frac {2 (d+e x)^{5/2} (A c e-b B e+B c d)}{5 c}}{c}+\frac {2 B (d+e x)^{7/2}}{7 c}\) |
| \(\Big \downarrow \) 1196 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {\sqrt {d+e x} \left (A c^3 d^3+\left (B (c d-b e)^3+A c e \left (3 c^2 d^2-3 b c e d+b^2 e^2\right )\right ) x\right )}{c x^2+b x}dx}{c}+\frac {2 (d+e x)^{3/2} \left (A c e (2 c d-b e)+B (c d-b e)^2\right )}{3 c}}{c}+\frac {2 (d+e x)^{5/2} (A c e-b B e+B c d)}{5 c}}{c}+\frac {2 B (d+e x)^{7/2}}{7 c}\) |
| \(\Big \downarrow \) 1196 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \frac {A c^4 d^4+\left (B (c d-b e)^4+A c e \left (4 c^3 d^3-6 b c^2 e d^2+4 b^2 c e^2 d-b^3 e^3\right )\right ) x}{\sqrt {d+e x} \left (c x^2+b x\right )}dx}{c}+\frac {2 \sqrt {d+e x} \left (A c e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )+B (c d-b e)^3\right )}{c}}{c}+\frac {2 (d+e x)^{3/2} \left (A c e (2 c d-b e)+B (c d-b e)^2\right )}{3 c}}{c}+\frac {2 (d+e x)^{5/2} (A c e-b B e+B c d)}{5 c}}{c}+\frac {2 B (d+e x)^{7/2}}{7 c}\) |
| \(\Big \downarrow \) 1197 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 \int -\frac {d \left (B (c d-b e)^4+A c e \left (3 c^3 d^3-6 b c^2 e d^2+4 b^2 c e^2 d-b^3 e^3\right )\right )-\left (B (c d-b e)^4+A c e \left (4 c^3 d^3-6 b c^2 e d^2+4 b^2 c e^2 d-b^3 e^3\right )\right ) (d+e x)}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{c}+\frac {2 \sqrt {d+e x} \left (A c e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )+B (c d-b e)^3\right )}{c}}{c}+\frac {2 (d+e x)^{3/2} \left (A c e (2 c d-b e)+B (c d-b e)^2\right )}{3 c}}{c}+\frac {2 (d+e x)^{5/2} (A c e-b B e+B c d)}{5 c}}{c}+\frac {2 B (d+e x)^{7/2}}{7 c}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 \sqrt {d+e x} \left (A c e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )+B (c d-b e)^3\right )}{c}-\frac {2 \int \frac {d \left (B (c d-b e)^4+A c e \left (3 c^3 d^3-6 b c^2 e d^2+4 b^2 c e^2 d-b^3 e^3\right )\right )-\left (B (c d-b e)^4+A c e \left (4 c^3 d^3-6 b c^2 e d^2+4 b^2 c e^2 d-b^3 e^3\right )\right ) (d+e x)}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{c}}{c}+\frac {2 (d+e x)^{3/2} \left (A c e (2 c d-b e)+B (c d-b e)^2\right )}{3 c}}{c}+\frac {2 (d+e x)^{5/2} (A c e-b B e+B c d)}{5 c}}{c}+\frac {2 B (d+e x)^{7/2}}{7 c}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 \left (\frac {(b B-A c) (c d-b e)^4 \int \frac {1}{-c d+b e+c (d+e x)}d\sqrt {d+e x}}{b}+\frac {A c^5 d^4 \int \frac {1}{c (d+e x)-c d}d\sqrt {d+e x}}{b}\right )}{c}+\frac {2 \sqrt {d+e x} \left (A c e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )+B (c d-b e)^3\right )}{c}}{c}+\frac {2 (d+e x)^{3/2} \left (A c e (2 c d-b e)+B (c d-b e)^2\right )}{3 c}}{c}+\frac {2 (d+e x)^{5/2} (A c e-b B e+B c d)}{5 c}}{c}+\frac {2 B (d+e x)^{7/2}}{7 c}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {\frac {2 \left (-\frac {(b B-A c) (c d-b e)^{7/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b \sqrt {c}}-\frac {A c^4 d^{7/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b}\right )}{c}+\frac {2 \sqrt {d+e x} \left (A c e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )+B (c d-b e)^3\right )}{c}}{c}+\frac {2 (d+e x)^{3/2} \left (A c e (2 c d-b e)+B (c d-b e)^2\right )}{3 c}}{c}+\frac {2 (d+e x)^{5/2} (A c e-b B e+B c d)}{5 c}}{c}+\frac {2 B (d+e x)^{7/2}}{7 c}\) |
(2*B*(d + e*x)^(7/2))/(7*c) + ((2*(B*c*d - b*B*e + A*c*e)*(d + e*x)^(5/2)) /(5*c) + ((2*(B*(c*d - b*e)^2 + A*c*e*(2*c*d - b*e))*(d + e*x)^(3/2))/(3*c ) + ((2*(B*(c*d - b*e)^3 + A*c*e*(3*c^2*d^2 - 3*b*c*d*e + b^2*e^2))*Sqrt[d + e*x])/c + (2*(-((A*c^4*d^(7/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/b) - ((b *B - A*c)*(c*d - b*e)^(7/2)*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e ]])/(b*Sqrt[c])))/c)/c)/c)/c
3.13.29.3.1 Defintions of rubi rules used
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_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[g*((d + e*x)^m/(c*m)), x] + Simp[1/c Int [(d + e*x)^(m - 1)*(Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x]/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && FractionQ[m] & & GtQ[m, 0]
Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)), x_Symbol] :> Simp[2 Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; Fr eeQ[{a, b, c, d, e, f, g}, x]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
Time = 0.68 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.98
| method | result | size |
| pseudoelliptic | \(\frac {-2 \left (b e -c d \right )^{4} \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (b e -c d \right ) c}}\right )+2 \sqrt {\left (b e -c d \right ) c}\, \left (-A \,d^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right ) c^{4}+\sqrt {e x +d}\, \left (-B \,b^{3} e^{3}+c \left (\left (\frac {B x}{3}+A \right ) e +\frac {10 B d}{3}\right ) e^{2} b^{2}-\frac {10 c^{2} e \left (\frac {x \left (\frac {3 B x}{5}+A \right ) e^{2}}{10}+d \left (\frac {8 B x}{25}+A \right ) e +\frac {29 B \,d^{2}}{25}\right ) b}{3}+\frac {58 c^{3} \left (\frac {3 x^{2} \left (\frac {5 B x}{7}+A \right ) e^{3}}{58}+\frac {8 \left (\frac {33 B x}{56}+A \right ) x d \,e^{2}}{29}+d^{2} \left (\frac {61 B x}{203}+A \right ) e +\frac {88 B \,d^{3}}{203}\right )}{15}\right ) b \right )}{c^{4} b \sqrt {\left (b e -c d \right ) c}}\) | \(224\) |
| derivativedivides | \(\frac {\frac {2 B \left (e x +d \right )^{\frac {7}{2}} c^{3}}{7}+\frac {2 A \,c^{3} e \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {2 B b \,c^{2} e \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 B \,c^{3} d \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {2 A b \,c^{2} e^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {4 A \,c^{3} d e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 B \,b^{2} c \,e^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}-\frac {4 B b \,c^{2} d e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 B \,c^{3} d^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}+2 A \,b^{2} c \,e^{3} \sqrt {e x +d}-6 A b \,c^{2} d \,e^{2} \sqrt {e x +d}+6 A \,c^{3} d^{2} e \sqrt {e x +d}-2 B \,b^{3} e^{3} \sqrt {e x +d}+6 B \,b^{2} c d \,e^{2} \sqrt {e x +d}-6 B b \,c^{2} d^{2} e \sqrt {e x +d}+2 B \,c^{3} d^{3} \sqrt {e x +d}}{c^{4}}+\frac {2 \left (-A \,b^{4} c \,e^{4}+4 A \,b^{3} c^{2} d \,e^{3}-6 A \,b^{2} c^{3} d^{2} e^{2}+4 A b \,c^{4} d^{3} e -d^{4} A \,c^{5}+b^{5} B \,e^{4}-4 B \,b^{4} c d \,e^{3}+6 B \,b^{3} c^{2} d^{2} e^{2}-4 B \,b^{2} c^{3} d^{3} e +B b \,c^{4} d^{4}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (b e -c d \right ) c}}\right )}{c^{4} b \sqrt {\left (b e -c d \right ) c}}-\frac {2 A \,d^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b}\) | \(441\) |
| default | \(\frac {\frac {2 B \left (e x +d \right )^{\frac {7}{2}} c^{3}}{7}+\frac {2 A \,c^{3} e \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {2 B b \,c^{2} e \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 B \,c^{3} d \left (e x +d \right )^{\frac {5}{2}}}{5}-\frac {2 A b \,c^{2} e^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {4 A \,c^{3} d e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 B \,b^{2} c \,e^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}-\frac {4 B b \,c^{2} d e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {2 B \,c^{3} d^{2} \left (e x +d \right )^{\frac {3}{2}}}{3}+2 A \,b^{2} c \,e^{3} \sqrt {e x +d}-6 A b \,c^{2} d \,e^{2} \sqrt {e x +d}+6 A \,c^{3} d^{2} e \sqrt {e x +d}-2 B \,b^{3} e^{3} \sqrt {e x +d}+6 B \,b^{2} c d \,e^{2} \sqrt {e x +d}-6 B b \,c^{2} d^{2} e \sqrt {e x +d}+2 B \,c^{3} d^{3} \sqrt {e x +d}}{c^{4}}+\frac {2 \left (-A \,b^{4} c \,e^{4}+4 A \,b^{3} c^{2} d \,e^{3}-6 A \,b^{2} c^{3} d^{2} e^{2}+4 A b \,c^{4} d^{3} e -d^{4} A \,c^{5}+b^{5} B \,e^{4}-4 B \,b^{4} c d \,e^{3}+6 B \,b^{3} c^{2} d^{2} e^{2}-4 B \,b^{2} c^{3} d^{3} e +B b \,c^{4} d^{4}\right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (b e -c d \right ) c}}\right )}{c^{4} b \sqrt {\left (b e -c d \right ) c}}-\frac {2 A \,d^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b}\) | \(441\) |
2*(-(b*e-c*d)^4*(A*c-B*b)*arctan(c*(e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2))+((b* e-c*d)*c)^(1/2)*(-A*d^(7/2)*arctanh((e*x+d)^(1/2)/d^(1/2))*c^4+(e*x+d)^(1/ 2)*(-B*b^3*e^3+c*((1/3*B*x+A)*e+10/3*B*d)*e^2*b^2-10/3*c^2*e*(1/10*x*(3/5* B*x+A)*e^2+d*(8/25*B*x+A)*e+29/25*B*d^2)*b+58/15*c^3*(3/58*x^2*(5/7*B*x+A) *e^3+8/29*(33/56*B*x+A)*x*d*e^2+d^2*(61/203*B*x+A)*e+88/203*B*d^3))*b))/(( b*e-c*d)*c)^(1/2)/c^4/b
Time = 12.75 (sec) , antiderivative size = 1474, normalized size of antiderivative = 6.46 \[ \int \frac {(A+B x) (d+e x)^{7/2}}{b x+c x^2} \, dx=\text {Too large to display} \]
[1/105*(105*A*c^4*d^(7/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 1 05*((B*b*c^3 - A*c^4)*d^3 - 3*(B*b^2*c^2 - A*b*c^3)*d^2*e + 3*(B*b^3*c - A *b^2*c^2)*d*e^2 - (B*b^4 - A*b^3*c)*e^3)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) + 2*(15*B* b*c^3*e^3*x^3 + 176*B*b*c^3*d^3 - 406*(B*b^2*c^2 - A*b*c^3)*d^2*e + 350*(B *b^3*c - A*b^2*c^2)*d*e^2 - 105*(B*b^4 - A*b^3*c)*e^3 + 3*(22*B*b*c^3*d*e^ 2 - 7*(B*b^2*c^2 - A*b*c^3)*e^3)*x^2 + (122*B*b*c^3*d^2*e - 112*(B*b^2*c^2 - A*b*c^3)*d*e^2 + 35*(B*b^3*c - A*b^2*c^2)*e^3)*x)*sqrt(e*x + d))/(b*c^4 ), 1/105*(105*A*c^4*d^(7/2)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - 210*((B*b*c^3 - A*c^4)*d^3 - 3*(B*b^2*c^2 - A*b*c^3)*d^2*e + 3*(B*b^3*c - A*b^2*c^2)*d*e^2 - (B*b^4 - A*b^3*c)*e^3)*sqrt(-(c*d - b*e)/c)*arctan(-sq rt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)) + 2*(15*B*b*c^3*e^3*x^3 + 176*B*b*c^3*d^3 - 406*(B*b^2*c^2 - A*b*c^3)*d^2*e + 350*(B*b^3*c - A*b^2*c ^2)*d*e^2 - 105*(B*b^4 - A*b^3*c)*e^3 + 3*(22*B*b*c^3*d*e^2 - 7*(B*b^2*c^2 - A*b*c^3)*e^3)*x^2 + (122*B*b*c^3*d^2*e - 112*(B*b^2*c^2 - A*b*c^3)*d*e^ 2 + 35*(B*b^3*c - A*b^2*c^2)*e^3)*x)*sqrt(e*x + d))/(b*c^4), 1/105*(210*A* c^4*sqrt(-d)*d^3*arctan(sqrt(e*x + d)*sqrt(-d)/d) - 105*((B*b*c^3 - A*c^4) *d^3 - 3*(B*b^2*c^2 - A*b*c^3)*d^2*e + 3*(B*b^3*c - A*b^2*c^2)*d*e^2 - (B* b^4 - A*b^3*c)*e^3)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d - b*e + 2*sqrt( e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)) + 2*(15*B*b*c^3*e^3*x^3 + 17...
Time = 9.33 (sec) , antiderivative size = 403, normalized size of antiderivative = 1.77 \[ \int \frac {(A+B x) (d+e x)^{7/2}}{b x+c x^2} \, dx=\begin {cases} \frac {2 \left (\frac {A d^{4} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b \sqrt {- d}} + \frac {B e \left (d + e x\right )^{\frac {7}{2}}}{7 c} + \frac {\left (d + e x\right )^{\frac {5}{2}} \left (A c e^{2} - B b e^{2} + B c d e\right )}{5 c^{2}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \left (- A b c e^{3} + 2 A c^{2} d e^{2} + B b^{2} e^{3} - 2 B b c d e^{2} + B c^{2} d^{2} e\right )}{3 c^{3}} + \frac {\sqrt {d + e x} \left (A b^{2} c e^{4} - 3 A b c^{2} d e^{3} + 3 A c^{3} d^{2} e^{2} - B b^{3} e^{4} + 3 B b^{2} c d e^{3} - 3 B b c^{2} d^{2} e^{2} + B c^{3} d^{3} e\right )}{c^{4}} + \frac {e \left (- A c + B b\right ) \left (b e - c d\right )^{4} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b c^{5} \sqrt {\frac {b e - c d}{c}}}\right )}{e} & \text {for}\: e \neq 0 \\d^{\frac {7}{2}} \left (\frac {B \log {\left (b x + c x^{2} \right )}}{2 c} + \left (A - \frac {B b}{2 c}\right ) \left (- \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\- \frac {\log {\left (b - 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b} - \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\\frac {\log {\left (b + 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b}\right )\right ) & \text {otherwise} \end {cases} \]
Piecewise((2*(A*d**4*e*atan(sqrt(d + e*x)/sqrt(-d))/(b*sqrt(-d)) + B*e*(d + e*x)**(7/2)/(7*c) + (d + e*x)**(5/2)*(A*c*e**2 - B*b*e**2 + B*c*d*e)/(5* c**2) + (d + e*x)**(3/2)*(-A*b*c*e**3 + 2*A*c**2*d*e**2 + B*b**2*e**3 - 2* B*b*c*d*e**2 + B*c**2*d**2*e)/(3*c**3) + sqrt(d + e*x)*(A*b**2*c*e**4 - 3* A*b*c**2*d*e**3 + 3*A*c**3*d**2*e**2 - B*b**3*e**4 + 3*B*b**2*c*d*e**3 - 3 *B*b*c**2*d**2*e**2 + B*c**3*d**3*e)/c**4 + e*(-A*c + B*b)*(b*e - c*d)**4* atan(sqrt(d + e*x)/sqrt((b*e - c*d)/c))/(b*c**5*sqrt((b*e - c*d)/c)))/e, N e(e, 0)), (d**(7/2)*(B*log(b*x + c*x**2)/(2*c) + (A - B*b/(2*c))*(-2*c*Pie cewise(((b/(2*c) + x)/b, Eq(c, 0)), (-log(b - 2*c*(b/(2*c) + x))/(2*c), Tr ue))/b - 2*c*Piecewise(((b/(2*c) + x)/b, Eq(c, 0)), (log(b + 2*c*(b/(2*c) + x))/(2*c), True))/b)), True))
Exception generated. \[ \int \frac {(A+B x) (d+e x)^{7/2}}{b x+c x^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b*e-c*d>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 460 vs. \(2 (198) = 396\).
Time = 0.28 (sec) , antiderivative size = 460, normalized size of antiderivative = 2.02 \[ \int \frac {(A+B x) (d+e x)^{7/2}}{b x+c x^2} \, dx=\frac {2 \, A d^{4} \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{b \sqrt {-d}} + \frac {2 \, {\left (B b c^{4} d^{4} - A c^{5} d^{4} - 4 \, B b^{2} c^{3} d^{3} e + 4 \, A b c^{4} d^{3} e + 6 \, B b^{3} c^{2} d^{2} e^{2} - 6 \, A b^{2} c^{3} d^{2} e^{2} - 4 \, B b^{4} c d e^{3} + 4 \, A b^{3} c^{2} d e^{3} + B b^{5} e^{4} - A b^{4} c e^{4}\right )} \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{\sqrt {-c^{2} d + b c e} b c^{4}} + \frac {2 \, {\left (15 \, {\left (e x + d\right )}^{\frac {7}{2}} B c^{6} + 21 \, {\left (e x + d\right )}^{\frac {5}{2}} B c^{6} d + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} B c^{6} d^{2} + 105 \, \sqrt {e x + d} B c^{6} d^{3} - 21 \, {\left (e x + d\right )}^{\frac {5}{2}} B b c^{5} e + 21 \, {\left (e x + d\right )}^{\frac {5}{2}} A c^{6} e - 70 \, {\left (e x + d\right )}^{\frac {3}{2}} B b c^{5} d e + 70 \, {\left (e x + d\right )}^{\frac {3}{2}} A c^{6} d e - 315 \, \sqrt {e x + d} B b c^{5} d^{2} e + 315 \, \sqrt {e x + d} A c^{6} d^{2} e + 35 \, {\left (e x + d\right )}^{\frac {3}{2}} B b^{2} c^{4} e^{2} - 35 \, {\left (e x + d\right )}^{\frac {3}{2}} A b c^{5} e^{2} + 315 \, \sqrt {e x + d} B b^{2} c^{4} d e^{2} - 315 \, \sqrt {e x + d} A b c^{5} d e^{2} - 105 \, \sqrt {e x + d} B b^{3} c^{3} e^{3} + 105 \, \sqrt {e x + d} A b^{2} c^{4} e^{3}\right )}}{105 \, c^{7}} \]
2*A*d^4*arctan(sqrt(e*x + d)/sqrt(-d))/(b*sqrt(-d)) + 2*(B*b*c^4*d^4 - A*c ^5*d^4 - 4*B*b^2*c^3*d^3*e + 4*A*b*c^4*d^3*e + 6*B*b^3*c^2*d^2*e^2 - 6*A*b ^2*c^3*d^2*e^2 - 4*B*b^4*c*d*e^3 + 4*A*b^3*c^2*d*e^3 + B*b^5*e^4 - A*b^4*c *e^4)*arctan(sqrt(e*x + d)*c/sqrt(-c^2*d + b*c*e))/(sqrt(-c^2*d + b*c*e)*b *c^4) + 2/105*(15*(e*x + d)^(7/2)*B*c^6 + 21*(e*x + d)^(5/2)*B*c^6*d + 35* (e*x + d)^(3/2)*B*c^6*d^2 + 105*sqrt(e*x + d)*B*c^6*d^3 - 21*(e*x + d)^(5/ 2)*B*b*c^5*e + 21*(e*x + d)^(5/2)*A*c^6*e - 70*(e*x + d)^(3/2)*B*b*c^5*d*e + 70*(e*x + d)^(3/2)*A*c^6*d*e - 315*sqrt(e*x + d)*B*b*c^5*d^2*e + 315*sq rt(e*x + d)*A*c^6*d^2*e + 35*(e*x + d)^(3/2)*B*b^2*c^4*e^2 - 35*(e*x + d)^ (3/2)*A*b*c^5*e^2 + 315*sqrt(e*x + d)*B*b^2*c^4*d*e^2 - 315*sqrt(e*x + d)* A*b*c^5*d*e^2 - 105*sqrt(e*x + d)*B*b^3*c^3*e^3 + 105*sqrt(e*x + d)*A*b^2* c^4*e^3)/c^7
Time = 11.00 (sec) , antiderivative size = 6515, normalized size of antiderivative = 28.57 \[ \int \frac {(A+B x) (d+e x)^{7/2}}{b x+c x^2} \, dx=\text {Too large to display} \]
((2*A*e - 2*B*d)/(5*c) - (2*B*(b*e - 2*c*d))/(5*c^2))*(d + e*x)^(5/2) - (( (c*d^2 - b*d*e)*((2*A*e - 2*B*d)/c - (2*B*(b*e - 2*c*d))/c^2))/c - ((b*e - 2*c*d)*(((b*e - 2*c*d)*((2*A*e - 2*B*d)/c - (2*B*(b*e - 2*c*d))/c^2))/c + (2*B*(c*d^2 - b*d*e))/c^2))/c)*(d + e*x)^(1/2) - (((b*e - 2*c*d)*((2*A*e - 2*B*d)/c - (2*B*(b*e - 2*c*d))/c^2))/(3*c) + (2*B*(c*d^2 - b*d*e))/(3*c^ 2))*(d + e*x)^(3/2) + (2*B*(d + e*x)^(7/2))/(7*c) - (A*atan(((A*((8*(d + e *x)^(1/2)*(B^2*b^10*e^10 + A^2*b^8*c^2*e^10 + 2*A^2*c^10*d^8*e^2 + 28*A^2* b^2*c^8*d^6*e^4 - 56*A^2*b^3*c^7*d^5*e^5 + 70*A^2*b^4*c^6*d^4*e^6 - 56*A^2 *b^5*c^5*d^3*e^7 + 28*A^2*b^6*c^4*d^2*e^8 + B^2*b^2*c^8*d^8*e^2 - 8*B^2*b^ 3*c^7*d^7*e^3 + 28*B^2*b^4*c^6*d^6*e^4 - 56*B^2*b^5*c^5*d^5*e^5 + 70*B^2*b ^6*c^4*d^4*e^6 - 56*B^2*b^7*c^3*d^3*e^7 + 28*B^2*b^8*c^2*d^2*e^8 - 8*B^2*b ^9*c*d*e^9 - 8*A^2*b*c^9*d^7*e^3 - 8*A^2*b^7*c^3*d*e^9 - 2*A*B*b^9*c*e^10 - 2*A*B*b*c^9*d^8*e^2 + 16*A*B*b^8*c^2*d*e^9 + 16*A*B*b^2*c^8*d^7*e^3 - 56 *A*B*b^3*c^7*d^6*e^4 + 112*A*B*b^4*c^6*d^5*e^5 - 140*A*B*b^5*c^5*d^4*e^6 + 112*A*B*b^6*c^4*d^3*e^7 - 56*A*B*b^7*c^3*d^2*e^8))/c^7 + (A*(d^7)^(1/2)*( (8*(B*b^6*c^5*d*e^6 - A*b^5*c^6*d*e^6 + 3*A*b^2*c^9*d^4*e^3 - 6*A*b^3*c^8* d^3*e^4 + 4*A*b^4*c^7*d^2*e^5 + B*b^2*c^9*d^5*e^2 - 4*B*b^3*c^8*d^4*e^3 + 6*B*b^4*c^7*d^3*e^4 - 4*B*b^5*c^6*d^2*e^5))/c^7 + (8*A*(b^3*c^9*e^3 - 2*b^ 2*c^10*d*e^2)*(d^7)^(1/2)*(d + e*x)^(1/2))/(b*c^7)))/b)*(d^7)^(1/2)*1i)/b + (A*((8*(d + e*x)^(1/2)*(B^2*b^10*e^10 + A^2*b^8*c^2*e^10 + 2*A^2*c^10...