Integrand size = 26, antiderivative size = 225 \[ \int \frac {A+B x}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx=\frac {2 (B d-A e)}{5 d (c d-b e) (d+e x)^{5/2}}+\frac {2 \left (B c d^2-A e (2 c d-b e)\right )}{3 d^2 (c d-b e)^2 (d+e x)^{3/2}}+\frac {2 \left (B c^2 d^3-A e \left (3 c^2 d^2-3 b c d e+b^2 e^2\right )\right )}{d^3 (c d-b e)^3 \sqrt {d+e x}}-\frac {2 A \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b d^{7/2}}-\frac {2 c^{5/2} (b B-A c) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b (c d-b e)^{7/2}} \]
2/5*(-A*e+B*d)/d/(-b*e+c*d)/(e*x+d)^(5/2)+2/3*(B*c*d^2-A*e*(-b*e+2*c*d))/d ^2/(-b*e+c*d)^2/(e*x+d)^(3/2)-2*A*arctanh((e*x+d)^(1/2)/d^(1/2))/b/d^(7/2) -2*c^(5/2)*(-A*c+B*b)*arctanh(c^(1/2)*(e*x+d)^(1/2)/(-b*e+c*d)^(1/2))/b/(- b*e+c*d)^(7/2)+2*(B*c^2*d^3-A*e*(b^2*e^2-3*b*c*d*e+3*c^2*d^2))/d^3/(-b*e+c *d)^3/(e*x+d)^(1/2)
Time = 0.59 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx=\frac {2 B d^3 \left (3 b^2 e^2-b c e (11 d+5 e x)+c^2 \left (23 d^2+35 d e x+15 e^2 x^2\right )\right )-2 A e \left (-3 b c d e \left (22 d^2+35 d e x+15 e^2 x^2\right )+b^2 e^2 \left (23 d^2+35 d e x+15 e^2 x^2\right )+c^2 d^2 \left (58 d^2+100 d e x+45 e^2 x^2\right )\right )}{15 d^3 (c d-b e)^3 (d+e x)^{5/2}}+\frac {2 c^{5/2} (-b B+A c) \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )}{b (-c d+b e)^{7/2}}-\frac {2 A \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b d^{7/2}} \]
(2*B*d^3*(3*b^2*e^2 - b*c*e*(11*d + 5*e*x) + c^2*(23*d^2 + 35*d*e*x + 15*e ^2*x^2)) - 2*A*e*(-3*b*c*d*e*(22*d^2 + 35*d*e*x + 15*e^2*x^2) + b^2*e^2*(2 3*d^2 + 35*d*e*x + 15*e^2*x^2) + c^2*d^2*(58*d^2 + 100*d*e*x + 45*e^2*x^2) ))/(15*d^3*(c*d - b*e)^3*(d + e*x)^(5/2)) + (2*c^(5/2)*(-(b*B) + A*c)*ArcT an[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[-(c*d) + b*e]])/(b*(-(c*d) + b*e)^(7/2)) - (2*A*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b*d^(7/2))
Time = 0.66 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.26, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {1198, 1198, 1198, 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.
| \(\displaystyle \int \frac {A+B x}{\left (b x+c x^2\right ) (d+e x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1198 |
| \(\displaystyle \frac {\int \frac {A (c d-b e)+c (B d-A e) x}{(d+e x)^{5/2} \left (c x^2+b x\right )}dx}{d (c d-b e)}+\frac {2 (B d-A e)}{5 d (d+e x)^{5/2} (c d-b e)}\) |
| \(\Big \downarrow \) 1198 |
| \(\displaystyle \frac {\frac {\int \frac {A (c d-b e)^2+c \left (B c d^2-A e (2 c d-b e)\right ) x}{(d+e x)^{3/2} \left (c x^2+b x\right )}dx}{d (c d-b e)}+\frac {2 \left (B c d^2-A e (2 c d-b e)\right )}{3 d (d+e x)^{3/2} (c d-b e)}}{d (c d-b e)}+\frac {2 (B d-A e)}{5 d (d+e x)^{5/2} (c d-b e)}\) |
| \(\Big \downarrow \) 1198 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {A (c d-b e)^3+c \left (B c^2 d^3-A e \left (3 c^2 d^2-3 b c e d+b^2 e^2\right )\right ) x}{\sqrt {d+e x} \left (c x^2+b x\right )}dx}{d (c d-b e)}+\frac {2 \left (B c^2 d^3-A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{d \sqrt {d+e x} (c d-b e)}}{d (c d-b e)}+\frac {2 \left (B c d^2-A e (2 c d-b e)\right )}{3 d (d+e x)^{3/2} (c d-b e)}}{d (c d-b e)}+\frac {2 (B d-A e)}{5 d (d+e x)^{5/2} (c d-b e)}\) |
| \(\Big \downarrow \) 1197 |
| \(\displaystyle \frac {\frac {\frac {2 \int -\frac {B c^3 d^4-A e \left (4 c^3 d^3-6 b c^2 e d^2+4 b^2 c e^2 d-b^3 e^3\right )-c \left (B c^2 d^3-A e \left (3 c^2 d^2-3 b c e d+b^2 e^2\right )\right ) (d+e x)}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{d (c d-b e)}+\frac {2 \left (B c^2 d^3-A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{d \sqrt {d+e x} (c d-b e)}}{d (c d-b e)}+\frac {2 \left (B c d^2-A e (2 c d-b e)\right )}{3 d (d+e x)^{3/2} (c d-b e)}}{d (c d-b e)}+\frac {2 (B d-A e)}{5 d (d+e x)^{5/2} (c d-b e)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\frac {2 \left (B c^2 d^3-A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{d \sqrt {d+e x} (c d-b e)}-\frac {2 \int \frac {B c^3 d^4-A e \left (4 c^3 d^3-6 b c^2 e d^2+4 b^2 c e^2 d-b^3 e^3\right )-c \left (B c^2 d^3-A e \left (3 c^2 d^2-3 b c e d+b^2 e^2\right )\right ) (d+e x)}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{d (c d-b e)}}{d (c d-b e)}+\frac {2 \left (B c d^2-A e (2 c d-b e)\right )}{3 d (d+e x)^{3/2} (c d-b e)}}{d (c d-b e)}+\frac {2 (B d-A e)}{5 d (d+e x)^{5/2} (c d-b e)}\) |
| \(\Big \downarrow \) 1480 |
| \(\displaystyle \frac {\frac {\frac {2 \left (\frac {c^3 d^3 (b B-A c) \int \frac {1}{-c d+b e+c (d+e x)}d\sqrt {d+e x}}{b}+\frac {A c (c d-b e)^3 \int \frac {1}{c (d+e x)-c d}d\sqrt {d+e x}}{b}\right )}{d (c d-b e)}+\frac {2 \left (B c^2 d^3-A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{d \sqrt {d+e x} (c d-b e)}}{d (c d-b e)}+\frac {2 \left (B c d^2-A e (2 c d-b e)\right )}{3 d (d+e x)^{3/2} (c d-b e)}}{d (c d-b e)}+\frac {2 (B d-A e)}{5 d (d+e x)^{5/2} (c d-b e)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {2 \left (-\frac {c^{5/2} d^3 (b B-A c) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b \sqrt {c d-b e}}-\frac {A \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) (c d-b e)^3}{b \sqrt {d}}\right )}{d (c d-b e)}+\frac {2 \left (B c^2 d^3-A e \left (b^2 e^2-3 b c d e+3 c^2 d^2\right )\right )}{d \sqrt {d+e x} (c d-b e)}}{d (c d-b e)}+\frac {2 \left (B c d^2-A e (2 c d-b e)\right )}{3 d (d+e x)^{3/2} (c d-b e)}}{d (c d-b e)}+\frac {2 (B d-A e)}{5 d (d+e x)^{5/2} (c d-b e)}\) |
(2*(B*d - A*e))/(5*d*(c*d - b*e)*(d + e*x)^(5/2)) + ((2*(B*c*d^2 - A*e*(2* c*d - b*e)))/(3*d*(c*d - b*e)*(d + e*x)^(3/2)) + ((2*(B*c^2*d^3 - A*e*(3*c ^2*d^2 - 3*b*c*d*e + b^2*e^2)))/(d*(c*d - b*e)*Sqrt[d + e*x]) + (2*(-((A*( c*d - b*e)^3*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b*Sqrt[d])) - (c^(5/2)*(b*B - A*c)*d^3*ArcTanh[(Sqrt[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*Sqrt[c*d - b*e])))/(d*(c*d - b*e)))/(d*(c*d - b*e)))/(d*(c*d - b*e))
3.13.36.3.1 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[((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_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*f - d*g)*((d + e*x)^(m + 1)/((m + 1)*(c *d^2 - b*d*e + a*e^2))), x] + Simp[1/(c*d^2 - b*d*e + a*e^2) Int[(d + e*x )^(m + 1)*(Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x]/(a + b*x + c*x^ 2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && FractionQ[m] && LtQ[m, -1 ]
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 = 0.42 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.95
| method | result | size |
| pseudoelliptic | \(\frac {\frac {2 A e}{5}-\frac {2 B d}{5}}{d \left (b e -c d \right ) \left (e x +d \right )^{\frac {5}{2}}}+\frac {\frac {2}{3} A b \,e^{2}-\frac {4}{3} A c d e +\frac {2}{3} B c \,d^{2}}{d^{2} \left (b e -c d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}+\frac {2 A \,b^{2} e^{3}-6 A b c d \,e^{2}+6 A \,c^{2} d^{2} e -2 B \,c^{2} d^{3}}{\sqrt {e x +d}\, d^{3} \left (b e -c d \right )^{3}}+\frac {2 c^{3} \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (b e -c d \right ) c}}\right )}{\left (b e -c d \right )^{3} b \sqrt {\left (b e -c d \right ) c}}-\frac {2 A \,\operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b \,d^{\frac {7}{2}}}\) | \(213\) |
| derivativedivides | \(-\frac {2 \left (-A e +B d \right )}{5 d \left (b e -c d \right ) \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 \left (-A b \,e^{2}+2 A c d e -B c \,d^{2}\right )}{3 d^{2} \left (b e -c d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 \left (-A \,b^{2} e^{3}+3 A b c d \,e^{2}-3 A \,c^{2} d^{2} e +B \,c^{2} d^{3}\right )}{d^{3} \left (b e -c d \right )^{3} \sqrt {e x +d}}+\frac {2 c^{3} \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (b e -c d \right ) c}}\right )}{\left (b e -c d \right )^{3} b \sqrt {\left (b e -c d \right ) c}}-\frac {2 A \,\operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b \,d^{\frac {7}{2}}}\) | \(215\) |
| default | \(-\frac {2 \left (-A e +B d \right )}{5 d \left (b e -c d \right ) \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 \left (-A b \,e^{2}+2 A c d e -B c \,d^{2}\right )}{3 d^{2} \left (b e -c d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 \left (-A \,b^{2} e^{3}+3 A b c d \,e^{2}-3 A \,c^{2} d^{2} e +B \,c^{2} d^{3}\right )}{d^{3} \left (b e -c d \right )^{3} \sqrt {e x +d}}+\frac {2 c^{3} \left (A c -B b \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {\left (b e -c d \right ) c}}\right )}{\left (b e -c d \right )^{3} b \sqrt {\left (b e -c d \right ) c}}-\frac {2 A \,\operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b \,d^{\frac {7}{2}}}\) | \(215\) |
2/5*(A*e-B*d)/d/(b*e-c*d)/(e*x+d)^(5/2)+2/3*(A*b*e^2-2*A*c*d*e+B*c*d^2)/d^ 2/(b*e-c*d)^2/(e*x+d)^(3/2)+(2*A*b^2*e^3-6*A*b*c*d*e^2+6*A*c^2*d^2*e-2*B*c ^2*d^3)/(e*x+d)^(1/2)/d^3/(b*e-c*d)^3+2/(b*e-c*d)^3*c^3*(A*c-B*b)/b/((b*e- c*d)*c)^(1/2)*arctan(c*(e*x+d)^(1/2)/((b*e-c*d)*c)^(1/2))-2*A*arctanh((e*x +d)^(1/2)/d^(1/2))/b/d^(7/2)
Leaf count of result is larger than twice the leaf count of optimal. 763 vs. \(2 (199) = 398\).
Time = 7.41 (sec) , antiderivative size = 3081, normalized size of antiderivative = 13.69 \[ \int \frac {A+B x}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx=\text {Too large to display} \]
[1/15*(15*((B*b*c^2 - A*c^3)*d^4*e^3*x^3 + 3*(B*b*c^2 - A*c^3)*d^5*e^2*x^2 + 3*(B*b*c^2 - A*c^3)*d^6*e*x + (B*b*c^2 - A*c^3)*d^7)*sqrt(c/(c*d - b*e) )*log((c*e*x + 2*c*d - b*e - 2*(c*d - b*e)*sqrt(e*x + d)*sqrt(c/(c*d - b*e )))/(c*x + b)) + 15*(A*c^3*d^6 - 3*A*b*c^2*d^5*e + 3*A*b^2*c*d^4*e^2 - A*b ^3*d^3*e^3 + (A*c^3*d^3*e^3 - 3*A*b*c^2*d^2*e^4 + 3*A*b^2*c*d*e^5 - A*b^3* e^6)*x^3 + 3*(A*c^3*d^4*e^2 - 3*A*b*c^2*d^3*e^3 + 3*A*b^2*c*d^2*e^4 - A*b^ 3*d*e^5)*x^2 + 3*(A*c^3*d^5*e - 3*A*b*c^2*d^4*e^2 + 3*A*b^2*c*d^3*e^3 - A* b^3*d^2*e^4)*x)*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*( 23*B*b*c^2*d^6 - 23*A*b^3*d^3*e^3 - (11*B*b^2*c + 58*A*b*c^2)*d^5*e + 3*(B *b^3 + 22*A*b^2*c)*d^4*e^2 + 15*(B*b*c^2*d^4*e^2 - 3*A*b*c^2*d^3*e^3 + 3*A *b^2*c*d^2*e^4 - A*b^3*d*e^5)*x^2 + 5*(7*B*b*c^2*d^5*e + 21*A*b^2*c*d^3*e^ 3 - 7*A*b^3*d^2*e^4 - (B*b^2*c + 20*A*b*c^2)*d^4*e^2)*x)*sqrt(e*x + d))/(b *c^3*d^10 - 3*b^2*c^2*d^9*e + 3*b^3*c*d^8*e^2 - b^4*d^7*e^3 + (b*c^3*d^7*e ^3 - 3*b^2*c^2*d^6*e^4 + 3*b^3*c*d^5*e^5 - b^4*d^4*e^6)*x^3 + 3*(b*c^3*d^8 *e^2 - 3*b^2*c^2*d^7*e^3 + 3*b^3*c*d^6*e^4 - b^4*d^5*e^5)*x^2 + 3*(b*c^3*d ^9*e - 3*b^2*c^2*d^8*e^2 + 3*b^3*c*d^7*e^3 - b^4*d^6*e^4)*x), -1/15*(30*(( B*b*c^2 - A*c^3)*d^4*e^3*x^3 + 3*(B*b*c^2 - A*c^3)*d^5*e^2*x^2 + 3*(B*b*c^ 2 - A*c^3)*d^6*e*x + (B*b*c^2 - A*c^3)*d^7)*sqrt(-c/(c*d - b*e))*arctan(-( c*d - b*e)*sqrt(e*x + d)*sqrt(-c/(c*d - b*e))/(c*e*x + c*d)) - 15*(A*c^3*d ^6 - 3*A*b*c^2*d^5*e + 3*A*b^2*c*d^4*e^2 - A*b^3*d^3*e^3 + (A*c^3*d^3*e...
Time = 10.61 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.45 \[ \int \frac {A+B x}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx=\begin {cases} \frac {2 \left (\frac {A e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {- d}} \right )}}{b d^{3} \sqrt {- d}} - \frac {e \left (- A e + B d\right )}{5 d \left (d + e x\right )^{\frac {5}{2}} \left (b e - c d\right )} + \frac {e \left (A b e^{2} - 2 A c d e + B c d^{2}\right )}{3 d^{2} \left (d + e x\right )^{\frac {3}{2}} \left (b e - c d\right )^{2}} + \frac {e \left (A b^{2} e^{3} - 3 A b c d e^{2} + 3 A c^{2} d^{2} e - B c^{2} d^{3}\right )}{d^{3} \sqrt {d + e x} \left (b e - c d\right )^{3}} - \frac {c^{2} e \left (- A c + B b\right ) \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {b e - c d}{c}}} \right )}}{b \sqrt {\frac {b e - c d}{c}} \left (b e - c d\right )^{3}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {\frac {B \log {\left (b x + c x^{2} \right )}}{2 c} + \left (A - \frac {B b}{2 c}\right ) \left (- \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\- \frac {\log {\left (b - 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b} - \frac {2 c \left (\begin {cases} \frac {\frac {b}{2 c} + x}{b} & \text {for}\: c = 0 \\\frac {\log {\left (b + 2 c \left (\frac {b}{2 c} + x\right ) \right )}}{2 c} & \text {otherwise} \end {cases}\right )}{b}\right )}{d^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((2*(A*e*atan(sqrt(d + e*x)/sqrt(-d))/(b*d**3*sqrt(-d)) - e*(-A*e + B*d)/(5*d*(d + e*x)**(5/2)*(b*e - c*d)) + e*(A*b*e**2 - 2*A*c*d*e + B*c *d**2)/(3*d**2*(d + e*x)**(3/2)*(b*e - c*d)**2) + e*(A*b**2*e**3 - 3*A*b*c *d*e**2 + 3*A*c**2*d**2*e - B*c**2*d**3)/(d**3*sqrt(d + e*x)*(b*e - c*d)** 3) - c**2*e*(-A*c + B*b)*atan(sqrt(d + e*x)/sqrt((b*e - c*d)/c))/(b*sqrt(( b*e - c*d)/c)*(b*e - c*d)**3))/e, Ne(e, 0)), ((B*log(b*x + c*x**2)/(2*c) + (A - B*b/(2*c))*(-2*c*Piecewise(((b/(2*c) + x)/b, Eq(c, 0)), (-log(b - 2* c*(b/(2*c) + x))/(2*c), True))/b - 2*c*Piecewise(((b/(2*c) + x)/b, Eq(c, 0 )), (log(b + 2*c*(b/(2*c) + x))/(2*c), True))/b))/d**(7/2), True))
Exception generated. \[ \int \frac {A+B x}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.29 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.67 \[ \int \frac {A+B x}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx=\frac {2 \, {\left (B b c^{3} - A c^{4}\right )} \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{{\left (b c^{3} d^{3} - 3 \, b^{2} c^{2} d^{2} e + 3 \, b^{3} c d e^{2} - b^{4} e^{3}\right )} \sqrt {-c^{2} d + b c e}} + \frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} B c^{2} d^{3} + 5 \, {\left (e x + d\right )} B c^{2} d^{4} + 3 \, B c^{2} d^{5} - 45 \, {\left (e x + d\right )}^{2} A c^{2} d^{2} e - 5 \, {\left (e x + d\right )} B b c d^{3} e - 10 \, {\left (e x + d\right )} A c^{2} d^{3} e - 6 \, B b c d^{4} e - 3 \, A c^{2} d^{4} e + 45 \, {\left (e x + d\right )}^{2} A b c d e^{2} + 15 \, {\left (e x + d\right )} A b c d^{2} e^{2} + 3 \, B b^{2} d^{3} e^{2} + 6 \, A b c d^{3} e^{2} - 15 \, {\left (e x + d\right )}^{2} A b^{2} e^{3} - 5 \, {\left (e x + d\right )} A b^{2} d e^{3} - 3 \, A b^{2} d^{2} e^{3}\right )}}{15 \, {\left (c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} - b^{3} d^{3} e^{3}\right )} {\left (e x + d\right )}^{\frac {5}{2}}} + \frac {2 \, A \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{b \sqrt {-d} d^{3}} \]
2*(B*b*c^3 - A*c^4)*arctan(sqrt(e*x + d)*c/sqrt(-c^2*d + b*c*e))/((b*c^3*d ^3 - 3*b^2*c^2*d^2*e + 3*b^3*c*d*e^2 - b^4*e^3)*sqrt(-c^2*d + b*c*e)) + 2/ 15*(15*(e*x + d)^2*B*c^2*d^3 + 5*(e*x + d)*B*c^2*d^4 + 3*B*c^2*d^5 - 45*(e *x + d)^2*A*c^2*d^2*e - 5*(e*x + d)*B*b*c*d^3*e - 10*(e*x + d)*A*c^2*d^3*e - 6*B*b*c*d^4*e - 3*A*c^2*d^4*e + 45*(e*x + d)^2*A*b*c*d*e^2 + 15*(e*x + d)*A*b*c*d^2*e^2 + 3*B*b^2*d^3*e^2 + 6*A*b*c*d^3*e^2 - 15*(e*x + d)^2*A*b^ 2*e^3 - 5*(e*x + d)*A*b^2*d*e^3 - 3*A*b^2*d^2*e^3)/((c^3*d^6 - 3*b*c^2*d^5 *e + 3*b^2*c*d^4*e^2 - b^3*d^3*e^3)*(e*x + d)^(5/2)) + 2*A*arctan(sqrt(e*x + d)/sqrt(-d))/(b*sqrt(-d)*d^3)
Time = 14.49 (sec) , antiderivative size = 13404, normalized size of antiderivative = 59.57 \[ \int \frac {A+B x}{(d+e x)^{7/2} \left (b x+c x^2\right )} \, dx=\text {Too large to display} \]
- ((2*(A*e - B*d))/(5*(c*d^2 - b*d*e)) + (2*(d + e*x)^2*(A*b^2*e^3 - B*c^2 *d^3 + 3*A*c^2*d^2*e - 3*A*b*c*d*e^2))/(c*d^2 - b*d*e)^3 - (2*(d + e*x)*(A *b*e^2 + B*c*d^2 - 2*A*c*d*e))/(3*(c*d^2 - b*d*e)^2))/(d + e*x)^(5/2) - (a tan((((-c^5*(b*e - c*d)^7)^(1/2)*(A*c - B*b)*((d + e*x)^(1/2)*(16*A^2*c^18 *d^24*e^2 + 1128*A^2*b^2*c^16*d^22*e^4 - 4312*A^2*b^3*c^15*d^21*e^5 + 1192 8*A^2*b^4*c^14*d^20*e^6 - 25032*A^2*b^5*c^13*d^19*e^7 + 40712*A^2*b^6*c^12 *d^18*e^8 - 51768*A^2*b^7*c^11*d^17*e^9 + 51552*A^2*b^8*c^10*d^16*e^10 - 4 0048*A^2*b^9*c^9*d^15*e^11 + 24024*A^2*b^10*c^8*d^14*e^12 - 10920*A^2*b^11 *c^7*d^13*e^13 + 3640*A^2*b^12*c^6*d^12*e^14 - 840*A^2*b^13*c^5*d^11*e^15 + 120*A^2*b^14*c^4*d^10*e^16 - 8*A^2*b^15*c^3*d^9*e^17 + 8*B^2*b^2*c^16*d^ 24*e^2 - 72*B^2*b^3*c^15*d^23*e^3 + 288*B^2*b^4*c^14*d^22*e^4 - 672*B^2*b^ 5*c^13*d^21*e^5 + 1008*B^2*b^6*c^12*d^20*e^6 - 1008*B^2*b^7*c^11*d^19*e^7 + 672*B^2*b^8*c^10*d^18*e^8 - 288*B^2*b^9*c^9*d^17*e^9 + 72*B^2*b^10*c^8*d ^16*e^10 - 8*B^2*b^11*c^7*d^15*e^11 - 192*A^2*b*c^17*d^23*e^3 - 16*A*B*b*c ^17*d^24*e^2 + 144*A*B*b^2*c^16*d^23*e^3 - 576*A*B*b^3*c^15*d^22*e^4 + 134 4*A*B*b^4*c^14*d^21*e^5 - 2016*A*B*b^5*c^13*d^20*e^6 + 2016*A*B*b^6*c^12*d ^19*e^7 - 1344*A*B*b^7*c^11*d^18*e^8 + 576*A*B*b^8*c^10*d^17*e^9 - 144*A*B *b^9*c^9*d^16*e^10 + 16*A*B*b^10*c^8*d^15*e^11) - ((-c^5*(b*e - c*d)^7)^(1 /2)*(A*c - B*b)*(((-c^5*(b*e - c*d)^7)^(1/2)*(A*c - B*b)*(d + e*x)^(1/2)*( 16*b^2*c^18*d^31*e^2 - 248*b^3*c^17*d^30*e^3 + 1800*b^4*c^16*d^29*e^4 -...