Integrand size = 26, antiderivative size = 351 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )^2} \, dx=-\frac {e \left (6 A c^2 d^2-b^2 e (2 B d-5 A e)-3 b c d (B d+2 A e)\right )}{3 b^2 d^2 (c d-b e)^2 (d+e x)^{3/2}}+\frac {c (b B d-2 A c d+A b e)}{b^2 d (c d-b e) (b+c x) (d+e x)^{3/2}}-\frac {A}{b d x (b+c x) (d+e x)^{3/2}}-\frac {e \left (2 A c^3 d^3-b^2 c d e (6 B d-11 A e)+b^3 e^2 (2 B d-5 A e)-b c^2 d^2 (B d+3 A e)\right )}{b^2 d^3 (c d-b e)^3 \sqrt {d+e x}}-\frac {(2 b B d-4 A c d-5 A b e) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{b^3 d^{7/2}}+\frac {c^{5/2} \left (2 b B c d-4 A c^2 d-7 b^2 B e+9 A b c e\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b^3 (c d-b e)^{7/2}} \] Output:
-1/3*e*(6*A*c^2*d^2-b^2*e*(-5*A*e+2*B*d)-3*b*c*d*(2*A*e+B*d))/b^2/d^2/(-b* e+c*d)^2/(e*x+d)^(3/2)+c*(A*b*e-2*A*c*d+B*b*d)/b^2/d/(-b*e+c*d)/(c*x+b)/(e *x+d)^(3/2)-A/b/d/x/(c*x+b)/(e*x+d)^(3/2)-e*(2*A*c^3*d^3-b^2*c*d*e*(-11*A* e+6*B*d)+b^3*e^2*(-5*A*e+2*B*d)-b*c^2*d^2*(3*A*e+B*d))/b^2/d^3/(-b*e+c*d)^ 3/(e*x+d)^(1/2)-(-5*A*b*e-4*A*c*d+2*B*b*d)*arctanh((e*x+d)^(1/2)/d^(1/2))/ b^3/d^(7/2)+c^(5/2)*(9*A*b*c*e-4*A*c^2*d-7*B*b^2*e+2*B*b*c*d)*arctanh(c^(1 /2)*(e*x+d)^(1/2)/(-b*e+c*d)^(1/2))/b^3/(-b*e+c*d)^(7/2)
Time = 2.63 (sec) , antiderivative size = 385, normalized size of antiderivative = 1.10 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )^2} \, dx=\frac {\frac {b \left (b B d x \left (3 c^3 d^2 (d+e x)^2-2 b^3 e^3 (4 d+3 e x)+2 b c^2 d e^2 x (10 d+9 e x)+2 b^2 c e^2 \left (10 d^2+5 d e x-3 e^2 x^2\right )\right )+A \left (-6 c^4 d^3 x (d+e x)^2-3 b c^3 d^2 (d-3 e x) (d+e x)^2+b^4 e^3 \left (3 d^2+20 d e x+15 e^2 x^2\right )+b^2 c^2 d e \left (9 d^3+9 d^2 e x-35 d e^2 x^2-33 e^3 x^3\right )+b^3 c e^2 \left (-9 d^3-41 d^2 e x-13 d e^2 x^2+15 e^3 x^3\right )\right )\right )}{d^3 (c d-b e)^3 x (b+c x) (d+e x)^{3/2}}-\frac {3 c^{5/2} \left (4 A c^2 d+7 b^2 B e-b c (2 B d+9 A e)\right ) \arctan \left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {-c d+b e}}\right )}{(-c d+b e)^{7/2}}+\frac {3 (-2 b B d+4 A c d+5 A b e) \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{d^{7/2}}}{3 b^3} \] Input:
Integrate[(A + B*x)/((d + e*x)^(5/2)*(b*x + c*x^2)^2),x]
Output:
((b*(b*B*d*x*(3*c^3*d^2*(d + e*x)^2 - 2*b^3*e^3*(4*d + 3*e*x) + 2*b*c^2*d* e^2*x*(10*d + 9*e*x) + 2*b^2*c*e^2*(10*d^2 + 5*d*e*x - 3*e^2*x^2)) + A*(-6 *c^4*d^3*x*(d + e*x)^2 - 3*b*c^3*d^2*(d - 3*e*x)*(d + e*x)^2 + b^4*e^3*(3* d^2 + 20*d*e*x + 15*e^2*x^2) + b^2*c^2*d*e*(9*d^3 + 9*d^2*e*x - 35*d*e^2*x ^2 - 33*e^3*x^3) + b^3*c*e^2*(-9*d^3 - 41*d^2*e*x - 13*d*e^2*x^2 + 15*e^3* x^3))))/(d^3*(c*d - b*e)^3*x*(b + c*x)*(d + e*x)^(3/2)) - (3*c^(5/2)*(4*A* c^2*d + 7*b^2*B*e - b*c*(2*B*d + 9*A*e))*ArcTan[(Sqrt[c]*Sqrt[d + e*x])/Sq rt[-(c*d) + b*e]])/(-(c*d) + b*e)^(7/2) + (3*(-2*b*B*d + 4*A*c*d + 5*A*b*e )*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/d^(7/2))/(3*b^3)
Time = 1.89 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {1235, 27, 1198, 1198, 1197, 25, 27, 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.
\(\displaystyle \int \frac {A+B x}{\left (b x+c x^2\right )^2 (d+e x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 1235 |
\(\displaystyle -\frac {\int -\frac {(c d-b e) (2 b B d-4 A c d-5 A b e)+5 c e (b B d-2 A c d+A b e) x}{2 (d+e x)^{5/2} \left (c x^2+b x\right )}dx}{b^2 d (c d-b e)}-\frac {c x (2 A c d-b (A e+B d))+A b (c d-b e)}{b^2 d \left (b x+c x^2\right ) (d+e x)^{3/2} (c d-b e)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(c d-b e) (2 b B d-4 A c d-5 A b e)+5 c e (b B d-2 A c d+A b e) x}{(d+e x)^{5/2} \left (c x^2+b x\right )}dx}{2 b^2 d (c d-b e)}-\frac {c x (2 A c d-b (A e+B d))+A b (c d-b e)}{b^2 d \left (b x+c x^2\right ) (d+e x)^{3/2} (c d-b e)}\) |
\(\Big \downarrow \) 1198 |
\(\displaystyle \frac {\frac {\int \frac {(c d-b e)^2 (2 b B d-4 A c d-5 A b e)-c e \left (-e (2 B d-5 A e) b^2-3 c d (B d+2 A e) b+6 A c^2 d^2\right ) x}{(d+e x)^{3/2} \left (c x^2+b x\right )}dx}{d (c d-b e)}-\frac {2 e \left (b^2 (-e) (2 B d-5 A e)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )}{3 d (d+e x)^{3/2} (c d-b e)}}{2 b^2 d (c d-b e)}-\frac {c x (2 A c d-b (A e+B d))+A b (c d-b e)}{b^2 d \left (b x+c x^2\right ) (d+e x)^{3/2} (c d-b e)}\) |
\(\Big \downarrow \) 1198 |
\(\displaystyle \frac {\frac {\frac {\int \frac {(c d-b e)^3 (2 b B d-4 A c d-5 A b e)-c e \left (e^2 (2 B d-5 A e) b^3-c d e (6 B d-11 A e) b^2-c^2 d^2 (B d+3 A e) b+2 A c^3 d^3\right ) x}{\sqrt {d+e x} \left (c x^2+b x\right )}dx}{d (c d-b e)}-\frac {2 e \left (b^3 e^2 (2 B d-5 A e)-b^2 c d e (6 B d-11 A e)-b c^2 d^2 (3 A e+B d)+2 A c^3 d^3\right )}{d \sqrt {d+e x} (c d-b e)}}{d (c d-b e)}-\frac {2 e \left (b^2 (-e) (2 B d-5 A e)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )}{3 d (d+e x)^{3/2} (c d-b e)}}{2 b^2 d (c d-b e)}-\frac {c x (2 A c d-b (A e+B d))+A b (c d-b e)}{b^2 d \left (b x+c x^2\right ) (d+e x)^{3/2} (c d-b e)}\) |
\(\Big \downarrow \) 1197 |
\(\displaystyle \frac {\frac {\frac {2 \int -\frac {e \left (e^3 (2 B d-5 A e) b^4-8 c d e^2 (B d-2 A e) b^3+2 c^2 d^2 e (6 B d-7 A e) b^2-c^3 d^3 (B d+4 A e) b+2 A c^4 d^4+c \left (e^2 (2 B d-5 A e) b^3-c d e (6 B d-11 A e) b^2-c^2 d^2 (B d+3 A e) b+2 A c^3 d^3\right ) (d+e x)\right )}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{d (c d-b e)}-\frac {2 e \left (b^3 e^2 (2 B d-5 A e)-b^2 c d e (6 B d-11 A e)-b c^2 d^2 (3 A e+B d)+2 A c^3 d^3\right )}{d \sqrt {d+e x} (c d-b e)}}{d (c d-b e)}-\frac {2 e \left (b^2 (-e) (2 B d-5 A e)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )}{3 d (d+e x)^{3/2} (c d-b e)}}{2 b^2 d (c d-b e)}-\frac {c x (2 A c d-b (A e+B d))+A b (c d-b e)}{b^2 d \left (b x+c x^2\right ) (d+e x)^{3/2} (c d-b e)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {-\frac {2 \int \frac {e \left (e^3 (2 B d-5 A e) b^4-8 c d e^2 (B d-2 A e) b^3+2 c^2 d^2 e (6 B d-7 A e) b^2-c^3 d^3 (B d+4 A e) b+2 A c^4 d^4+c \left (e^2 (2 B d-5 A e) b^3-c d e (6 B d-11 A e) b^2-c^2 d^2 (B d+3 A e) b+2 A c^3 d^3\right ) (d+e x)\right )}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{d (c d-b e)}-\frac {2 e \left (b^3 e^2 (2 B d-5 A e)-b^2 c d e (6 B d-11 A e)-b c^2 d^2 (3 A e+B d)+2 A c^3 d^3\right )}{d \sqrt {d+e x} (c d-b e)}}{d (c d-b e)}-\frac {2 e \left (b^2 (-e) (2 B d-5 A e)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )}{3 d (d+e x)^{3/2} (c d-b e)}}{2 b^2 d (c d-b e)}-\frac {c x (2 A c d-b (A e+B d))+A b (c d-b e)}{b^2 d \left (b x+c x^2\right ) (d+e x)^{3/2} (c d-b e)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {-\frac {2 e \int \frac {e^3 (2 B d-5 A e) b^4-8 c d e^2 (B d-2 A e) b^3+2 c^2 d^2 e (6 B d-7 A e) b^2-c^3 d^3 (B d+4 A e) b+2 A c^4 d^4+c \left (e^2 (2 B d-5 A e) b^3-c d e (6 B d-11 A e) b^2-c^2 d^2 (B d+3 A e) b+2 A c^3 d^3\right ) (d+e x)}{c (d+e x)^2-(2 c d-b e) (d+e x)+d (c d-b e)}d\sqrt {d+e x}}{d (c d-b e)}-\frac {2 e \left (b^3 e^2 (2 B d-5 A e)-b^2 c d e (6 B d-11 A e)-b c^2 d^2 (3 A e+B d)+2 A c^3 d^3\right )}{d \sqrt {d+e x} (c d-b e)}}{d (c d-b e)}-\frac {2 e \left (b^2 (-e) (2 B d-5 A e)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )}{3 d (d+e x)^{3/2} (c d-b e)}}{2 b^2 d (c d-b e)}-\frac {c x (2 A c d-b (A e+B d))+A b (c d-b e)}{b^2 d \left (b x+c x^2\right ) (d+e x)^{3/2} (c d-b e)}\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle \frac {\frac {-\frac {2 e \left (\frac {c^3 d^3 \left (9 A b c e-4 A c^2 d-7 b^2 B e+2 b B c d\right ) \int \frac {1}{-c d+b e+c (d+e x)}d\sqrt {d+e x}}{b e}-\frac {c (c d-b e)^3 (-5 A b e-4 A c d+2 b B d) \int \frac {1}{c (d+e x)-c d}d\sqrt {d+e x}}{b e}\right )}{d (c d-b e)}-\frac {2 e \left (b^3 e^2 (2 B d-5 A e)-b^2 c d e (6 B d-11 A e)-b c^2 d^2 (3 A e+B d)+2 A c^3 d^3\right )}{d \sqrt {d+e x} (c d-b e)}}{d (c d-b e)}-\frac {2 e \left (b^2 (-e) (2 B d-5 A e)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )}{3 d (d+e x)^{3/2} (c d-b e)}}{2 b^2 d (c d-b e)}-\frac {c x (2 A c d-b (A e+B d))+A b (c d-b e)}{b^2 d \left (b x+c x^2\right ) (d+e x)^{3/2} (c d-b e)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {-\frac {2 e \left (\frac {\text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) (c d-b e)^3 (-5 A b e-4 A c d+2 b B d)}{b \sqrt {d} e}-\frac {c^{5/2} d^3 \left (9 A b c e-4 A c^2 d-7 b^2 B e+2 b B c d\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d+e x}}{\sqrt {c d-b e}}\right )}{b e \sqrt {c d-b e}}\right )}{d (c d-b e)}-\frac {2 e \left (b^3 e^2 (2 B d-5 A e)-b^2 c d e (6 B d-11 A e)-b c^2 d^2 (3 A e+B d)+2 A c^3 d^3\right )}{d \sqrt {d+e x} (c d-b e)}}{d (c d-b e)}-\frac {2 e \left (b^2 (-e) (2 B d-5 A e)-3 b c d (2 A e+B d)+6 A c^2 d^2\right )}{3 d (d+e x)^{3/2} (c d-b e)}}{2 b^2 d (c d-b e)}-\frac {c x (2 A c d-b (A e+B d))+A b (c d-b e)}{b^2 d \left (b x+c x^2\right ) (d+e x)^{3/2} (c d-b e)}\) |
Input:
Int[(A + B*x)/((d + e*x)^(5/2)*(b*x + c*x^2)^2),x]
Output:
-((A*b*(c*d - b*e) + c*(2*A*c*d - b*(B*d + A*e))*x)/(b^2*d*(c*d - b*e)*(d + e*x)^(3/2)*(b*x + c*x^2))) + ((-2*e*(6*A*c^2*d^2 - b^2*e*(2*B*d - 5*A*e) - 3*b*c*d*(B*d + 2*A*e)))/(3*d*(c*d - b*e)*(d + e*x)^(3/2)) + ((-2*e*(2*A *c^3*d^3 - b^2*c*d*e*(6*B*d - 11*A*e) + b^3*e^2*(2*B*d - 5*A*e) - b*c^2*d^ 2*(B*d + 3*A*e)))/(d*(c*d - b*e)*Sqrt[d + e*x]) - (2*e*(((c*d - b*e)^3*(2* b*B*d - 4*A*c*d - 5*A*b*e)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(b*Sqrt[d]*e) - (c^(5/2)*d^3*(2*b*B*c*d - 4*A*c^2*d - 7*b^2*B*e + 9*A*b*c*e)*ArcTanh[(Sqr t[c]*Sqrt[d + e*x])/Sqrt[c*d - b*e]])/(b*e*Sqrt[c*d - b*e])))/(d*(c*d - b* e)))/(d*(c*d - b*e)))/(2*b^2*d*(c*d - b*e))
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_)^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_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2 *a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^m *(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p] )
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 = 1.68 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(2 e^{2} \left (\frac {c^{3} \left (\frac {\left (\frac {1}{2} A c e b -\frac {1}{2} b^{2} B e \right ) \sqrt {e x +d}}{\left (e x +d \right ) c +b e -c d}+\frac {\left (9 A c e b -4 A \,c^{2} d -7 b^{2} B e +2 B b c d \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{2 \sqrt {c \left (b e -c d \right )}}\right )}{\left (b e -c d \right )^{3} e^{2} b^{3}}+\frac {-\frac {A b \sqrt {e x +d}}{2 x}+\frac {\left (5 A b e +4 A c d -2 B b d \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{2 \sqrt {d}}}{b^{3} e^{2} d^{3}}-\frac {A e -B d}{3 d^{2} \left (b e -c d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 A b \,e^{2}-4 A c d e -B b d e +3 B c \,d^{2}}{d^{3} \left (b e -c d \right )^{3} \sqrt {e x +d}}\right )\) | \(267\) |
default | \(2 e^{2} \left (\frac {c^{3} \left (\frac {\left (\frac {1}{2} A c e b -\frac {1}{2} b^{2} B e \right ) \sqrt {e x +d}}{\left (e x +d \right ) c +b e -c d}+\frac {\left (9 A c e b -4 A \,c^{2} d -7 b^{2} B e +2 B b c d \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{2 \sqrt {c \left (b e -c d \right )}}\right )}{\left (b e -c d \right )^{3} e^{2} b^{3}}+\frac {-\frac {A b \sqrt {e x +d}}{2 x}+\frac {\left (5 A b e +4 A c d -2 B b d \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{2 \sqrt {d}}}{b^{3} e^{2} d^{3}}-\frac {A e -B d}{3 d^{2} \left (b e -c d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 A b \,e^{2}-4 A c d e -B b d e +3 B c \,d^{2}}{d^{3} \left (b e -c d \right )^{3} \sqrt {e x +d}}\right )\) | \(267\) |
risch | \(-\frac {A \sqrt {e x +d}}{d^{3} b^{2} x}-\frac {e \left (-\frac {\left (5 A b e +4 A c d -2 B b d \right ) \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )}{b e \sqrt {d}}-\frac {2 d^{3} c^{3} \left (\frac {\left (\frac {1}{2} A c e b -\frac {1}{2} b^{2} B e \right ) \sqrt {e x +d}}{\left (e x +d \right ) c +b e -c d}+\frac {\left (9 A c e b -4 A \,c^{2} d -7 b^{2} B e +2 B b c d \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right )}{2 \sqrt {c \left (b e -c d \right )}}\right )}{\left (b e -c d \right )^{3} b e}+\frac {2 e \,b^{2} \left (2 A b \,e^{2}-4 A c d e -B b d e +3 B c \,d^{2}\right )}{\left (b e -c d \right )^{3} \sqrt {e x +d}}+\frac {2 d e \,b^{2} \left (A e -B d \right )}{3 \left (b e -c d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}\right )}{b^{2} d^{3}}\) | \(279\) |
pseudoelliptic | \(\frac {\left (e x +d \right )^{\frac {3}{2}} \left (c x +b \right ) \left (9 A c e b -4 A \,c^{2} d -7 b^{2} B e +2 B b c d \right ) \arctan \left (\frac {c \sqrt {e x +d}}{\sqrt {c \left (b e -c d \right )}}\right ) c^{3} d^{\frac {7}{2}} x +\left (\left (5 A b e +4 A c d -2 B b d \right ) x \left (c x +b \right ) \left (e x +d \right )^{\frac {3}{2}} \left (b e -c d \right )^{3} \operatorname {arctanh}\left (\frac {\sqrt {e x +d}}{\sqrt {d}}\right )+\sqrt {d}\, b \left (2 A \,d^{3} x \left (e x +d \right )^{2} c^{4}+\left (d \left (-B x +A \right )-3 A e x \right ) d^{2} \left (e x +d \right )^{2} b \,c^{3}-3 e d \left (A \,d^{3}+e x \left (\frac {20 B x}{9}+A \right ) d^{2}-\frac {35 e^{2} \left (-\frac {18 B x}{35}+A \right ) x^{2} d}{9}-\frac {11 A \,e^{3} x^{3}}{3}\right ) b^{2} c^{2}+3 e^{2} b^{3} \left (\left (-\frac {20 B x}{9}+A \right ) d^{3}+\frac {41 \left (-\frac {10 B x}{41}+A \right ) e x \,d^{2}}{9}+\frac {13 e^{2} x^{2} \left (\frac {6 B x}{13}+A \right ) d}{9}-\frac {5 A \,e^{3} x^{3}}{3}\right ) c -e^{3} \left (\left (-\frac {8 B x}{3}+A \right ) d^{2}+\frac {20 e x \left (-\frac {3 B x}{10}+A \right ) d}{3}+5 A \,e^{2} x^{2}\right ) b^{4}\right )\right ) \sqrt {c \left (b e -c d \right )}}{\sqrt {c \left (b e -c d \right )}\, \left (e x +d \right )^{\frac {3}{2}} x \,d^{\frac {7}{2}} b^{3} \left (c x +b \right ) \left (b e -c d \right )^{3}}\) | \(383\) |
Input:
int((B*x+A)/(e*x+d)^(5/2)/(c*x^2+b*x)^2,x,method=_RETURNVERBOSE)
Output:
2*e^2*(c^3/(b*e-c*d)^3/e^2/b^3*((1/2*A*c*e*b-1/2*b^2*B*e)*(e*x+d)^(1/2)/(( e*x+d)*c+b*e-c*d)+1/2*(9*A*b*c*e-4*A*c^2*d-7*B*b^2*e+2*B*b*c*d)/(c*(b*e-c* d))^(1/2)*arctan(c*(e*x+d)^(1/2)/(c*(b*e-c*d))^(1/2)))+1/b^3/e^2/d^3*(-1/2 *A*b*(e*x+d)^(1/2)/x+1/2*(5*A*b*e+4*A*c*d-2*B*b*d)/d^(1/2)*arctanh((e*x+d) ^(1/2)/d^(1/2)))-1/3*(A*e-B*d)/d^2/(b*e-c*d)^2/(e*x+d)^(3/2)-(2*A*b*e^2-4* A*c*d*e-B*b*d*e+3*B*c*d^2)/d^3/(b*e-c*d)^3/(e*x+d)^(1/2))
Leaf count of result is larger than twice the leaf count of optimal. 1386 vs. \(2 (325) = 650\).
Time = 25.46 (sec) , antiderivative size = 5620, normalized size of antiderivative = 16.01 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:
integrate((B*x+A)/(e*x+d)^(5/2)/(c*x^2+b*x)^2,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )^2} \, dx=\text {Timed out} \] Input:
integrate((B*x+A)/(e*x+d)**(5/2)/(c*x**2+b*x)**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((B*x+A)/(e*x+d)^(5/2)/(c*x^2+b*x)^2,x, algorithm="maxima")
Output:
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 = 0.27 (sec) , antiderivative size = 589, normalized size of antiderivative = 1.68 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )^2} \, dx=-\frac {{\left (2 \, B b c^{4} d - 4 \, A c^{5} d - 7 \, B b^{2} c^{3} e + 9 \, A b c^{4} e\right )} \arctan \left (\frac {\sqrt {e x + d} c}{\sqrt {-c^{2} d + b c e}}\right )}{{\left (b^{3} c^{3} d^{3} - 3 \, b^{4} c^{2} d^{2} e + 3 \, b^{5} c d e^{2} - b^{6} e^{3}\right )} \sqrt {-c^{2} d + b c e}} + \frac {{\left (e x + d\right )}^{\frac {3}{2}} B b c^{3} d^{3} e - 2 \, {\left (e x + d\right )}^{\frac {3}{2}} A c^{4} d^{3} e - \sqrt {e x + d} B b c^{3} d^{4} e + 2 \, \sqrt {e x + d} A c^{4} d^{4} e + 3 \, {\left (e x + d\right )}^{\frac {3}{2}} A b c^{3} d^{2} e^{2} - 4 \, \sqrt {e x + d} A b c^{3} d^{3} e^{2} - 3 \, {\left (e x + d\right )}^{\frac {3}{2}} A b^{2} c^{2} d e^{3} + 6 \, \sqrt {e x + d} A b^{2} c^{2} d^{2} e^{3} + {\left (e x + d\right )}^{\frac {3}{2}} A b^{3} c e^{4} - 4 \, \sqrt {e x + d} A b^{3} c d e^{4} + \sqrt {e x + d} A b^{4} e^{5}}{{\left (b^{2} c^{3} d^{6} - 3 \, b^{3} c^{2} d^{5} e + 3 \, b^{4} c d^{4} e^{2} - b^{5} d^{3} e^{3}\right )} {\left ({\left (e x + d\right )}^{2} c - 2 \, {\left (e x + d\right )} c d + c d^{2} + {\left (e x + d\right )} b e - b d e\right )}} + \frac {2 \, {\left (9 \, {\left (e x + d\right )} B c d^{2} e^{2} + B c d^{3} e^{2} - 3 \, {\left (e x + d\right )} B b d e^{3} - 12 \, {\left (e x + d\right )} A c d e^{3} - B b d^{2} e^{3} - A c d^{2} e^{3} + 6 \, {\left (e x + d\right )} A b e^{4} + A b d e^{4}\right )}}{3 \, {\left (c^{3} d^{6} - 3 \, b c^{2} d^{5} e + 3 \, b^{2} c d^{4} e^{2} - b^{3} d^{3} e^{3}\right )} {\left (e x + d\right )}^{\frac {3}{2}}} + \frac {{\left (2 \, B b d - 4 \, A c d - 5 \, A b e\right )} \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{b^{3} \sqrt {-d} d^{3}} \] Input:
integrate((B*x+A)/(e*x+d)^(5/2)/(c*x^2+b*x)^2,x, algorithm="giac")
Output:
-(2*B*b*c^4*d - 4*A*c^5*d - 7*B*b^2*c^3*e + 9*A*b*c^4*e)*arctan(sqrt(e*x + d)*c/sqrt(-c^2*d + b*c*e))/((b^3*c^3*d^3 - 3*b^4*c^2*d^2*e + 3*b^5*c*d*e^ 2 - b^6*e^3)*sqrt(-c^2*d + b*c*e)) + ((e*x + d)^(3/2)*B*b*c^3*d^3*e - 2*(e *x + d)^(3/2)*A*c^4*d^3*e - sqrt(e*x + d)*B*b*c^3*d^4*e + 2*sqrt(e*x + d)* A*c^4*d^4*e + 3*(e*x + d)^(3/2)*A*b*c^3*d^2*e^2 - 4*sqrt(e*x + d)*A*b*c^3* d^3*e^2 - 3*(e*x + d)^(3/2)*A*b^2*c^2*d*e^3 + 6*sqrt(e*x + d)*A*b^2*c^2*d^ 2*e^3 + (e*x + d)^(3/2)*A*b^3*c*e^4 - 4*sqrt(e*x + d)*A*b^3*c*d*e^4 + sqrt (e*x + d)*A*b^4*e^5)/((b^2*c^3*d^6 - 3*b^3*c^2*d^5*e + 3*b^4*c*d^4*e^2 - b ^5*d^3*e^3)*((e*x + d)^2*c - 2*(e*x + d)*c*d + c*d^2 + (e*x + d)*b*e - b*d *e)) + 2/3*(9*(e*x + d)*B*c*d^2*e^2 + B*c*d^3*e^2 - 3*(e*x + d)*B*b*d*e^3 - 12*(e*x + d)*A*c*d*e^3 - B*b*d^2*e^3 - A*c*d^2*e^3 + 6*(e*x + d)*A*b*e^4 + A*b*d*e^4)/((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)^(3/2)) + (2*B*b*d - 4*A*c*d - 5*A*b*e)*arctan(sqrt(e*x + d)/sqrt( -d))/(b^3*sqrt(-d)*d^3)
Time = 15.37 (sec) , antiderivative size = 18450, normalized size of antiderivative = 52.56 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:
int((A + B*x)/((b*x + c*x^2)^2*(d + e*x)^(5/2)),x)
Output:
atan((A^2*c^13*d^12*(-(16*A^2*c^9*d^2 + 81*A^2*b^2*c^7*e^2 + 4*B^2*b^2*c^7 *d^2 + 49*B^2*b^4*c^5*e^2 - 126*A*B*b^3*c^6*e^2 - 28*B^2*b^3*c^6*d*e - 16* A*B*b*c^8*d^2 - 72*A^2*b*c^8*d*e + 92*A*B*b^2*c^7*d*e)/(b^13*e^7 - b^6*c^7 *d^7 + 7*b^7*c^6*d^6*e - 21*b^8*c^5*d^5*e^2 + 35*b^9*c^4*d^4*e^3 - 35*b^10 *c^3*d^3*e^4 + 21*b^11*c^2*d^2*e^5 - 7*b^12*c*d*e^6))^(1/2)*(d + e*x)^(1/2 )*32i - b^6*c^11*d^17*(-(16*A^2*c^9*d^2 + 81*A^2*b^2*c^7*e^2 + 4*B^2*b^2*c ^7*d^2 + 49*B^2*b^4*c^5*e^2 - 126*A*B*b^3*c^6*e^2 - 28*B^2*b^3*c^6*d*e - 1 6*A*B*b*c^8*d^2 - 72*A^2*b*c^8*d*e + 92*A*B*b^2*c^7*d*e)/(b^13*e^7 - b^6*c ^7*d^7 + 7*b^7*c^6*d^6*e - 21*b^8*c^5*d^5*e^2 + 35*b^9*c^4*d^4*e^3 - 35*b^ 10*c^3*d^3*e^4 + 21*b^11*c^2*d^2*e^5 - 7*b^12*c*d*e^6))^(3/2)*(d + e*x)^(1 /2)*2i + b^17*d^6*e^11*(-(16*A^2*c^9*d^2 + 81*A^2*b^2*c^7*e^2 + 4*B^2*b^2* c^7*d^2 + 49*B^2*b^4*c^5*e^2 - 126*A*B*b^3*c^6*e^2 - 28*B^2*b^3*c^6*d*e - 16*A*B*b*c^8*d^2 - 72*A^2*b*c^8*d*e + 92*A*B*b^2*c^7*d*e)/(b^13*e^7 - b^6* c^7*d^7 + 7*b^7*c^6*d^6*e - 21*b^8*c^5*d^5*e^2 + 35*b^9*c^4*d^4*e^3 - 35*b ^10*c^3*d^3*e^4 + 21*b^11*c^2*d^2*e^5 - 7*b^12*c*d*e^6))^(3/2)*(d + e*x)^( 1/2)*1i + B^2*b^2*c^11*d^12*(-(16*A^2*c^9*d^2 + 81*A^2*b^2*c^7*e^2 + 4*B^2 *b^2*c^7*d^2 + 49*B^2*b^4*c^5*e^2 - 126*A*B*b^3*c^6*e^2 - 28*B^2*b^3*c^6*d *e - 16*A*B*b*c^8*d^2 - 72*A^2*b*c^8*d*e + 92*A*B*b^2*c^7*d*e)/(b^13*e^7 - b^6*c^7*d^7 + 7*b^7*c^6*d^6*e - 21*b^8*c^5*d^5*e^2 + 35*b^9*c^4*d^4*e^3 - 35*b^10*c^3*d^3*e^4 + 21*b^11*c^2*d^2*e^5 - 7*b^12*c*d*e^6))^(1/2)*(d ...
Time = 0.28 (sec) , antiderivative size = 3890, normalized size of antiderivative = 11.08 \[ \int \frac {A+B x}{(d+e x)^{5/2} \left (b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:
int((B*x+A)/(e*x+d)^(5/2)/(c*x^2+b*x)^2,x)
Output:
(54*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)* sqrt(b*e - c*d)))*a*b**2*c**3*d**5*e*x + 54*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*a*b**2*c**3*d**4 *e**2*x**2 - 24*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)* c)/(sqrt(c)*sqrt(b*e - c*d)))*a*b*c**4*d**6*x + 30*sqrt(c)*sqrt(d + e*x)*s qrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*a*b*c**4* d**5*e*x**2 + 54*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x) *c)/(sqrt(c)*sqrt(b*e - c*d)))*a*b*c**4*d**4*e**2*x**3 - 24*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))* a*c**5*d**6*x**2 - 24*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*a*c**5*d**5*e*x**3 - 42*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))* b**4*c**2*d**5*e*x - 42*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b**4*c**2*d**4*e**2*x**2 + 12*sqrt(c )*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b**3*c**3*d**6*x - 30*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan(( sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b**3*c**3*d**5*e*x**2 - 42*sqr t(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b* e - c*d)))*b**3*c**3*d**4*e**2*x**3 + 12*sqrt(c)*sqrt(d + e*x)*sqrt(b*e - c*d)*atan((sqrt(d + e*x)*c)/(sqrt(c)*sqrt(b*e - c*d)))*b**2*c**4*d**6*x...