Integrand size = 21, antiderivative size = 236 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2}{3 b d \left (b x+c x^2\right )^{3/2}}+\frac {2 (2 c d+b e) x}{b^2 d^2 \left (b x+c x^2\right )^{3/2}}+\frac {2 c \left (8 c^2 d^2-4 b c d e-3 b^2 e^2\right ) x^2}{3 b^3 d^2 (c d-b e) \left (b x+c x^2\right )^{3/2}}+\frac {2 c (2 c d-b e) \left (8 c^2 d^2-8 b c d e-3 b^2 e^2\right ) x}{3 b^4 d^2 (c d-b e)^2 \sqrt {b x+c x^2}}+\frac {2 e^4 \text {arctanh}\left (\frac {\sqrt {c d-b e} x}{\sqrt {d} \sqrt {b x+c x^2}}\right )}{d^{5/2} (c d-b e)^{5/2}} \] Output:
-2/3/b/d/(c*x^2+b*x)^(3/2)+2*(b*e+2*c*d)*x/b^2/d^2/(c*x^2+b*x)^(3/2)+2/3*c *(-3*b^2*e^2-4*b*c*d*e+8*c^2*d^2)*x^2/b^3/d^2/(-b*e+c*d)/(c*x^2+b*x)^(3/2) +2/3*c*(-b*e+2*c*d)*(-3*b^2*e^2-8*b*c*d*e+8*c^2*d^2)*x/b^4/d^2/(-b*e+c*d)^ 2/(c*x^2+b*x)^(1/2)+2*e^4*arctanh((-b*e+c*d)^(1/2)*x/d^(1/2)/(c*x^2+b*x)^( 1/2))/d^(5/2)/(-b*e+c*d)^(5/2)
Time = 0.83 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{5/2}} \, dx=\frac {2 \sqrt {d} \sqrt {-c d+b e} \left (16 c^5 d^3 x^3+24 b c^4 d^2 x^2 (d-e x)+b^5 e^2 (-d+3 e x)+2 b^2 c^3 d x \left (3 d^2-18 d e x+e^2 x^2\right )+2 b^4 c e \left (d^2+3 e^2 x^2\right )-b^3 c^2 \left (d^3+9 d^2 e x-3 d e^2 x^2-3 e^3 x^3\right )\right )-6 b^4 e^4 x^{3/2} (b+c x)^{3/2} \arctan \left (\frac {-e \sqrt {x} \sqrt {b+c x}+\sqrt {c} (d+e x)}{\sqrt {d} \sqrt {-c d+b e}}\right )}{3 b^4 d^{5/2} (-c d+b e)^{5/2} (x (b+c x))^{3/2}} \] Input:
Integrate[1/((d + e*x)*(b*x + c*x^2)^(5/2)),x]
Output:
(2*Sqrt[d]*Sqrt[-(c*d) + b*e]*(16*c^5*d^3*x^3 + 24*b*c^4*d^2*x^2*(d - e*x) + b^5*e^2*(-d + 3*e*x) + 2*b^2*c^3*d*x*(3*d^2 - 18*d*e*x + e^2*x^2) + 2*b ^4*c*e*(d^2 + 3*e^2*x^2) - b^3*c^2*(d^3 + 9*d^2*e*x - 3*d*e^2*x^2 - 3*e^3* x^3)) - 6*b^4*e^4*x^(3/2)*(b + c*x)^(3/2)*ArcTan[(-(e*Sqrt[x]*Sqrt[b + c*x ]) + Sqrt[c]*(d + e*x))/(Sqrt[d]*Sqrt[-(c*d) + b*e])])/(3*b^4*d^(5/2)*(-(c *d) + b*e)^(5/2)*(x*(b + c*x))^(3/2))
Time = 0.73 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1165, 27, 1235, 27, 1154, 219}
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.
| \(\displaystyle \int \frac {1}{\left (b x+c x^2\right )^{5/2} (d+e x)} \, dx\) |
| \(\Big \downarrow \) 1165 |
| \(\displaystyle -\frac {2 \int \frac {8 c^2 d^2-4 b c e d-3 b^2 e^2+4 c e (2 c d-b e) x}{2 (d+e x) \left (c x^2+b x\right )^{3/2}}dx}{3 b^2 d (c d-b e)}-\frac {2 (c x (2 c d-b e)+b (c d-b e))}{3 b^2 d \left (b x+c x^2\right )^{3/2} (c d-b e)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {8 c^2 d^2-4 b c e d-3 b^2 e^2+4 c e (2 c d-b e) x}{(d+e x) \left (c x^2+b x\right )^{3/2}}dx}{3 b^2 d (c d-b e)}-\frac {2 (c x (2 c d-b e)+b (c d-b e))}{3 b^2 d \left (b x+c x^2\right )^{3/2} (c d-b e)}\) |
| \(\Big \downarrow \) 1235 |
| \(\displaystyle -\frac {-\frac {2 \int \frac {3 b^4 e^4}{2 (d+e x) \sqrt {c x^2+b x}}dx}{b^2 d (c d-b e)}-\frac {2 \left (c x (2 c d-b e) \left (-3 b^2 e^2-8 b c d e+8 c^2 d^2\right )+b (c d-b e) \left (-3 b^2 e^2-4 b c d e+8 c^2 d^2\right )\right )}{b^2 d \sqrt {b x+c x^2} (c d-b e)}}{3 b^2 d (c d-b e)}-\frac {2 (c x (2 c d-b e)+b (c d-b e))}{3 b^2 d \left (b x+c x^2\right )^{3/2} (c d-b e)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {3 b^2 e^4 \int \frac {1}{(d+e x) \sqrt {c x^2+b x}}dx}{d (c d-b e)}-\frac {2 \left (c x (2 c d-b e) \left (-3 b^2 e^2-8 b c d e+8 c^2 d^2\right )+b (c d-b e) \left (-3 b^2 e^2-4 b c d e+8 c^2 d^2\right )\right )}{b^2 d \sqrt {b x+c x^2} (c d-b e)}}{3 b^2 d (c d-b e)}-\frac {2 (c x (2 c d-b e)+b (c d-b e))}{3 b^2 d \left (b x+c x^2\right )^{3/2} (c d-b e)}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle -\frac {\frac {6 b^2 e^4 \int \frac {1}{4 d (c d-b e)-\frac {(b d+(2 c d-b e) x)^2}{c x^2+b x}}d\left (-\frac {b d+(2 c d-b e) x}{\sqrt {c x^2+b x}}\right )}{d (c d-b e)}-\frac {2 \left (c x (2 c d-b e) \left (-3 b^2 e^2-8 b c d e+8 c^2 d^2\right )+b (c d-b e) \left (-3 b^2 e^2-4 b c d e+8 c^2 d^2\right )\right )}{b^2 d \sqrt {b x+c x^2} (c d-b e)}}{3 b^2 d (c d-b e)}-\frac {2 (c x (2 c d-b e)+b (c d-b e))}{3 b^2 d \left (b x+c x^2\right )^{3/2} (c d-b e)}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {-\frac {3 b^2 e^4 \text {arctanh}\left (\frac {x (2 c d-b e)+b d}{2 \sqrt {d} \sqrt {b x+c x^2} \sqrt {c d-b e}}\right )}{d^{3/2} (c d-b e)^{3/2}}-\frac {2 \left (c x (2 c d-b e) \left (-3 b^2 e^2-8 b c d e+8 c^2 d^2\right )+b (c d-b e) \left (-3 b^2 e^2-4 b c d e+8 c^2 d^2\right )\right )}{b^2 d \sqrt {b x+c x^2} (c d-b e)}}{3 b^2 d (c d-b e)}-\frac {2 (c x (2 c d-b e)+b (c d-b e))}{3 b^2 d \left (b x+c x^2\right )^{3/2} (c d-b e)}\) |
Input:
Int[1/((d + e*x)*(b*x + c*x^2)^(5/2)),x]
Output:
(-2*(b*(c*d - b*e) + c*(2*c*d - b*e)*x))/(3*b^2*d*(c*d - b*e)*(b*x + c*x^2 )^(3/2)) - ((-2*(b*(c*d - b*e)*(8*c^2*d^2 - 4*b*c*d*e - 3*b^2*e^2) + c*(2* c*d - b*e)*(8*c^2*d^2 - 8*b*c*d*e - 3*b^2*e^2)*x))/(b^2*d*(c*d - b*e)*Sqrt [b*x + c*x^2]) - (3*b^2*e^4*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqrt[d]*Sqr t[c*d - b*e]*Sqrt[b*x + c*x^2])])/(d^(3/2)*(c*d - b*e)^(3/2)))/(3*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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*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*Simp[b*c*d*e*(2*p - m + 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]
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] )
Time = 0.93 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.92
| method | result | size |
| pseudoelliptic | \(-\frac {2 \left (3 b^{4} e^{4} x \sqrt {x \left (c x +b \right )}\, \left (c x +b \right ) \arctan \left (\frac {\sqrt {x \left (c x +b \right )}\, d}{x \sqrt {d \left (b e -c d \right )}}\right )+\sqrt {d \left (b e -c d \right )}\, \left (c^{2} \left (2 c x +b \right ) \left (-8 c^{2} x^{2}-8 c b x +b^{2}\right ) d^{3}-2 e c b \left (-12 c^{3} x^{3}-18 b \,c^{2} x^{2}-\frac {9}{2} b^{2} c x +b^{3}\right ) d^{2}+b^{2} e^{2} \left (-2 c x +b \right ) \left (c x +b \right )^{2} d -3 b^{3} e^{3} x \left (c x +b \right )^{2}\right )\right )}{3 \sqrt {x \left (c x +b \right )}\, \sqrt {d \left (b e -c d \right )}\, x \,d^{2} \left (b e -c d \right )^{2} \left (c x +b \right ) b^{4}}\) | \(218\) |
| risch | \(-\frac {2 \left (c x +b \right ) \left (-3 b e x -8 c d x +b d \right )}{3 b^{4} d^{2} \sqrt {x \left (c x +b \right )}\, x}+\frac {-\frac {e^{3} b^{3} \ln \left (\frac {-\frac {2 d \left (b e -c d \right )}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (b e -c d \right )^{2} \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}}-\frac {b c \,d^{2} \left (\frac {2 \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}}{3 b \left (\frac {b}{c}+x \right )^{2}}+\frac {4 c \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}}{3 b^{2} \left (\frac {b}{c}+x \right )}\right )}{b e -c d}-\frac {2 c^{2} d^{2} \left (3 b e -2 c d \right ) \sqrt {c \left (\frac {b}{c}+x \right )^{2}-\left (\frac {b}{c}+x \right ) b}}{\left (b e -c d \right )^{2} b \left (\frac {b}{c}+x \right )}}{b^{3} d^{2}}\) | \(351\) |
| default | \(\frac {-\frac {e^{2}}{3 d \left (b e -c d \right ) \left (c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}\right )^{\frac {3}{2}}}+\frac {\left (b e -2 c d \right ) e \left (\frac {\frac {4 c \left (x +\frac {d}{e}\right )}{3}+\frac {2 \left (b e -2 c d \right )}{3 e}}{\left (-\frac {4 c d \left (b e -c d \right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \left (c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}\right )^{\frac {3}{2}}}+\frac {16 c \left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right )}{3 \left (-\frac {4 c d \left (b e -c d \right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right )^{2} \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}\right )}{2 d \left (b e -c d \right )}-\frac {e^{2} \left (-\frac {e^{2}}{d \left (b e -c d \right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}+\frac {\left (b e -2 c d \right ) e \left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right )}{d \left (b e -c d \right ) \left (-\frac {4 c d \left (b e -c d \right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}+\frac {e^{2} \ln \left (\frac {-\frac {2 d \left (b e -c d \right )}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}-\frac {d \left (b e -c d \right )}{e^{2}}}}{x +\frac {d}{e}}\right )}{d \left (b e -c d \right ) \sqrt {-\frac {d \left (b e -c d \right )}{e^{2}}}}\right )}{d \left (b e -c d \right )}}{e}\) | \(660\) |
Input:
int(1/(e*x+d)/(c*x^2+b*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/3/(x*(c*x+b))^(1/2)/(d*(b*e-c*d))^(1/2)*(3*b^4*e^4*x*(x*(c*x+b))^(1/2)* (c*x+b)*arctan((x*(c*x+b))^(1/2)/x*d/(d*(b*e-c*d))^(1/2))+(d*(b*e-c*d))^(1 /2)*(c^2*(2*c*x+b)*(-8*c^2*x^2-8*b*c*x+b^2)*d^3-2*e*c*b*(-12*c^3*x^3-18*b* c^2*x^2-9/2*b^2*c*x+b^3)*d^2+b^2*e^2*(-2*c*x+b)*(c*x+b)^2*d-3*b^3*e^3*x*(c *x+b)^2))/x/d^2/(b*e-c*d)^2/(c*x+b)/b^4
Leaf count of result is larger than twice the leaf count of optimal. 498 vs. \(2 (212) = 424\).
Time = 0.15 (sec) , antiderivative size = 1012, normalized size of antiderivative = 4.29 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate(1/(e*x+d)/(c*x^2+b*x)^(5/2),x, algorithm="fricas")
Output:
[1/3*(3*(b^4*c^2*e^4*x^4 + 2*b^5*c*e^4*x^3 + b^6*e^4*x^2)*sqrt(c*d^2 - b*d *e)*log((b*d + (2*c*d - b*e)*x + 2*sqrt(c*d^2 - b*d*e)*sqrt(c*x^2 + b*x))/ (e*x + d)) - 2*(b^3*c^3*d^5 - 3*b^4*c^2*d^4*e + 3*b^5*c*d^3*e^2 - b^6*d^2* e^3 - (16*c^6*d^5 - 40*b*c^5*d^4*e + 26*b^2*c^4*d^3*e^2 + b^3*c^3*d^2*e^3 - 3*b^4*c^2*d*e^4)*x^3 - 3*(8*b*c^5*d^5 - 20*b^2*c^4*d^4*e + 13*b^3*c^3*d^ 3*e^2 + b^4*c^2*d^2*e^3 - 2*b^5*c*d*e^4)*x^2 - 3*(2*b^2*c^4*d^5 - 5*b^3*c^ 3*d^4*e + 3*b^4*c^2*d^3*e^2 + b^5*c*d^2*e^3 - b^6*d*e^4)*x)*sqrt(c*x^2 + b *x))/((b^4*c^5*d^6 - 3*b^5*c^4*d^5*e + 3*b^6*c^3*d^4*e^2 - b^7*c^2*d^3*e^3 )*x^4 + 2*(b^5*c^4*d^6 - 3*b^6*c^3*d^5*e + 3*b^7*c^2*d^4*e^2 - b^8*c*d^3*e ^3)*x^3 + (b^6*c^3*d^6 - 3*b^7*c^2*d^5*e + 3*b^8*c*d^4*e^2 - b^9*d^3*e^3)* x^2), -2/3*(3*(b^4*c^2*e^4*x^4 + 2*b^5*c*e^4*x^3 + b^6*e^4*x^2)*sqrt(-c*d^ 2 + b*d*e)*arctan(sqrt(-c*d^2 + b*d*e)*sqrt(c*x^2 + b*x)/(c*d*x + b*d)) + (b^3*c^3*d^5 - 3*b^4*c^2*d^4*e + 3*b^5*c*d^3*e^2 - b^6*d^2*e^3 - (16*c^6*d ^5 - 40*b*c^5*d^4*e + 26*b^2*c^4*d^3*e^2 + b^3*c^3*d^2*e^3 - 3*b^4*c^2*d*e ^4)*x^3 - 3*(8*b*c^5*d^5 - 20*b^2*c^4*d^4*e + 13*b^3*c^3*d^3*e^2 + b^4*c^2 *d^2*e^3 - 2*b^5*c*d*e^4)*x^2 - 3*(2*b^2*c^4*d^5 - 5*b^3*c^3*d^4*e + 3*b^4 *c^2*d^3*e^2 + b^5*c*d^2*e^3 - b^6*d*e^4)*x)*sqrt(c*x^2 + b*x))/((b^4*c^5* d^6 - 3*b^5*c^4*d^5*e + 3*b^6*c^3*d^4*e^2 - b^7*c^2*d^3*e^3)*x^4 + 2*(b^5* c^4*d^6 - 3*b^6*c^3*d^5*e + 3*b^7*c^2*d^4*e^2 - b^8*c*d^3*e^3)*x^3 + (b^6* c^3*d^6 - 3*b^7*c^2*d^5*e + 3*b^8*c*d^4*e^2 - b^9*d^3*e^3)*x^2)]
\[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {1}{\left (x \left (b + c x\right )\right )^{\frac {5}{2}} \left (d + e x\right )}\, dx \] Input:
integrate(1/(e*x+d)/(c*x**2+b*x)**(5/2),x)
Output:
Integral(1/((x*(b + c*x))**(5/2)*(d + e*x)), x)
Exception generated. \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(e*x+d)/(c*x^2+b*x)^(5/2),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
Leaf count of result is larger than twice the leaf count of optimal. 645 vs. \(2 (212) = 424\).
Time = 0.13 (sec) , antiderivative size = 645, normalized size of antiderivative = 2.73 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{5/2}} \, dx=-\frac {2 \, e^{4} \arctan \left (\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} + b d e}}\right )}{{\left (c^{2} d^{4} - 2 \, b c d^{3} e + b^{2} d^{2} e^{2}\right )} \sqrt {-c d^{2} + b d e}} + \frac {2 \, {\left ({\left ({\left (\frac {{\left (16 \, c^{7} d^{10} - 56 \, b c^{6} d^{9} e + 66 \, b^{2} c^{5} d^{8} e^{2} - 25 \, b^{3} c^{4} d^{7} e^{3} - 4 \, b^{4} c^{3} d^{6} e^{4} + 3 \, b^{5} c^{2} d^{5} e^{5}\right )} x}{b^{4} c^{4} d^{11} - 4 \, b^{5} c^{3} d^{10} e + 6 \, b^{6} c^{2} d^{9} e^{2} - 4 \, b^{7} c d^{8} e^{3} + b^{8} d^{7} e^{4}} + \frac {3 \, {\left (8 \, b c^{6} d^{10} - 28 \, b^{2} c^{5} d^{9} e + 33 \, b^{3} c^{4} d^{8} e^{2} - 12 \, b^{4} c^{3} d^{7} e^{3} - 3 \, b^{5} c^{2} d^{6} e^{4} + 2 \, b^{6} c d^{5} e^{5}\right )}}{b^{4} c^{4} d^{11} - 4 \, b^{5} c^{3} d^{10} e + 6 \, b^{6} c^{2} d^{9} e^{2} - 4 \, b^{7} c d^{8} e^{3} + b^{8} d^{7} e^{4}}\right )} x + \frac {3 \, {\left (2 \, b^{2} c^{5} d^{10} - 7 \, b^{3} c^{4} d^{9} e + 8 \, b^{4} c^{3} d^{8} e^{2} - 2 \, b^{5} c^{2} d^{7} e^{3} - 2 \, b^{6} c d^{6} e^{4} + b^{7} d^{5} e^{5}\right )}}{b^{4} c^{4} d^{11} - 4 \, b^{5} c^{3} d^{10} e + 6 \, b^{6} c^{2} d^{9} e^{2} - 4 \, b^{7} c d^{8} e^{3} + b^{8} d^{7} e^{4}}\right )} x - \frac {b^{3} c^{4} d^{10} - 4 \, b^{4} c^{3} d^{9} e + 6 \, b^{5} c^{2} d^{8} e^{2} - 4 \, b^{6} c d^{7} e^{3} + b^{7} d^{6} e^{4}}{b^{4} c^{4} d^{11} - 4 \, b^{5} c^{3} d^{10} e + 6 \, b^{6} c^{2} d^{9} e^{2} - 4 \, b^{7} c d^{8} e^{3} + b^{8} d^{7} e^{4}}\right )}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \] Input:
integrate(1/(e*x+d)/(c*x^2+b*x)^(5/2),x, algorithm="giac")
Output:
-2*e^4*arctan(((sqrt(c)*x - sqrt(c*x^2 + b*x))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e))/((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2)*sqrt(-c*d^2 + b*d*e)) + 2 /3*((((16*c^7*d^10 - 56*b*c^6*d^9*e + 66*b^2*c^5*d^8*e^2 - 25*b^3*c^4*d^7* e^3 - 4*b^4*c^3*d^6*e^4 + 3*b^5*c^2*d^5*e^5)*x/(b^4*c^4*d^11 - 4*b^5*c^3*d ^10*e + 6*b^6*c^2*d^9*e^2 - 4*b^7*c*d^8*e^3 + b^8*d^7*e^4) + 3*(8*b*c^6*d^ 10 - 28*b^2*c^5*d^9*e + 33*b^3*c^4*d^8*e^2 - 12*b^4*c^3*d^7*e^3 - 3*b^5*c^ 2*d^6*e^4 + 2*b^6*c*d^5*e^5)/(b^4*c^4*d^11 - 4*b^5*c^3*d^10*e + 6*b^6*c^2* d^9*e^2 - 4*b^7*c*d^8*e^3 + b^8*d^7*e^4))*x + 3*(2*b^2*c^5*d^10 - 7*b^3*c^ 4*d^9*e + 8*b^4*c^3*d^8*e^2 - 2*b^5*c^2*d^7*e^3 - 2*b^6*c*d^6*e^4 + b^7*d^ 5*e^5)/(b^4*c^4*d^11 - 4*b^5*c^3*d^10*e + 6*b^6*c^2*d^9*e^2 - 4*b^7*c*d^8* e^3 + b^8*d^7*e^4))*x - (b^3*c^4*d^10 - 4*b^4*c^3*d^9*e + 6*b^5*c^2*d^8*e^ 2 - 4*b^6*c*d^7*e^3 + b^7*d^6*e^4)/(b^4*c^4*d^11 - 4*b^5*c^3*d^10*e + 6*b^ 6*c^2*d^9*e^2 - 4*b^7*c*d^8*e^3 + b^8*d^7*e^4))/(c*x^2 + b*x)^(3/2)
Timed out. \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{5/2}} \, dx=\int \frac {1}{{\left (c\,x^2+b\,x\right )}^{5/2}\,\left (d+e\,x\right )} \,d x \] Input:
int(1/((b*x + c*x^2)^(5/2)*(d + e*x)),x)
Output:
int(1/((b*x + c*x^2)^(5/2)*(d + e*x)), x)
Time = 0.40 (sec) , antiderivative size = 880, normalized size of antiderivative = 3.73 \[ \int \frac {1}{(d+e x) \left (b x+c x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:
int(1/(e*x+d)/(c*x^2+b*x)^(5/2),x)
Output:
(2*( - 3*sqrt(d)*sqrt(b + c*x)*sqrt(b*e - c*d)*atan((sqrt(b*e - c*d) - sqr t(e)*sqrt(b + c*x) - sqrt(x)*sqrt(e)*sqrt(c))/(sqrt(d)*sqrt(c)))*b**5*e**4 *x**2 - 3*sqrt(d)*sqrt(b + c*x)*sqrt(b*e - c*d)*atan((sqrt(b*e - c*d) - sq rt(e)*sqrt(b + c*x) - sqrt(x)*sqrt(e)*sqrt(c))/(sqrt(d)*sqrt(c)))*b**4*c*e **4*x**3 - 3*sqrt(d)*sqrt(b + c*x)*sqrt(b*e - c*d)*atan((sqrt(b*e - c*d) + sqrt(e)*sqrt(b + c*x) + sqrt(x)*sqrt(e)*sqrt(c))/(sqrt(d)*sqrt(c)))*b**5* e**4*x**2 - 3*sqrt(d)*sqrt(b + c*x)*sqrt(b*e - c*d)*atan((sqrt(b*e - c*d) + sqrt(e)*sqrt(b + c*x) + sqrt(x)*sqrt(e)*sqrt(c))/(sqrt(d)*sqrt(c)))*b**4 *c*e**4*x**3 - sqrt(c)*sqrt(b + c*x)*b**5*d*e**4*x**2 - 5*sqrt(c)*sqrt(b + c*x)*b**4*c*d**2*e**3*x**2 - sqrt(c)*sqrt(b + c*x)*b**4*c*d*e**4*x**3 + 3 0*sqrt(c)*sqrt(b + c*x)*b**3*c**2*d**3*e**2*x**2 - 5*sqrt(c)*sqrt(b + c*x) *b**3*c**2*d**2*e**3*x**3 - 40*sqrt(c)*sqrt(b + c*x)*b**2*c**3*d**4*e*x**2 + 30*sqrt(c)*sqrt(b + c*x)*b**2*c**3*d**3*e**2*x**3 + 16*sqrt(c)*sqrt(b + c*x)*b*c**4*d**5*x**2 - 40*sqrt(c)*sqrt(b + c*x)*b*c**4*d**4*e*x**3 + 16* sqrt(c)*sqrt(b + c*x)*c**5*d**5*x**3 - sqrt(x)*b**6*d**2*e**3 + 3*sqrt(x)* b**6*d*e**4*x + 3*sqrt(x)*b**5*c*d**3*e**2 - 3*sqrt(x)*b**5*c*d**2*e**3*x + 6*sqrt(x)*b**5*c*d*e**4*x**2 - 3*sqrt(x)*b**4*c**2*d**4*e - 9*sqrt(x)*b* *4*c**2*d**3*e**2*x - 3*sqrt(x)*b**4*c**2*d**2*e**3*x**2 + 3*sqrt(x)*b**4* c**2*d*e**4*x**3 + sqrt(x)*b**3*c**3*d**5 + 15*sqrt(x)*b**3*c**3*d**4*e*x - 39*sqrt(x)*b**3*c**3*d**3*e**2*x**2 - sqrt(x)*b**3*c**3*d**2*e**3*x**...