Integrand size = 44, antiderivative size = 259 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^7} \, dx=-\frac {2 c^2 g \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e^2 (d+e x)}+\frac {2 c g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 e^2 (d+e x)^3}-\frac {2 g \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{5 e^2 (d+e x)^5}-\frac {2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{7 e^2 (2 c d-b e) (d+e x)^7}-\frac {2 c^{5/2} g \arctan \left (\frac {\sqrt {c} (d+e x)}{\sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{e^2} \] Output:
-2*c^2*g*(d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2)/e^2/(e*x+d)+2/3*c*g*(d*(-b *e+c*d)-b*e^2*x-c*e^2*x^2)^(3/2)/e^2/(e*x+d)^3-2/5*g*(d*(-b*e+c*d)-b*e^2*x -c*e^2*x^2)^(5/2)/e^2/(e*x+d)^5-2/7*(-d*g+e*f)*(d*(-b*e+c*d)-b*e^2*x-c*e^2 *x^2)^(7/2)/e^2/(-b*e+2*c*d)/(e*x+d)^7-2*c^(5/2)*g*arctan(c^(1/2)*(e*x+d)/ (d*(-b*e+c*d)-b*e^2*x-c*e^2*x^2)^(1/2))/e^2
Time = 1.17 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.15 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^7} \, dx=\frac {2 ((d+e x) (-b e+c (d-e x)))^{5/2} \left (-\frac {-210 c^3 d g (d+e x)^3+105 b c^2 e g (d+e x)^3+35 b c e g (d+e x)^2 (-c d+b e+c e x)-42 c d g (d+e x) (-c d+b e+c e x)^2+21 b e g (d+e x) (-c d+b e+c e x)^2+15 e f (-c d+b e+c e x)^3+70 c^2 d g (d+e x)^2 (-b e+c (d-e x))+15 d g (-b e+c (d-e x))^3}{(-2 c d+b e) (d+e x)^6 (-c d+b e+c e x)^2}+\frac {105 c^{5/2} g \arctan \left (\frac {\sqrt {c d-b e-c e x}}{\sqrt {c} \sqrt {d+e x}}\right )}{(d+e x)^{5/2} (-b e+c (d-e x))^{5/2}}\right )}{105 e^2} \] Input:
Integrate[((f + g*x)*(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(5/2))/(d + e*x )^7,x]
Output:
(2*((d + e*x)*(-(b*e) + c*(d - e*x)))^(5/2)*(-((-210*c^3*d*g*(d + e*x)^3 + 105*b*c^2*e*g*(d + e*x)^3 + 35*b*c*e*g*(d + e*x)^2*(-(c*d) + b*e + c*e*x) - 42*c*d*g*(d + e*x)*(-(c*d) + b*e + c*e*x)^2 + 21*b*e*g*(d + e*x)*(-(c*d ) + b*e + c*e*x)^2 + 15*e*f*(-(c*d) + b*e + c*e*x)^3 + 70*c^2*d*g*(d + e*x )^2*(-(b*e) + c*(d - e*x)) + 15*d*g*(-(b*e) + c*(d - e*x))^3)/((-2*c*d + b *e)*(d + e*x)^6*(-(c*d) + b*e + c*e*x)^2)) + (105*c^(5/2)*g*ArcTan[Sqrt[c* d - b*e - c*e*x]/(Sqrt[c]*Sqrt[d + e*x])])/((d + e*x)^(5/2)*(-(b*e) + c*(d - e*x))^(5/2))))/(105*e^2)
Time = 0.90 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.159, Rules used = {1220, 1130, 1130, 1125, 27, 1092, 217}
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.
\(\displaystyle \int \frac {(f+g x) \left (-b d e-b e^2 x+c d^2-c e^2 x^2\right )^{5/2}}{(d+e x)^7} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {g \int \frac {\left (-c x^2 e^2-b x e^2+d (c d-b e)\right )^{5/2}}{(d+e x)^6}dx}{e}-\frac {2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{7 e^2 (d+e x)^7 (2 c d-b e)}\) |
\(\Big \downarrow \) 1130 |
\(\displaystyle \frac {g \left (-c \int \frac {\left (-c x^2 e^2-b x e^2+d (c d-b e)\right )^{3/2}}{(d+e x)^4}dx-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{5 e (d+e x)^5}\right )}{e}-\frac {2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{7 e^2 (d+e x)^7 (2 c d-b e)}\) |
\(\Big \downarrow \) 1130 |
\(\displaystyle \frac {g \left (-c \left (-c \int \frac {\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}{(d+e x)^2}dx-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^3}\right )-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{5 e (d+e x)^5}\right )}{e}-\frac {2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{7 e^2 (d+e x)^7 (2 c d-b e)}\) |
\(\Big \downarrow \) 1125 |
\(\displaystyle \frac {g \left (-c \left (-c \left (-\frac {\int \frac {c e^2}{\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx}{e^2}-\frac {2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e (d+e x)}\right )-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^3}\right )-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{5 e (d+e x)^5}\right )}{e}-\frac {2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{7 e^2 (d+e x)^7 (2 c d-b e)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {g \left (-c \left (-c \left (-c \int \frac {1}{\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}dx-\frac {2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e (d+e x)}\right )-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^3}\right )-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{5 e (d+e x)^5}\right )}{e}-\frac {2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{7 e^2 (d+e x)^7 (2 c d-b e)}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {g \left (-c \left (-c \left (-2 c \int \frac {1}{-\frac {(b+2 c x)^2 e^4}{-c x^2 e^2-b x e^2+d (c d-b e)}-4 c e^2}d\left (-\frac {e^2 (b+2 c x)}{\sqrt {-c x^2 e^2-b x e^2+d (c d-b e)}}\right )-\frac {2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e (d+e x)}\right )-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^3}\right )-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{5 e (d+e x)^5}\right )}{e}-\frac {2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{7 e^2 (d+e x)^7 (2 c d-b e)}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {g \left (-c \left (-c \left (-\frac {\sqrt {c} \arctan \left (\frac {e (b+2 c x)}{2 \sqrt {c} \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}\right )}{e}-\frac {2 \sqrt {d (c d-b e)-b e^2 x-c e^2 x^2}}{e (d+e x)}\right )-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^3}\right )-\frac {2 \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{5/2}}{5 e (d+e x)^5}\right )}{e}-\frac {2 (e f-d g) \left (d (c d-b e)-b e^2 x-c e^2 x^2\right )^{7/2}}{7 e^2 (d+e x)^7 (2 c d-b e)}\) |
Input:
Int[((f + g*x)*(c*d^2 - b*d*e - b*e^2*x - c*e^2*x^2)^(5/2))/(d + e*x)^7,x]
Output:
(-2*(e*f - d*g)*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(7/2))/(7*e^2*(2*c*d - b*e)*(d + e*x)^7) + (g*((-2*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(5/2) )/(5*e*(d + e*x)^5) - c*((-2*(d*(c*d - b*e) - b*e^2*x - c*e^2*x^2)^(3/2))/ (3*e*(d + e*x)^3) - c*((-2*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])/(e*( d + e*x)) - (Sqrt[c]*ArcTan[(e*(b + 2*c*x))/(2*Sqrt[c]*Sqrt[d*(c*d - b*e) - b*e^2*x - c*e^2*x^2])])/e))))/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, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ Symbol] :> Simp[-2*e^(2*m + 3)*(Sqrt[a + b*x + c*x^2]/((-2*c*d + b*e)^(m + 2)*(d + e*x))), x] - Simp[e^(2*m + 2) Int[(1/Sqrt[a + b*x + c*x^2])*Expan dToSum[((-2*c*d + b*e)^(-m - 1) - ((-c)*d + b*e + c*e*x)^(-m - 1))/(d + e*x ), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ILtQ[m, 0] && EqQ[m + p, -3/2]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + p + 1))), x] - Simp[c*(p/(e^2*(m + p + 1))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] & & IntegerQ[2*p]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Leaf count of result is larger than twice the leaf count of optimal. \(824\) vs. \(2(239)=478\).
Time = 9.43 (sec) , antiderivative size = 825, normalized size of antiderivative = 3.19
method | result | size |
default | \(\frac {g \left (-\frac {2 \left (-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{5 \left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )^{6}}-\frac {2 c \,e^{2} \left (-\frac {2 \left (-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{3 \left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )^{5}}-\frac {4 c \,e^{2} \left (-\frac {2 \left (-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{\left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )^{4}}-\frac {6 c \,e^{2} \left (\frac {2 \left (-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{\left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )^{3}}+\frac {8 c \,e^{2} \left (\frac {2 \left (-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{3 \left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )^{2}}+\frac {10 c \,e^{2} \left (\frac {\left (-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+\frac {\left (-b \,e^{2}+2 d e c \right ) \left (-\frac {\left (-2 c \,e^{2} \left (x +\frac {d}{e}\right )-b \,e^{2}+2 d e c \right ) \left (-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 c \,e^{2}}+\frac {3 \left (-b \,e^{2}+2 d e c \right )^{2} \left (-\frac {\left (-2 c \,e^{2} \left (x +\frac {d}{e}\right )-b \,e^{2}+2 d e c \right ) \sqrt {-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )}}{4 c \,e^{2}}+\frac {\left (-b \,e^{2}+2 d e c \right )^{2} \arctan \left (\frac {\sqrt {c \,e^{2}}\, \left (x +\frac {d}{e}-\frac {-b \,e^{2}+2 d e c}{2 c \,e^{2}}\right )}{\sqrt {-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )}}\right )}{8 c \,e^{2} \sqrt {c \,e^{2}}}\right )}{16 c \,e^{2}}\right )}{2}\right )}{3 \left (-b \,e^{2}+2 d e c \right )}\right )}{-b \,e^{2}+2 d e c}\right )}{-b \,e^{2}+2 d e c}\right )}{3 \left (-b \,e^{2}+2 d e c \right )}\right )}{5 \left (-b \,e^{2}+2 d e c \right )}\right )}{e^{7}}+\frac {2 \left (d g -e f \right ) \left (-c \,e^{2} \left (x +\frac {d}{e}\right )^{2}+\left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{7 e^{8} \left (-b \,e^{2}+2 d e c \right ) \left (x +\frac {d}{e}\right )^{7}}\) | \(825\) |
Input:
int((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2)/(e*x+d)^7,x,method=_RET URNVERBOSE)
Output:
g/e^7*(-2/5/(-b*e^2+2*c*d*e)/(x+d/e)^6*(-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)* (x+d/e))^(7/2)-2/5*c*e^2/(-b*e^2+2*c*d*e)*(-2/3/(-b*e^2+2*c*d*e)/(x+d/e)^5 *(-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(7/2)-4/3*c*e^2/(-b*e^2+2*c*d *e)*(-2/(-b*e^2+2*c*d*e)/(x+d/e)^4*(-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d /e))^(7/2)-6*c*e^2/(-b*e^2+2*c*d*e)*(2/(-b*e^2+2*c*d*e)/(x+d/e)^3*(-c*e^2* (x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(7/2)+8*c*e^2/(-b*e^2+2*c*d*e)*(2/3/(- b*e^2+2*c*d*e)/(x+d/e)^2*(-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(7/2) +10/3*c*e^2/(-b*e^2+2*c*d*e)*(1/5*(-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/ e))^(5/2)+1/2*(-b*e^2+2*c*d*e)*(-1/8*(-2*c*e^2*(x+d/e)-b*e^2+2*d*e*c)/c/e^ 2*(-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(3/2)+3/16*(-b*e^2+2*c*d*e)^ 2/c/e^2*(-1/4*(-2*c*e^2*(x+d/e)-b*e^2+2*d*e*c)/c/e^2*(-c*e^2*(x+d/e)^2+(-b *e^2+2*c*d*e)*(x+d/e))^(1/2)+1/8*(-b*e^2+2*c*d*e)^2/c/e^2/(c*e^2)^(1/2)*ar ctan((c*e^2)^(1/2)*(x+d/e-1/2*(-b*e^2+2*c*d*e)/c/e^2)/(-c*e^2*(x+d/e)^2+(- b*e^2+2*c*d*e)*(x+d/e))^(1/2))))))))))+2/7*(d*g-e*f)/e^8/(-b*e^2+2*c*d*e)/ (x+d/e)^7*(-c*e^2*(x+d/e)^2+(-b*e^2+2*c*d*e)*(x+d/e))^(7/2)
Leaf count of result is larger than twice the leaf count of optimal. 613 vs. \(2 (239) = 478\).
Time = 34.44 (sec) , antiderivative size = 1239, normalized size of antiderivative = 4.78 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^7} \, dx=\text {Too large to display} \] Input:
integrate((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2)/(e*x+d)^7,x, algo rithm="fricas")
Output:
[1/210*(105*((2*c^3*d*e^4 - b*c^2*e^5)*g*x^4 + 4*(2*c^3*d^2*e^3 - b*c^2*d* e^4)*g*x^3 + 6*(2*c^3*d^3*e^2 - b*c^2*d^2*e^3)*g*x^2 + 4*(2*c^3*d^4*e - b* c^2*d^3*e^2)*g*x + (2*c^3*d^5 - b*c^2*d^4*e)*g)*sqrt(-c)*log(8*c^2*e^2*x^2 + 8*b*c*e^2*x - 4*c^2*d^2 + 4*b*c*d*e + b^2*e^2 - 4*sqrt(-c*e^2*x^2 - b*e ^2*x + c*d^2 - b*d*e)*(2*c*e*x + b*e)*sqrt(-c)) + 4*sqrt(-c*e^2*x^2 - b*e^ 2*x + c*d^2 - b*d*e)*((15*c^3*e^4*f - (337*c^3*d*e^3 - 161*b*c^2*e^4)*g)*x ^3 - (45*(c^3*d*e^3 - b*c^2*e^4)*f + (613*c^3*d^2*e^2 - 130*b*c^2*d*e^3 - 77*b^2*c*e^4)*g)*x^2 - 15*(c^3*d^3*e - 3*b*c^2*d^2*e^2 + 3*b^2*c*d*e^3 - b ^3*e^4)*f - (167*c^3*d^4 - 60*b*c^2*d^3*e + 4*b^2*c*d^2*e^2 - 6*b^3*d*e^3) *g + (45*(c^3*d^2*e^2 - 2*b*c^2*d*e^3 + b^2*c*e^4)*f - (563*c^3*d^3*e - 20 9*b*c^2*d^2*e^2 + 17*b^2*c*d*e^3 - 21*b^3*e^4)*g)*x))/(2*c*d^5*e^2 - b*d^4 *e^3 + (2*c*d*e^6 - b*e^7)*x^4 + 4*(2*c*d^2*e^5 - b*d*e^6)*x^3 + 6*(2*c*d^ 3*e^4 - b*d^2*e^5)*x^2 + 4*(2*c*d^4*e^3 - b*d^3*e^4)*x), 1/105*(105*((2*c^ 3*d*e^4 - b*c^2*e^5)*g*x^4 + 4*(2*c^3*d^2*e^3 - b*c^2*d*e^4)*g*x^3 + 6*(2* c^3*d^3*e^2 - b*c^2*d^2*e^3)*g*x^2 + 4*(2*c^3*d^4*e - b*c^2*d^3*e^2)*g*x + (2*c^3*d^5 - b*c^2*d^4*e)*g)*sqrt(c)*arctan(1/2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*(2*c*e*x + b*e)*sqrt(c)/(c^2*e^2*x^2 + b*c*e^2*x - c^2*d ^2 + b*c*d*e)) + 2*sqrt(-c*e^2*x^2 - b*e^2*x + c*d^2 - b*d*e)*((15*c^3*e^4 *f - (337*c^3*d*e^3 - 161*b*c^2*e^4)*g)*x^3 - (45*(c^3*d*e^3 - b*c^2*e^4)* f + (613*c^3*d^2*e^2 - 130*b*c^2*d*e^3 - 77*b^2*c*e^4)*g)*x^2 - 15*(c^3...
\[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^7} \, dx=\int \frac {\left (- \left (d + e x\right ) \left (b e - c d + c e x\right )\right )^{\frac {5}{2}} \left (f + g x\right )}{\left (d + e x\right )^{7}}\, dx \] Input:
integrate((g*x+f)*(-c*e**2*x**2-b*e**2*x-b*d*e+c*d**2)**(5/2)/(e*x+d)**7,x )
Output:
Integral((-(d + e*x)*(b*e - c*d + c*e*x))**(5/2)*(f + g*x)/(d + e*x)**7, x )
Exception generated. \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^7} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2)/(e*x+d)^7,x, algo rithm="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-2*c*d>0)', see `assume?` for more deta
Timed out. \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^7} \, dx=\text {Timed out} \] Input:
integrate((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2)/(e*x+d)^7,x, algo rithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^7} \, dx=\int \frac {\left (f+g\,x\right )\,{\left (c\,d^2-b\,d\,e-c\,e^2\,x^2-b\,e^2\,x\right )}^{5/2}}{{\left (d+e\,x\right )}^7} \,d x \] Input:
int(((f + g*x)*(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(5/2))/(d + e*x)^7,x)
Output:
int(((f + g*x)*(c*d^2 - c*e^2*x^2 - b*d*e - b*e^2*x)^(5/2))/(d + e*x)^7, x )
Time = 0.98 (sec) , antiderivative size = 2439, normalized size of antiderivative = 9.42 \[ \int \frac {(f+g x) \left (c d^2-b d e-b e^2 x-c e^2 x^2\right )^{5/2}}{(d+e x)^7} \, dx =\text {Too large to display} \] Input:
int((g*x+f)*(-c*e^2*x^2-b*e^2*x-b*d*e+c*d^2)^(5/2)/(e*x+d)^7,x)
Output:
(2*i*( - 105*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2* c*d))*b**2*c**2*d**4*e**2*g - 420*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x )*i)/sqrt( - b*e + 2*c*d))*b**2*c**2*d**3*e**3*g*x - 630*sqrt(c)*asinh((sq rt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d))*b**2*c**2*d**2*e**4*g*x* *2 - 420*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d) )*b**2*c**2*d*e**5*g*x**3 - 105*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)* i)/sqrt( - b*e + 2*c*d))*b**2*c**2*e**6*g*x**4 + 420*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d))*b*c**3*d**5*e*g + 1680*sqrt( c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d))*b*c**3*d**4* e**2*g*x + 2520*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d))*b*c**3*d**3*e**3*g*x**2 + 1680*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d))*b*c**3*d**2*e**4*g*x**3 + 420*sqrt(c)*asi nh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d))*b*c**3*d*e**5*g*x* *4 - 420*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d) )*c**4*d**6*g - 1680*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d))*c**4*d**5*e*g*x - 2520*sqrt(c)*asinh((sqrt( - b*e + c*d - c* e*x)*i)/sqrt( - b*e + 2*c*d))*c**4*d**4*e**2*g*x**2 - 1680*sqrt(c)*asinh(( sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d))*c**4*d**3*e**3*g*x**3 - 420*sqrt(c)*asinh((sqrt( - b*e + c*d - c*e*x)*i)/sqrt( - b*e + 2*c*d))*c **4*d**2*e**4*g*x**4 - 6*sqrt(d + e*x)*sqrt(b*e - 2*c*d)*sqrt( - b*e + ...