3.2363 \(\int \frac {(a+b x+c x^2)^{5/2}}{(d+e x)^5} \, dx\)

Optimal. Leaf size=492 \[ \frac {5 \left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{128 e^6 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac {5 \left (a+b x+c x^2\right )^{3/2} \left (3 e x \left (-4 c e (2 b d-a e)+b^2 e^2+8 c^2 d^2\right )-4 c d e (3 b d-a e)-b e^2 (b d-4 a e)+16 c^2 d^3\right )}{96 e^3 (d+e x)^3 \left (a e^2-b d e+c d^2\right )}+\frac {5 \sqrt {a+b x+c x^2} \left (2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (5 b d-4 a e)+4 b c e^2 (4 b d-5 a e)+b^3 e^3+64 c^3 d^3\right )}{64 e^5 (d+e x) \left (a e^2-b d e+c d^2\right )}-\frac {5 c^{3/2} (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 e^6}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 e (d+e x)^4} \]

[Out]

-5/96*(16*c^2*d^3-b*e^2*(-4*a*e+b*d)-4*c*d*e*(-a*e+3*b*d)+3*e*(8*c^2*d^2+b^2*e^2-4*c*e*(-a*e+2*b*d))*x)*(c*x^2
+b*x+a)^(3/2)/e^3/(a*e^2-b*d*e+c*d^2)/(e*x+d)^3-1/4*(c*x^2+b*x+a)^(5/2)/e/(e*x+d)^4-5/2*c^(3/2)*(-b*e+2*c*d)*a
rctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/e^6+5/128*(128*c^4*d^4-b^4*e^4-8*b^2*c*e^3*(-3*a*e+2*b*d)-64
*c^3*d^2*e*(-3*a*e+4*b*d)+48*c^2*e^2*(a^2*e^2-4*a*b*d*e+3*b^2*d^2))*arctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d)*x)/(a*
e^2-b*d*e+c*d^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/e^6/(a*e^2-b*d*e+c*d^2)^(3/2)+5/64*(64*c^3*d^3+b^3*e^3+4*b*c*e^2*(
-5*a*e+4*b*d)-16*c^2*d*e*(-4*a*e+5*b*d)+2*c*e*(16*c^2*d^2+b^2*e^2-4*c*e*(-3*a*e+4*b*d))*x)*(c*x^2+b*x+a)^(1/2)
/e^5/(a*e^2-b*d*e+c*d^2)/(e*x+d)

________________________________________________________________________________________

Rubi [A]  time = 0.73, antiderivative size = 492, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {732, 810, 812, 843, 621, 206, 724} \[ \frac {5 \left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{128 e^6 \left (a e^2-b d e+c d^2\right )^{3/2}}-\frac {5 \left (a+b x+c x^2\right )^{3/2} \left (3 e x \left (-4 c e (2 b d-a e)+b^2 e^2+8 c^2 d^2\right )-4 c d e (3 b d-a e)-b e^2 (b d-4 a e)+16 c^2 d^3\right )}{96 e^3 (d+e x)^3 \left (a e^2-b d e+c d^2\right )}+\frac {5 \sqrt {a+b x+c x^2} \left (2 c e x \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right )-16 c^2 d e (5 b d-4 a e)+4 b c e^2 (4 b d-5 a e)+b^3 e^3+64 c^3 d^3\right )}{64 e^5 (d+e x) \left (a e^2-b d e+c d^2\right )}-\frac {5 c^{3/2} (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 e^6}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 e (d+e x)^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(5/2)/(d + e*x)^5,x]

[Out]

(5*(64*c^3*d^3 + b^3*e^3 + 4*b*c*e^2*(4*b*d - 5*a*e) - 16*c^2*d*e*(5*b*d - 4*a*e) + 2*c*e*(16*c^2*d^2 + b^2*e^
2 - 4*c*e*(4*b*d - 3*a*e))*x)*Sqrt[a + b*x + c*x^2])/(64*e^5*(c*d^2 - b*d*e + a*e^2)*(d + e*x)) - (5*(16*c^2*d
^3 - b*e^2*(b*d - 4*a*e) - 4*c*d*e*(3*b*d - a*e) + 3*e*(8*c^2*d^2 + b^2*e^2 - 4*c*e*(2*b*d - a*e))*x)*(a + b*x
+ c*x^2)^(3/2))/(96*e^3*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^3) - (a + b*x + c*x^2)^(5/2)/(4*e*(d + e*x)^4) - (5
*c^(3/2)*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*e^6) + (5*(128*c^4*d^4 - b^4
*e^4 - 8*b^2*c*e^3*(2*b*d - 3*a*e) - 64*c^3*d^2*e*(4*b*d - 3*a*e) + 48*c^2*e^2*(3*b^2*d^2 - 4*a*b*d*e + a^2*e^
2))*ArcTanh[(b*d - 2*a*e + (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])])/(128*e^6*(
c*d^2 - b*d*e + a*e^2)^(3/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-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] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 732

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ
[2*c*d - b*e, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*
c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + b*x + c*x^2)^p)/(e^2*(m + 1)*(m
+ 2*p + 2)), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x+c x^2\right )^{5/2}}{(d+e x)^5} \, dx &=-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 e (d+e x)^4}+\frac {5 \int \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{(d+e x)^4} \, dx}{8 e}\\ &=-\frac {5 \left (16 c^2 d^3-b e^2 (b d-4 a e)-4 c d e (3 b d-a e)+3 e \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{96 e^3 \left (c d^2-b d e+a e^2\right ) (d+e x)^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 e (d+e x)^4}-\frac {5 \int \frac {\left (\frac {1}{2} \left (12 b^2 c d e+16 a c^2 d e+b^3 e^2-4 b c \left (4 c d^2+5 a e^2\right )\right )-c \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{(d+e x)^2} \, dx}{32 e^3 \left (c d^2-b d e+a e^2\right )}\\ &=\frac {5 \left (64 c^3 d^3+b^3 e^3+4 b c e^2 (4 b d-5 a e)-16 c^2 d e (5 b d-4 a e)+2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{64 e^5 \left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac {5 \left (16 c^2 d^3-b e^2 (b d-4 a e)-4 c d e (3 b d-a e)+3 e \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{96 e^3 \left (c d^2-b d e+a e^2\right ) (d+e x)^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 e (d+e x)^4}+\frac {5 \int \frac {\frac {1}{2} \left (-16 b^3 c d e^2-b^4 e^3-64 b c^2 d \left (c d^2+2 a e^2\right )+16 a c^2 e \left (4 c d^2+3 a e^2\right )+8 b^2 c e \left (10 c d^2+3 a e^2\right )\right )-32 c^2 (2 c d-b e) \left (c d^2-b d e+a e^2\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{64 e^5 \left (c d^2-b d e+a e^2\right )}\\ &=\frac {5 \left (64 c^3 d^3+b^3 e^3+4 b c e^2 (4 b d-5 a e)-16 c^2 d e (5 b d-4 a e)+2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{64 e^5 \left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac {5 \left (16 c^2 d^3-b e^2 (b d-4 a e)-4 c d e (3 b d-a e)+3 e \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{96 e^3 \left (c d^2-b d e+a e^2\right ) (d+e x)^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 e (d+e x)^4}-\frac {\left (5 c^2 (2 c d-b e)\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 e^6}+\frac {\left (5 \left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{128 e^6 \left (c d^2-b d e+a e^2\right )}\\ &=\frac {5 \left (64 c^3 d^3+b^3 e^3+4 b c e^2 (4 b d-5 a e)-16 c^2 d e (5 b d-4 a e)+2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{64 e^5 \left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac {5 \left (16 c^2 d^3-b e^2 (b d-4 a e)-4 c d e (3 b d-a e)+3 e \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{96 e^3 \left (c d^2-b d e+a e^2\right ) (d+e x)^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 e (d+e x)^4}-\frac {\left (5 c^2 (2 c d-b e)\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{e^6}-\frac {\left (5 \left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c d^2-4 b d e+4 a e^2-x^2} \, dx,x,\frac {-b d+2 a e-(2 c d-b e) x}{\sqrt {a+b x+c x^2}}\right )}{64 e^6 \left (c d^2-b d e+a e^2\right )}\\ &=\frac {5 \left (64 c^3 d^3+b^3 e^3+4 b c e^2 (4 b d-5 a e)-16 c^2 d e (5 b d-4 a e)+2 c e \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{64 e^5 \left (c d^2-b d e+a e^2\right ) (d+e x)}-\frac {5 \left (16 c^2 d^3-b e^2 (b d-4 a e)-4 c d e (3 b d-a e)+3 e \left (8 c^2 d^2+b^2 e^2-4 c e (2 b d-a e)\right ) x\right ) \left (a+b x+c x^2\right )^{3/2}}{96 e^3 \left (c d^2-b d e+a e^2\right ) (d+e x)^3}-\frac {\left (a+b x+c x^2\right )^{5/2}}{4 e (d+e x)^4}-\frac {5 c^{3/2} (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 e^6}+\frac {5 \left (128 c^4 d^4-b^4 e^4-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)+48 c^2 e^2 \left (3 b^2 d^2-4 a b d e+a^2 e^2\right )\right ) \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{128 e^6 \left (c d^2-b d e+a e^2\right )^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 3.67, size = 593, normalized size = 1.21 \[ \frac {-\frac {15 \left (48 c^2 e^2 \left (a^2 e^2-4 a b d e+3 b^2 d^2\right )-8 b^2 c e^3 (2 b d-3 a e)-64 c^3 d^2 e (4 b d-3 a e)-b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac {2 a e-b d+b e x-2 c d x}{2 \sqrt {a+x (b+c x)} \sqrt {e (a e-b d)+c d^2}}\right )}{\left (e (a e-b d)+c d^2\right )^{3/2}}+\frac {2 e \sqrt {a+x (b+c x)} \left (2 c e^2 \left (-4 a^2 e^2 \left (11 d^2+20 d e x+27 e^2 x^2\right )-2 a b e \left (45 d^3+169 d^2 e x+191 d e^2 x^2+139 e^3 x^3\right )+b^2 d \left (120 d^3+435 d^2 e x+566 d e^2 x^2+323 e^3 x^3\right )\right )+e^3 \left (-48 a^3 e^3+8 a^2 b e^2 (d-17 e x)+2 a b^2 e \left (5 d^2+18 d e x-59 e^2 x^2\right )+b^3 \left (15 d^3+55 d^2 e x+73 d e^2 x^2-15 e^3 x^3\right )\right )-8 c^2 e \left (b d \left (150 d^4+530 d^3 e x+665 d^2 e^2 x^2+327 d e^3 x^3+24 e^4 x^4\right )-a e \left (100 d^4+355 d^3 e x+448 d^2 e^2 x^2+235 d e^3 x^3+24 e^4 x^4\right )\right )+16 c^3 d^2 \left (60 d^4+210 d^3 e x+260 d^2 e^2 x^2+125 d e^3 x^3+12 e^4 x^4\right )\right )}{(d+e x)^4 \left (e (a e-b d)+c d^2\right )}-960 c^{3/2} (2 c d-b e) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{384 e^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(5/2)/(d + e*x)^5,x]

[Out]

((2*e*Sqrt[a + x*(b + c*x)]*(16*c^3*d^2*(60*d^4 + 210*d^3*e*x + 260*d^2*e^2*x^2 + 125*d*e^3*x^3 + 12*e^4*x^4)
+ e^3*(-48*a^3*e^3 + 8*a^2*b*e^2*(d - 17*e*x) + 2*a*b^2*e*(5*d^2 + 18*d*e*x - 59*e^2*x^2) + b^3*(15*d^3 + 55*d
^2*e*x + 73*d*e^2*x^2 - 15*e^3*x^3)) + 2*c*e^2*(-4*a^2*e^2*(11*d^2 + 20*d*e*x + 27*e^2*x^2) - 2*a*b*e*(45*d^3
+ 169*d^2*e*x + 191*d*e^2*x^2 + 139*e^3*x^3) + b^2*d*(120*d^3 + 435*d^2*e*x + 566*d*e^2*x^2 + 323*e^3*x^3)) -
8*c^2*e*(-(a*e*(100*d^4 + 355*d^3*e*x + 448*d^2*e^2*x^2 + 235*d*e^3*x^3 + 24*e^4*x^4)) + b*d*(150*d^4 + 530*d^
3*e*x + 665*d^2*e^2*x^2 + 327*d*e^3*x^3 + 24*e^4*x^4))))/((c*d^2 + e*(-(b*d) + a*e))*(d + e*x)^4) - 960*c^(3/2
)*(2*c*d - b*e)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] - (15*(128*c^4*d^4 - b^4*e^4 - 8*b^2*c*
e^3*(2*b*d - 3*a*e) - 64*c^3*d^2*e*(4*b*d - 3*a*e) + 48*c^2*e^2*(3*b^2*d^2 - 4*a*b*d*e + a^2*e^2))*ArcTanh[(-(
b*d) + 2*a*e - 2*c*d*x + b*e*x)/(2*Sqrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b + c*x)])])/(c*d^2 + e*(-(b*d)
+ a*e))^(3/2))/(384*e^6)

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(e*x+d)^5,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(e*x+d)^5,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Evaluation time: 1.14Unable to divide, perhaps due to rounding error%%%{%%%{1,[0,8,0,0,0]%%%},[4,0
]%%%}+%%%{%%%{-2,[0,8,0,0,1]%%%},[2,0]%%%}+%%%{%%%{1,[0,8,0,0,2]%%%},[0,0]%%%} / %%%{%%%{1,[1,2,0,0,0]%%%}+%%%
{-1,[0,1,1,1,0]%%%}+%%%{1,[0,0,0,2,1]%%%},[4,0]%%%}+%%%{%%%{-2,[1,2,0,0,1]%%%}+%%%{2,[0,1,1,1,1]%%%}+%%%{-2,[0
,0,0,2,2]%%%},[2,0]%%%}+%%%{%%%{1,[1,2,0,0,2]%%%}+%%%{-1,[0,1,1,1,2]%%%}+%%%{1,[0,0,0,2,3]%%%},[0,0]%%%} Error
: Bad Argument Value

________________________________________________________________________________________

maple [B]  time = 0.34, size = 28635, normalized size = 58.20 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(5/2)/(e*x+d)^5,x)

[Out]

result too large to display

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/2)/(e*x+d)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e^2-b*d*e>0)', see `assume?`
for more details)Is a*e^2-b*d*e                            +c*d^2 zero or nonzero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (d+e\,x\right )}^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(5/2)/(d + e*x)^5,x)

[Out]

int((a + b*x + c*x^2)^(5/2)/(d + e*x)^5, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b x + c x^{2}\right )^{\frac {5}{2}}}{\left (d + e x\right )^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(5/2)/(e*x+d)**5,x)

[Out]

Integral((a + b*x + c*x**2)**(5/2)/(d + e*x)**5, x)

________________________________________________________________________________________