3.21.31 \(\int \frac {(a+b x+c x^2)^{5/2}}{(d+e x)^4} \, dx\)
Optimal. Leaf size=337 \[ \frac {5 \sqrt {c} \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 e^6}-\frac {5 (2 c d-b e) \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{16 e^6 \sqrt {a e^2-b d e+c d^2}}-\frac {5 \sqrt {a+b x+c x^2} \left (-4 c e (3 b d-a e)+b^2 e^2+4 c e x (2 c d-b e)+16 c^2 d^2\right )}{8 e^5 (d+e x)}+\frac {5 \left (a+b x+c x^2\right )^{3/2} (-b e+4 c d+2 c e x)}{12 e^3 (d+e x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{3 e (d+e x)^3} \]
________________________________________________________________________________________
Rubi [A] time = 0.43, antiderivative size = 337, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used =
{732, 812, 843, 621, 206, 724} \begin {gather*} -\frac {5 \sqrt {a+b x+c x^2} \left (-4 c e (3 b d-a e)+b^2 e^2+4 c e x (2 c d-b e)+16 c^2 d^2\right )}{8 e^5 (d+e x)}+\frac {5 \sqrt {c} \left (-4 c e (4 b d-a e)+3 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 e^6}-\frac {5 (2 c d-b e) \left (-4 c e (4 b d-3 a e)+b^2 e^2+16 c^2 d^2\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 )}{16 e^6 \sqrt {a e^2-b d e+c d^2}}+\frac {5 \left (a+b x+c x^2\right )^{3/2} (-b e+4 c d+2 c e x)}{12 e^3 (d+e x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{3 e (d+e x)^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + b*x + c*x^2)^(5/2)/(d + e*x)^4,x]
[Out]
(-5*(16*c^2*d^2 + b^2*e^2 - 4*c*e*(3*b*d - a*e) + 4*c*e*(2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/(8*e^5*(d + e*
x)) + (5*(4*c*d - b*e + 2*c*e*x)*(a + b*x + c*x^2)^(3/2))/(12*e^3*(d + e*x)^2) - (a + b*x + c*x^2)^(5/2)/(3*e*
(d + e*x)^3) + (5*Sqrt[c]*(16*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(4*b*d - a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a
+ b*x + c*x^2])])/(8*e^6) - (5*(2*c*d - b*e)*(16*c^2*d^2 + b^2*e^2 - 4*c*e*(4*b*d - 3*a*e))*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])])/(16*e^6*Sqrt[c*d^2 - b*d*e + a*
e^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 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)^4} \, dx &=-\frac {\left (a+b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}+\frac {5 \int \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/2}}{(d+e x)^3} \, dx}{6 e}\\ &=\frac {5 (4 c d-b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}-\frac {5 \int \frac {\left (2 \left (4 b c d-b^2 e-4 a c e\right )+8 c (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{(d+e x)^2} \, dx}{16 e^3}\\ &=-\frac {5 \left (16 c^2 d^2+b^2 e^2-4 c e (3 b d-a e)+4 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{8 e^5 (d+e x)}+\frac {5 (4 c d-b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}+\frac {5 \int \frac {2 \left (8 c (b d-a e) (2 c d-b e)-b e \left (4 b c d-b^2 e-4 a c e\right )\right )+4 c \left (16 c^2 d^2+3 b^2 e^2-4 c e (4 b d-a e)\right ) x}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{32 e^5}\\ &=-\frac {5 \left (16 c^2 d^2+b^2 e^2-4 c e (3 b d-a e)+4 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{8 e^5 (d+e x)}+\frac {5 (4 c d-b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}-\frac {\left (5 (2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+b x+c x^2}} \, dx}{16 e^6}+\frac {\left (5 c \left (16 c^2 d^2+3 b^2 e^2-4 c e (4 b d-a e)\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{8 e^6}\\ &=-\frac {5 \left (16 c^2 d^2+b^2 e^2-4 c e (3 b d-a e)+4 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{8 e^5 (d+e x)}+\frac {5 (4 c d-b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}+\frac {\left (5 (2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\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 )}{8 e^6}+\frac {\left (5 c \left (16 c^2 d^2+3 b^2 e^2-4 c e (4 b d-a e)\right )\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 )}{4 e^6}\\ &=-\frac {5 \left (16 c^2 d^2+b^2 e^2-4 c e (3 b d-a e)+4 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{8 e^5 (d+e x)}+\frac {5 (4 c d-b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}+\frac {5 \sqrt {c} \left (16 c^2 d^2+3 b^2 e^2-4 c e (4 b d-a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 e^6}-\frac {5 (2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\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 )}{16 e^6 \sqrt {c d^2-b d e+a e^2}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 3.01, size = 559, normalized size = 1.66 \begin {gather*} \frac {\frac {5 \left (\frac {(a+x (b+c x))^{3/2} \left (2 c^2 e (2 a e (2 e x-3 d)+3 b d (5 d-2 e x))+b c e^2 (10 a e-15 b d+b e x)+b^3 e^3-4 c^3 d^2 (4 d-3 e x)\right )}{2 e^2}-\frac {(a+x (b+c x))^{5/2} \left (4 c e (2 a e-3 b d)+b^2 e^2+12 c^2 d^2\right )}{2 (d+e x)}+\frac {3 \left (2 e \sqrt {a+x (b+c x)} \left (e (a e-b d)+c d^2\right ) \left (4 c^2 e (a e (e x-3 d)+b d (7 d-2 e x))+b c e^2 (8 a e-13 b d+b e x)+b^3 e^3+8 c^3 d^2 (e x-2 d)\right )+2 \sqrt {c} \left (4 c e (a e-4 b d)+3 b^2 e^2+16 c^2 d^2\right ) \left (e (a e-b d)+c d^2\right )^2 \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )+(2 c d-b e) \left (4 c e (3 a e-4 b d)+b^2 e^2+16 c^2 d^2\right ) \left (e (a e-b d)+c d^2\right )^{3/2} \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 )\right )}{4 e^5}\right )}{2 \left (e (a e-b d)+c d^2\right )^2}+\frac {5 (a+x (b+c x))^{5/2} (2 c d-b e)}{2 (d+e x)^2 \left (e (a e-b d)+c d^2\right )}-\frac {2 (a+x (b+c x))^{5/2}}{(d+e x)^3}}{6 e} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a + b*x + c*x^2)^(5/2)/(d + e*x)^4,x]
[Out]
((-2*(a + x*(b + c*x))^(5/2))/(d + e*x)^3 + (5*(2*c*d - b*e)*(a + x*(b + c*x))^(5/2))/(2*(c*d^2 + e*(-(b*d) +
a*e))*(d + e*x)^2) + (5*(-1/2*((12*c^2*d^2 + b^2*e^2 + 4*c*e*(-3*b*d + 2*a*e))*(a + x*(b + c*x))^(5/2))/(d + e
*x) + ((a + x*(b + c*x))^(3/2)*(b^3*e^3 - 4*c^3*d^2*(4*d - 3*e*x) + b*c*e^2*(-15*b*d + 10*a*e + b*e*x) + 2*c^2
*e*(3*b*d*(5*d - 2*e*x) + 2*a*e*(-3*d + 2*e*x))))/(2*e^2) + (3*(2*e*(c*d^2 + e*(-(b*d) + a*e))*Sqrt[a + x*(b +
c*x)]*(b^3*e^3 + 8*c^3*d^2*(-2*d + e*x) + b*c*e^2*(-13*b*d + 8*a*e + b*e*x) + 4*c^2*e*(b*d*(7*d - 2*e*x) + a*
e*(-3*d + e*x))) + 2*Sqrt[c]*(16*c^2*d^2 + 3*b^2*e^2 + 4*c*e*(-4*b*d + a*e))*(c*d^2 + e*(-(b*d) + a*e))^2*ArcT
anh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])] + (2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e))^(3/2)*(16*c^2*d^
2 + b^2*e^2 + 4*c*e*(-4*b*d + 3*a*e))*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)])]))/(4*e^5)))/(2*(c*d^2 + e*(-(b*d) + a*e))^2))/(6*e)
________________________________________________________________________________________
IntegrateAlgebraic [F] time = 180.13, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}
Verification is not applicable to the result.
[In]
IntegrateAlgebraic[(a + b*x + c*x^2)^(5/2)/(d + e*x)^4,x]
[Out]
$Aborted
________________________________________________________________________________________
fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*x^2+b*x+a)^(5/2)/(e*x+d)^4,x, algorithm="fricas")
[Out]
Timed out
________________________________________________________________________________________
giac [B] time = 24.64, size = 2120, normalized size = 6.29
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*x^2+b*x+a)^(5/2)/(e*x+d)^4,x, algorithm="giac")
[Out]
-5/8*(32*c^3*d^3 - 48*b*c^2*d^2*e + 18*b^2*c*d*e^2 + 24*a*c^2*d*e^2 - b^3*e^3 - 12*a*b*c*e^3)*arctan(-((sqrt(c
)*x - sqrt(c*x^2 + b*x + a))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e - a*e^2))*e^(-6)/sqrt(-c*d^2 + b*d*e - a*e^2)
- 5/8*(16*c^3*d^2 - 16*b*c^2*d*e + 3*b^2*c*e^2 + 4*a*c^2*e^2)*e^(-6)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))*sqrt(c) + b))/sqrt(c) + 1/4*(2*c^2*x*e^(-4) - (16*c^3*d*e^10 - 9*b*c^2*e^11)*e^(-15)/c)*sqrt(c*x^2 + b*x
+ a) - 1/24*(1680*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*c^4*d^4*e + 1504*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3
*c^(9/2)*d^5 + 480*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*c^(7/2)*d^3*e^2 - 400*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))^3*b*c^(7/2)*d^4*e + 2256*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b*c^4*d^5 - 2160*(sqrt(c)*x - sqrt(c*x^2 +
b*x + a))^4*b*c^3*d^3*e^2 - 2412*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b^2*c^3*d^4*e - 2832*(sqrt(c)*x - sqrt
(c*x^2 + b*x + a))^2*a*c^4*d^4*e + 1128*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b^2*c^(7/2)*d^5 - 720*(sqrt(c)*x -
sqrt(c*x^2 + b*x + a))^5*b*c^(5/2)*d^2*e^3 - 1308*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^2*c^(5/2)*d^3*e^2 -
1808*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*c^(7/2)*d^3*e^2 - 1272*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b^3*c
^(5/2)*d^4*e - 2832*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*b*c^(7/2)*d^4*e + 188*b^3*c^3*d^5 + 666*(sqrt(c)*x -
sqrt(c*x^2 + b*x + a))^4*b^2*c^2*d^2*e^3 + 216*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a*c^3*d^2*e^3 + 462*(sqr
t(c)*x - sqrt(c*x^2 + b*x + a))^2*b^3*c^2*d^3*e^2 + 2952*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*b*c^3*d^3*e^2
- 188*b^4*c^2*d^4*e - 708*a*b^2*c^3*d^4*e + 306*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*b^2*c^(3/2)*d*e^4 + 216
*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*a*c^(5/2)*d*e^4 + 574*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^3*c^(3/2)
*d^2*e^3 + 3144*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*b*c^(5/2)*d^2*e^3 + 324*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))*b^4*c^(3/2)*d^3*e^2 + 3420*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*b^2*c^(5/2)*d^3*e^2 + 1776*(sqrt(c)*x -
sqrt(c*x^2 + b*x + a))*a^2*c^(7/2)*d^3*e^2 - 21*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*b^3*c*d*e^4 + 324*(sqrt
(c)*x - sqrt(c*x^2 + b*x + a))^4*a*b*c^2*d*e^4 + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b^4*c*d^2*e^3 + 144*
(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a*b^2*c^2*d^2*e^3 + 33*b^5*c*d^3*e^2 + 746*a*b^3*c^2*d^3*e^2 + 888*a^2*b
*c^3*d^3*e^2 - 33*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*b^3*sqrt(c)*e^5 - 108*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))^5*a*b*c^(3/2)*e^5 - 40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b^4*sqrt(c)*d*e^4 - 912*(sqrt(c)*x - sqrt(c*x
^2 + b*x + a))^3*a*b^2*c^(3/2)*d*e^4 - 672*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^2*c^(5/2)*d*e^4 - 15*(sqrt(
c)*x - sqrt(c*x^2 + b*x + a))*b^5*sqrt(c)*d^2*e^3 - 774*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*b^3*c^(3/2)*d^2*
e^3 - 2664*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^2*b*c^(5/2)*d^2*e^3 - 144*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))
^4*a*b^2*c*e^5 - 144*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*a^2*c^2*e^5 - 168*(sqrt(c)*x - sqrt(c*x^2 + b*x + a
))^2*a*b^3*c*d*e^4 - 1008*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^2*b*c^2*d*e^4 - 114*a*b^4*c*d^2*e^3 - 1050*a
^2*b^2*c^2*d^2*e^3 - 376*a^3*c^3*d^2*e^3 + 40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*a*b^3*sqrt(c)*e^5 + 48*(sq
rt(c)*x - sqrt(c*x^2 + b*x + a))^3*a^2*b*c^(3/2)*e^5 + 30*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*b^4*sqrt(c)*d*
e^4 + 486*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^2*b^2*c^(3/2)*d*e^4 + 456*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*
a^3*c^(5/2)*d*e^4 + 144*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*a^2*b^2*c*e^5 + 192*(sqrt(c)*x - sqrt(c*x^2 + b*
x + a))^2*a^3*c^2*e^5 + 129*a^2*b^3*c*d*e^4 + 604*a^3*b*c^2*d*e^4 - 15*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^2
*b^3*sqrt(c)*e^5 - 36*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a^3*b*c^(3/2)*e^5 - 48*a^3*b^2*c*e^5 - 112*a^4*c^2*e
^5)*e^(-6)/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*e + 2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c)*d + b*d -
a*e)^3*sqrt(c))
________________________________________________________________________________________
maple [B] time = 0.10, size = 18718, normalized size = 55.54 \begin {gather*} \text {output too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c*x^2+b*x+a)^(5/2)/(e*x+d)^4,x)
[Out]
result too large to display
________________________________________________________________________________________
maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*x^2+b*x+a)^(5/2)/(e*x+d)^4,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 \begin {gather*} \int \frac {{\left (c\,x^2+b\,x+a\right )}^{5/2}}{{\left (d+e\,x\right )}^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*x + c*x^2)^(5/2)/(d + e*x)^4,x)
[Out]
int((a + b*x + c*x^2)^(5/2)/(d + e*x)^4, x)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x + c x^{2}\right )^{\frac {5}{2}}}{\left (d + e x\right )^{4}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c*x**2+b*x+a)**(5/2)/(e*x+d)**4,x)
[Out]
Integral((a + b*x + c*x**2)**(5/2)/(d + e*x)**4, x)
________________________________________________________________________________________