3.24.62 \(\int \frac {(a+b x+c x^2)^{5/2}}{(d+e x)^4} \, dx\) [2362]

Optimal. Leaf size=337 \[ -\frac {5 \left (16 c^2 d^2+b^2 e^2-4 c e (3 b d-a e)+4 c e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2}}{8 e^5 (d+e x)}+\frac {5 (4 c d-b e+2 c e x) \left (a+b x+c x^2\right )^{3/2}}{12 e^3 (d+e x)^2}-\frac {\left (a+b x+c x^2\right )^{5/2}}{3 e (d+e x)^3}+\frac {5 \sqrt {c} \left (16 c^2 d^2+3 b^2 e^2-4 c e (4 b d-a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 e^6}-\frac {5 (2 c d-b e) \left (16 c^2 d^2+b^2 e^2-4 c e (4 b d-3 a e)\right ) \tanh ^{-1}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{16 e^6 \sqrt {c d^2-b d e+a e^2}} \]

[Out]

5/12*(2*c*e*x-b*e+4*c*d)*(c*x^2+b*x+a)^(3/2)/e^3/(e*x+d)^2-1/3*(c*x^2+b*x+a)^(5/2)/e/(e*x+d)^3+5/8*(16*c^2*d^2
+3*b^2*e^2-4*c*e*(-a*e+4*b*d))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))*c^(1/2)/e^6-5/16*(-b*e+2*c*d
)*(16*c^2*d^2+b^2*e^2-4*c*e*(-3*a*e+4*b*d))*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)^(1/2)-5/8*(16*c^2*d^2+b^2*e^2-4*c*e*(-a*e+3*b*d)+4*c*e*(-b*e+2*c*d
)*x)*(c*x^2+b*x+a)^(1/2)/e^5/(e*x+d)

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Rubi [A]
time = 0.28, 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 = {746, 826, 857, 635, 212, 738} \begin {gather*} \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} \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 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 738

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 746

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 826

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 857

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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 11.99, size = 559, normalized size = 1.66 \begin {gather*} \frac {-\frac {2 (a+x (b+c x))^{5/2}}{(d+e x)^3}+\frac {5 (2 c d-b e) (a+x (b+c x))^{5/2}}{2 \left (c d^2+e (-b d+a e)\right ) (d+e x)^2}+\frac {5 \left (-\frac {\left (12 c^2 d^2+b^2 e^2+4 c e (-3 b d+2 a e)\right ) (a+x (b+c x))^{5/2}}{2 (d+e x)}+\frac {(a+x (b+c x))^{3/2} \left (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))\right )}{2 e^2}+\frac {3 \left (2 e \left (c d^2+e (-b d+a e)\right ) \sqrt {a+x (b+c x)} \left (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))\right )+2 \sqrt {c} \left (16 c^2 d^2+3 b^2 e^2+4 c e (-4 b d+a e)\right ) \left (c d^2+e (-b d+a e)\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 (c d^2+e (-b d+a e)\right )^{3/2} \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 x+b e x}{2 \sqrt {c d^2+e (-b d+a e)} \sqrt {a+x (b+c x)}}\right )\right )}{4 e^5}\right )}{2 \left (c d^2+e (-b d+a e)\right )^2}}{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)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(4597\) vs. \(2(305)=610\).
time = 1.07, size = 4598, normalized size = 13.64

method	result	size
default	\(\text {Expression too large to display}\)	\(4598\)
risch	\(\text {Expression too large to display}\)	\(16782\)

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,method=_RETURNVERBOSE)

[Out]

1/e^4*(-1/3/(a*e^2-b*d*e+c*d^2)*e^2/(x+d/e)^3*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(7
/2)+1/6*e*(b*e-2*c*d)/(a*e^2-b*d*e+c*d^2)*(-1/2/(a*e^2-b*d*e+c*d^2)*e^2/(x+d/e)^2*(c*(x+d/e)^2+1/e*(b*e-2*c*d)
*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(7/2)+3/4*e*(b*e-2*c*d)/(a*e^2-b*d*e+c*d^2)*(-1/(a*e^2-b*d*e+c*d^2)*e^2/(x+d
/e)*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(7/2)+5/2*e*(b*e-2*c*d)/(a*e^2-b*d*e+c*d^2)*
(1/5*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(5/2)+1/2/e*(b*e-2*c*d)*(1/8*(2*c*(x+d/e)+1
/e*(b*e-2*c*d))/c*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)+3/16*(4*c*(a*e^2-b*d*e+c
*d^2)/e^2-1/e^2*(b*e-2*c*d)^2)/c*(1/4*(2*c*(x+d/e)+1/e*(b*e-2*c*d))/c*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*
e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/8*(4*c*(a*e^2-b*d*e+c*d^2)/e^2-1/e^2*(b*e-2*c*d)^2)/c^(3/2)*ln((1/2/e*(b*e-2*c*d
)+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))))+(a*e^2-b*d*e+c*d^2
)/e^2*(1/3*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)+1/2/e*(b*e-2*c*d)*(1/4*(2*c*(x+
d/e)+1/e*(b*e-2*c*d))/c*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/8*(4*c*(a*e^2-b*
d*e+c*d^2)/e^2-1/e^2*(b*e-2*c*d)^2)/c^(3/2)*ln((1/2/e*(b*e-2*c*d)+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+1/e*(b*e-2*c
*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)))+(a*e^2-b*d*e+c*d^2)/e^2*((c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*
e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/2/e*(b*e-2*c*d)*ln((1/2/e*(b*e-2*c*d)+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+1/e*(b*e-2
*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/c^(1/2)-(a*e^2-b*d*e+c*d^2)/e^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*
ln((2*(a*e^2-b*d*e+c*d^2)/e^2+1/e*(b*e-2*c*d)*(x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2+1/e*(b*e-
2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e)))))+6*c/(a*e^2-b*d*e+c*d^2)*e^2*(1/12*(2*c*(x+d/e)+1/e*
(b*e-2*c*d))/c*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(5/2)+5/24*(4*c*(a*e^2-b*d*e+c*d^
2)/e^2-1/e^2*(b*e-2*c*d)^2)/c*(1/8*(2*c*(x+d/e)+1/e*(b*e-2*c*d))/c*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2
-b*d*e+c*d^2)/e^2)^(3/2)+3/16*(4*c*(a*e^2-b*d*e+c*d^2)/e^2-1/e^2*(b*e-2*c*d)^2)/c*(1/4*(2*c*(x+d/e)+1/e*(b*e-2
*c*d))/c*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/8*(4*c*(a*e^2-b*d*e+c*d^2)/e^2-
1/e^2*(b*e-2*c*d)^2)/c^(3/2)*ln((1/2/e*(b*e-2*c*d)+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*
e^2-b*d*e+c*d^2)/e^2)^(1/2))))))+5/2*c/(a*e^2-b*d*e+c*d^2)*e^2*(1/5*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^
2-b*d*e+c*d^2)/e^2)^(5/2)+1/2/e*(b*e-2*c*d)*(1/8*(2*c*(x+d/e)+1/e*(b*e-2*c*d))/c*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*
(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)+3/16*(4*c*(a*e^2-b*d*e+c*d^2)/e^2-1/e^2*(b*e-2*c*d)^2)/c*(1/4*(2*c*(x+d
/e)+1/e*(b*e-2*c*d))/c*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/8*(4*c*(a*e^2-b*d
*e+c*d^2)/e^2-1/e^2*(b*e-2*c*d)^2)/c^(3/2)*ln((1/2/e*(b*e-2*c*d)+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+1/e*(b*e-2*c*
d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))))+(a*e^2-b*d*e+c*d^2)/e^2*(1/3*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)
+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)+1/2/e*(b*e-2*c*d)*(1/4*(2*c*(x+d/e)+1/e*(b*e-2*c*d))/c*(c*(x+d/e)^2+1/e*(b*e-2
*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/8*(4*c*(a*e^2-b*d*e+c*d^2)/e^2-1/e^2*(b*e-2*c*d)^2)/c^(3/2)*ln(
(1/2/e*(b*e-2*c*d)+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)))+(a
*e^2-b*d*e+c*d^2)/e^2*((c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/2/e*(b*e-2*c*d)*l
n((1/2/e*(b*e-2*c*d)+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/c
^(1/2)-(a*e^2-b*d*e+c*d^2)/e^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+1/e*(b*e-2*c*d)*(
x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/
(x+d/e))))))+4/3*c/(a*e^2-b*d*e+c*d^2)*e^2*(-1/(a*e^2-b*d*e+c*d^2)*e^2/(x+d/e)*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x
+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(7/2)+5/2*e*(b*e-2*c*d)/(a*e^2-b*d*e+c*d^2)*(1/5*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(
x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(5/2)+1/2/e*(b*e-2*c*d)*(1/8*(2*c*(x+d/e)+1/e*(b*e-2*c*d))/c*(c*(x+d/e)^2+1/e*
(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)+3/16*(4*c*(a*e^2-b*d*e+c*d^2)/e^2-1/e^2*(b*e-2*c*d)^2)/c*(1
/4*(2*c*(x+d/e)+1/e*(b*e-2*c*d))/c*(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/8*(4*
c*(a*e^2-b*d*e+c*d^2)/e^2-1/e^2*(b*e-2*c*d)^2)/c^(3/2)*ln((1/2/e*(b*e-2*c*d)+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+1
/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))))+(a*e^2-b*d*e+c*d^2)/e^2*(1/3*(c*(x+d/e)^2+1/e*(b*e-2*
c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)+1/2/e*(b*e-2*c*d)*(1/4*(2*c*(x+d/e)+1/e*(b*e-2*c*d))/c*(c*(x+d/e)^
2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/8*(4*c*(a*e^2-b*d*e+c*d^2)/e^2-1/e^2*(b*e-2*c*d)^2)
/c^(3/2)*ln((1/2/e*(b*e-2*c*d)+c*(x+d/e))/c^(1/2)+(c*(x+d/e)^2+1/e*(b*e-2*c*d)*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2
)^(1/2)))+(a*e^2-b*d*e+c*d^2)/e^2*((c*(x+d/e)^2...

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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(c*d^2-%e*b*d+%e^2*a>0)', see `
assume?` for

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Fricas [F(-1)] Timed out
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

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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)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 2120 vs. \(2 (313) = 626\).
time = 12.99, size = 2120, normalized size = 6.29 \begin {gather*} \text {Too large to display} \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="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))

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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)

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