3.2359 \(\int \frac{(a+b x+c x^2)^{5/2}}{d+e x} \, dx\)

Optimal. Leaf size=459 \[ \frac{\sqrt{a+b x+c x^2} \left (8 c^2 e^2 \left (16 a^2 e^2-39 a b d e+22 b^2 d^2\right )-2 c e x (2 c d-b e) \left (-4 c e (4 b d-7 a e)-3 b^2 e^2+16 c^2 d^2\right )-2 b^2 c e^3 (5 b d-14 a e)-32 c^3 d^2 e (9 b d-8 a e)-3 b^4 e^4+128 c^4 d^4\right )}{128 c^2 e^5}-\frac{(2 c d-b e) \left (16 c^2 e^2 \left (15 a^2 e^2-20 a b d e+7 b^2 d^2\right )+8 b^2 c e^3 (2 b d-5 a e)-64 c^3 d^2 e (4 b d-5 a e)+3 b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{256 c^{5/2} e^6}+\frac{\left (a+b x+c x^2\right )^{3/2} \left (-2 c e (11 b d-8 a e)+3 b^2 e^2-6 c e x (2 c d-b e)+16 c^2 d^2\right )}{48 c e^3}+\frac{\left (a e^2-b d e+c d^2\right )^{5/2} \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{e^6}+\frac{\left (a+b x+c x^2\right )^{5/2}}{5 e} \]

[Out]

((128*c^4*d^4 - 3*b^4*e^4 - 2*b^2*c*e^3*(5*b*d - 14*a*e) - 32*c^3*d^2*e*(9*b*d - 8*a*e) + 8*c^2*e^2*(22*b^2*d^
2 - 39*a*b*d*e + 16*a^2*e^2) - 2*c*e*(2*c*d - b*e)*(16*c^2*d^2 - 3*b^2*e^2 - 4*c*e*(4*b*d - 7*a*e))*x)*Sqrt[a
+ b*x + c*x^2])/(128*c^2*e^5) + ((16*c^2*d^2 + 3*b^2*e^2 - 2*c*e*(11*b*d - 8*a*e) - 6*c*e*(2*c*d - b*e)*x)*(a
+ b*x + c*x^2)^(3/2))/(48*c*e^3) + (a + b*x + c*x^2)^(5/2)/(5*e) - ((2*c*d - b*e)*(128*c^4*d^4 + 3*b^4*e^4 + 8
*b^2*c*e^3*(2*b*d - 5*a*e) - 64*c^3*d^2*e*(4*b*d - 5*a*e) + 16*c^2*e^2*(7*b^2*d^2 - 20*a*b*d*e + 15*a^2*e^2))*
ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(256*c^(5/2)*e^6) + ((c*d^2 - b*d*e + a*e^2)^(5/2)*Arc
Tanh[(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])])/e^6

________________________________________________________________________________________

Rubi [A]  time = 0.749641, antiderivative size = 459, normalized size of antiderivative = 1., 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 = {734, 814, 843, 621, 206, 724} \[ \frac{\sqrt{a+b x+c x^2} \left (8 c^2 e^2 \left (16 a^2 e^2-39 a b d e+22 b^2 d^2\right )-2 c e x (2 c d-b e) \left (-4 c e (4 b d-7 a e)-3 b^2 e^2+16 c^2 d^2\right )-2 b^2 c e^3 (5 b d-14 a e)-32 c^3 d^2 e (9 b d-8 a e)-3 b^4 e^4+128 c^4 d^4\right )}{128 c^2 e^5}-\frac{(2 c d-b e) \left (16 c^2 e^2 \left (15 a^2 e^2-20 a b d e+7 b^2 d^2\right )+8 b^2 c e^3 (2 b d-5 a e)-64 c^3 d^2 e (4 b d-5 a e)+3 b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{256 c^{5/2} e^6}+\frac{\left (a+b x+c x^2\right )^{3/2} \left (-2 c e (11 b d-8 a e)+3 b^2 e^2-6 c e x (2 c d-b e)+16 c^2 d^2\right )}{48 c e^3}+\frac{\left (a e^2-b d e+c d^2\right )^{5/2} \tanh ^{-1}\left (\frac{-2 a e+x (2 c d-b e)+b d}{2 \sqrt{a+b x+c x^2} \sqrt{a e^2-b d e+c d^2}}\right )}{e^6}+\frac{\left (a+b x+c x^2\right )^{5/2}}{5 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((128*c^4*d^4 - 3*b^4*e^4 - 2*b^2*c*e^3*(5*b*d - 14*a*e) - 32*c^3*d^2*e*(9*b*d - 8*a*e) + 8*c^2*e^2*(22*b^2*d^
2 - 39*a*b*d*e + 16*a^2*e^2) - 2*c*e*(2*c*d - b*e)*(16*c^2*d^2 - 3*b^2*e^2 - 4*c*e*(4*b*d - 7*a*e))*x)*Sqrt[a
+ b*x + c*x^2])/(128*c^2*e^5) + ((16*c^2*d^2 + 3*b^2*e^2 - 2*c*e*(11*b*d - 8*a*e) - 6*c*e*(2*c*d - b*e)*x)*(a
+ b*x + c*x^2)^(3/2))/(48*c*e^3) + (a + b*x + c*x^2)^(5/2)/(5*e) - ((2*c*d - b*e)*(128*c^4*d^4 + 3*b^4*e^4 + 8
*b^2*c*e^3*(2*b*d - 5*a*e) - 64*c^3*d^2*e*(4*b*d - 5*a*e) + 16*c^2*e^2*(7*b^2*d^2 - 20*a*b*d*e + 15*a^2*e^2))*
ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(256*c^(5/2)*e^6) + ((c*d^2 - b*d*e + a*e^2)^(5/2)*Arc
Tanh[(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])])/e^6

Rule 734

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 + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, 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] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 814

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

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 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 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]

Rubi steps

\begin{align*} \int \frac{\left (a+b x+c x^2\right )^{5/2}}{d+e x} \, dx &=\frac{\left (a+b x+c x^2\right )^{5/2}}{5 e}-\frac{\int \frac{(b d-2 a e+(2 c d-b e) x) \left (a+b x+c x^2\right )^{3/2}}{d+e x} \, dx}{2 e}\\ &=\frac{\left (16 c^2 d^2+3 b^2 e^2-2 c e (11 b d-8 a e)-6 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{48 c e^3}+\frac{\left (a+b x+c x^2\right )^{5/2}}{5 e}+\frac{\int \frac{\left (\frac{1}{2} \left (8 c e (b d-2 a e)^2+2 (2 c d-b e) \left (2 a c d e-b d \left (4 c d-\frac{3 b e}{2}\right )\right )\right )-\frac{1}{2} (2 c d-b e) \left (16 c^2 d^2-3 b^2 e^2-4 c e (4 b d-7 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{d+e x} \, dx}{16 c e^3}\\ &=\frac{\left (128 c^4 d^4-3 b^4 e^4-2 b^2 c e^3 (5 b d-14 a e)-32 c^3 d^2 e (9 b d-8 a e)+8 c^2 e^2 \left (22 b^2 d^2-39 a b d e+16 a^2 e^2\right )-2 c e (2 c d-b e) \left (16 c^2 d^2-3 b^2 e^2-4 c e (4 b d-7 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{128 c^2 e^5}+\frac{\left (16 c^2 d^2+3 b^2 e^2-2 c e (11 b d-8 a e)-6 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{48 c e^3}+\frac{\left (a+b x+c x^2\right )^{5/2}}{5 e}-\frac{\int \frac{\frac{1}{4} \left (d (2 c d-b e) \left (4 b c d-b^2 e-4 a c e\right ) \left (16 c^2 d^2-3 b^2 e^2-4 c e (4 b d-7 a e)\right )+4 c e (b d-2 a e) \left (8 c e (b d-2 a e)^2-d (2 c d-b e) \left (8 b c d-3 b^2 e-4 a c e\right )\right )\right )+\frac{1}{4} (2 c d-b e) \left (128 c^4 d^4+3 b^4 e^4+8 b^2 c e^3 (2 b d-5 a e)-64 c^3 d^2 e (4 b d-5 a e)+16 c^2 e^2 \left (7 b^2 d^2-20 a b d e+15 a^2 e^2\right )\right ) x}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{64 c^2 e^5}\\ &=\frac{\left (128 c^4 d^4-3 b^4 e^4-2 b^2 c e^3 (5 b d-14 a e)-32 c^3 d^2 e (9 b d-8 a e)+8 c^2 e^2 \left (22 b^2 d^2-39 a b d e+16 a^2 e^2\right )-2 c e (2 c d-b e) \left (16 c^2 d^2-3 b^2 e^2-4 c e (4 b d-7 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{128 c^2 e^5}+\frac{\left (16 c^2 d^2+3 b^2 e^2-2 c e (11 b d-8 a e)-6 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{48 c e^3}+\frac{\left (a+b x+c x^2\right )^{5/2}}{5 e}+\frac{\left (c d^2-b d e+a e^2\right )^3 \int \frac{1}{(d+e x) \sqrt{a+b x+c x^2}} \, dx}{e^6}-\frac{\left ((2 c d-b e) \left (128 c^4 d^4+3 b^4 e^4+8 b^2 c e^3 (2 b d-5 a e)-64 c^3 d^2 e (4 b d-5 a e)+16 c^2 e^2 \left (7 b^2 d^2-20 a b d e+15 a^2 e^2\right )\right )\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{256 c^2 e^6}\\ &=\frac{\left (128 c^4 d^4-3 b^4 e^4-2 b^2 c e^3 (5 b d-14 a e)-32 c^3 d^2 e (9 b d-8 a e)+8 c^2 e^2 \left (22 b^2 d^2-39 a b d e+16 a^2 e^2\right )-2 c e (2 c d-b e) \left (16 c^2 d^2-3 b^2 e^2-4 c e (4 b d-7 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{128 c^2 e^5}+\frac{\left (16 c^2 d^2+3 b^2 e^2-2 c e (11 b d-8 a e)-6 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{48 c e^3}+\frac{\left (a+b x+c x^2\right )^{5/2}}{5 e}-\frac{\left (2 \left (c d^2-b d e+a e^2\right )^3\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 )}{e^6}-\frac{\left ((2 c d-b e) \left (128 c^4 d^4+3 b^4 e^4+8 b^2 c e^3 (2 b d-5 a e)-64 c^3 d^2 e (4 b d-5 a e)+16 c^2 e^2 \left (7 b^2 d^2-20 a b d e+15 a^2 e^2\right )\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 )}{128 c^2 e^6}\\ &=\frac{\left (128 c^4 d^4-3 b^4 e^4-2 b^2 c e^3 (5 b d-14 a e)-32 c^3 d^2 e (9 b d-8 a e)+8 c^2 e^2 \left (22 b^2 d^2-39 a b d e+16 a^2 e^2\right )-2 c e (2 c d-b e) \left (16 c^2 d^2-3 b^2 e^2-4 c e (4 b d-7 a e)\right ) x\right ) \sqrt{a+b x+c x^2}}{128 c^2 e^5}+\frac{\left (16 c^2 d^2+3 b^2 e^2-2 c e (11 b d-8 a e)-6 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{48 c e^3}+\frac{\left (a+b x+c x^2\right )^{5/2}}{5 e}-\frac{(2 c d-b e) \left (128 c^4 d^4+3 b^4 e^4+8 b^2 c e^3 (2 b d-5 a e)-64 c^3 d^2 e (4 b d-5 a e)+16 c^2 e^2 \left (7 b^2 d^2-20 a b d e+15 a^2 e^2\right )\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{256 c^{5/2} e^6}+\frac{\left (c d^2-b d e+a e^2\right )^{5/2} \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 )}{e^6}\\ \end{align*}

Mathematica [A]  time = 1.2842, size = 440, normalized size = 0.96 \[ -\frac{\frac{2 e \sqrt{a+x (b+c x)} \left (-4 c^2 e^2 \left (32 a^2 e^2+2 a b e (7 e x-39 d)+b^2 d (44 d-5 e x)\right )+2 b^2 c e^3 (-14 a e+5 b d+3 b e x)+16 c^3 d e (a e (7 e x-16 d)+6 b d (3 d-e x))+3 b^4 e^4-64 c^4 d^3 (2 d-e x)\right )}{c^2}+\frac{(2 c d-b e) \left (16 c^2 e^2 \left (15 a^2 e^2-20 a b d e+7 b^2 d^2\right )+8 b^2 c e^3 (2 b d-5 a e)-64 c^3 d^2 e (4 b d-5 a e)+3 b^4 e^4+128 c^4 d^4\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{c^{5/2}}+256 \left (e (a e-b d)+c d^2\right )^{5/2} \tanh ^{-1}\left (\frac{2 a e-b d+b e x-2 c d x}{2 \sqrt{a+x (b+c x)} \sqrt{e (a e-b d)+c d^2}}\right )}{256 e^6}+\frac{(a+x (b+c x))^{3/2} \left (2 c e (8 a e-11 b d+3 b e x)+3 b^2 e^2+4 c^2 d (4 d-3 e x)\right )}{48 c e^3}+\frac{(a+x (b+c x))^{5/2}}{5 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + x*(b + c*x))^(5/2)/(5*e) + ((a + x*(b + c*x))^(3/2)*(3*b^2*e^2 + 4*c^2*d*(4*d - 3*e*x) + 2*c*e*(-11*b*d +
8*a*e + 3*b*e*x)))/(48*c*e^3) - ((2*e*Sqrt[a + x*(b + c*x)]*(3*b^4*e^4 - 64*c^4*d^3*(2*d - e*x) + 2*b^2*c*e^3
*(5*b*d - 14*a*e + 3*b*e*x) - 4*c^2*e^2*(32*a^2*e^2 + b^2*d*(44*d - 5*e*x) + 2*a*b*e*(-39*d + 7*e*x)) + 16*c^3
*d*e*(6*b*d*(3*d - e*x) + a*e*(-16*d + 7*e*x))))/c^2 + ((2*c*d - b*e)*(128*c^4*d^4 + 3*b^4*e^4 + 8*b^2*c*e^3*(
2*b*d - 5*a*e) - 64*c^3*d^2*e*(4*b*d - 5*a*e) + 16*c^2*e^2*(7*b^2*d^2 - 20*a*b*d*e + 15*a^2*e^2))*ArcTanh[(b +
2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/c^(5/2) + 256*(c*d^2 + e*(-(b*d) + a*e))^(5/2)*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)])])/(256*e^6)

________________________________________________________________________________________

Maple [B]  time = 0.23, size = 4126, normalized size = 9. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/e*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(5/2)+1/3/e*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+
x)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)*a+1/e*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*a^2+
5/16/e^3*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^
(1/2))/c^(1/2)*b^3*d^2+6/e^4/((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+(b*e-2*c*d)/e*(d/e+
x)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(d/e+x
))*a*b*d^3*c-1/e^7/((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+(b*e-2*c*d)/e*(d/e+x)+2*((a*e
^2-b*d*e+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(d/e+x))*c^3*d^6
-1/2/e^4*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*c^2*d^3-9/4/e^4*((d/e+x)^2*c+(b*e
-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b*d^3*c+2/e^3*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e
+c*d^2)/e^2)^(1/2)*a*c*d^2-15/8/e^2*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x
)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(1/2)*d*a^2+5/128/e^2/c^(3/2)*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+((d
/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*b^4*d-5/2/e^4*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)
/c^(1/2)+((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(3/2)*d^3*a+5/2/e^5*ln((1/2*(b*e
-2*c*d)/e+(d/e+x)*c)/c^(1/2)+((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(3/2)*d^4*b-
15/8/e^4*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^
(1/2))*c^(1/2)*d^3*b^2+15/16/e/c^(1/2)*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+((d/e+x)^2*c+(b*e-2*c*d)/e*(d/
e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*a^2*b-5/32/e/c^(3/2)*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+((d/e+x)^2*
c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*b^3*a-1/4/e^2*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^
2-b*d*e+c*d^2)/e^2)^(3/2)*x*c*d+7/16/e*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*a*b
+1/e^4/((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+(b*e-2*c*d)/e*(d/e+x)+2*((a*e^2-b*d*e+c*d
^2)/e^2)^(1/2)*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(d/e+x))*b^3*d^3-3/64/e/c*((
d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*b^3-5/32/e^2*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e
+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*b^2*d+7/32/e/c*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2
)^(1/2)*b^2*a-39/16/e^2*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b*a*d-5/64/e^2/c*((d
/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b^3*d+3/256/e/c^(5/2)*ln((1/2*(b*e-2*c*d)/e+(d/
e+x)*c)/c^(1/2)+((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*b^5+1/e^5*((d/e+x)^2*c+(b*e
-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*c^2*d^4-1/e/((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+(b*e-2*c*d)/e*(d/e+x)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2
-b*d*e+c*d^2)/e^2)^(1/2))/(d/e+x))*a^3-1/e^6*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+((d/e+x)^2*c+(b*e-2*c*d)
/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(5/2)*d^5+1/3/e^3*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e
+c*d^2)/e^2)^(3/2)*c*d^2-7/8/e^2*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*a*c*d+3/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+(b*e-2*c*d)/e*(d/e+x)+2*((a*e^2-b*d*e+c*d^2)/
e^2)^(1/2)*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(d/e+x))*a^2*b*d+3/e^6/((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+(b*e-2*c*d)/e*(d/e+x)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(
(d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(d/e+x))*b*d^5*c^2-3/e^5/((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+(b*e-2*c*d)/e*(d/e+x)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((d/e+x)^2*
c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(d/e+x))*a*c^2*d^4+1/8/e*((d/e+x)^2*c+(b*e-2*c*d)/e*(d
/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)*x*b+1/16/e/c*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^
(3/2)*b^2-11/24/e^2*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)*b*d-3/128/e/c^2*((d/e+x)
^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*b^4+11/8/e^3*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e
^2-b*d*e+c*d^2)/e^2)^(1/2)*b^2*d^2-3/e^3/((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+(b*e-2*
c*d)/e*(d/e+x)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(
1/2))/(d/e+x))*a^2*c*d^2+3/4/e^3*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*x*b*c*d^2-1
5/16/e^2/c^(1/2)*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^
2)/e^2)^(1/2))*a*b^2*d+15/4/e^3*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a
*e^2-b*d*e+c*d^2)/e^2)^(1/2))*c^(1/2)*d^2*a*b-3/e^5/((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+(b*e-2*c*d)/e*(d/e+x)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*
d^2)/e^2)^(1/2))/(d/e+x))*b^2*d^4*c-3/e^3/((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+(b*e-2
*c*d)/e*(d/e+x)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((d/e+x)^2*c+(b*e-2*c*d)/e*(d/e+x)+(a*e^2-b*d*e+c*d^2)/e^2)^
(1/2))/(d/e+x))*a*b^2*d^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError