3.24.59 \(\int \frac {(a+b x+c x^2)^{5/2}}{d+e x} \, dx\) [2359]

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

[Out]

1/48*(16*c^2*d^2+3*b^2*e^2-2*c*e*(-8*a*e+11*b*d)-6*c*e*(-b*e+2*c*d)*x)*(c*x^2+b*x+a)^(3/2)/c/e^3+1/5*(c*x^2+b*
x+a)^(5/2)/e-1/256*(-b*e+2*c*d)*(128*c^4*d^4+3*b^4*e^4+8*b^2*c*e^3*(-5*a*e+2*b*d)-64*c^3*d^2*e*(-5*a*e+4*b*d)+
16*c^2*e^2*(15*a^2*e^2-20*a*b*d*e+7*b^2*d^2))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(5/2)/e^6+(
a*e^2-b*d*e+c*d^2)^(5/2)*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+1/128*(128*c^4*d^4-3*b^4*e^4-2*b^2*c*e^3*(-14*a*e+5*b*d)-32*c^3*d^2*e*(-8*a*e+9*b*d)+8*c^2*e^2*(16*a^2*e^
2-39*a*b*d*e+22*b^2*d^2)-2*c*e*(-b*e+2*c*d)*(16*c^2*d^2-3*b^2*e^2-4*c*e*(-7*a*e+4*b*d))*x)*(c*x^2+b*x+a)^(1/2)
/c^2/e^5

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Rubi [A]
time = 0.48, antiderivative size = 459, 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 = {748, 828, 857, 635, 212, 738} \begin {gather*} \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} \end {gather*}

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 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 748

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 828

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 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} \, 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 ) \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 )}{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 ) \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 )}{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*}

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Mathematica [A]
time = 10.81, size = 440, normalized size = 0.96 \begin {gather*} \frac {(a+x (b+c x))^{5/2}}{5 e}+\frac {(a+x (b+c x))^{3/2} \left (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)\right )}{48 c e^3}-\frac {\frac {2 e \sqrt {a+x (b+c x)} \left (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 \left (32 a^2 e^2+b^2 d (44 d-5 e x)+2 a b e (-39 d+7 e x)\right )+16 c^3 d e (6 b d (3 d-e x)+a e (-16 d+7 e x))\right )}{c^2}+\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+x (b+c x)}}\right )}{c^{5/2}}+256 \left (c d^2+e (-b d+a e)\right )^{5/2} \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 )}{256 e^6} \end {gather*}

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)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1045\) vs. \(2(429)=858\).
time = 0.88, size = 1046, normalized size = 2.28

method	result	size
default	\(\frac {\frac {\left (c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}\right )^{\frac {5}{2}}}{5}+\frac {\left (b e -2 c d \right ) \left (\frac {\left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right ) \left (c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (\frac {4 c \left (e^{2} a -b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \left (\frac {\left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}{4 c}+\frac {\left (\frac {4 c \left (e^{2} a -b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \ln \left (\frac {\frac {b e -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )}{2 e}+\frac {\left (e^{2} a -b d e +c \,d^{2}\right ) \left (\frac {\left (c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}}}{3}+\frac {\left (b e -2 c d \right ) \left (\frac {\left (2 c \left (x +\frac {d}{e}\right )+\frac {b e -2 c d}{e}\right ) \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}{4 c}+\frac {\left (\frac {4 c \left (e^{2} a -b d e +c \,d^{2}\right )}{e^{2}}-\frac {\left (b e -2 c d \right )^{2}}{e^{2}}\right ) \ln \left (\frac {\frac {b e -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\right )}{8 c^{\frac {3}{2}}}\right )}{2 e}+\frac {\left (e^{2} a -b d e +c \,d^{2}\right ) \left (\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}+\frac {\left (b e -2 c d \right ) \ln \left (\frac {\frac {b e -2 c d}{2 e}+c \left (x +\frac {d}{e}\right )}{\sqrt {c}}+\sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\right )}{2 e \sqrt {c}}-\frac {\left (e^{2} a -b d e +c \,d^{2}\right ) \ln \left (\frac {\frac {2 e^{2} a -2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{e^{2} \sqrt {\frac {e^{2} a -b d e +c \,d^{2}}{e^{2}}}}\right )}{e^{2}}\right )}{e^{2}}}{e}\)	\(1046\)
risch	\(\text {Expression too large to display}\)	\(2341\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e*(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)))))

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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),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),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}}}{d + e x}\, 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),x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

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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}}{d+e\,x} \,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),x)

[Out]

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

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