3.2333 \(\int (d+e x)^3 \sqrt {a+b x+c x^2} \, dx\)

Optimal. Leaf size=248 \[ -\frac {\left (b^2-4 a c\right ) (2 c d-b e) \left (-4 c e (3 a e+4 b d)+7 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{9/2}}+\frac {(b+2 c x) \sqrt {a+b x+c x^2} (2 c d-b e) \left (-4 c e (3 a e+4 b d)+7 b^2 e^2+16 c^2 d^2\right )}{128 c^4}+\frac {e \left (a+b x+c x^2\right )^{3/2} \left (-2 c e (16 a e+75 b d)+35 b^2 e^2+42 c e x (2 c d-b e)+192 c^2 d^2\right )}{240 c^3}+\frac {e (d+e x)^2 \left (a+b x+c x^2\right )^{3/2}}{5 c} \]

[Out]

1/5*e*(e*x+d)^2*(c*x^2+b*x+a)^(3/2)/c+1/240*e*(192*c^2*d^2+35*b^2*e^2-2*c*e*(16*a*e+75*b*d)+42*c*e*(-b*e+2*c*d
)*x)*(c*x^2+b*x+a)^(3/2)/c^3-1/256*(-4*a*c+b^2)*(-b*e+2*c*d)*(16*c^2*d^2+7*b^2*e^2-4*c*e*(3*a*e+4*b*d))*arctan
h(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(9/2)+1/128*(-b*e+2*c*d)*(16*c^2*d^2+7*b^2*e^2-4*c*e*(3*a*e+4*b
*d))*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^4

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Rubi [A]  time = 0.25, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {742, 779, 612, 621, 206} \[ \frac {e \left (a+b x+c x^2\right )^{3/2} \left (-2 c e (16 a e+75 b d)+35 b^2 e^2+42 c e x (2 c d-b e)+192 c^2 d^2\right )}{240 c^3}+\frac {(b+2 c x) \sqrt {a+b x+c x^2} (2 c d-b e) \left (-4 c e (3 a e+4 b d)+7 b^2 e^2+16 c^2 d^2\right )}{128 c^4}-\frac {\left (b^2-4 a c\right ) (2 c d-b e) \left (-4 c e (3 a e+4 b d)+7 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{9/2}}+\frac {e (d+e x)^2 \left (a+b x+c x^2\right )^{3/2}}{5 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3*Sqrt[a + b*x + c*x^2],x]

[Out]

((2*c*d - b*e)*(16*c^2*d^2 + 7*b^2*e^2 - 4*c*e*(4*b*d + 3*a*e))*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(128*c^4) +
(e*(d + e*x)^2*(a + b*x + c*x^2)^(3/2))/(5*c) + (e*(192*c^2*d^2 + 35*b^2*e^2 - 2*c*e*(75*b*d + 16*a*e) + 42*c
*e*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(3/2))/(240*c^3) - ((b^2 - 4*a*c)*(2*c*d - b*e)*(16*c^2*d^2 + 7*b^2*e^2
- 4*c*e*(4*b*d + 3*a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(256*c^(9/2))

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 612

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

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 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, 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]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rule 779

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

Rubi steps

\begin {align*} \int (d+e x)^3 \sqrt {a+b x+c x^2} \, dx &=\frac {e (d+e x)^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac {\int (d+e x) \left (\frac {1}{2} \left (10 c d^2-e (3 b d+4 a e)\right )+\frac {7}{2} e (2 c d-b e) x\right ) \sqrt {a+b x+c x^2} \, dx}{5 c}\\ &=\frac {e (d+e x)^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac {e \left (192 c^2 d^2+35 b^2 e^2-2 c e (75 b d+16 a e)+42 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3}+\frac {\left ((2 c d-b e) \left (16 c^2 d^2+7 b^2 e^2-4 c e (4 b d+3 a e)\right )\right ) \int \sqrt {a+b x+c x^2} \, dx}{32 c^3}\\ &=\frac {(2 c d-b e) \left (16 c^2 d^2+7 b^2 e^2-4 c e (4 b d+3 a e)\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^4}+\frac {e (d+e x)^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac {e \left (192 c^2 d^2+35 b^2 e^2-2 c e (75 b d+16 a e)+42 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3}-\frac {\left (\left (b^2-4 a c\right ) (2 c d-b e) \left (16 c^2 d^2+7 b^2 e^2-4 c e (4 b d+3 a e)\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{256 c^4}\\ &=\frac {(2 c d-b e) \left (16 c^2 d^2+7 b^2 e^2-4 c e (4 b d+3 a e)\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^4}+\frac {e (d+e x)^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac {e \left (192 c^2 d^2+35 b^2 e^2-2 c e (75 b d+16 a e)+42 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3}-\frac {\left (\left (b^2-4 a c\right ) (2 c d-b e) \left (16 c^2 d^2+7 b^2 e^2-4 c e (4 b d+3 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 )}{128 c^4}\\ &=\frac {(2 c d-b e) \left (16 c^2 d^2+7 b^2 e^2-4 c e (4 b d+3 a e)\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^4}+\frac {e (d+e x)^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac {e \left (192 c^2 d^2+35 b^2 e^2-2 c e (75 b d+16 a e)+42 c e (2 c d-b e) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3}-\frac {\left (b^2-4 a c\right ) (2 c d-b e) \left (16 c^2 d^2+7 b^2 e^2-4 c e (4 b d+3 a e)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 206, normalized size = 0.83 \[ \frac {\frac {e (a+x (b+c x))^{3/2} \left (-2 c e (16 a e+75 b d+21 b e x)+35 b^2 e^2+12 c^2 d (16 d+7 e x)\right )}{48 c^2}+\frac {5 (2 c d-b e) \left (-4 c e (3 a e+4 b d)+7 b^2 e^2+16 c^2 d^2\right ) \left (2 \sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)}-\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )}{256 c^{7/2}}+e (d+e x)^2 (a+x (b+c x))^{3/2}}{5 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3*Sqrt[a + b*x + c*x^2],x]

[Out]

(e*(d + e*x)^2*(a + x*(b + c*x))^(3/2) + (e*(a + x*(b + c*x))^(3/2)*(35*b^2*e^2 + 12*c^2*d*(16*d + 7*e*x) - 2*
c*e*(75*b*d + 16*a*e + 21*b*e*x)))/(48*c^2) + (5*(2*c*d - b*e)*(16*c^2*d^2 + 7*b^2*e^2 - 4*c*e*(4*b*d + 3*a*e)
)*(2*Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b + c*x)] - (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b +
c*x)])]))/(256*c^(7/2)))/(5*c)

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fricas [A]  time = 1.02, size = 787, normalized size = 3.17 \[ \left [-\frac {15 \, {\left (32 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} d^{3} - 48 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} d^{2} e + 6 \, {\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d e^{2} - {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} e^{3}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (384 \, c^{5} e^{3} x^{4} + 480 \, b c^{4} d^{3} - 240 \, {\left (3 \, b^{2} c^{3} - 8 \, a c^{4}\right )} d^{2} e + 30 \, {\left (15 \, b^{3} c^{2} - 52 \, a b c^{3}\right )} d e^{2} - {\left (105 \, b^{4} c - 460 \, a b^{2} c^{2} + 256 \, a^{2} c^{3}\right )} e^{3} + 48 \, {\left (30 \, c^{5} d e^{2} + b c^{4} e^{3}\right )} x^{3} + 8 \, {\left (240 \, c^{5} d^{2} e + 30 \, b c^{4} d e^{2} - {\left (7 \, b^{2} c^{3} - 16 \, a c^{4}\right )} e^{3}\right )} x^{2} + 2 \, {\left (480 \, c^{5} d^{3} + 240 \, b c^{4} d^{2} e - 30 \, {\left (5 \, b^{2} c^{3} - 12 \, a c^{4}\right )} d e^{2} + {\left (35 \, b^{3} c^{2} - 116 \, a b c^{3}\right )} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{7680 \, c^{5}}, \frac {15 \, {\left (32 \, {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} d^{3} - 48 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} d^{2} e + 6 \, {\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} d e^{2} - {\left (7 \, b^{5} - 40 \, a b^{3} c + 48 \, a^{2} b c^{2}\right )} e^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (384 \, c^{5} e^{3} x^{4} + 480 \, b c^{4} d^{3} - 240 \, {\left (3 \, b^{2} c^{3} - 8 \, a c^{4}\right )} d^{2} e + 30 \, {\left (15 \, b^{3} c^{2} - 52 \, a b c^{3}\right )} d e^{2} - {\left (105 \, b^{4} c - 460 \, a b^{2} c^{2} + 256 \, a^{2} c^{3}\right )} e^{3} + 48 \, {\left (30 \, c^{5} d e^{2} + b c^{4} e^{3}\right )} x^{3} + 8 \, {\left (240 \, c^{5} d^{2} e + 30 \, b c^{4} d e^{2} - {\left (7 \, b^{2} c^{3} - 16 \, a c^{4}\right )} e^{3}\right )} x^{2} + 2 \, {\left (480 \, c^{5} d^{3} + 240 \, b c^{4} d^{2} e - 30 \, {\left (5 \, b^{2} c^{3} - 12 \, a c^{4}\right )} d e^{2} + {\left (35 \, b^{3} c^{2} - 116 \, a b c^{3}\right )} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3840 \, c^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/7680*(15*(32*(b^2*c^3 - 4*a*c^4)*d^3 - 48*(b^3*c^2 - 4*a*b*c^3)*d^2*e + 6*(5*b^4*c - 24*a*b^2*c^2 + 16*a^2
*c^3)*d*e^2 - (7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*e^3)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 +
b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(384*c^5*e^3*x^4 + 480*b*c^4*d^3 - 240*(3*b^2*c^3 - 8*a*c^4)*d^2*e
+ 30*(15*b^3*c^2 - 52*a*b*c^3)*d*e^2 - (105*b^4*c - 460*a*b^2*c^2 + 256*a^2*c^3)*e^3 + 48*(30*c^5*d*e^2 + b*c^
4*e^3)*x^3 + 8*(240*c^5*d^2*e + 30*b*c^4*d*e^2 - (7*b^2*c^3 - 16*a*c^4)*e^3)*x^2 + 2*(480*c^5*d^3 + 240*b*c^4*
d^2*e - 30*(5*b^2*c^3 - 12*a*c^4)*d*e^2 + (35*b^3*c^2 - 116*a*b*c^3)*e^3)*x)*sqrt(c*x^2 + b*x + a))/c^5, 1/384
0*(15*(32*(b^2*c^3 - 4*a*c^4)*d^3 - 48*(b^3*c^2 - 4*a*b*c^3)*d^2*e + 6*(5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*d
*e^2 - (7*b^5 - 40*a*b^3*c + 48*a^2*b*c^2)*e^3)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)
/(c^2*x^2 + b*c*x + a*c)) + 2*(384*c^5*e^3*x^4 + 480*b*c^4*d^3 - 240*(3*b^2*c^3 - 8*a*c^4)*d^2*e + 30*(15*b^3*
c^2 - 52*a*b*c^3)*d*e^2 - (105*b^4*c - 460*a*b^2*c^2 + 256*a^2*c^3)*e^3 + 48*(30*c^5*d*e^2 + b*c^4*e^3)*x^3 +
8*(240*c^5*d^2*e + 30*b*c^4*d*e^2 - (7*b^2*c^3 - 16*a*c^4)*e^3)*x^2 + 2*(480*c^5*d^3 + 240*b*c^4*d^2*e - 30*(5
*b^2*c^3 - 12*a*c^4)*d*e^2 + (35*b^3*c^2 - 116*a*b*c^3)*e^3)*x)*sqrt(c*x^2 + b*x + a))/c^5]

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giac [A]  time = 0.23, size = 380, normalized size = 1.53 \[ \frac {1}{1920} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, x e^{3} + \frac {30 \, c^{4} d e^{2} + b c^{3} e^{3}}{c^{4}}\right )} x + \frac {240 \, c^{4} d^{2} e + 30 \, b c^{3} d e^{2} - 7 \, b^{2} c^{2} e^{3} + 16 \, a c^{3} e^{3}}{c^{4}}\right )} x + \frac {480 \, c^{4} d^{3} + 240 \, b c^{3} d^{2} e - 150 \, b^{2} c^{2} d e^{2} + 360 \, a c^{3} d e^{2} + 35 \, b^{3} c e^{3} - 116 \, a b c^{2} e^{3}}{c^{4}}\right )} x + \frac {480 \, b c^{3} d^{3} - 720 \, b^{2} c^{2} d^{2} e + 1920 \, a c^{3} d^{2} e + 450 \, b^{3} c d e^{2} - 1560 \, a b c^{2} d e^{2} - 105 \, b^{4} e^{3} + 460 \, a b^{2} c e^{3} - 256 \, a^{2} c^{2} e^{3}}{c^{4}}\right )} + \frac {{\left (32 \, b^{2} c^{3} d^{3} - 128 \, a c^{4} d^{3} - 48 \, b^{3} c^{2} d^{2} e + 192 \, a b c^{3} d^{2} e + 30 \, b^{4} c d e^{2} - 144 \, a b^{2} c^{2} d e^{2} + 96 \, a^{2} c^{3} d e^{2} - 7 \, b^{5} e^{3} + 40 \, a b^{3} c e^{3} - 48 \, a^{2} b c^{2} e^{3}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{256 \, c^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/1920*sqrt(c*x^2 + b*x + a)*(2*(4*(6*(8*x*e^3 + (30*c^4*d*e^2 + b*c^3*e^3)/c^4)*x + (240*c^4*d^2*e + 30*b*c^3
*d*e^2 - 7*b^2*c^2*e^3 + 16*a*c^3*e^3)/c^4)*x + (480*c^4*d^3 + 240*b*c^3*d^2*e - 150*b^2*c^2*d*e^2 + 360*a*c^3
*d*e^2 + 35*b^3*c*e^3 - 116*a*b*c^2*e^3)/c^4)*x + (480*b*c^3*d^3 - 720*b^2*c^2*d^2*e + 1920*a*c^3*d^2*e + 450*
b^3*c*d*e^2 - 1560*a*b*c^2*d*e^2 - 105*b^4*e^3 + 460*a*b^2*c*e^3 - 256*a^2*c^2*e^3)/c^4) + 1/256*(32*b^2*c^3*d
^3 - 128*a*c^4*d^3 - 48*b^3*c^2*d^2*e + 192*a*b*c^3*d^2*e + 30*b^4*c*d*e^2 - 144*a*b^2*c^2*d*e^2 + 96*a^2*c^3*
d*e^2 - 7*b^5*e^3 + 40*a*b^3*c*e^3 - 48*a^2*b*c^2*e^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c)
- b))/c^(9/2)

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maple [B]  time = 0.10, size = 795, normalized size = 3.21 \[ \frac {3 a^{2} b \,e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {5}{2}}}-\frac {3 a^{2} d \,e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}-\frac {5 a \,b^{3} e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{32 c^{\frac {7}{2}}}+\frac {9 a \,b^{2} d \,e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {5}{2}}}-\frac {3 a b \,d^{2} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {3}{2}}}+\frac {a \,d^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}+\frac {7 b^{5} e^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{256 c^{\frac {9}{2}}}-\frac {15 b^{4} d \,e^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {7}{2}}}+\frac {3 b^{3} d^{2} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {5}{2}}}-\frac {b^{2} d^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, a b \,e^{3} x}{16 c^{2}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, a d \,e^{2} x}{8 c}-\frac {7 \sqrt {c \,x^{2}+b x +a}\, b^{3} e^{3} x}{64 c^{3}}+\frac {15 \sqrt {c \,x^{2}+b x +a}\, b^{2} d \,e^{2} x}{32 c^{2}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, b \,d^{2} e x}{4 c}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} e^{3} x^{2}}{5 c}+\frac {\sqrt {c \,x^{2}+b x +a}\, d^{3} x}{2}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, a \,b^{2} e^{3}}{32 c^{3}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, a b d \,e^{2}}{16 c^{2}}-\frac {7 \sqrt {c \,x^{2}+b x +a}\, b^{4} e^{3}}{128 c^{4}}+\frac {15 \sqrt {c \,x^{2}+b x +a}\, b^{3} d \,e^{2}}{64 c^{3}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, b^{2} d^{2} e}{8 c^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, b \,d^{3}}{4 c}-\frac {7 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b \,e^{3} x}{40 c^{2}}+\frac {3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} d \,e^{2} x}{4 c}-\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a \,e^{3}}{15 c^{2}}+\frac {7 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{2} e^{3}}{48 c^{3}}-\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b d \,e^{2}}{8 c^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} d^{2} e}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(c*x^2+b*x+a)^(1/2),x)

[Out]

-7/40*e^3*b/c^2*x*(c*x^2+b*x+a)^(3/2)-3/8*d^2*e*b^2/c^2*(c*x^2+b*x+a)^(1/2)+3/16*d^2*e*b^3/c^(5/2)*ln((c*x+1/2
*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+7/256*e^3*b^5/c^(9/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-2/15*e^3*a/
c^2*(c*x^2+b*x+a)^(3/2)+1/5*e^3*x^2*(c*x^2+b*x+a)^(3/2)/c+7/48*e^3*b^2/c^3*(c*x^2+b*x+a)^(3/2)+1/2*d^3/c^(1/2)
*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a+d^2*e*(c*x^2+b*x+a)^(3/2)/c-7/128*e^3*b^4/c^4*(c*x^2+b*x+a)^(1/
2)-1/8*d^3/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^2+1/4*d^3/c*(c*x^2+b*x+a)^(1/2)*b+15/32*d*e^2
*b^2/c^2*x*(c*x^2+b*x+a)^(1/2)+9/16*d*e^2*b^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a+3/16*e^3*b
/c^2*a*x*(c*x^2+b*x+a)^(1/2)-3/4*d^2*e*b/c*x*(c*x^2+b*x+a)^(1/2)-3/4*d^2*e*b/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c
*x^2+b*x+a)^(1/2))*a-3/16*d*e^2*a/c^2*(c*x^2+b*x+a)^(1/2)*b-3/8*d*e^2*a/c*x*(c*x^2+b*x+a)^(1/2)+1/2*d^3*x*(c*x
^2+b*x+a)^(1/2)-7/64*e^3*b^3/c^3*x*(c*x^2+b*x+a)^(1/2)-5/32*e^3*b^3/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+
a)^(1/2))*a+3/16*e^3*b/c^(5/2)*a^2*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+3/4*d*e^2*x*(c*x^2+b*x+a)^(3/2)
/c-5/8*d*e^2*b/c^2*(c*x^2+b*x+a)^(3/2)+3/32*e^3*b^2/c^3*a*(c*x^2+b*x+a)^(1/2)+15/64*d*e^2*b^3/c^3*(c*x^2+b*x+a
)^(1/2)-15/128*d*e^2*b^4/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-3/8*d*e^2*a^2/c^(3/2)*ln((c*x+1/2
*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))

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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((e*x+d)^3*(c*x^2+b*x+a)^(1/2),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(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [B]  time = 2.25, size = 632, normalized size = 2.55 \[ d^3\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {7\,b\,e^3\,\left (\frac {5\,b\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}-\frac {x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c}+\frac {a\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{4\,c}\right )}{10\,c}+\frac {e^3\,x^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{5\,c}+\frac {d^3\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}-\frac {2\,a\,e^3\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{5\,c}-\frac {3\,a\,d\,e^2\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{4\,c}+\frac {3\,d^2\,e\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}-\frac {15\,b\,d\,e^2\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}+\frac {d^2\,e\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{8\,c^2}+\frac {3\,d\,e^2\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3*(a + b*x + c*x^2)^(1/2),x)

[Out]

d^3*(x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (7*b*e^3*((5*b*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(
1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2)))
/(8*c) - (x*(a + b*x + c*x^2)^(3/2))/(4*c) + (a*((x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (log((b/2 + c*x)/c^
(1/2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/(4*c)))/(10*c) + (e^3*x^2*(a + b*x + c*x^2)^(3/2
))/(5*c) + (d^3*log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2)) - (2*a*e^3*((log
((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 +
2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2)))/(5*c) - (3*a*d*e^2*((x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (lo
g((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/(4*c) + (3*d^2*e*log((b + 2*c*x)
/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) - (15*b*d*e^2*((log((b + 2*c*x)/c^(1/2) +
2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x
^2)^(1/2))/(24*c^2)))/(8*c) + (d^2*e*(8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(8*c^2) + (3
*d*e^2*x*(a + b*x + c*x^2)^(3/2))/(4*c)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d + e x\right )^{3} \sqrt {a + b x + c x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((d + e*x)**3*sqrt(a + b*x + c*x**2), x)

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