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

Optimal. Leaf size=317 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 d x+e) \sqrt {c+d x^2+e x} \left (2 a d \left (4 c d-5 e^2\right )-b \left (12 c d e-7 e^3\right )\right )}{128 d^4 (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2+e x\right )^{3/2} \left (-6 d x (10 a d-7 b e)+50 a d e+32 b c d-35 b e^2\right )}{240 d^3 (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (4 c d-e^2\right ) \left (8 a c d^2-10 a d e^2-12 b c d e+7 b e^3\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{256 d^{9/2} (a+b x)}+\frac {b x^2 \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2+e x\right )^{3/2}}{5 d (a+b x)} \]

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Rubi [A]  time = 0.33, antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1000, 832, 779, 612, 621, 206} \begin {gather*} -\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2+e x\right )^{3/2} \left (-6 d x (10 a d-7 b e)+50 a d e+32 b c d-35 b e^2\right )}{240 d^3 (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (2 d x+e) \sqrt {c+d x^2+e x} \left (2 a d \left (4 c d-5 e^2\right )-b \left (12 c d e-7 e^3\right )\right )}{128 d^4 (a+b x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (4 c d-e^2\right ) \left (8 a c d^2-10 a d e^2-12 b c d e+7 b e^3\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{256 d^{9/2} (a+b x)}+\frac {b x^2 \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2+e x\right )^{3/2}}{5 d (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a*d*(4*c*d - 5*e^2) - b*(12*c*d*e - 7*e^3))*(e + 2*d*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + e*x + d*x^
2])/(128*d^4*(a + b*x)) + (b*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(c + e*x + d*x^2)^(3/2))/(5*d*(a + b*x)) - ((32
*b*c*d + 50*a*d*e - 35*b*e^2 - 6*d*(10*a*d - 7*b*e)*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(c + e*x + d*x^2)^(3/2))/
(240*d^3*(a + b*x)) - ((4*c*d - e^2)*(8*a*c*d^2 - 12*b*c*d*e - 10*a*d*e^2 + 7*b*e^3)*Sqrt[a^2 + 2*a*b*x + b^2*
x^2]*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + e*x + d*x^2])])/(256*d^(9/2)*(a + b*x))

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 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]

Rule 832

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

Rule 1000

Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_)
, x_Symbol] :> Dist[(a + b*x + c*x^2)^FracPart[p]/((4*c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p])), Int[(g + h*x
)^m*(b + 2*c*x)^(2*p)*(d + e*x + f*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q}, x] && EqQ[b^2 -
4*a*c, 0]

Rubi steps

\begin {align*} \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^2 \left (2 a b+2 b^2 x\right ) \sqrt {c+e x+d x^2} \, dx}{2 a b+2 b^2 x}\\ &=\frac {b x^2 \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{5 d (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x \left (-4 b^2 c+b (10 a d-7 b e) x\right ) \sqrt {c+e x+d x^2} \, dx}{5 d \left (2 a b+2 b^2 x\right )}\\ &=\frac {b x^2 \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{5 d (a+b x)}-\frac {\left (32 b c d+50 a d e-35 b e^2-6 d (10 a d-7 b e) x\right ) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{240 d^3 (a+b x)}+\frac {\left (\left (16 b^2 c d e-2 b c d (10 a d-7 b e)+\frac {5}{2} b e^2 (10 a d-7 b e)\right ) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \sqrt {c+e x+d x^2} \, dx}{40 d^3 \left (2 a b+2 b^2 x\right )}\\ &=-\frac {\left (8 a c d^2-12 b c d e-10 a d e^2+7 b e^3\right ) (e+2 d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{128 d^4 (a+b x)}+\frac {b x^2 \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{5 d (a+b x)}-\frac {\left (32 b c d+50 a d e-35 b e^2-6 d (10 a d-7 b e) x\right ) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{240 d^3 (a+b x)}+\frac {\left (\left (4 c d-e^2\right ) \left (16 b^2 c d e-2 b c d (10 a d-7 b e)+\frac {5}{2} b e^2 (10 a d-7 b e)\right ) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {1}{\sqrt {c+e x+d x^2}} \, dx}{320 d^4 \left (2 a b+2 b^2 x\right )}\\ &=-\frac {\left (8 a c d^2-12 b c d e-10 a d e^2+7 b e^3\right ) (e+2 d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{128 d^4 (a+b x)}+\frac {b x^2 \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{5 d (a+b x)}-\frac {\left (32 b c d+50 a d e-35 b e^2-6 d (10 a d-7 b e) x\right ) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{240 d^3 (a+b x)}+\frac {\left (\left (4 c d-e^2\right ) \left (16 b^2 c d e-2 b c d (10 a d-7 b e)+\frac {5}{2} b e^2 (10 a d-7 b e)\right ) \sqrt {a^2+2 a b x+b^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{4 d-x^2} \, dx,x,\frac {e+2 d x}{\sqrt {c+e x+d x^2}}\right )}{160 d^4 \left (2 a b+2 b^2 x\right )}\\ &=-\frac {\left (8 a c d^2-12 b c d e-10 a d e^2+7 b e^3\right ) (e+2 d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{128 d^4 (a+b x)}+\frac {b x^2 \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{5 d (a+b x)}-\frac {\left (32 b c d+50 a d e-35 b e^2-6 d (10 a d-7 b e) x\right ) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+e x+d x^2\right )^{3/2}}{240 d^3 (a+b x)}-\frac {\left (4 c d-e^2\right ) \left (8 a c d^2-12 b c d e-10 a d e^2+7 b e^3\right ) \sqrt {a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{256 d^{9/2} (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 198, normalized size = 0.62 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (-\frac {5 \left (2 a d \left (4 c d-5 e^2\right )+b \left (7 e^3-12 c d e\right )\right ) \left (\left (4 c d-e^2\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+x (d x+e)}}\right )+2 \sqrt {d} (2 d x+e) \sqrt {c+x (d x+e)}\right )}{256 d^{7/2}}+\frac {(c+x (d x+e))^{3/2} (10 a d (6 d x-5 e)-32 b c d+7 b e (5 e-6 d x))}{48 d^2}+b x^2 (c+x (d x+e))^{3/2}\right )}{5 d (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a + b*x)^2]*(b*x^2*(c + x*(e + d*x))^(3/2) + ((c + x*(e + d*x))^(3/2)*(-32*b*c*d + 7*b*e*(5*e - 6*d*x)
+ 10*a*d*(-5*e + 6*d*x)))/(48*d^2) - (5*(2*a*d*(4*c*d - 5*e^2) + b*(-12*c*d*e + 7*e^3))*(2*Sqrt[d]*(e + 2*d*x)
*Sqrt[c + x*(e + d*x)] + (4*c*d - e^2)*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + x*(e + d*x)])]))/(256*d^(7/2)))
)/(5*d*(a + b*x))

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IntegrateAlgebraic [A]  time = 1.00, size = 263, normalized size = 0.83 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (\frac {\left (32 a c^2 d^3-48 a c d^2 e^2+10 a d e^4-48 b c^2 d^2 e+40 b c d e^3-7 b e^5\right ) \log \left (-2 \sqrt {d} \sqrt {c+d x^2+e x}+2 d x+e\right )}{256 d^{9/2}}+\frac {\sqrt {c+d x^2+e x} \left (240 a c d^3 x-520 a c d^2 e+480 a d^4 x^3+80 a d^3 e x^2-100 a d^2 e^2 x+150 a d e^3-256 b c^2 d^2+128 b c d^3 x^2-232 b c d^2 e x+460 b c d e^2+384 b d^4 x^4+48 b d^3 e x^3-56 b d^2 e^2 x^2+70 b d e^3 x-105 b e^4\right )}{1920 d^4}\right )}{a+b x} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + e*x + d*x^2],x]

[Out]

(Sqrt[(a + b*x)^2]*((Sqrt[c + e*x + d*x^2]*(-256*b*c^2*d^2 - 520*a*c*d^2*e + 460*b*c*d*e^2 + 150*a*d*e^3 - 105
*b*e^4 + 240*a*c*d^3*x - 232*b*c*d^2*e*x - 100*a*d^2*e^2*x + 70*b*d*e^3*x + 128*b*c*d^3*x^2 + 80*a*d^3*e*x^2 -
56*b*d^2*e^2*x^2 + 480*a*d^4*x^3 + 48*b*d^3*e*x^3 + 384*b*d^4*x^4))/(1920*d^4) + ((32*a*c^2*d^3 - 48*b*c^2*d^
2*e - 48*a*c*d^2*e^2 + 40*b*c*d*e^3 + 10*a*d*e^4 - 7*b*e^5)*Log[e + 2*d*x - 2*Sqrt[d]*Sqrt[c + e*x + d*x^2]])/
(256*d^(9/2))))/(a + b*x)

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fricas [A]  time = 0.47, size = 517, normalized size = 1.63 \begin {gather*} \left [-\frac {15 \, {\left (32 \, a c^{2} d^{3} - 48 \, b c^{2} d^{2} e - 48 \, a c d^{2} e^{2} + 40 \, b c d e^{3} + 10 \, a d e^{4} - 7 \, b e^{5}\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d e x + 4 \, \sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) - 4 \, {\left (384 \, b d^{5} x^{4} - 256 \, b c^{2} d^{3} - 520 \, a c d^{3} e + 460 \, b c d^{2} e^{2} + 150 \, a d^{2} e^{3} - 105 \, b d e^{4} + 48 \, {\left (10 \, a d^{5} + b d^{4} e\right )} x^{3} + 8 \, {\left (16 \, b c d^{4} + 10 \, a d^{4} e - 7 \, b d^{3} e^{2}\right )} x^{2} + 2 \, {\left (120 \, a c d^{4} - 116 \, b c d^{3} e - 50 \, a d^{3} e^{2} + 35 \, b d^{2} e^{3}\right )} x\right )} \sqrt {d x^{2} + e x + c}}{7680 \, d^{5}}, \frac {15 \, {\left (32 \, a c^{2} d^{3} - 48 \, b c^{2} d^{2} e - 48 \, a c d^{2} e^{2} + 40 \, b c d e^{3} + 10 \, a d e^{4} - 7 \, b e^{5}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d e x + c d\right )}}\right ) + 2 \, {\left (384 \, b d^{5} x^{4} - 256 \, b c^{2} d^{3} - 520 \, a c d^{3} e + 460 \, b c d^{2} e^{2} + 150 \, a d^{2} e^{3} - 105 \, b d e^{4} + 48 \, {\left (10 \, a d^{5} + b d^{4} e\right )} x^{3} + 8 \, {\left (16 \, b c d^{4} + 10 \, a d^{4} e - 7 \, b d^{3} e^{2}\right )} x^{2} + 2 \, {\left (120 \, a c d^{4} - 116 \, b c d^{3} e - 50 \, a d^{3} e^{2} + 35 \, b d^{2} e^{3}\right )} x\right )} \sqrt {d x^{2} + e x + c}}{3840 \, d^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/7680*(15*(32*a*c^2*d^3 - 48*b*c^2*d^2*e - 48*a*c*d^2*e^2 + 40*b*c*d*e^3 + 10*a*d*e^4 - 7*b*e^5)*sqrt(d)*lo
g(8*d^2*x^2 + 8*d*e*x + 4*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(d) + 4*c*d + e^2) - 4*(384*b*d^5*x^4 - 256*b*
c^2*d^3 - 520*a*c*d^3*e + 460*b*c*d^2*e^2 + 150*a*d^2*e^3 - 105*b*d*e^4 + 48*(10*a*d^5 + b*d^4*e)*x^3 + 8*(16*
b*c*d^4 + 10*a*d^4*e - 7*b*d^3*e^2)*x^2 + 2*(120*a*c*d^4 - 116*b*c*d^3*e - 50*a*d^3*e^2 + 35*b*d^2*e^3)*x)*sqr
t(d*x^2 + e*x + c))/d^5, 1/3840*(15*(32*a*c^2*d^3 - 48*b*c^2*d^2*e - 48*a*c*d^2*e^2 + 40*b*c*d*e^3 + 10*a*d*e^
4 - 7*b*e^5)*sqrt(-d)*arctan(1/2*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2 + d*e*x + c*d)) + 2*(384*
b*d^5*x^4 - 256*b*c^2*d^3 - 520*a*c*d^3*e + 460*b*c*d^2*e^2 + 150*a*d^2*e^3 - 105*b*d*e^4 + 48*(10*a*d^5 + b*d
^4*e)*x^3 + 8*(16*b*c*d^4 + 10*a*d^4*e - 7*b*d^3*e^2)*x^2 + 2*(120*a*c*d^4 - 116*b*c*d^3*e - 50*a*d^3*e^2 + 35
*b*d^2*e^3)*x)*sqrt(d*x^2 + e*x + c))/d^5]

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giac [A]  time = 0.31, size = 368, normalized size = 1.16 \begin {gather*} \frac {1}{1920} \, \sqrt {d x^{2} + x e + c} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, b x \mathrm {sgn}\left (b x + a\right ) + \frac {10 \, a d^{4} \mathrm {sgn}\left (b x + a\right ) + b d^{3} e \mathrm {sgn}\left (b x + a\right )}{d^{4}}\right )} x + \frac {16 \, b c d^{3} \mathrm {sgn}\left (b x + a\right ) + 10 \, a d^{3} e \mathrm {sgn}\left (b x + a\right ) - 7 \, b d^{2} e^{2} \mathrm {sgn}\left (b x + a\right )}{d^{4}}\right )} x + \frac {120 \, a c d^{3} \mathrm {sgn}\left (b x + a\right ) - 116 \, b c d^{2} e \mathrm {sgn}\left (b x + a\right ) - 50 \, a d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 35 \, b d e^{3} \mathrm {sgn}\left (b x + a\right )}{d^{4}}\right )} x - \frac {256 \, b c^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) + 520 \, a c d^{2} e \mathrm {sgn}\left (b x + a\right ) - 460 \, b c d e^{2} \mathrm {sgn}\left (b x + a\right ) - 150 \, a d e^{3} \mathrm {sgn}\left (b x + a\right ) + 105 \, b e^{4} \mathrm {sgn}\left (b x + a\right )}{d^{4}}\right )} + \frac {{\left (32 \, a c^{2} d^{3} \mathrm {sgn}\left (b x + a\right ) - 48 \, b c^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 48 \, a c d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 40 \, b c d e^{3} \mathrm {sgn}\left (b x + a\right ) + 10 \, a d e^{4} \mathrm {sgn}\left (b x + a\right ) - 7 \, b e^{5} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | -2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )} \sqrt {d} - e \right |}\right )}{256 \, d^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*sqrt(d*x^2 + x*e + c)*(2*(4*(6*(8*b*x*sgn(b*x + a) + (10*a*d^4*sgn(b*x + a) + b*d^3*e*sgn(b*x + a))/d^4
)*x + (16*b*c*d^3*sgn(b*x + a) + 10*a*d^3*e*sgn(b*x + a) - 7*b*d^2*e^2*sgn(b*x + a))/d^4)*x + (120*a*c*d^3*sgn
(b*x + a) - 116*b*c*d^2*e*sgn(b*x + a) - 50*a*d^2*e^2*sgn(b*x + a) + 35*b*d*e^3*sgn(b*x + a))/d^4)*x - (256*b*
c^2*d^2*sgn(b*x + a) + 520*a*c*d^2*e*sgn(b*x + a) - 460*b*c*d*e^2*sgn(b*x + a) - 150*a*d*e^3*sgn(b*x + a) + 10
5*b*e^4*sgn(b*x + a))/d^4) + 1/256*(32*a*c^2*d^3*sgn(b*x + a) - 48*b*c^2*d^2*e*sgn(b*x + a) - 48*a*c*d^2*e^2*s
gn(b*x + a) + 40*b*c*d*e^3*sgn(b*x + a) + 10*a*d*e^4*sgn(b*x + a) - 7*b*e^5*sgn(b*x + a))*log(abs(-2*(sqrt(d)*
x - sqrt(d*x^2 + x*e + c))*sqrt(d) - e))/d^(9/2)

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maple [C]  time = 0.02, size = 530, normalized size = 1.67 \begin {gather*} \frac {\left (-480 a \,c^{2} d^{4} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )+720 a c \,d^{3} e^{2} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-150 a \,d^{2} e^{4} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )+720 b \,c^{2} d^{3} e \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-600 b c \,d^{2} e^{3} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )+105 b d \,e^{5} \ln \left (\frac {2 d x +e +2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}}{2 \sqrt {d}}\right )-480 \sqrt {d \,x^{2}+e x +c}\, a c \,d^{\frac {9}{2}} x +600 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {7}{2}} e^{2} x +720 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{\frac {7}{2}} e x +768 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b \,d^{\frac {9}{2}} x^{2}-420 \sqrt {d \,x^{2}+e x +c}\, b \,d^{\frac {5}{2}} e^{3} x -240 \sqrt {d \,x^{2}+e x +c}\, a c \,d^{\frac {7}{2}} e +960 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a \,d^{\frac {9}{2}} x +300 \sqrt {d \,x^{2}+e x +c}\, a \,d^{\frac {5}{2}} e^{3}+360 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{\frac {5}{2}} e^{2}-672 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b \,d^{\frac {7}{2}} e x -210 \sqrt {d \,x^{2}+e x +c}\, b \,d^{\frac {3}{2}} e^{4}-800 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a \,d^{\frac {7}{2}} e -512 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b c \,d^{\frac {7}{2}}+560 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b \,d^{\frac {5}{2}} e^{2}\right ) \mathrm {csgn}\left (b x +a \right )}{3840 d^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3840*csgn(b*x+a)*(768*(d*x^2+e*x+c)^(3/2)*d^(9/2)*x^2*b+960*(d*x^2+e*x+c)^(3/2)*d^(9/2)*x*a-672*(d*x^2+e*x+c
)^(3/2)*d^(7/2)*x*b*e-800*(d*x^2+e*x+c)^(3/2)*d^(7/2)*a*e-512*(d*x^2+e*x+c)^(3/2)*d^(7/2)*b*c+560*(d*x^2+e*x+c
)^(3/2)*d^(5/2)*b*e^2-480*(d*x^2+e*x+c)^(1/2)*d^(9/2)*x*a*c+600*(d*x^2+e*x+c)^(1/2)*d^(7/2)*x*a*e^2+720*(d*x^2
+e*x+c)^(1/2)*d^(7/2)*x*b*c*e-420*(d*x^2+e*x+c)^(1/2)*d^(5/2)*x*b*e^3-240*(d*x^2+e*x+c)^(1/2)*d^(7/2)*a*c*e+30
0*(d*x^2+e*x+c)^(1/2)*d^(5/2)*a*e^3+360*(d*x^2+e*x+c)^(1/2)*d^(5/2)*b*c*e^2-210*(d*x^2+e*x+c)^(1/2)*d^(3/2)*b*
e^4-480*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*a*c^2*d^4+720*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^
(1/2)+2*d*x+e)/d^(1/2))*a*c*d^3*e^2-150*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*a*d^2*e^4+720*
ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*b*c^2*d^3*e-600*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+
2*d*x+e)/d^(1/2))*b*c*d^2*e^3+105*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*b*d*e^5)/d^(11/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {d x^{2} + e x + c} \sqrt {{\left (b x + a\right )}^{2}} x^{2}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(d*x^2 + e*x + c)*sqrt((b*x + a)^2)*x^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+e\,x+c} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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