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

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

[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))

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Rubi [A]  time = 0.333016, antiderivative size = 317, normalized size of antiderivative = 1., 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} \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2+e x\right )^{3/2} \left (-6 d x (10 a d-7 b e)+50 a d e+32 b c d-35 b e^2\right )}{240 d^3 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (2 d x+e) \sqrt{c+d x^2+e x} \left (2 a d \left (4 c d-5 e^2\right )-b \left (12 c d e-7 e^3\right )\right )}{128 d^4 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} \left (4 c d-e^2\right ) \left (8 a c d^2-10 a d e^2-12 b c d e+7 b e^3\right ) \tanh ^{-1}\left (\frac{2 d x+e}{2 \sqrt{d} \sqrt{c+d x^2+e x}}\right )}{256 d^{9/2} (a+b x)}+\frac{b x^2 \sqrt{a^2+2 a b x+b^2 x^2} \left (c+d x^2+e x\right )^{3/2}}{5 d (a+b x)} \]

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 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]

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 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 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 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])

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*}

Mathematica [A]  time = 0.281873, size = 198, normalized size = 0.62 \[ \frac{\sqrt{(a+b x)^2} \left (-\frac{5 \left (2 a d \left (4 c d-5 e^2\right )+b \left (7 e^3-12 c d e\right )\right ) \left (\left (4 c d-e^2\right ) \tanh ^{-1}\left (\frac{2 d x+e}{2 \sqrt{d} \sqrt{c+x (d x+e)}}\right )+2 \sqrt{d} (2 d x+e) \sqrt{c+x (d x+e)}\right )}{256 d^{7/2}}+\frac{(c+x (d x+e))^{3/2} (10 a d (6 d x-5 e)-32 b c d+7 b e (5 e-6 d x))}{48 d^2}+b x^2 (c+x (d x+e))^{3/2}\right )}{5 d (a+b x)} \]

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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Maple [C]  time = 0.214, size = 530, normalized size = 1.7 \begin{align*}{\frac{{\it csgn} \left ( bx+a \right ) }{3840} \left ( 768\,{d}^{9/2} \left ( d{x}^{2}+ex+c \right ) ^{3/2}{x}^{2}b+960\,{d}^{9/2} \left ( d{x}^{2}+ex+c \right ) ^{3/2}xa-672\,{d}^{7/2} \left ( d{x}^{2}+ex+c \right ) ^{3/2}xbe-800\,{d}^{7/2} \left ( d{x}^{2}+ex+c \right ) ^{3/2}ae-512\,{d}^{7/2} \left ( d{x}^{2}+ex+c \right ) ^{3/2}bc+560\,{d}^{5/2} \left ( d{x}^{2}+ex+c \right ) ^{3/2}b{e}^{2}-480\,{d}^{9/2}\sqrt{d{x}^{2}+ex+c}xac+600\,{d}^{7/2}\sqrt{d{x}^{2}+ex+c}xa{e}^{2}+720\,{d}^{7/2}\sqrt{d{x}^{2}+ex+c}xbce-420\,{d}^{5/2}\sqrt{d{x}^{2}+ex+c}xb{e}^{3}-240\,{d}^{7/2}\sqrt{d{x}^{2}+ex+c}ace+300\,{d}^{5/2}\sqrt{d{x}^{2}+ex+c}a{e}^{3}+360\,{d}^{5/2}\sqrt{d{x}^{2}+ex+c}bc{e}^{2}-210\,{d}^{3/2}\sqrt{d{x}^{2}+ex+c}b{e}^{4}-480\,\ln \left ( 1/2\,{\frac{2\,\sqrt{d{x}^{2}+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) a{c}^{2}{d}^{4}+720\,\ln \left ( 1/2\,{\frac{2\,\sqrt{d{x}^{2}+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) ac{d}^{3}{e}^{2}-150\,\ln \left ( 1/2\,{\frac{2\,\sqrt{d{x}^{2}+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) a{d}^{2}{e}^{4}+720\,\ln \left ( 1/2\,{\frac{2\,\sqrt{d{x}^{2}+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) b{c}^{2}{d}^{3}e-600\,\ln \left ( 1/2\,{\frac{2\,\sqrt{d{x}^{2}+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) bc{d}^{2}{e}^{3}+105\,\ln \left ( 1/2\,{\frac{2\,\sqrt{d{x}^{2}+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) bd{e}^{5} \right ){d}^{-{\frac{11}{2}}}} \end{align*}

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^(9/2)*(d*x^2+e*x+c)^(3/2)*x^2*b+960*d^(9/2)*(d*x^2+e*x+c)^(3/2)*x*a-672*d^(7/2)*(d*x
^2+e*x+c)^(3/2)*x*b*e-800*d^(7/2)*(d*x^2+e*x+c)^(3/2)*a*e-512*d^(7/2)*(d*x^2+e*x+c)^(3/2)*b*c+560*d^(5/2)*(d*x
^2+e*x+c)^(3/2)*b*e^2-480*d^(9/2)*(d*x^2+e*x+c)^(1/2)*x*a*c+600*d^(7/2)*(d*x^2+e*x+c)^(1/2)*x*a*e^2+720*d^(7/2
)*(d*x^2+e*x+c)^(1/2)*x*b*c*e-420*d^(5/2)*(d*x^2+e*x+c)^(1/2)*x*b*e^3-240*d^(7/2)*(d*x^2+e*x+c)^(1/2)*a*c*e+30
0*d^(5/2)*(d*x^2+e*x+c)^(1/2)*a*e^3+360*d^(5/2)*(d*x^2+e*x+c)^(1/2)*b*c*e^2-210*d^(3/2)*(d*x^2+e*x+c)^(1/2)*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., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d x^{2} + e x + c} \sqrt{{\left (b x + a\right )}^{2}} x^{2}\,{d x} \end{align*}

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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Fricas [A]  time = 1.84483, size = 1250, normalized size = 3.94 \begin{align*} \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{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.18472, size = 497, normalized size = 1.57 \begin{align*} \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{align*}

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)