3.1.38 \(\int \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx\) [38]

Optimal. Leaf size=283 \[ -\frac {\left (4 b c d-16 a d^2-5 b e^2\right ) (e+2 d x) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{64 d^3 \left (a+b x^2\right )}-\frac {5 b e \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{24 d^2 \left (a+b x^2\right )}+\frac {b x \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 d \left (a+b x^2\right )}-\frac {\left (4 c d-e^2\right ) \left (4 b c d-16 a d^2-5 b e^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{128 d^{7/2} \left (a+b x^2\right )} \]

[Out]

-5/24*b*e*(d*x^2+e*x+c)^(3/2)*((b*x^2+a)^2)^(1/2)/d^2/(b*x^2+a)+1/4*b*x*(d*x^2+e*x+c)^(3/2)*((b*x^2+a)^2)^(1/2
)/d/(b*x^2+a)-1/128*(4*c*d-e^2)*(-16*a*d^2+4*b*c*d-5*b*e^2)*arctanh(1/2*(2*d*x+e)/d^(1/2)/(d*x^2+e*x+c)^(1/2))
*((b*x^2+a)^2)^(1/2)/d^(7/2)/(b*x^2+a)-1/64*(-16*a*d^2+4*b*c*d-5*b*e^2)*(2*d*x+e)*(d*x^2+e*x+c)^(1/2)*((b*x^2+
a)^2)^(1/2)/d^3/(b*x^2+a)

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Rubi [A]
time = 0.23, antiderivative size = 283, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {6876, 1675, 654, 626, 635, 212} \begin {gather*} -\frac {5 b e \sqrt {a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{24 d^2 \left (a+b x^2\right )}-\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (4 c d-e^2\right ) \left (-16 a d^2+4 b c d-5 b e^2\right ) \tanh ^{-1}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{128 d^{7/2} \left (a+b x^2\right )}-\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} (2 d x+e) \sqrt {c+d x^2+e x} \left (-16 a d^2+4 b c d-5 b e^2\right )}{64 d^3 \left (a+b x^2\right )}+\frac {b x \sqrt {a^2+2 a b x^2+b^2 x^4} \left (c+d x^2+e x\right )^{3/2}}{4 d \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

-1/64*((4*b*c*d - 16*a*d^2 - 5*b*e^2)*(e + 2*d*x)*Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(d^3*
(a + b*x^2)) - (5*b*e*(c + e*x + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(24*d^2*(a + b*x^2)) + (b*x*(c
+ e*x + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*d*(a + b*x^2)) - ((4*c*d - e^2)*(4*b*c*d - 16*a*d^2 -
5*b*e^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + e*x + d*x^2])])/(128*d^(7/2)
*(a + b*x^2))

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 626

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 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 654

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

Rule 1675

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 6876

Int[(u_)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[Sqrt[a + b*x^n + c*x^(2*n)]/((4
*c)^(p - 1/2)*(b + 2*c*x^n)), Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] &
& EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx &=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (2 a b+2 b^2 x^2\right ) \sqrt {c+e x+d x^2} \, dx}{2 a b+2 b^2 x^2}\\ &=\frac {b x \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 d \left (a+b x^2\right )}+\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (-2 b (b c-4 a d)-5 b^2 e x\right ) \sqrt {c+e x+d x^2} \, dx}{4 d \left (2 a b+2 b^2 x^2\right )}\\ &=-\frac {5 b e \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{24 d^2 \left (a+b x^2\right )}+\frac {b x \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 d \left (a+b x^2\right )}+\frac {\left (\left (-4 b d (b c-4 a d)+5 b^2 e^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}\right ) \int \sqrt {c+e x+d x^2} \, dx}{8 d^2 \left (2 a b+2 b^2 x^2\right )}\\ &=-\frac {\left (4 b c d-16 a d^2-5 b e^2\right ) (e+2 d x) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{64 d^3 \left (a+b x^2\right )}-\frac {5 b e \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{24 d^2 \left (a+b x^2\right )}+\frac {b x \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 d \left (a+b x^2\right )}+\frac {\left (\left (4 c d-e^2\right ) \left (-4 b d (b c-4 a d)+5 b^2 e^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}\right ) \int \frac {1}{\sqrt {c+e x+d x^2}} \, dx}{64 d^3 \left (2 a b+2 b^2 x^2\right )}\\ &=-\frac {\left (4 b c d-16 a d^2-5 b e^2\right ) (e+2 d x) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{64 d^3 \left (a+b x^2\right )}-\frac {5 b e \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{24 d^2 \left (a+b x^2\right )}+\frac {b x \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 d \left (a+b x^2\right )}+\frac {\left (\left (4 c d-e^2\right ) \left (-4 b d (b c-4 a d)+5 b^2 e^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}\right ) \text {Subst}\left (\int \frac {1}{4 d-x^2} \, dx,x,\frac {e+2 d x}{\sqrt {c+e x+d x^2}}\right )}{32 d^3 \left (2 a b+2 b^2 x^2\right )}\\ &=-\frac {\left (4 b c d-16 a d^2-5 b e^2\right ) (e+2 d x) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{64 d^3 \left (a+b x^2\right )}-\frac {5 b e \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{24 d^2 \left (a+b x^2\right )}+\frac {b x \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 d \left (a+b x^2\right )}-\frac {\left (4 c d-e^2\right ) \left (4 b c d-16 a d^2-5 b e^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{128 d^{7/2} \left (a+b x^2\right )}\\ \end {align*}

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Mathematica [A]
time = 0.41, size = 166, normalized size = 0.59 \begin {gather*} \frac {\sqrt {\left (a+b x^2\right )^2} \left (2 \sqrt {d} \sqrt {c+x (e+d x)} \left (48 a d^2 (e+2 d x)+b \left (15 e^3-10 d e^2 x+8 d^2 e x^2+48 d^3 x^3+4 c d (-13 e+6 d x)\right )\right )+3 \left (4 c d-e^2\right ) \left (4 b c d-16 a d^2-5 b e^2\right ) \log \left (e+2 d x-2 \sqrt {d} \sqrt {c+x (e+d x)}\right )\right )}{384 d^{7/2} \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

(Sqrt[(a + b*x^2)^2]*(2*Sqrt[d]*Sqrt[c + x*(e + d*x)]*(48*a*d^2*(e + 2*d*x) + b*(15*e^3 - 10*d*e^2*x + 8*d^2*e
*x^2 + 48*d^3*x^3 + 4*c*d*(-13*e + 6*d*x))) + 3*(4*c*d - e^2)*(4*b*c*d - 16*a*d^2 - 5*b*e^2)*Log[e + 2*d*x - 2
*Sqrt[d]*Sqrt[c + x*(e + d*x)]]))/(384*d^(7/2)*(a + b*x^2))

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Maple [A]
time = 0.19, size = 373, normalized size = 1.32

method	result	size
risch	\(\frac {\left (48 x^{3} d^{3} b +8 b e \,x^{2} d^{2}+96 a \,d^{3} x +24 b c \,d^{2} x -10 b d \,e^{2} x +48 a \,d^{2} e -52 b c d e +15 e^{3} b \right ) \sqrt {d \,x^{2}+e x +c}\, \sqrt {\left (b \,x^{2}+a \right )^{2}}}{192 d^{3} \left (b \,x^{2}+a \right )}+\frac {\left (\frac {\ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right ) a c}{2 \sqrt {d}}-\frac {\ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right ) e^{2} a}{8 d^{\frac {3}{2}}}-\frac {\ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right ) b \,c^{2}}{8 d^{\frac {3}{2}}}+\frac {3 \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right ) b c \,e^{2}}{16 d^{\frac {5}{2}}}-\frac {5 \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right ) b \,e^{4}}{128 d^{\frac {7}{2}}}\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\)	\(292\)
default	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (96 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {7}{2}} b x -80 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} d^{\frac {5}{2}} b e +192 d^{\frac {9}{2}} \sqrt {d \,x^{2}+e x +c}\, a x -48 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {7}{2}} b c x +60 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} b \,e^{2} x +96 d^{\frac {7}{2}} \sqrt {d \,x^{2}+e x +c}\, a e -24 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} b c e +30 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {3}{2}} b \,e^{3}+192 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a c \,d^{4}-48 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a \,d^{3} e^{2}-48 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b \,c^{2} d^{3}+72 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b c \,d^{2} e^{2}-15 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b d \,e^{4}\right )}{384 \left (b \,x^{2}+a \right ) d^{\frac {9}{2}}}\)	\(373\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*((b*x^2+a)^2)^(1/2)*(96*(d*x^2+e*x+c)^(3/2)*d^(7/2)*b*x-80*(d*x^2+e*x+c)^(3/2)*d^(5/2)*b*e+192*d^(9/2)*(
d*x^2+e*x+c)^(1/2)*a*x-48*(d*x^2+e*x+c)^(1/2)*d^(7/2)*b*c*x+60*(d*x^2+e*x+c)^(1/2)*d^(5/2)*b*e^2*x+96*d^(7/2)*
(d*x^2+e*x+c)^(1/2)*a*e-24*(d*x^2+e*x+c)^(1/2)*d^(5/2)*b*c*e+30*(d*x^2+e*x+c)^(1/2)*d^(3/2)*b*e^3+192*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^4-48*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1
/2))*a*d^3*e^2-48*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+72*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^2-15*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^
4)/(b*x^2+a)/d^(9/2)

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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((d*x^2+e*x+c)^(1/2)*((b*x^2+a)^2)^(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*c*d-%e^2>0)', see `assume?`
for more det

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Fricas [A]
time = 0.38, size = 360, normalized size = 1.27 \begin {gather*} \left [\frac {3 \, {\left (16 \, b c^{2} d^{2} - 64 \, a c d^{3} + 5 \, b e^{4} - 8 \, {\left (3 \, b c d - 2 \, a d^{2}\right )} e^{2}\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d x e - 4 \, \sqrt {d x^{2} + x e + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + 4 \, {\left (48 \, b d^{4} x^{3} - 10 \, b d^{2} x e^{2} + 15 \, b d e^{3} + 24 \, {\left (b c d^{3} + 4 \, a d^{4}\right )} x + 4 \, {\left (2 \, b d^{3} x^{2} - 13 \, b c d^{2} + 12 \, a d^{3}\right )} e\right )} \sqrt {d x^{2} + x e + c}}{768 \, d^{4}}, \frac {3 \, {\left (16 \, b c^{2} d^{2} - 64 \, a c d^{3} + 5 \, b e^{4} - 8 \, {\left (3 \, b c d - 2 \, a d^{2}\right )} e^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + x e + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d x e + c d\right )}}\right ) + 2 \, {\left (48 \, b d^{4} x^{3} - 10 \, b d^{2} x e^{2} + 15 \, b d e^{3} + 24 \, {\left (b c d^{3} + 4 \, a d^{4}\right )} x + 4 \, {\left (2 \, b d^{3} x^{2} - 13 \, b c d^{2} + 12 \, a d^{3}\right )} e\right )} \sqrt {d x^{2} + x e + c}}{384 \, d^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/768*(3*(16*b*c^2*d^2 - 64*a*c*d^3 + 5*b*e^4 - 8*(3*b*c*d - 2*a*d^2)*e^2)*sqrt(d)*log(8*d^2*x^2 + 8*d*x*e -
4*sqrt(d*x^2 + x*e + c)*(2*d*x + e)*sqrt(d) + 4*c*d + e^2) + 4*(48*b*d^4*x^3 - 10*b*d^2*x*e^2 + 15*b*d*e^3 + 2
4*(b*c*d^3 + 4*a*d^4)*x + 4*(2*b*d^3*x^2 - 13*b*c*d^2 + 12*a*d^3)*e)*sqrt(d*x^2 + x*e + c))/d^4, 1/384*(3*(16*
b*c^2*d^2 - 64*a*c*d^3 + 5*b*e^4 - 8*(3*b*c*d - 2*a*d^2)*e^2)*sqrt(-d)*arctan(1/2*sqrt(d*x^2 + x*e + c)*(2*d*x
+ e)*sqrt(-d)/(d^2*x^2 + d*x*e + c*d)) + 2*(48*b*d^4*x^3 - 10*b*d^2*x*e^2 + 15*b*d*e^3 + 24*(b*c*d^3 + 4*a*d^
4)*x + 4*(2*b*d^3*x^2 - 13*b*c*d^2 + 12*a*d^3)*e)*sqrt(d*x^2 + x*e + c))/d^4]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.71, size = 265, normalized size = 0.94 \begin {gather*} \frac {1}{192} \, \sqrt {d x^{2} + x e + c} {\left (2 \, {\left (4 \, {\left (6 \, b x \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {b e \mathrm {sgn}\left (b x^{2} + a\right )}{d}\right )} x + \frac {12 \, b c d^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 48 \, a d^{3} \mathrm {sgn}\left (b x^{2} + a\right ) - 5 \, b d e^{2} \mathrm {sgn}\left (b x^{2} + a\right )}{d^{3}}\right )} x - \frac {52 \, b c d e \mathrm {sgn}\left (b x^{2} + a\right ) - 48 \, a d^{2} e \mathrm {sgn}\left (b x^{2} + a\right ) - 15 \, b e^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{d^{3}}\right )} + \frac {{\left (16 \, b c^{2} d^{2} \mathrm {sgn}\left (b x^{2} + a\right ) - 64 \, a c d^{3} \mathrm {sgn}\left (b x^{2} + a\right ) - 24 \, b c d e^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 16 \, a d^{2} e^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, b e^{4} \mathrm {sgn}\left (b x^{2} + a\right )\right )} \log \left ({\left | -2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + x e + c}\right )} \sqrt {d} - e \right |}\right )}{128 \, d^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(d*x^2 + x*e + c)*(2*(4*(6*b*x*sgn(b*x^2 + a) + b*e*sgn(b*x^2 + a)/d)*x + (12*b*c*d^2*sgn(b*x^2 + a)
+ 48*a*d^3*sgn(b*x^2 + a) - 5*b*d*e^2*sgn(b*x^2 + a))/d^3)*x - (52*b*c*d*e*sgn(b*x^2 + a) - 48*a*d^2*e*sgn(b*
x^2 + a) - 15*b*e^3*sgn(b*x^2 + a))/d^3) + 1/128*(16*b*c^2*d^2*sgn(b*x^2 + a) - 64*a*c*d^3*sgn(b*x^2 + a) - 24
*b*c*d*e^2*sgn(b*x^2 + a) + 16*a*d^2*e^2*sgn(b*x^2 + a) + 5*b*e^4*sgn(b*x^2 + a))*log(abs(-2*(sqrt(d)*x - sqrt
(d*x^2 + x*e + c))*sqrt(d) - e))/d^(7/2)

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

[Out]

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

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