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

Optimal. Leaf size=210 \[ -\frac {\left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{16 c^{5/2} d^{5/2} e^{3/2}}+\frac {\left (c d^2-a e^2\right ) \left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 c^2 d^2 e}+\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 c d} \]

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Rubi [A]  time = 0.10, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {640, 612, 621, 206} \begin {gather*} \frac {\left (c d^2-a e^2\right ) \left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{8 c^2 d^2 e}-\frac {\left (c d^2-a e^2\right )^3 \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{16 c^{5/2} d^{5/2} e^{3/2}}+\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((c*d^2 - a*e^2)*(c*d^2 + a*e^2 + 2*c*d*e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(8*c^2*d^2*e) + (a*d
*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(3*c*d) - ((c*d^2 - a*e^2)^3*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*
Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(16*c^(5/2)*d^(5/2)*e^(3/2))

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 640

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]

Rubi steps

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

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Mathematica [A]  time = 0.71, size = 262, normalized size = 1.25 \begin {gather*} \frac {\sqrt {c d} \left (\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {c d} (d+e x) \left (-3 a^3 e^5+a^2 c d e^3 (8 d-e x)+a c^2 d^2 e \left (3 d^2+22 d e x+10 e^2 x^2\right )+c^3 d^3 x \left (3 d^2+14 d e x+8 e^2 x^2\right )\right )-3 \left (c d^2-a e^2\right )^{7/2} \sqrt {a e+c d x} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d} \sqrt {c d^2-a e^2}}\right )\right )}{24 c^{7/2} d^{7/2} e^{3/2} \sqrt {(d+e x) (a e+c d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c*d]*(Sqrt[c]*Sqrt[d]*Sqrt[c*d]*Sqrt[e]*(d + e*x)*(-3*a^3*e^5 + a^2*c*d*e^3*(8*d - e*x) + c^3*d^3*x*(3*d
^2 + 14*d*e*x + 8*e^2*x^2) + a*c^2*d^2*e*(3*d^2 + 22*d*e*x + 10*e^2*x^2)) - 3*(c*d^2 - a*e^2)^(7/2)*Sqrt[a*e +
c*d*x]*Sqrt[(c*d*(d + e*x))/(c*d^2 - a*e^2)]*ArcSinh[(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d*x])/(Sqrt[c*d]*S
qrt[c*d^2 - a*e^2])]))/(24*c^(7/2)*d^(7/2)*e^(3/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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IntegrateAlgebraic [B]  time = 9.95, size = 13210, normalized size = 62.90 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

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fricas [A]  time = 0.44, size = 532, normalized size = 2.53 \begin {gather*} \left [-\frac {3 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt {c d e} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {c d e} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right ) - 4 \, {\left (8 \, c^{3} d^{3} e^{3} x^{2} + 3 \, c^{3} d^{5} e + 8 \, a c^{2} d^{3} e^{3} - 3 \, a^{2} c d e^{5} + 2 \, {\left (7 \, c^{3} d^{4} e^{2} + a c^{2} d^{2} e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{96 \, c^{3} d^{3} e^{2}}, \frac {3 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt {-c d e} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-c d e}}{2 \, {\left (c^{2} d^{2} e^{2} x^{2} + a c d^{2} e^{2} + {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right )}}\right ) + 2 \, {\left (8 \, c^{3} d^{3} e^{3} x^{2} + 3 \, c^{3} d^{5} e + 8 \, a c^{2} d^{3} e^{3} - 3 \, a^{2} c d e^{5} + 2 \, {\left (7 \, c^{3} d^{4} e^{2} + a c^{2} d^{2} e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{48 \, c^{3} d^{3} e^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(3*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(c*d*e)*log(8*c^2*d^2*e^2*x^2 + c^2*d^4
+ 6*a*c*d^2*e^2 + a^2*e^4 + 4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(c*d
*e) + 8*(c^2*d^3*e + a*c*d*e^3)*x) - 4*(8*c^3*d^3*e^3*x^2 + 3*c^3*d^5*e + 8*a*c^2*d^3*e^3 - 3*a^2*c*d*e^5 + 2*
(7*c^3*d^4*e^2 + a*c^2*d^2*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^3*d^3*e^2), 1/48*(3*(c^3*d^
6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(-c*d*e)*arctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e
^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(-c*d*e)/(c^2*d^2*e^2*x^2 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)) +
2*(8*c^3*d^3*e^3*x^2 + 3*c^3*d^5*e + 8*a*c^2*d^3*e^3 - 3*a^2*c*d*e^5 + 2*(7*c^3*d^4*e^2 + a*c^2*d^2*e^4)*x)*s
qrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^3*d^3*e^2)]

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giac [A]  time = 0.42, size = 216, normalized size = 1.03 \begin {gather*} \frac {1}{24} \, \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e} {\left (2 \, {\left (4 \, x e + \frac {{\left (7 \, c^{2} d^{3} e^{2} + a c d e^{4}\right )} e^{\left (-2\right )}}{c^{2} d^{2}}\right )} x + \frac {{\left (3 \, c^{2} d^{4} e + 8 \, a c d^{2} e^{3} - 3 \, a^{2} e^{5}\right )} e^{\left (-2\right )}}{c^{2} d^{2}}\right )} + \frac {{\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} e^{\left (-\frac {3}{2}\right )} \log \left ({\left | -c d^{2} - 2 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}\right )} \sqrt {c d} e^{\frac {1}{2}} - a e^{2} \right |}\right )}{16 \, \sqrt {c d} c^{2} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)*(2*(4*x*e + (7*c^2*d^3*e^2 + a*c*d*e^4)*e^(-2)/(c^2*d^2))*x +
(3*c^2*d^4*e + 8*a*c*d^2*e^3 - 3*a^2*e^5)*e^(-2)/(c^2*d^2)) + 1/16*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e
^4 - a^3*e^6)*e^(-3/2)*log(abs(-c*d^2 - 2*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e))*
sqrt(c*d)*e^(1/2) - a*e^2))/(sqrt(c*d)*c^2*d^2)

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maple [B]  time = 0.08, size = 465, normalized size = 2.21 \begin {gather*} \frac {a^{3} e^{5} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{16 \sqrt {c d e}\, c^{2} d^{2}}-\frac {3 a^{2} e^{3} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{16 \sqrt {c d e}\, c}+\frac {3 a \,d^{2} e \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{16 \sqrt {c d e}}-\frac {c \,d^{4} \ln \left (\frac {c d e x +\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}}{\sqrt {c d e}}+\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\right )}{16 \sqrt {c d e}\, e}-\frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a \,e^{2} x}{4 c d}+\frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, d x}{4}-\frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, a^{2} e^{3}}{8 c^{2} d^{2}}+\frac {\sqrt {c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x}\, d^{2}}{8 e}+\frac {\left (c d e \,x^{2}+a d e +\left (a \,e^{2}+c \,d^{2}\right ) x \right )^{\frac {3}{2}}}{3 c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(3/2)/c/d-1/4/c/d*e^2*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x*a+1/4*d*
(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*x-1/8/c^2/d^2*e^3*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)*a^2+1/8*d^2/
e*(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2)+1/16/c^2/d^2*e^5*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/2)+(c*d
*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)*a^3-3/16/c*e^3*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d*e)^(1/
2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)*a^2+3/16*e*d^2*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^2)/(c*d
*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)*a-1/16*c*d^4/e*ln((c*d*e*x+1/2*a*e^2+1/2*c*d^
2)/(c*d*e)^(1/2)+(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(1/2))/(c*d*e)^(1/2)

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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((e*x+d)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^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(a*e^2-c*d^2>0)', see `assume?`
for more details)Is a*e^2-c*d^2 zero or nonzero?

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mupad [B]  time = 1.04, size = 307, normalized size = 1.46 \begin {gather*} d\,\left (\frac {x}{2}+\frac {c\,d^2+a\,e^2}{4\,c\,d\,e}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}-\frac {d\,\ln \left (2\,\sqrt {\left (a\,e+c\,d\,x\right )\,\left (d+e\,x\right )}\,\sqrt {c\,d\,e}+a\,e^2+c\,d^2+2\,c\,d\,e\,x\right )\,\left (\frac {{\left (c\,d^2+a\,e^2\right )}^2}{4}-a\,c\,d^2\,e^2\right )}{2\,{\left (c\,d\,e\right )}^{3/2}}+\frac {e\,\ln \left (2\,\sqrt {\left (a\,e+c\,d\,x\right )\,\left (d+e\,x\right )}\,\sqrt {c\,d\,e}+a\,e^2+c\,d^2+2\,c\,d\,e\,x\right )\,\left ({\left (c\,d^2+a\,e^2\right )}^3-4\,a\,c\,d^2\,e^2\,\left (c\,d^2+a\,e^2\right )\right )}{16\,{\left (c\,d\,e\right )}^{5/2}}+\frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}\,\left (8\,c\,d\,e\,\left (c\,d\,e\,x^2+a\,d\,e\right )-3\,{\left (c\,d^2+a\,e^2\right )}^2+2\,c\,d\,e\,x\,\left (c\,d^2+a\,e^2\right )\right )}{24\,c^2\,d^2\,e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*(x/2 + (a*e^2 + c*d^2)/(4*c*d*e))*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2) - (d*log(2*((a*e + c*d*x)*(d
+ e*x))^(1/2)*(c*d*e)^(1/2) + a*e^2 + c*d^2 + 2*c*d*e*x)*((a*e^2 + c*d^2)^2/4 - a*c*d^2*e^2))/(2*(c*d*e)^(3/2
)) + (e*log(2*((a*e + c*d*x)*(d + e*x))^(1/2)*(c*d*e)^(1/2) + a*e^2 + c*d^2 + 2*c*d*e*x)*((a*e^2 + c*d^2)^3 -
4*a*c*d^2*e^2*(a*e^2 + c*d^2)))/(16*(c*d*e)^(5/2)) + ((x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)*(8*c*d*e*(
a*d*e + c*d*e*x^2) - 3*(a*e^2 + c*d^2)^2 + 2*c*d*e*x*(a*e^2 + c*d^2)))/(24*c^2*d^2*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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