3.20.8 \(\int (d+e x)^3 \sqrt {a d e+(c d^2+a e^2) x+c d e x^2} \, dx\) [1908]

Optimal. Leaf size=328 \[ \frac {7 \left (c d^2-a e^2\right )^3 \left (c d^2+a e^2+2 c d e x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{128 c^4 d^4 e}+\frac {7 \left (c d^2-a e^2\right )^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{48 c^3 d^3}+\frac {7 \left (c d^2-a e^2\right ) (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{40 c^2 d^2}+\frac {(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 c d}-\frac {7 \left (c d^2-a e^2\right )^5 \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{256 c^{9/2} d^{9/2} e^{3/2}} \]

[Out]

7/48*(-a*e^2+c*d^2)^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/c^3/d^3+7/40*(-a*e^2+c*d^2)*(e*x+d)*(a*d*e+(a*e^
2+c*d^2)*x+c*d*e*x^2)^(3/2)/c^2/d^2+1/5*(e*x+d)^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/c/d-7/256*(-a*e^2+c*
d^2)^5*arctanh(1/2*(2*c*d*e*x+a*e^2+c*d^2)/c^(1/2)/d^(1/2)/e^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/c^
(9/2)/d^(9/2)/e^(3/2)+7/128*(-a*e^2+c*d^2)^3*(2*c*d*e*x+a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c
^4/d^4/e

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Rubi [A]
time = 0.18, antiderivative size = 328, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {684, 654, 626, 635, 212} \begin {gather*} -\frac {7 \left (c d^2-a e^2\right )^5 \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 )}{256 c^{9/2} d^{9/2} e^{3/2}}+\frac {7 \left (c d^2-a e^2\right )^3 \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}}{128 c^4 d^4 e}+\frac {7 \left (c d^2-a e^2\right )^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{48 c^3 d^3}+\frac {7 (d+e x) \left (c d^2-a e^2\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{40 c^2 d^2}+\frac {(d+e x)^2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{5 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*(c*d^2 - a*e^2)^3*(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])/(128*c^4*d^4*e)
+ (7*(c*d^2 - a*e^2)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(48*c^3*d^3) + (7*(c*d^2 - a*e^2)*(d + e
*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2))/(40*c^2*d^2) + ((d + e*x)^2*(a*d*e + (c*d^2 + a*e^2)*x + c*
d*e*x^2)^(3/2))/(5*c*d) - (7*(c*d^2 - a*e^2)^5*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])])/(256*c^(9/2)*d^(9/2)*e^(3/2))

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 684

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[(m + p)*((2*c*d - b*e)/(c*(m + 2*p + 1))), Int[(d + e
*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rubi steps

\begin {align*} \int (d+e x)^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx &=\frac {(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 c d}+\frac {\left (7 \left (d^2-\frac {a e^2}{c}\right )\right ) \int (d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx}{10 d}\\ &=\frac {7 \left (c d^2-a e^2\right ) (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{40 c^2 d^2}+\frac {(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 c d}+\frac {\left (7 \left (d^2-\frac {a e^2}{c}\right )^2\right ) \int (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx}{16 d^2}\\ &=\frac {7 \left (c d^2-a e^2\right )^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{48 c^3 d^3}+\frac {7 \left (c d^2-a e^2\right ) (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{40 c^2 d^2}+\frac {(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 c d}+\frac {\left (7 \left (d^2-\frac {a e^2}{c}\right )^3\right ) \int \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx}{32 d^3}\\ &=\frac {7 \left (c d^2-a e^2\right )^3 \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}}{128 c^4 d^4 e}+\frac {7 \left (c d^2-a e^2\right )^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{48 c^3 d^3}+\frac {7 \left (c d^2-a e^2\right ) (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{40 c^2 d^2}+\frac {(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 c d}-\frac {\left (7 \left (c d^2-a e^2\right )^5\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{256 c^4 d^4 e}\\ &=\frac {7 \left (c d^2-a e^2\right )^3 \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}}{128 c^4 d^4 e}+\frac {7 \left (c d^2-a e^2\right )^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{48 c^3 d^3}+\frac {7 \left (c d^2-a e^2\right ) (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{40 c^2 d^2}+\frac {(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 c d}-\frac {\left (7 \left (c d^2-a e^2\right )^5\right ) \text {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 )}{128 c^4 d^4 e}\\ &=\frac {7 \left (c d^2-a e^2\right )^3 \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}}{128 c^4 d^4 e}+\frac {7 \left (c d^2-a e^2\right )^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{48 c^3 d^3}+\frac {7 \left (c d^2-a e^2\right ) (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{40 c^2 d^2}+\frac {(d+e x)^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 c d}-\frac {7 \left (c d^2-a e^2\right )^5 \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 )}{256 c^{9/2} d^{9/2} e^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.49, size = 257, normalized size = 0.78 \begin {gather*} \frac {\left (c d^2-a e^2\right )^5 \sqrt {(a e+c d x) (d+e x)} \left (\frac {\sqrt {c} \sqrt {d} \sqrt {e} (d+e x)^4 \left (105 c^4 d^4-\frac {105 e^4 (a e+c d x)^4}{(d+e x)^4}+\frac {490 c d e^3 (a e+c d x)^3}{(d+e x)^3}-\frac {896 c^2 d^2 e^2 (a e+c d x)^2}{(d+e x)^2}+\frac {790 c^3 d^3 e (a e+c d x)}{d+e x}\right )}{\left (c d^2-a e^2\right )^5}-\frac {105 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{\sqrt {a e+c d x} \sqrt {d+e x}}\right )}{1920 c^{9/2} d^{9/2} e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((c*d^2 - a*e^2)^5*Sqrt[(a*e + c*d*x)*(d + e*x)]*((Sqrt[c]*Sqrt[d]*Sqrt[e]*(d + e*x)^4*(105*c^4*d^4 - (105*e^4
*(a*e + c*d*x)^4)/(d + e*x)^4 + (490*c*d*e^3*(a*e + c*d*x)^3)/(d + e*x)^3 - (896*c^2*d^2*e^2*(a*e + c*d*x)^2)/
(d + e*x)^2 + (790*c^3*d^3*e*(a*e + c*d*x))/(d + e*x)))/(c*d^2 - a*e^2)^5 - (105*ArcTanh[(Sqrt[e]*Sqrt[a*e + c
*d*x])/(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/(Sqrt[a*e + c*d*x]*Sqrt[d + e*x])))/(1920*c^(9/2)*d^(9/2)*e^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1542\) vs. \(2(294)=588\).
time = 0.71, size = 1543, normalized size = 4.70

method	result	size
default	\(\text {Expression too large to display}\)	\(1543\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^3*(1/5*x^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/c/d/e-7/10*(a*e^2+c*d^2)/c/d/e*(1/4*x*(a*d*e+(a*e^2+c*d^2
)*x+c*d*e*x^2)^(3/2)/c/d/e-5/8*(a*e^2+c*d^2)/c/d/e*(1/3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/c/d/e-1/2*(a*e
^2+c*d^2)/c/d/e*(1/4*(2*c*d*e*x+a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c/d/e+1/8*(4*a*c*d^2*e^2-
(a*e^2+c*d^2)^2)/c/d/e*ln((1/2*e^2*a+1/2*c*d^2+c*d*e*x)/(c*d*e)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))
/(c*d*e)^(1/2)))-1/4*a/c*(1/4*(2*c*d*e*x+a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c/d/e+1/8*(4*a*c
*d^2*e^2-(a*e^2+c*d^2)^2)/c/d/e*ln((1/2*e^2*a+1/2*c*d^2+c*d*e*x)/(c*d*e)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^
2)^(1/2))/(c*d*e)^(1/2)))-2/5*a/c*(1/3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/c/d/e-1/2*(a*e^2+c*d^2)/c/d/e*(
1/4*(2*c*d*e*x+a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c/d/e+1/8*(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/
c/d/e*ln((1/2*e^2*a+1/2*c*d^2+c*d*e*x)/(c*d*e)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(c*d*e)^(1/2))))
+3*d*e^2*(1/4*x*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/c/d/e-5/8*(a*e^2+c*d^2)/c/d/e*(1/3*(a*d*e+(a*e^2+c*d^2
)*x+c*d*e*x^2)^(3/2)/c/d/e-1/2*(a*e^2+c*d^2)/c/d/e*(1/4*(2*c*d*e*x+a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x
^2)^(1/2)/c/d/e+1/8*(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/c/d/e*ln((1/2*e^2*a+1/2*c*d^2+c*d*e*x)/(c*d*e)^(1/2)+(a*d*
e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(c*d*e)^(1/2)))-1/4*a/c*(1/4*(2*c*d*e*x+a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*
x+c*d*e*x^2)^(1/2)/c/d/e+1/8*(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/c/d/e*ln((1/2*e^2*a+1/2*c*d^2+c*d*e*x)/(c*d*e)^(1
/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(c*d*e)^(1/2)))+3*d^2*e*(1/3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3
/2)/c/d/e-1/2*(a*e^2+c*d^2)/c/d/e*(1/4*(2*c*d*e*x+a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c/d/e+1
/8*(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/c/d/e*ln((1/2*e^2*a+1/2*c*d^2+c*d*e*x)/(c*d*e)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x
+c*d*e*x^2)^(1/2))/(c*d*e)^(1/2)))+d^3*(1/4*(2*c*d*e*x+a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c/
d/e+1/8*(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/c/d/e*ln((1/2*e^2*a+1/2*c*d^2+c*d*e*x)/(c*d*e)^(1/2)+(a*d*e+(a*e^2+c*d
^2)*x+c*d*e*x^2)^(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)^3*(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(c*d^2-%e^2*a>0)', see `assume?
` for more d

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Fricas [A]
time = 2.80, size = 819, normalized size = 2.50 \begin {gather*} \left [\frac {{\left (105 \, {\left (c^{5} d^{10} - 5 \, a c^{4} d^{8} e^{2} + 10 \, a^{2} c^{3} d^{6} e^{4} - 10 \, a^{3} c^{2} d^{4} e^{6} + 5 \, a^{4} c d^{2} e^{8} - a^{5} e^{10}\right )} \sqrt {c d} e^{\frac {1}{2}} \log \left (8 \, c^{2} d^{3} x e + c^{2} d^{4} + 8 \, a c d x e^{3} + a^{2} e^{4} - 4 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (2 \, c d x e + c d^{2} + a e^{2}\right )} \sqrt {c d} e^{\frac {1}{2}} + 2 \, {\left (4 \, c^{2} d^{2} x^{2} + 3 \, a c d^{2}\right )} e^{2}\right ) + 4 \, {\left (1210 \, c^{5} d^{8} x e^{2} + 105 \, c^{5} d^{9} e + 70 \, a^{3} c^{2} d^{2} x e^{8} - 105 \, a^{4} c d e^{9} - 14 \, {\left (4 \, a^{2} c^{3} d^{3} x^{2} - 35 \, a^{3} c^{2} d^{3}\right )} e^{7} + 2 \, {\left (24 \, a c^{4} d^{4} x^{3} - 161 \, a^{2} c^{3} d^{4} x\right )} e^{6} + 128 \, {\left (3 \, c^{5} d^{5} x^{4} + 2 \, a c^{4} d^{5} x^{2} - 7 \, a^{2} c^{3} d^{5}\right )} e^{5} + 2 \, {\left (744 \, c^{5} d^{6} x^{3} + 289 \, a c^{4} d^{6} x\right )} e^{4} + 2 \, {\left (1052 \, c^{5} d^{7} x^{2} + 395 \, a c^{4} d^{7}\right )} e^{3}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}\right )} e^{\left (-2\right )}}{7680 \, c^{5} d^{5}}, \frac {{\left (105 \, {\left (c^{5} d^{10} - 5 \, a c^{4} d^{8} e^{2} + 10 \, a^{2} c^{3} d^{6} e^{4} - 10 \, a^{3} c^{2} d^{4} e^{6} + 5 \, a^{4} c d^{2} e^{8} - a^{5} e^{10}\right )} \sqrt {-c d e} \arctan \left (\frac {\sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (2 \, c d x e + c d^{2} + a e^{2}\right )} \sqrt {-c d e}}{2 \, {\left (c^{2} d^{3} x e + a c d x e^{3} + {\left (c^{2} d^{2} x^{2} + a c d^{2}\right )} e^{2}\right )}}\right ) + 2 \, {\left (1210 \, c^{5} d^{8} x e^{2} + 105 \, c^{5} d^{9} e + 70 \, a^{3} c^{2} d^{2} x e^{8} - 105 \, a^{4} c d e^{9} - 14 \, {\left (4 \, a^{2} c^{3} d^{3} x^{2} - 35 \, a^{3} c^{2} d^{3}\right )} e^{7} + 2 \, {\left (24 \, a c^{4} d^{4} x^{3} - 161 \, a^{2} c^{3} d^{4} x\right )} e^{6} + 128 \, {\left (3 \, c^{5} d^{5} x^{4} + 2 \, a c^{4} d^{5} x^{2} - 7 \, a^{2} c^{3} d^{5}\right )} e^{5} + 2 \, {\left (744 \, c^{5} d^{6} x^{3} + 289 \, a c^{4} d^{6} x\right )} e^{4} + 2 \, {\left (1052 \, c^{5} d^{7} x^{2} + 395 \, a c^{4} d^{7}\right )} e^{3}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}\right )} e^{\left (-2\right )}}{3840 \, c^{5} d^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/7680*(105*(c^5*d^10 - 5*a*c^4*d^8*e^2 + 10*a^2*c^3*d^6*e^4 - 10*a^3*c^2*d^4*e^6 + 5*a^4*c*d^2*e^8 - a^5*e^1
0)*sqrt(c*d)*e^(1/2)*log(8*c^2*d^3*x*e + c^2*d^4 + 8*a*c*d*x*e^3 + a^2*e^4 - 4*sqrt(c*d^2*x + a*x*e^2 + (c*d*x
^2 + a*d)*e)*(2*c*d*x*e + c*d^2 + a*e^2)*sqrt(c*d)*e^(1/2) + 2*(4*c^2*d^2*x^2 + 3*a*c*d^2)*e^2) + 4*(1210*c^5*
d^8*x*e^2 + 105*c^5*d^9*e + 70*a^3*c^2*d^2*x*e^8 - 105*a^4*c*d*e^9 - 14*(4*a^2*c^3*d^3*x^2 - 35*a^3*c^2*d^3)*e
^7 + 2*(24*a*c^4*d^4*x^3 - 161*a^2*c^3*d^4*x)*e^6 + 128*(3*c^5*d^5*x^4 + 2*a*c^4*d^5*x^2 - 7*a^2*c^3*d^5)*e^5
+ 2*(744*c^5*d^6*x^3 + 289*a*c^4*d^6*x)*e^4 + 2*(1052*c^5*d^7*x^2 + 395*a*c^4*d^7)*e^3)*sqrt(c*d^2*x + a*x*e^2
+ (c*d*x^2 + a*d)*e))*e^(-2)/(c^5*d^5), 1/3840*(105*(c^5*d^10 - 5*a*c^4*d^8*e^2 + 10*a^2*c^3*d^6*e^4 - 10*a^3
*c^2*d^4*e^6 + 5*a^4*c*d^2*e^8 - a^5*e^10)*sqrt(-c*d*e)*arctan(1/2*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)
*(2*c*d*x*e + c*d^2 + a*e^2)*sqrt(-c*d*e)/(c^2*d^3*x*e + a*c*d*x*e^3 + (c^2*d^2*x^2 + a*c*d^2)*e^2)) + 2*(1210
*c^5*d^8*x*e^2 + 105*c^5*d^9*e + 70*a^3*c^2*d^2*x*e^8 - 105*a^4*c*d*e^9 - 14*(4*a^2*c^3*d^3*x^2 - 35*a^3*c^2*d
^3)*e^7 + 2*(24*a*c^4*d^4*x^3 - 161*a^2*c^3*d^4*x)*e^6 + 128*(3*c^5*d^5*x^4 + 2*a*c^4*d^5*x^2 - 7*a^2*c^3*d^5)
*e^5 + 2*(744*c^5*d^6*x^3 + 289*a*c^4*d^6*x)*e^4 + 2*(1052*c^5*d^7*x^2 + 395*a*c^4*d^7)*e^3)*sqrt(c*d^2*x + a*
x*e^2 + (c*d*x^2 + a*d)*e))*e^(-2)/(c^5*d^5)]

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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 )^{3}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(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)**3, x)

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Giac [A]
time = 0.73, size = 377, normalized size = 1.15 \begin {gather*} \frac {1}{1920} \, \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, x e^{3} + \frac {{\left (31 \, c^{4} d^{5} e^{6} + a c^{3} d^{3} e^{8}\right )} e^{\left (-4\right )}}{c^{4} d^{4}}\right )} x + \frac {{\left (263 \, c^{4} d^{6} e^{5} + 32 \, a c^{3} d^{4} e^{7} - 7 \, a^{2} c^{2} d^{2} e^{9}\right )} e^{\left (-4\right )}}{c^{4} d^{4}}\right )} x + \frac {{\left (605 \, c^{4} d^{7} e^{4} + 289 \, a c^{3} d^{5} e^{6} - 161 \, a^{2} c^{2} d^{3} e^{8} + 35 \, a^{3} c d e^{10}\right )} e^{\left (-4\right )}}{c^{4} d^{4}}\right )} x + \frac {{\left (105 \, c^{4} d^{8} e^{3} + 790 \, a c^{3} d^{6} e^{5} - 896 \, a^{2} c^{2} d^{4} e^{7} + 490 \, a^{3} c d^{2} e^{9} - 105 \, a^{4} e^{11}\right )} e^{\left (-4\right )}}{c^{4} d^{4}}\right )} + \frac {7 \, {\left (c^{5} d^{10} - 5 \, a c^{4} d^{8} e^{2} + 10 \, a^{2} c^{3} d^{6} e^{4} - 10 \, a^{3} c^{2} d^{4} e^{6} + 5 \, a^{4} c d^{2} e^{8} - a^{5} e^{10}\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 )}{256 \, \sqrt {c d} c^{4} d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)*(2*(4*(6*(8*x*e^3 + (31*c^4*d^5*e^6 + a*c^3*d^3*e^8)*e^(-4)
/(c^4*d^4))*x + (263*c^4*d^6*e^5 + 32*a*c^3*d^4*e^7 - 7*a^2*c^2*d^2*e^9)*e^(-4)/(c^4*d^4))*x + (605*c^4*d^7*e^
4 + 289*a*c^3*d^5*e^6 - 161*a^2*c^2*d^3*e^8 + 35*a^3*c*d*e^10)*e^(-4)/(c^4*d^4))*x + (105*c^4*d^8*e^3 + 790*a*
c^3*d^6*e^5 - 896*a^2*c^2*d^4*e^7 + 490*a^3*c*d^2*e^9 - 105*a^4*e^11)*e^(-4)/(c^4*d^4)) + 7/256*(c^5*d^10 - 5*
a*c^4*d^8*e^2 + 10*a^2*c^3*d^6*e^4 - 10*a^3*c^2*d^4*e^6 + 5*a^4*c*d^2*e^8 - a^5*e^10)*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^4*d^4)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (d+e\,x\right )}^3\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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