3.20.55 \(\int \frac {(d+e x)^4}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\) [1955]

Optimal. Leaf size=241 \[ -\frac {2 (d+e x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {15 e \left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c^3 d^3}+\frac {5 e (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c^2 d^2}+\frac {15 \sqrt {e} \left (c d^2-a e^2\right )^2 \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{8 c^{7/2} d^{7/2}} \]

[Out]

15/8*(-a*e^2+c*d^2)^2*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))*e^(1/2)/c^(7/2)/d^(7/2)-2*(e*x+d)^3/c/d/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+15/4*e*(-a*e^2+c*
d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c^3/d^3+5/2*e*(e*x+d)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/c^2
/d^2

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Rubi [A]
time = 0.12, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {682, 684, 654, 635, 212} \begin {gather*} \frac {15 \sqrt {e} \left (c d^2-a e^2\right )^2 \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 )}{8 c^{7/2} d^{7/2}}+\frac {15 e \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 c^3 d^3}+\frac {5 e (d+e x) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{2 c^2 d^2}-\frac {2 (d+e x)^3}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^4/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2),x]

[Out]

(-2*(d + e*x)^3)/(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (15*e*(c*d^2 - a*e^2)*Sqrt[a*d*e + (c*d^2
+ a*e^2)*x + c*d*e*x^2])/(4*c^3*d^3) + (5*e*(d + e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(2*c^2*d^2
) + (15*Sqrt[e]*(c*d^2 - a*e^2)^2*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])])/(8*c^(7/2)*d^(7/2))

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 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 682

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

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 \frac {(d+e x)^4}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=-\frac {2 (d+e x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {(5 e) \int \frac {(d+e x)^2}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{c d}\\ &=-\frac {2 (d+e x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 e (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c^2 d^2}+\frac {\left (15 e \left (c d^2-a e^2\right )\right ) \int \frac {d+e x}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{4 c^2 d^2}\\ &=-\frac {2 (d+e x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {15 e \left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c^3 d^3}+\frac {5 e (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c^2 d^2}+\frac {\left (15 e \left (c d^2-a e^2\right )^2\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 c^3 d^3}\\ &=-\frac {2 (d+e x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {15 e \left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c^3 d^3}+\frac {5 e (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c^2 d^2}+\frac {\left (15 e \left (c d^2-a e^2\right )^2\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 )}{4 c^3 d^3}\\ &=-\frac {2 (d+e x)^3}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {15 e \left (c d^2-a e^2\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 c^3 d^3}+\frac {5 e (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 c^2 d^2}+\frac {15 \sqrt {e} \left (c d^2-a e^2\right )^2 \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 )}{8 c^{7/2} d^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.30, size = 180, normalized size = 0.75 \begin {gather*} \frac {-\sqrt {c} \sqrt {d} (d+e x) \left (15 a^2 e^4+5 a c d e^2 (-5 d+e x)+c^2 d^2 \left (8 d^2-9 d e x-2 e^2 x^2\right )\right )+15 \sqrt {e} \left (c d^2-a e^2\right )^2 \sqrt {a e+c d x} \sqrt {d+e x} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {e} \sqrt {a e+c d x}}\right )}{4 c^{7/2} d^{7/2} \sqrt {(a e+c d x) (d+e x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^4/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2),x]

[Out]

(-(Sqrt[c]*Sqrt[d]*(d + e*x)*(15*a^2*e^4 + 5*a*c*d*e^2*(-5*d + e*x) + c^2*d^2*(8*d^2 - 9*d*e*x - 2*e^2*x^2)))
+ 15*Sqrt[e]*(c*d^2 - a*e^2)^2*Sqrt[a*e + c*d*x]*Sqrt[d + e*x]*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/(Sqrt[e
]*Sqrt[a*e + c*d*x])])/(4*c^(7/2)*d^(7/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1769\) vs. \(2(213)=426\).
time = 0.69, size = 1770, normalized size = 7.34

method	result	size
default	\(\text {Expression too large to display}\)	\(1770\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

e^4*(1/2*x^3/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-5/4*(a*e^2+c*d^2)/c/d/e*(x^2/c/d/e/(a*d*e+(a*e^2+c*
d^2)*x+c*d*e*x^2)^(1/2)-3/2*(a*e^2+c*d^2)/c/d/e*(-x/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-1/2*(a*e^2+c
*d^2)/c/d/e*(-1/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-(a*e^2+c*d^2)/c/d/e*(2*c*d*e*x+a*e^2+c*d^2)/(4*a
*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))+1/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*a/c*(-1/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c
*d*e*x^2)^(1/2)-(a*e^2+c*d^2)/c/d/e*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^
2)*x+c*d*e*x^2)^(1/2)))-3/2*a/c*(-x/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-1/2*(a*e^2+c*d^2)/c/d/e*(-1/
c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-(a*e^2+c*d^2)/c/d/e*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^
2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))+1/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)))+4*d*e^3*(x^2/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1
/2)-3/2*(a*e^2+c*d^2)/c/d/e*(-x/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-1/2*(a*e^2+c*d^2)/c/d/e*(-1/c/d/
e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-(a*e^2+c*d^2)/c/d/e*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*
d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))+1/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*a/c*(-1/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-(a*e
^2+c*d^2)/c/d/e*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2
)))+6*d^2*e^2*(-x/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-1/2*(a*e^2+c*d^2)/c/d/e*(-1/c/d/e/(a*d*e+(a*e^
2+c*d^2)*x+c*d*e*x^2)^(1/2)-(a*e^2+c*d^2)/c/d/e*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e
+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))+1/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))+4*d^3*e*(-1/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-(a*e^2+c*d^2)/c/
d/e*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))+2*d^4*(2*
c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(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)^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),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 = 2.72, size = 572, normalized size = 2.37 \begin {gather*} \left [\frac {15 \, {\left (c^{3} d^{5} x - 2 \, a c^{2} d^{3} x e^{2} + a c^{2} d^{4} e + a^{2} c d x e^{4} - 2 \, a^{2} c d^{2} e^{3} + a^{3} e^{5}\right )} \sqrt {\frac {1}{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 \, {\left (2 \, c^{2} d^{2} x e + c^{2} d^{3} + a c d e^{2}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} \sqrt {\frac {1}{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 (9 \, c^{2} d^{3} x e - 8 \, c^{2} d^{4} - 5 \, a c d x e^{3} - 15 \, a^{2} e^{4} + {\left (2 \, c^{2} d^{2} x^{2} + 25 \, a c d^{2}\right )} e^{2}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}}{16 \, {\left (c^{4} d^{4} x + a c^{3} d^{3} e\right )}}, -\frac {15 \, {\left (c^{3} d^{5} x - 2 \, a c^{2} d^{3} x e^{2} + a c^{2} d^{4} e + a^{2} c d x e^{4} - 2 \, a^{2} c d^{2} e^{3} + a^{3} e^{5}\right )} \sqrt {-\frac {e}{c d}} \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 {-\frac {e}{c d}}}{2 \, {\left (c d^{2} x e + a x e^{3} + {\left (c d x^{2} + a d\right )} e^{2}\right )}}\right ) - 2 \, {\left (9 \, c^{2} d^{3} x e - 8 \, c^{2} d^{4} - 5 \, a c d x e^{3} - 15 \, a^{2} e^{4} + {\left (2 \, c^{2} d^{2} x^{2} + 25 \, a c d^{2}\right )} e^{2}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}}{8 \, {\left (c^{4} d^{4} x + a c^{3} d^{3} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(15*(c^3*d^5*x - 2*a*c^2*d^3*x*e^2 + a*c^2*d^4*e + a^2*c*d*x*e^4 - 2*a^2*c*d^2*e^3 + a^3*e^5)*sqrt(1/(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*(2*c^2*d^2*x*e + c^2*d^3 + a*c*d*e^2)*sq
rt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*sqrt(1/(c*d))*e^(1/2) + 2*(4*c^2*d^2*x^2 + 3*a*c*d^2)*e^2) + 4*(9*c^
2*d^3*x*e - 8*c^2*d^4 - 5*a*c*d*x*e^3 - 15*a^2*e^4 + (2*c^2*d^2*x^2 + 25*a*c*d^2)*e^2)*sqrt(c*d^2*x + a*x*e^2
+ (c*d*x^2 + a*d)*e))/(c^4*d^4*x + a*c^3*d^3*e), -1/8*(15*(c^3*d^5*x - 2*a*c^2*d^3*x*e^2 + a*c^2*d^4*e + a^2*c
*d*x*e^4 - 2*a^2*c*d^2*e^3 + a^3*e^5)*sqrt(-e/(c*d))*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(-e/(c*d))/(c*d^2*x*e + a*x*e^3 + (c*d*x^2 + a*d)*e^2)) - 2*(9*c^2*d^3*x*e - 8*c
^2*d^4 - 5*a*c*d*x*e^3 - 15*a^2*e^4 + (2*c^2*d^2*x^2 + 25*a*c*d^2)*e^2)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*
d)*e))/(c^4*d^4*x + a*c^3*d^3*e)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{4}}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**4/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)

[Out]

Integral((d + e*x)**4/((d + e*x)*(a*e + c*d*x))**(3/2), x)

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Giac [A]
time = 2.23, size = 293, normalized size = 1.22 \begin {gather*} \frac {1}{4} \, \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e} {\left (\frac {2 \, x e^{2}}{c^{2} d^{2}} + \frac {{\left (9 \, c^{6} d^{7} e^{2} - 7 \, a c^{5} d^{5} e^{4}\right )} e^{\left (-1\right )}}{c^{8} d^{8}}\right )} - \frac {15 \, {\left (\sqrt {c d} c^{2} d^{4} e^{\frac {3}{2}} - 2 \, \sqrt {c d} a c d^{2} e^{\frac {7}{2}} + \sqrt {c d} a^{2} e^{\frac {11}{2}}\right )} e^{\left (-1\right )} \log \left ({\left | -\sqrt {c d} c d^{2} e^{\frac {1}{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 )} c d e - \sqrt {c d} a e^{\frac {5}{2}} \right |}\right )}{8 \, c^{4} d^{4}} + \frac {2 \, {\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 )}}{{\left ({\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 )} c d + \sqrt {c d} a e^{\frac {3}{2}}\right )} c^{3} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

1/4*sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e)*(2*x*e^2/(c^2*d^2) + (9*c^6*d^7*e^2 - 7*a*c^5*d^5*e^4)*e^(-1)/
(c^8*d^8)) - 15/8*(sqrt(c*d)*c^2*d^4*e^(3/2) - 2*sqrt(c*d)*a*c*d^2*e^(7/2) + sqrt(c*d)*a^2*e^(11/2))*e^(-1)*lo
g(abs(-sqrt(c*d)*c*d^2*e^(1/2) - 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))*c*d*e -
sqrt(c*d)*a*e^(5/2)))/(c^4*d^4) + 2*(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)*x*e^
(1/2) - sqrt(c*d*x^2*e + c*d^2*x + a*x*e^2 + a*d*e))*c*d + sqrt(c*d)*a*e^(3/2))*c^3*d^3)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^4}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^4/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(3/2),x)

[Out]

int((d + e*x)^4/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(3/2), x)

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