3.20.66 \(\int \frac {(d+e x)^4}{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx\) [1966]

Optimal. Leaf size=172 \[ -\frac {2 (d+e x)^3}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {2 e (d+e x)}{c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {e^{3/2} \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{c^{5/2} d^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {682, 666, 635, 212} \begin {gather*} \frac {e^{3/2} \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 )}{c^{5/2} d^{5/2}}-\frac {2 e (d+e x)}{c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 (d+e x)^3}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \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)^(5/2),x]

[Out]

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

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 666

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

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

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Mathematica [A]
time = 0.24, size = 137, normalized size = 0.80 \begin {gather*} \frac {2 \sqrt {(a e+c d x) (d+e x)} \left (-\frac {\sqrt {c} \sqrt {d} \left (3 a e^2+c d (d+4 e x)\right )}{(a e+c d x)^2}+\frac {3 e^{3/2} \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 )}{3 c^{5/2} d^{5/2}} \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)^(5/2),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2598\) vs. \(2(150)=300\).
time = 0.72, size = 2599, normalized size = 15.11

method	result	size
default	\(\text {Expression too large to display}\)	\(2599\)

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)^(5/2),x,method=_RETURNVERBOSE)

[Out]

e^4*(-1/3*x^3/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(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)^(3/2)+1/2*(a*e^2+c*d^2)/c/d/e*(-1/2*x/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/4*(a
*e^2+c*d^2)/c/d/e*(-1/3/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(a*e^2+c*d^2)/c/d/e*(2/3*(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)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2
-(a*e^2+c*d^2)^2)^2*(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)))+1/2*a/c*(2/3*(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)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2
-(a*e^2+c*d^2)^2)^2*(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)))+2*a/c*(-1/3/c/d/e/(a*d*e
+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(a*e^2+c*d^2)/c/d/e*(2/3*(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)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(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))))+1/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)))+4*d*e^3*(-x^2/c/d/e/(a*d*e+(
a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)+1/2*(a*e^2+c*d^2)/c/d/e*(-1/2*x/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-
1/4*(a*e^2+c*d^2)/c/d/e*(-1/3/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(a*e^2+c*d^2)/c/d/e*(2/3*(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)^(3/2)+16/3*c*d*e/(4*a*c*d
^2*e^2-(a*e^2+c*d^2)^2)^2*(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)))+1/2*a/c*(2/3*(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)^(3/2)+16/3*c*d*e/(4*a*c*d
^2*e^2-(a*e^2+c*d^2)^2)^2*(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)))+2*a/c*(-1/3/c/d/e/
(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(a*e^2+c*d^2)/c/d/e*(2/3*(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)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(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))))+6*d^2*e^2*(-1/2*x/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*
x^2)^(3/2)-1/4*(a*e^2+c*d^2)/c/d/e*(-1/3/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(a*e^2+c*d^2)/c/d/e
*(2/3*(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)^(3/2)+16/3*c*d
*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(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)))+1/2*a/c
*(2/3*(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)^(3/2)+16/3*c*d
*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(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)))+4*d^3*e
*(-1/3/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-1/2*(a*e^2+c*d^2)/c/d/e*(2/3*(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)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)
^2*(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)))+d^4*(2/3*(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)^(3/2)+16/3*c*d*e/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^2*(
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))

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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)^(5/2),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 = 3.68, size = 466, normalized size = 2.71 \begin {gather*} \left [\frac {3 \, {\left (c^{2} d^{2} x^{2} e + 2 \, a c d x e^{2} + a^{2} e^{3}\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 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (4 \, c d x e + c d^{2} + 3 \, a e^{2}\right )}}{6 \, {\left (c^{4} d^{4} x^{2} + 2 \, a c^{3} d^{3} x e + a^{2} c^{2} d^{2} e^{2}\right )}}, -\frac {3 \, {\left (c^{2} d^{2} x^{2} e + 2 \, a c d x e^{2} + a^{2} e^{3}\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 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (4 \, c d x e + c d^{2} + 3 \, a e^{2}\right )}}{3 \, {\left (c^{4} d^{4} x^{2} + 2 \, a c^{3} d^{3} x e + a^{2} c^{2} d^{2} e^{2}\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)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(c^2*d^2*x^2*e + 2*a*c*d*x*e^2 + a^2*e^3)*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)*sqrt(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*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*(4*c*d*x*e
+ c*d^2 + 3*a*e^2))/(c^4*d^4*x^2 + 2*a*c^3*d^3*x*e + a^2*c^2*d^2*e^2), -1/3*(3*(c^2*d^2*x^2*e + 2*a*c*d*x*e^2
+ a^2*e^3)*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)*s
qrt(-e/(c*d))/(c*d^2*x*e + a*x*e^3 + (c*d*x^2 + a*d)*e^2)) + 2*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*d)*e)*(4*
c*d*x*e + c*d^2 + 3*a*e^2))/(c^4*d^4*x^2 + 2*a*c^3*d^3*x*e + a^2*c^2*d^2*e^2)]

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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 {5}{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)**(5/2),x)

[Out]

Integral((d + e*x)**4/((d + e*x)*(a*e + c*d*x))**(5/2), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{%%%{1,[2,2,8]%%%},[4,4]%%%}+%%%{%%%{-4,[3,4,6]%%%},[4,3]
%%%}+%%%{%%

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \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 )}^{5/2}} \,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)^(5/2),x)

[Out]

int((d + e*x)^4/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2), x)

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