3.20.26 \(\int \frac {(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}}{(d+e x)^4} \, dx\) [1926]

Optimal. Leaf size=169 \[ -\frac {2 c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e^2 (d+e x)}-\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e (d+e x)^3}+\frac {c^{3/2} d^{3/2} \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{e^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {676, 635, 212} \begin {gather*} \frac {c^{3/2} d^{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 )}{e^{5/2}}-\frac {2 c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e^2 (d+e x)}-\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(e^2*(d + e*x)) - (2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x
^2)^(3/2))/(3*e*(d + e*x)^3) + (c^(3/2)*d^(3/2)*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])])/e^(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 676

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + p + 1))), x] - Dist[c*(p/(e^2*(m + 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] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rubi steps

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

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Mathematica [A]
time = 0.25, size = 134, normalized size = 0.79 \begin {gather*} \frac {2 ((a e+c d x) (d+e x))^{3/2} \left (-\frac {\sqrt {e} \left (e+\frac {3 c d (d+e x)}{a e+c d x}\right )}{(d+e x)^3}+\frac {3 c^{3/2} d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {e} \sqrt {a e+c d x}}\right )}{(a e+c d x)^{3/2} (d+e x)^{3/2}}\right )}{3 e^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(472\) vs. \(2(147)=294\).
time = 0.70, size = 473, normalized size = 2.80

method	result	size
default	\(\frac {-\frac {2 \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{3 \left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{4}}+\frac {2 c d e \left (-\frac {2 \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{3}}+\frac {4 c d e \left (\frac {2 \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{2}}-\frac {6 c d e \left (\frac {\left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{3}+\frac {\left (e^{2} a -c \,d^{2}\right ) \left (\frac {\left (2 c d e \left (x +\frac {d}{e}\right )+e^{2} a -c \,d^{2}\right ) \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{4 c d e}-\frac {\left (e^{2} a -c \,d^{2}\right )^{2} \ln \left (\frac {\frac {e^{2} a}{2}-\frac {c \,d^{2}}{2}+c d e \left (x +\frac {d}{e}\right )}{\sqrt {c d e}}+\sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{8 c d e \sqrt {c d e}}\right )}{2}\right )}{e^{2} a -c \,d^{2}}\right )}{e^{2} a -c \,d^{2}}\right )}{3 \left (e^{2} a -c \,d^{2}\right )}}{e^{4}}\)	\(473\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e^4*(-2/3/(a*e^2-c*d^2)/(x+d/e)^4*(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x+d/e))^(5/2)+2/3*c*d*e/(a*e^2-c*d^2)*(-2/
(a*e^2-c*d^2)/(x+d/e)^3*(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x+d/e))^(5/2)+4*c*d*e/(a*e^2-c*d^2)*(2/(a*e^2-c*d^2)/(
x+d/e)^2*(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x+d/e))^(5/2)-6*c*d*e/(a*e^2-c*d^2)*(1/3*(c*d*e*(x+d/e)^2+(a*e^2-c*d^
2)*(x+d/e))^(3/2)+1/2*(a*e^2-c*d^2)*(1/4*(2*c*d*e*(x+d/e)+e^2*a-c*d^2)/c/d/e*(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x
+d/e))^(1/2)-1/8*(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+d/e))/(c*d*e)^(1/2)+(c*d*e*(x+d/e)^2+(
a*e^2-c*d^2)*(x+d/e))^(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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^4,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 = 4.52, size = 413, normalized size = 2.44 \begin {gather*} \left [\frac {3 \, {\left (c d x^{2} e^{2} + 2 \, c d^{2} x e + c d^{3}\right )} \sqrt {c d} e^{\left (-\frac {1}{2}\right )} \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^{2} + c d^{2} e + a e^{3}\right )} \sqrt {c d} e^{\left (-\frac {1}{2}\right )} + 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 + 3 \, c d^{2} + a e^{2}\right )}}{6 \, {\left (x^{2} e^{4} + 2 \, d x e^{3} + d^{2} e^{2}\right )}}, -\frac {3 \, {\left (c d x^{2} e^{2} + 2 \, c d^{2} x e + c d^{3}\right )} \sqrt {-c d e^{\left (-1\right )}} \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^{\left (-1\right )}}}{2 \, {\left (c^{2} d^{3} x + a c d x e^{2} + {\left (c^{2} d^{2} x^{2} + a c d^{2}\right )} e\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 + 3 \, c d^{2} + a e^{2}\right )}}{3 \, {\left (x^{2} e^{4} + 2 \, d x e^{3} + d^{2} e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/(e*x+d)^4,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,[0,0,10]%%%},[4,4]%%%}+%%%{%%%{-4,[1,2,8]%%%},[4,3
]%%%}+%%%{%

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}}{{\left (d+e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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