3.20.15 \(\int \frac {\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}}{(d+e x)^4} \, dx\) [1915]

Optimal. Leaf size=111 \[ \frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{5 \left (c d^2-a e^2\right ) (d+e x)^4}+\frac {4 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{15 \left (c d^2-a e^2\right )^2 (d+e x)^3} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {672, 664} \begin {gather*} \frac {4 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{15 (d+e x)^3 \left (c d^2-a e^2\right )^2}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{5 (d+e x)^4 \left (c d^2-a e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^m*((a +
b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b*e))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] &&
EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rule 672

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-e)*(d + e*x)^m*((a
+ b*x + c*x^2)^(p + 1)/((m + p + 1)*(2*c*d - b*e))), x] + Dist[c*(Simplify[m + 2*p + 2]/((m + p + 1)*(2*c*d -
b*e))), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a
*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 61, normalized size = 0.55 \begin {gather*} \frac {2 ((a e+c d x) (d+e x))^{3/2} \left (-3 a e^2+c d (5 d+2 e x)\right )}{15 \left (c d^2-a e^2\right )^2 (d+e x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.73, size = 131, normalized size = 1.18

method	result	size
gosper	\(-\frac {2 \left (c d x +a e \right ) \left (-2 c d e x +3 e^{2} a -5 c \,d^{2}\right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{15 \left (e x +d \right )^{3} \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right )}\)	\(90\)
trager	\(-\frac {2 \left (-2 e \,c^{2} d^{2} x^{2}+a c d \,e^{2} x -5 c^{2} d^{3} x +3 a^{2} e^{3}-5 d^{2} e a c \right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{15 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \left (e x +d \right )^{3}}\)	\(109\)
default	\(\frac {-\frac {2 \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{5 \left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{4}}+\frac {4 c d e \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{15 \left (e^{2} a -c \,d^{2}\right )^{2} \left (x +\frac {d}{e}\right )^{3}}}{e^{4}}\)	\(131\)

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)^(1/2)/(e*x+d)^4,x,method=_RETURNVERBOSE)

[Out]

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

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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)^(1/2)/(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 = 4.57, size = 208, normalized size = 1.87 \begin {gather*} \frac {2 \, {\left (5 \, c^{2} d^{3} x - a c d x e^{2} - 3 \, a^{2} e^{3} + {\left (2 \, c^{2} d^{2} x^{2} + 5 \, a c d^{2}\right )} e\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}}{15 \, {\left (3 \, c^{2} d^{6} x e + c^{2} d^{7} + a^{2} x^{3} e^{7} + 3 \, a^{2} d x^{2} e^{6} - {\left (2 \, a c d^{2} x^{3} - 3 \, a^{2} d^{2} x\right )} e^{5} - {\left (6 \, a c d^{3} x^{2} - a^{2} d^{3}\right )} e^{4} + {\left (c^{2} d^{4} x^{3} - 6 \, a c d^{4} x\right )} e^{3} + {\left (3 \, c^{2} d^{5} x^{2} - 2 \, a c d^{5}\right )} e^{2}\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)^(1/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

2/15*(5*c^2*d^3*x - a*c*d*x*e^2 - 3*a^2*e^3 + (2*c^2*d^2*x^2 + 5*a*c*d^2)*e)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2
+ a*d)*e)/(3*c^2*d^6*x*e + c^2*d^7 + a^2*x^3*e^7 + 3*a^2*d*x^2*e^6 - (2*a*c*d^2*x^3 - 3*a^2*d^2*x)*e^5 - (6*a
*c*d^3*x^2 - a^2*d^3)*e^4 + (c^2*d^4*x^3 - 6*a*c*d^4*x)*e^3 + (3*c^2*d^5*x^2 - 2*a*c*d^5)*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)**(1/2)/(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)^(1/2)/(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,[0,0,3]%%%},[6]%%%}+%%%{%%{[%%%{-6,[0,1,2]%%%},0]:
[1,0,%%%{-1

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Mupad [B]
time = 1.40, size = 562, normalized size = 5.06 \begin {gather*} \frac {\left (\frac {4\,c^2\,d^3}{5\,\left (a\,e^2-c\,d^2\right )\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )}-\frac {4\,a\,c\,d\,e^2}{5\,\left (a\,e^2-c\,d^2\right )\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{{\left (d+e\,x\right )}^2}-\frac {\left (\frac {2\,a\,e^2}{5\,a\,e^3-5\,c\,d^2\,e}-\frac {2\,c\,d^2}{5\,a\,e^3-5\,c\,d^2\,e}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{{\left (d+e\,x\right )}^3}+\frac {\left (\frac {4\,c^3\,d^4+4\,a\,c^2\,d^2\,e^2}{15\,e\,{\left (a\,e^2-c\,d^2\right )}^3}-\frac {8\,c^3\,d^4}{15\,e\,{\left (a\,e^2-c\,d^2\right )}^3}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{d+e\,x}+\frac {\left (\frac {2\,c^2\,d^3+2\,a\,c\,d\,e^2}{5\,\left (a\,e^2-c\,d^2\right )\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )}-\frac {4\,c^2\,d^3}{5\,\left (a\,e^2-c\,d^2\right )\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{{\left (d+e\,x\right )}^2}+\frac {\left (\frac {8\,c^3\,d^4}{15\,e\,{\left (a\,e^2-c\,d^2\right )}^3}-\frac {8\,a\,c^2\,d^2\,e}{15\,{\left (a\,e^2-c\,d^2\right )}^3}\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{d+e\,x}+\frac {8\,c^2\,d^2\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{15\,e\,{\left (a\,e^2-c\,d^2\right )}^2\,\left (d+e\,x\right )} \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)^(1/2)/(d + e*x)^4,x)

[Out]

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

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