3.20.53 \(\int \frac {1}{(d+e x)^4 \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [1953]

Optimal. Leaf size=231 \[ \frac {2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{7 \left (c d^2-a e^2\right ) (d+e x)^4}+\frac {12 c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{35 \left (c d^2-a e^2\right )^2 (d+e x)^3}+\frac {16 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{35 \left (c d^2-a e^2\right )^3 (d+e x)^2}+\frac {32 c^3 d^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{35 \left (c d^2-a e^2\right )^4 (d+e x)} \]

[Out]

2/7*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(-a*e^2+c*d^2)/(e*x+d)^4+12/35*c*d*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^
2)^(1/2)/(-a*e^2+c*d^2)^2/(e*x+d)^3+16/35*c^2*d^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(-a*e^2+c*d^2)^3/(e*
x+d)^2+32/35*c^3*d^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(-a*e^2+c*d^2)^4/(e*x+d)

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Rubi [A]
time = 0.09, antiderivative size = 231, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {32 c^3 d^3 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{35 (d+e x) \left (c d^2-a e^2\right )^4}+\frac {16 c^2 d^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{35 (d+e x)^2 \left (c d^2-a e^2\right )^3}+\frac {12 c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{35 (d+e x)^3 \left (c d^2-a e^2\right )^2}+\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{7 (d+e x)^4 \left (c d^2-a e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^4*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]),x]

[Out]

(2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(7*(c*d^2 - a*e^2)*(d + e*x)^4) + (12*c*d*Sqrt[a*d*e + (c*d^2
+ a*e^2)*x + c*d*e*x^2])/(35*(c*d^2 - a*e^2)^2*(d + e*x)^3) + (16*c^2*d^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d
*e*x^2])/(35*(c*d^2 - a*e^2)^3*(d + e*x)^2) + (32*c^3*d^3*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(35*(c*
d^2 - a*e^2)^4*(d + e*x))

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 {1}{(d+e x)^4 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx &=\frac {2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{7 \left (c d^2-a e^2\right ) (d+e x)^4}+\frac {(6 c d) \int \frac {1}{(d+e x)^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{7 \left (c d^2-a e^2\right )}\\ &=\frac {2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{7 \left (c d^2-a e^2\right ) (d+e x)^4}+\frac {12 c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{35 \left (c d^2-a e^2\right )^2 (d+e x)^3}+\frac {\left (24 c^2 d^2\right ) \int \frac {1}{(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{35 \left (c d^2-a e^2\right )^2}\\ &=\frac {2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{7 \left (c d^2-a e^2\right ) (d+e x)^4}+\frac {12 c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{35 \left (c d^2-a e^2\right )^2 (d+e x)^3}+\frac {16 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{35 \left (c d^2-a e^2\right )^3 (d+e x)^2}+\frac {\left (16 c^3 d^3\right ) \int \frac {1}{(d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{35 \left (c d^2-a e^2\right )^3}\\ &=\frac {2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{7 \left (c d^2-a e^2\right ) (d+e x)^4}+\frac {12 c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{35 \left (c d^2-a e^2\right )^2 (d+e x)^3}+\frac {16 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{35 \left (c d^2-a e^2\right )^3 (d+e x)^2}+\frac {32 c^3 d^3 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{35 \left (c d^2-a e^2\right )^4 (d+e x)}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 138, normalized size = 0.60 \begin {gather*} \frac {2 \sqrt {(a e+c d x) (d+e x)} \left (-5 a^3 e^6+3 a^2 c d e^4 (7 d+2 e x)-a c^2 d^2 e^2 \left (35 d^2+28 d e x+8 e^2 x^2\right )+c^3 d^3 \left (35 d^3+70 d^2 e x+56 d e^2 x^2+16 e^3 x^3\right )\right )}{35 \left (c d^2-a e^2\right )^4 (d+e x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^4*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]),x]

[Out]

(2*Sqrt[(a*e + c*d*x)*(d + e*x)]*(-5*a^3*e^6 + 3*a^2*c*d*e^4*(7*d + 2*e*x) - a*c^2*d^2*e^2*(35*d^2 + 28*d*e*x
+ 8*e^2*x^2) + c^3*d^3*(35*d^3 + 70*d^2*e*x + 56*d*e^2*x^2 + 16*e^3*x^3)))/(35*(c*d^2 - a*e^2)^4*(d + e*x)^4)

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Maple [A]
time = 0.86, size = 293, normalized size = 1.27

method	result	size
trager	\(-\frac {2 \left (-16 c^{3} d^{3} e^{3} x^{3}+8 a \,c^{2} d^{2} e^{4} x^{2}-56 c^{3} d^{4} e^{2} x^{2}-6 a^{2} c d \,e^{5} x +28 a \,c^{2} d^{3} e^{3} x -70 c^{3} d^{5} e x +5 e^{6} a^{3}-21 e^{4} d^{2} a^{2} c +35 d^{4} e^{2} c^{2} a -35 d^{6} c^{3}\right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{35 \left (a^{4} e^{8}-4 a^{3} c \,d^{2} e^{6}+6 a^{2} c^{2} d^{4} e^{4}-4 a \,c^{3} d^{6} e^{2}+c^{4} d^{8}\right ) \left (e x +d \right )^{4}}\)	\(209\)
gosper	\(-\frac {2 \left (c d x +a e \right ) \left (-16 c^{3} d^{3} e^{3} x^{3}+8 a \,c^{2} d^{2} e^{4} x^{2}-56 c^{3} d^{4} e^{2} x^{2}-6 a^{2} c d \,e^{5} x +28 a \,c^{2} d^{3} e^{3} x -70 c^{3} d^{5} e x +5 e^{6} a^{3}-21 e^{4} d^{2} a^{2} c +35 d^{4} e^{2} c^{2} a -35 d^{6} c^{3}\right )}{35 \left (e x +d \right )^{3} \left (a^{4} e^{8}-4 a^{3} c \,d^{2} e^{6}+6 a^{2} c^{2} d^{4} e^{4}-4 a \,c^{3} d^{6} e^{2}+c^{4} d^{8}\right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}\)	\(217\)
default	\(\frac {-\frac {2 \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{7 \left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{4}}-\frac {6 c d e \left (-\frac {2 \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{5 \left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{3}}-\frac {4 c d e \left (-\frac {2 \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{3 \left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{2}}+\frac {4 c d e \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{3 \left (e^{2} a -c \,d^{2}\right )^{2} \left (x +\frac {d}{e}\right )}\right )}{5 \left (e^{2} a -c \,d^{2}\right )}\right )}{7 \left (e^{2} a -c \,d^{2}\right )}}{e^{4}}\)	\(293\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e^4*(-2/7/(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))^(1/2)-6/7*c*d*e/(a*e^2-c*d^2)*(-2/
5/(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))^(1/2)-4/5*c*d*e/(a*e^2-c*d^2)*(-2/3/(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))^(1/2)+4/3*c*d*e/(a*e^2-c*d^2)^2/(x+d/e)*(c*d*e*(x+d/e)
^2+(a*e^2-c*d^2)*(x+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(1/(e*x+d)^4/(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 [B] Leaf count of result is larger than twice the leaf count of optimal. 458 vs. \(2 (219) = 438\).
time = 9.55, size = 458, normalized size = 1.98 \begin {gather*} \frac {2 \, {\left (70 \, c^{3} d^{5} x e + 35 \, c^{3} d^{6} + 6 \, a^{2} c d x e^{5} - 5 \, a^{3} e^{6} - {\left (8 \, a c^{2} d^{2} x^{2} - 21 \, a^{2} c d^{2}\right )} e^{4} + 4 \, {\left (4 \, c^{3} d^{3} x^{3} - 7 \, a c^{2} d^{3} x\right )} e^{3} + 7 \, {\left (8 \, c^{3} d^{4} x^{2} - 5 \, a c^{2} d^{4}\right )} e^{2}\right )} \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}}{35 \, {\left (4 \, c^{4} d^{11} x e + c^{4} d^{12} + a^{4} x^{4} e^{12} + 4 \, a^{4} d x^{3} e^{11} - 2 \, {\left (2 \, a^{3} c d^{2} x^{4} - 3 \, a^{4} d^{2} x^{2}\right )} e^{10} - 4 \, {\left (4 \, a^{3} c d^{3} x^{3} - a^{4} d^{3} x\right )} e^{9} + {\left (6 \, a^{2} c^{2} d^{4} x^{4} - 24 \, a^{3} c d^{4} x^{2} + a^{4} d^{4}\right )} e^{8} + 8 \, {\left (3 \, a^{2} c^{2} d^{5} x^{3} - 2 \, a^{3} c d^{5} x\right )} e^{7} - 4 \, {\left (a c^{3} d^{6} x^{4} - 9 \, a^{2} c^{2} d^{6} x^{2} + a^{3} c d^{6}\right )} e^{6} - 8 \, {\left (2 \, a c^{3} d^{7} x^{3} - 3 \, a^{2} c^{2} d^{7} x\right )} e^{5} + {\left (c^{4} d^{8} x^{4} - 24 \, a c^{3} d^{8} x^{2} + 6 \, a^{2} c^{2} d^{8}\right )} e^{4} + 4 \, {\left (c^{4} d^{9} x^{3} - 4 \, a c^{3} d^{9} x\right )} e^{3} + 2 \, {\left (3 \, c^{4} d^{10} x^{2} - 2 \, a c^{3} d^{10}\right )} e^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^4/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),x, algorithm="fricas")

[Out]

2/35*(70*c^3*d^5*x*e + 35*c^3*d^6 + 6*a^2*c*d*x*e^5 - 5*a^3*e^6 - (8*a*c^2*d^2*x^2 - 21*a^2*c*d^2)*e^4 + 4*(4*
c^3*d^3*x^3 - 7*a*c^2*d^3*x)*e^3 + 7*(8*c^3*d^4*x^2 - 5*a*c^2*d^4)*e^2)*sqrt(c*d^2*x + a*x*e^2 + (c*d*x^2 + a*
d)*e)/(4*c^4*d^11*x*e + c^4*d^12 + a^4*x^4*e^12 + 4*a^4*d*x^3*e^11 - 2*(2*a^3*c*d^2*x^4 - 3*a^4*d^2*x^2)*e^10
- 4*(4*a^3*c*d^3*x^3 - a^4*d^3*x)*e^9 + (6*a^2*c^2*d^4*x^4 - 24*a^3*c*d^4*x^2 + a^4*d^4)*e^8 + 8*(3*a^2*c^2*d^
5*x^3 - 2*a^3*c*d^5*x)*e^7 - 4*(a*c^3*d^6*x^4 - 9*a^2*c^2*d^6*x^2 + a^3*c*d^6)*e^6 - 8*(2*a*c^3*d^7*x^3 - 3*a^
2*c^2*d^7*x)*e^5 + (c^4*d^8*x^4 - 24*a*c^3*d^8*x^2 + 6*a^2*c^2*d^8)*e^4 + 4*(c^4*d^9*x^3 - 4*a*c^3*d^9*x)*e^3
+ 2*(3*c^4*d^10*x^2 - 2*a*c^3*d^10)*e^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {\left (d + e x\right ) \left (a e + c d x\right )} \left (d + e x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)**4/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2),x)

[Out]

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

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Mupad [B]
time = 0.81, size = 252, normalized size = 1.09 \begin {gather*} \frac {32\,c^3\,d^3\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{35\,{\left (a\,e^2-c\,d^2\right )}^4\,\left (d+e\,x\right )}-\frac {2\,e\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{\left (7\,a\,e^3-7\,c\,d^2\,e\right )\,{\left (d+e\,x\right )}^4}-\frac {48\,c^2\,d^2\,e\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{35\,{\left (a\,e^2-c\,d^2\right )}^2\,\left (3\,a\,e^3-3\,c\,d^2\,e\right )\,{\left (d+e\,x\right )}^2}+\frac {12\,c\,d\,e\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{7\,\left (a\,e^2-c\,d^2\right )\,\left (5\,a\,e^3-5\,c\,d^2\,e\right )\,{\left (d+e\,x\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((d + e*x)^4*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2)),x)

[Out]

(32*c^3*d^3*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/(35*(a*e^2 - c*d^2)^4*(d + e*x)) - (2*e*(x*(a*e^2 +
c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/((7*a*e^3 - 7*c*d^2*e)*(d + e*x)^4) - (48*c^2*d^2*e*(x*(a*e^2 + c*d^2) + a
*d*e + c*d*e*x^2)^(1/2))/(35*(a*e^2 - c*d^2)^2*(3*a*e^3 - 3*c*d^2*e)*(d + e*x)^2) + (12*c*d*e*(x*(a*e^2 + c*d^
2) + a*d*e + c*d*e*x^2)^(1/2))/(7*(a*e^2 - c*d^2)*(5*a*e^3 - 5*c*d^2*e)*(d + e*x)^3)

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