\(\int \frac {(d+e x)^4}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx\) [1081]

Optimal result

Integrand size = 32, antiderivative size = 42 \[ \int \frac {(d+e x)^4}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\frac {(d+e x) \log (d+e x)}{c^2 e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

[Out]

(e*x+d)*ln(e*x+d)/c^2/e/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {656, 622, 31} \[ \int \frac {(d+e x)^4}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\frac {(d+e x) \log (d+e x)}{c^2 e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

[In]

Int[(d + e*x)^4/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2),x]

[Out]

((d + e*x)*Log[d + e*x])/(c^2*e*Sqrt[c*d^2 + 2*c*d*e*x + c*e^2*x^2])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 622

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(b/2 + c*x)/Sqrt[a + b*x + c*x^2], Int[1/(b/2
+ c*x), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 656

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {1}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx}{c^2} \\ & = \frac {\left (c d e+c e^2 x\right ) \int \frac {1}{c d e+c e^2 x} \, dx}{c^2 \sqrt {c d^2+2 c d e x+c e^2 x^2}} \\ & = \frac {(d+e x) \log (d+e x)}{c^2 e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.74 \[ \int \frac {(d+e x)^4}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\frac {(d+e x) \log (d+e x)}{c^2 e \sqrt {c (d+e x)^2}} \]

[In]

Integrate[(d + e*x)^4/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(5/2),x]

[Out]

((d + e*x)*Log[d + e*x])/(c^2*e*Sqrt[c*(d + e*x)^2])

Maple [A] (verified)

Time = 2.70 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.71

method	result	size
risch	\(\frac {\left (e x +d \right ) \ln \left (e x +d \right )}{c^{2} \sqrt {c \left (e x +d \right )^{2}}\, e}\)	\(30\)
default	\(\frac {\left (e x +d \right )^{5} \ln \left (e x +d \right )}{\left (c \,x^{2} e^{2}+2 x c d e +c \,d^{2}\right )^{\frac {5}{2}} e}\)	\(40\)

[In]

int((e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/c^2/(c*(e*x+d)^2)^(1/2)*(e*x+d)*ln(e*x+d)/e

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.10 \[ \int \frac {(d+e x)^4}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}} \log \left (e x + d\right )}{c^{3} e^{2} x + c^{3} d e} \]

[In]

integrate((e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="fricas")

[Out]

sqrt(c*e^2*x^2 + 2*c*d*e*x + c*d^2)*log(e*x + d)/(c^3*e^2*x + c^3*d*e)

Sympy [A] (verification not implemented)

Time = 3.54 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.69 \[ \int \frac {(d+e x)^4}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\frac {\left (\frac {d}{e} + x\right ) \log {\left (\frac {d}{e} + x \right )}}{c^{2} \sqrt {c e^{2} \left (\frac {d}{e} + x\right )^{2}}} \]

[In]

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

[Out]

(d/e + x)*log(d/e + x)/(c**2*sqrt(c*e**2*(d/e + x)**2))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 449 vs. \(2 (40) = 80\).

Time = 0.24 (sec) , antiderivative size = 449, normalized size of antiderivative = 10.69 \[ \int \frac {(d+e x)^4}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\frac {1}{12} \, e^{4} {\left (\frac {48 \, \sqrt {c} d e^{3} x^{3} + 108 \, \sqrt {c} d^{2} e^{2} x^{2} + 88 \, \sqrt {c} d^{3} e x + 25 \, \sqrt {c} d^{4}}{c^{3} e^{9} x^{4} + 4 \, c^{3} d e^{8} x^{3} + 6 \, c^{3} d^{2} e^{7} x^{2} + 4 \, c^{3} d^{3} e^{6} x + c^{3} d^{4} e^{5}} + \frac {12 \, \log \left (e x + d\right )}{c^{\frac {5}{2}} e^{5}}\right )} - \frac {1}{3} \, d e^{3} {\left (\frac {3 \, c^{2} d^{3} e}{\left (c e^{2}\right )^{\frac {9}{2}} {\left (x + \frac {d}{e}\right )}^{4}} + \frac {12 \, x^{2}}{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} c e^{2}} - \frac {8 \, c d^{2}}{\left (c e^{2}\right )^{\frac {7}{2}} {\left (x + \frac {d}{e}\right )}^{3}} + \frac {8 \, d^{2}}{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} c e^{4}} + \frac {6 \, d}{\left (c e^{2}\right )^{\frac {5}{2}} e {\left (x + \frac {d}{e}\right )}^{2}} - \frac {6 \, d^{3}}{\left (c e^{2}\right )^{\frac {5}{2}} e^{3} {\left (x + \frac {d}{e}\right )}^{4}}\right )} - \frac {1}{2} \, d^{2} e^{2} {\left (\frac {3 \, c^{2} d^{2} e^{2}}{\left (c e^{2}\right )^{\frac {9}{2}} {\left (x + \frac {d}{e}\right )}^{4}} - \frac {8 \, c d e}{\left (c e^{2}\right )^{\frac {7}{2}} {\left (x + \frac {d}{e}\right )}^{3}} + \frac {6}{\left (c e^{2}\right )^{\frac {5}{2}} {\left (x + \frac {d}{e}\right )}^{2}}\right )} - \frac {1}{3} \, d^{3} e {\left (\frac {4}{{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {3}{2}} c e^{2}} - \frac {3 \, d}{\left (c e^{2}\right )^{\frac {5}{2}} e {\left (x + \frac {d}{e}\right )}^{4}}\right )} - \frac {d^{4}}{4 \, \left (c e^{2}\right )^{\frac {5}{2}} {\left (x + \frac {d}{e}\right )}^{4}} \]

[In]

integrate((e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="maxima")

[Out]

1/12*e^4*((48*sqrt(c)*d*e^3*x^3 + 108*sqrt(c)*d^2*e^2*x^2 + 88*sqrt(c)*d^3*e*x + 25*sqrt(c)*d^4)/(c^3*e^9*x^4
+ 4*c^3*d*e^8*x^3 + 6*c^3*d^2*e^7*x^2 + 4*c^3*d^3*e^6*x + c^3*d^4*e^5) + 12*log(e*x + d)/(c^(5/2)*e^5)) - 1/3*
d*e^3*(3*c^2*d^3*e/((c*e^2)^(9/2)*(x + d/e)^4) + 12*x^2/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c*e^2) - 8*c*d^
2/((c*e^2)^(7/2)*(x + d/e)^3) + 8*d^2/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c*e^4) + 6*d/((c*e^2)^(5/2)*e*(x
+ d/e)^2) - 6*d^3/((c*e^2)^(5/2)*e^3*(x + d/e)^4)) - 1/2*d^2*e^2*(3*c^2*d^2*e^2/((c*e^2)^(9/2)*(x + d/e)^4) -
8*c*d*e/((c*e^2)^(7/2)*(x + d/e)^3) + 6/((c*e^2)^(5/2)*(x + d/e)^2)) - 1/3*d^3*e*(4/((c*e^2*x^2 + 2*c*d*e*x +
c*d^2)^(3/2)*c*e^2) - 3*d/((c*e^2)^(5/2)*e*(x + d/e)^4)) - 1/4*d^4/((c*e^2)^(5/2)*(x + d/e)^4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.52 \[ \int \frac {(d+e x)^4}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\frac {\log \left ({\left | e x + d \right |}\right )}{c^{\frac {5}{2}} e \mathrm {sgn}\left (e x + d\right )} \]

[In]

integrate((e*x+d)^4/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2),x, algorithm="giac")

[Out]

log(abs(e*x + d))/(c^(5/2)*e*sgn(e*x + d))

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^4}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^4}{{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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