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

Optimal. Leaf size=42 \[ \frac {(d+e x) \log (d+e x)}{c^2 e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

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Rubi [A]  time = 0.03, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {642, 608, 31} \begin {gather*} \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 {gather*}

Antiderivative was successfully verified.

[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 608

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 642

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*} \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 {\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*}

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Mathematica [A]  time = 0.01, size = 31, normalized size = 0.74 \begin {gather*} \frac {(d+e x) \log (d+e x)}{c^2 e \sqrt {c (d+e x)^2}} \end {gather*}

Antiderivative was successfully verified.

[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])

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IntegrateAlgebraic [B]  time = 0.41, size = 136, normalized size = 3.24 \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {x \sqrt {c e^2}}{\sqrt {c} d}-\frac {\sqrt {c d^2+2 c d e x+c e^2 x^2}}{\sqrt {c} d}\right )}{c^{5/2} e}-\frac {\sqrt {c e^2} \log \left (x \left (c d e+c e^2 x\right )-x \sqrt {c e^2} \sqrt {c d^2+2 c d e x+c e^2 x^2}\right )}{2 c^3 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 46, normalized size = 1.10 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.62, size = 110, normalized size = 2.62 \begin {gather*} \frac {2 \, {\left (C_{0} d^{3} e^{\left (-3\right )} + {\left (3 \, C_{0} d^{2} e^{\left (-2\right )} + {\left (3 \, C_{0} d e^{\left (-1\right )} + C_{0} x\right )} x\right )} x\right )}}{{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {3}{2}}} - \frac {e^{\left (-1\right )} \log \left ({\left | -\sqrt {c} d e^{2} - {\left (\sqrt {c} x e - \sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}}\right )} e^{2} \right |}\right )}{c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

2*(C_0*d^3*e^(-3) + (3*C_0*d^2*e^(-2) + (3*C_0*d*e^(-1) + C_0*x)*x)*x)/(c*x^2*e^2 + 2*c*d*x*e + c*d^2)^(3/2) -
e^(-1)*log(abs(-sqrt(c)*d*e^2 - (sqrt(c)*x*e - sqrt(c*x^2*e^2 + 2*c*d*x*e + c*d^2))*e^2))/c^(5/2)

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maple [A]  time = 0.05, size = 40, normalized size = 0.95 \begin {gather*} \frac {\left (e x +d \right )^{5} \ln \left (e x +d \right )}{\left (c \,e^{2} x^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2)*(e*x+d)^5*ln(e*x+d)/e

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maxima [B]  time = 1.90, size = 398, normalized size = 9.48 \begin {gather*} \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 {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 \, 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}{c^{\frac {5}{2}} e^{6} {\left (x + \frac {d}{e}\right )}^{2}} - \frac {8 \, d^{2}}{c^{\frac {5}{2}} e^{7} {\left (x + \frac {d}{e}\right )}^{3}} - \frac {3 \, d^{3}}{c^{\frac {5}{2}} e^{8} {\left (x + \frac {d}{e}\right )}^{4}}\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}{c^{\frac {5}{2}} e^{6} {\left (x + \frac {d}{e}\right )}^{4}}\right )} - \frac {1}{2} \, d^{2} e^{2} {\left (\frac {6}{c^{\frac {5}{2}} e^{5} {\left (x + \frac {d}{e}\right )}^{2}} - \frac {8 \, d}{c^{\frac {5}{2}} e^{6} {\left (x + \frac {d}{e}\right )}^{3}} + \frac {3 \, d^{2}}{c^{\frac {5}{2}} e^{7} {\left (x + \frac {d}{e}\right )}^{4}}\right )} - \frac {d^{4}}{4 \, c^{\frac {5}{2}} e^{5} {\left (x + \frac {d}{e}\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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*(12*x^2/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c*e^2) + 8*d^2/((c*e^2*x^2 + 2*c*d*e*x + c*d^2)^(3/2)*c*e
^4) + 6*d/(c^(5/2)*e^6*(x + d/e)^2) - 8*d^2/(c^(5/2)*e^7*(x + d/e)^3) - 3*d^3/(c^(5/2)*e^8*(x + d/e)^4)) - 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^(5/2)*e^6*(x + d/e)^4)) - 1/2*d^2*e^2*(6/(c^(
5/2)*e^5*(x + d/e)^2) - 8*d/(c^(5/2)*e^6*(x + d/e)^3) + 3*d^2/(c^(5/2)*e^7*(x + d/e)^4)) - 1/4*d^4/(c^(5/2)*e^
5*(x + d/e)^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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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 (c \left (d + e x\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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