### 3.1081 $$\int \frac{(d+e x)^4}{(c d^2+2 c d e x+c e^2 x^2)^{5/2}} \, dx$$

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

[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])

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Rubi [A]  time = 0.0258827, antiderivative size = 42, normalized size of antiderivative = 1., 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} $\frac{(d+e x) \log (d+e x)}{c^2 e \sqrt{c d^2+2 c d e x+c e^2 x^2}}$

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 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]

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 31

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

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*}

Mathematica [A]  time = 0.0070995, size = 31, normalized size = 0.74 $\frac{(d+e x) \log (d+e x)}{c^2 e \sqrt{c (d+e x)^2}}$

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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Maple [A]  time = 0.042, size = 40, normalized size = 1. \begin{align*}{\frac{ \left ( ex+d \right ) ^{5}\ln \left ( ex+d \right ) }{e} \left ( c{e}^{2}{x}^{2}+2\,cdex+c{d}^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

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 = 1.57926, size = 606, normalized size = 14.43 \begin{align*} \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}} \end{align*}

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*(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)

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Fricas [A]  time = 2.32264, size = 97, normalized size = 2.31 \begin{align*} \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{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{4}}{\left (c \left (d + e x\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

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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Giac [B]  time = 1.55926, size = 149, normalized size = 3.55 \begin{align*} \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{align*}

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)