3.11.73 \(\int \frac {(d+e x)^2}{(c d^2+2 c d e x+c e^2 x^2)^{3/2}} \, dx\) [1073]

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

[Out]

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

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Rubi [A]
time = 0.02, 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 = {656, 622, 31} \begin {gather*} \frac {(d+e x) \log (d+e x)}{c 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)^2/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2),x]

[Out]

((d + e*x)*Log[d + e*x])/(c*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*} \int \frac {(d+e x)^2}{\left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}} \, dx &=\frac {\int \frac {1}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx}{c}\\ &=\frac {\left (c d e+c e^2 x\right ) \int \frac {1}{c d e+c e^2 x} \, dx}{c \sqrt {c d^2+2 c d e x+c e^2 x^2}}\\ &=\frac {(d+e x) \log (d+e x)}{c e \sqrt {c d^2+2 c d e x+c e^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 31, normalized size = 0.74 \begin {gather*} \frac {(d+e x) \log (d+e x)}{c e \sqrt {c (d+e x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2/(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2),x]

[Out]

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

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Maple [A]
time = 0.58, size = 40, normalized size = 0.95

method	result	size
risch	\(\frac {\left (e x +d \right ) \ln \left (e x +d \right )}{c \sqrt {\left (e x +d \right )^{2} c}\, e}\)	\(30\)
default	\(\frac {\left (e x +d \right )^{3} \ln \left (e x +d \right )}{\left (x^{2} c \,e^{2}+2 c d e x +c \,d^{2}\right )^{\frac {3}{2}} e}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2)*(e*x+d)^3*ln(e*x+d)/e

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Maxima [A]
time = 0.28, size = 79, normalized size = 1.88 \begin {gather*} -\frac {2 \, d e^{\left (-1\right )}}{\sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}} c} + \frac {e^{\left (-1\right )} \log \left (d e^{\left (-1\right )} + x\right )}{c^{\frac {3}{2}}} + \frac {2 \, d x e^{\left (-2\right )}}{{\left (d e^{\left (-1\right )} + x\right )}^{2} c^{\frac {3}{2}}} + \frac {2 \, d^{2} e^{\left (-3\right )}}{{\left (d e^{\left (-1\right )} + x\right )}^{2} c^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.22, size = 47, normalized size = 1.12 \begin {gather*} \frac {\sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}} \log \left (x e + d\right )}{c^{2} x e^{2} + c^{2} d e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{2}}{\left (c \left (d + e x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/(c*e**2*x**2+2*c*d*e*x+c*d**2)**(3/2),x)

[Out]

Integral((d + e*x)**2/(c*(d + e*x)**2)**(3/2), x)

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Giac [A]
time = 1.32, size = 23, normalized size = 0.55 \begin {gather*} \frac {e^{\left (-1\right )} \log \left ({\left | x e + d \right |}\right )}{c^{\frac {3}{2}} \mathrm {sgn}\left (x e + d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(c*e^2*x^2+2*c*d*e*x+c*d^2)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^2}{{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(3/2),x)

[Out]

int((d + e*x)^2/(c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(3/2), x)

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