3.11.34 \(\int \frac {\sqrt {c d^2+2 c d e x+c e^2 x^2}}{(d+e x)^2} \, dx\) [1034]

Optimal. Leaf size=40 \[ \frac {c (d+e x) \log (d+e x)}{e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

[Out]

c*(e*x+d)*ln(e*x+d)/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 = 40, 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 {c (d+e x) \log (d+e x)}{e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(d + e*x)*Log[d + e*x])/(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 {\sqrt {c d^2+2 c d e x+c e^2 x^2}}{(d+e x)^2} \, dx &=c \int \frac {1}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx\\ &=\frac {\left (c \left (c d e+c e^2 x\right )\right ) \int \frac {1}{c d e+c e^2 x} \, dx}{\sqrt {c d^2+2 c d e x+c e^2 x^2}}\\ &=\frac {c (d+e x) \log (d+e x)}{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 = 29, normalized size = 0.72 \begin {gather*} \frac {c (d+e x) \log (d+e x)}{e \sqrt {c (d+e x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.62, size = 40, normalized size = 1.00

method	result	size
risch	\(\frac {\sqrt {\left (e x +d \right )^{2} c}\, \ln \left (e x +d \right )}{\left (e x +d \right ) e}\)	\(29\)
default	\(\frac {\sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}\, \ln \left (e x +d \right )}{\left (e x +d \right ) e}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [A]
time = 3.24, size = 41, normalized size = 1.02 \begin {gather*} \frac {\sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}} \log \left (x e + d\right )}{x e^{2} + d e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.86, size = 21, normalized size = 0.52 \begin {gather*} \sqrt {c} e^{\left (-1\right )} \log \left ({\left | x e + d \right |}\right ) \mathrm {sgn}\left (x e + d\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c)*e^(-1)*log(abs(x*e + d))*sgn(x*e + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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