3.1046 \(\int \frac {(c d^2+2 c d e x+c e^2 x^2)^{3/2}}{(d+e x)^4} \, dx\)

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

[Out]

c^2*(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 = 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} \[ \frac {c^2 (d+e x) \log (d+e x)}{e \sqrt {c d^2+2 c d e x+c e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2)/(d + e*x)^4,x]

[Out]

(c^2*(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 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 {\left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{(d+e x)^4} \, dx &=c^2 \int \frac {1}{\sqrt {c d^2+2 c d e x+c e^2 x^2}} \, dx\\ &=\frac {\left (c^2 \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^2 (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 = 0.00, size = 31, normalized size = 0.74 \[ \frac {c^2 (d+e x) \log (d+e x)}{e \sqrt {c (d+e x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*d^2 + 2*c*d*e*x + c*e^2*x^2)^(3/2)/(d + e*x)^4,x]

[Out]

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

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fricas [A]  time = 0.95, size = 41, normalized size = 0.98 \[ \frac {\sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}} c \log \left (e x + d\right )}{e^{2} x + d e} \]

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)^(3/2)/(e*x+d)^4,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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)^(3/2)/(e*x+d)^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 2*(-c*sqrt(c*exp(2))/2/exp(2)*ln(abs(exp
(2)*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)-sqrt(c*exp(2))*exp(1)*d))+(c^2*exp(1)^4*d+c^2*e
xp(2)*exp(1)^2*d-2*c^2*exp(2)^2*d)/2/exp(1)^4/d/sqrt(c*exp(1)^2-c*exp(2))*atan((-d*sqrt(c*exp(2))+(sqrt(c*d^2+
2*c*d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)*exp(1))/d/sqrt(c*exp(1)^2-c*exp(2)))+(3*c^2*(sqrt(c*d^2+2*c*d*x
*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)^5*exp(1)^6*d-21*c^2*exp(2)*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))-sq
rt(c*exp(2))*x)^5*exp(1)^4*d+18*c^2*exp(2)^2*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)^5*exp(
1)^2*d+9*c^2*sqrt(c*exp(2))*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)^4*exp(1)^5*d^2+45*c^2*e
xp(2)*sqrt(c*exp(2))*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)^4*exp(1)^3*d^2-54*c^2*exp(2)^2
*sqrt(c*exp(2))*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)^4*exp(1)*d^2-8*c^3*(sqrt(c*d^2+2*c*
d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)^3*exp(1)^6*d^3-42*c^3*exp(2)*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2
))-sqrt(c*exp(2))*x)^3*exp(1)^4*d^3+6*c^3*exp(2)^2*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)^
3*exp(1)^2*d^3+44*c^3*exp(2)^3*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)^3*d^3+24*c^3*sqrt(c*
exp(2))*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)^2*exp(1)^5*d^4+30*c^3*exp(2)*sqrt(c*exp(2))
*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)^2*exp(1)^3*d^4-54*c^3*exp(2)^2*sqrt(c*exp(2))*(sqr
t(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)^2*exp(1)*d^4-3*c^4*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp
(2))-sqrt(c*exp(2))*x)*exp(1)^6*d^5-21*c^4*exp(2)*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)*e
xp(1)^4*d^5+24*c^4*exp(2)^2*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x)*exp(1)^2*d^5+3*c^4*sqrt
(c*exp(2))*exp(1)^5*d^6-3*c^4*exp(2)*sqrt(c*exp(2))*exp(1)^3*d^6)/6/exp(1)^4/((sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2
*exp(2))-sqrt(c*exp(2))*x)^2*exp(1)-2*sqrt(c*exp(2))*(sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))-sqrt(c*exp(2))*x
)*d+c*exp(1)*d^2)^3)

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maple [A]  time = 0.05, size = 40, normalized size = 0.95 \[ \frac {\left (c \,e^{2} x^{2}+2 c d e x +c \,d^{2}\right )^{\frac {3}{2}} \ln \left (e x +d \right )}{\left (e x +d \right )^{3} e} \]

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)^(3/2)/(e*x+d)^4,x)

[Out]

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

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)^(3/2)/(e*x+d)^4,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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^4} \,d x \]

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)^(3/2)/(d + e*x)^4,x)

[Out]

int((c*d^2 + c*e^2*x^2 + 2*c*d*e*x)^(3/2)/(d + e*x)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \left (d + e x\right )^{2}\right )^{\frac {3}{2}}}{\left (d + e x\right )^{4}}\, dx \]

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)**(3/2)/(e*x+d)**4,x)

[Out]

Integral((c*(d + e*x)**2)**(3/2)/(d + e*x)**4, x)

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