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

Optimal. Leaf size=31 \[ \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{5 e} \]

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Rubi [A]  time = 0.02, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {643, 629} \begin {gather*} \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{5 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 629

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d*(a + b*x + c*x^2)^(p +
1))/(b*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 643

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[e^(m - 1)/c^((m - 1)/2
), Int[(d + e*x)*(a + b*x + c*x^2)^(p + (m - 1)/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 - 1)/2]

Rubi steps

\begin {align*} \int \frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{d+e x} \, dx &=c \int (d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx\\ &=\frac {\left (c d^2+2 c d e x+c e^2 x^2\right )^{5/2}}{5 e}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.65 \begin {gather*} \frac {\left (c (d+e x)^2\right )^{5/2}}{5 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(d + e*x)^2)^(5/2)/(5*e)

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IntegrateAlgebraic [A]  time = 0.04, size = 20, normalized size = 0.65 \begin {gather*} \frac {\left (c (d+e x)^2\right )^{5/2}}{5 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(d + e*x)^2)^(5/2)/(5*e)

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fricas [B]  time = 0.40, size = 89, normalized size = 2.87 \begin {gather*} \frac {{\left (c^{2} e^{4} x^{5} + 5 \, c^{2} d e^{3} x^{4} + 10 \, c^{2} d^{2} e^{2} x^{3} + 10 \, c^{2} d^{3} e x^{2} + 5 \, c^{2} d^{4} x\right )} \sqrt {c e^{2} x^{2} + 2 \, c d e x + c d^{2}}}{5 \, {\left (e x + 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)^(5/2)/(e*x+d),x, algorithm="fricas")

[Out]

1/5*(c^2*e^4*x^5 + 5*c^2*d*e^3*x^4 + 10*c^2*d^2*e^2*x^3 + 10*c^2*d^3*e*x^2 + 5*c^2*d^4*x)*sqrt(c*e^2*x^2 + 2*c
*d*e*x + c*d^2)/(e*x + d)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \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)^(5/2)/(e*x+d),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 2*((((192*c^2*exp(1)^10*1/1920/exp(1)^7*
x+768*c^2*exp(1)^9*d*1/1920/exp(1)^7)*x+1152*c^2*exp(1)^8*d^2*1/1920/exp(1)^7)*x+768*c^2*exp(1)^7*d^3*1/1920/e
xp(1)^7)*x+192*c^2*exp(1)^6*d^4*1/1920/exp(1)^7)*sqrt(c*d^2+2*c*d*x*exp(1)+c*x^2*exp(2))+2*(-c^3*d^6*exp(1)^6+
3*c^3*d^6*exp(1)^4*exp(2)-3*c^3*d^6*exp(1)^2*exp(2)^2+c^3*d^6*exp(2)^3)*2/2/exp(1)^6/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)))

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maple [B]  time = 0.05, size = 73, normalized size = 2.35 \begin {gather*} \frac {\left (e^{4} x^{4}+5 d \,e^{3} x^{3}+10 d^{2} e^{2} x^{2}+10 d^{3} e x +5 d^{4}\right ) \left (c \,e^{2} x^{2}+2 c d e x +c \,d^{2}\right )^{\frac {5}{2}} x}{5 \left (e x +d \right )^{5}} \end {gather*}

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

[Out]

1/5*x*(e^4*x^4+5*d*e^3*x^3+10*d^2*e^2*x^2+10*d^3*e*x+5*d^4)*(c*e^2*x^2+2*c*d*e*x+c*d^2)^(5/2)/(e*x+d)^5

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maxima [A]  time = 1.34, size = 27, normalized size = 0.87 \begin {gather*} \frac {{\left (c e^{2} x^{2} + 2 \, c d e x + c d^{2}\right )}^{\frac {5}{2}}}{5 \, 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)^(5/2)/(e*x+d),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.48, size = 16, normalized size = 0.52 \begin {gather*} \frac {{\left (c\,{\left (d+e\,x\right )}^2\right )}^{5/2}}{5\,e} \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)^(5/2)/(d + e*x),x)

[Out]

(c*(d + e*x)^2)^(5/2)/(5*e)

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sympy [A]  time = 5.09, size = 39, normalized size = 1.26 \begin {gather*} \begin {cases} \frac {\left (c d^{2} + 2 c d e x + c e^{2} x^{2}\right )^{\frac {5}{2}}}{5 e} & \text {for}\: e \neq 0 \\\frac {x \left (c d^{2}\right )^{\frac {5}{2}}}{d} & \text {otherwise} \end {cases} \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)**(5/2)/(e*x+d),x)

[Out]

Piecewise(((c*d**2 + 2*c*d*e*x + c*e**2*x**2)**(5/2)/(5*e), Ne(e, 0)), (x*(c*d**2)**(5/2)/d, True))

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