3.11.30 \(\int (d+e x)^2 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx\) [1030]

Optimal. Leaf size=39 \[ \frac {(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 c e} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 39, 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 = {656, 623} \begin {gather*} \frac {(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 c e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 623

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1)
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

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 (d+e x)^2 \sqrt {c d^2+2 c d e x+c e^2 x^2} \, dx &=\frac {\int \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2} \, dx}{c}\\ &=\frac {(d+e x) \left (c d^2+2 c d e x+c e^2 x^2\right )^{3/2}}{4 c e}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 28, normalized size = 0.72 \begin {gather*} \frac {(d+e x) \left (c (d+e x)^2\right )^{3/2}}{4 c e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.54, size = 35, normalized size = 0.90

method	result	size
risch	\(\frac {\left (e x +d \right )^{3} \sqrt {\left (e x +d \right )^{2} c}}{4 e}\)	\(24\)
default	\(\frac {\left (e x +d \right )^{3} \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}{4 e}\)	\(35\)
gosper	\(\frac {x \left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}{4 e x +4 d}\)	\(62\)
trager	\(\frac {x \left (e^{3} x^{3}+4 d \,e^{2} x^{2}+6 d^{2} e x +4 d^{3}\right ) \sqrt {x^{2} c \,e^{2}+2 c d e x +c \,d^{2}}}{4 e x +4 d}\)	\(62\)

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)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 59, normalized size = 1.51 \begin {gather*} \frac {{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {3}{2}} d e^{\left (-1\right )}}{4 \, c} + \frac {{\left (c x^{2} e^{2} + 2 \, c d x e + c d^{2}\right )}^{\frac {3}{2}} x}{4 \, c} \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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.61, size = 63, normalized size = 1.62 \begin {gather*} \frac {{\left (x^{4} e^{3} + 4 \, d x^{3} e^{2} + 6 \, d^{2} x^{2} e + 4 \, d^{3} x\right )} \sqrt {c x^{2} e^{2} + 2 \, c d x e + c d^{2}}}{4 \, {\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)^(1/2),x, algorithm="fricas")

[Out]

1/4*(x^4*e^3 + 4*d*x^3*e^2 + 6*d^2*x^2*e + 4*d^3*x)*sqrt(c*x^2*e^2 + 2*c*d*x*e + c*d^2)/(x*e + d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \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((e*x+d)**2*(c*e**2*x**2+2*c*d*e*x+c*d**2)**(1/2),x)

[Out]

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

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Giac [A]
time = 1.53, size = 22, normalized size = 0.56 \begin {gather*} \frac {1}{4} \, {\left (x e + d\right )}^{4} \sqrt {c} e^{\left (-1\right )} \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)^(1/2),x, algorithm="giac")

[Out]

1/4*(x*e + d)^4*sqrt(c)*e^(-1)*sgn(x*e + d)

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Mupad [B]
time = 0.70, size = 76, normalized size = 1.95 \begin {gather*} \frac {\sqrt {c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2}\,\left (c\,d^3+e\,x\,\left (c\,d^2+2\,c\,d\,e\,x+c\,e^2\,x^2\right )+2\,c\,d^2\,e\,x+c\,d\,e^2\,x^2\right )}{4\,c\,e} \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)^(1/2),x)

[Out]

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

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