3.16.74 \(\int (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\) [1574]

Optimal. Leaf size=32 \[ \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b} \]

[Out]

1/6*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/b

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Rubi [A]
time = 0.00, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {623} \begin {gather*} \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(6*b)

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)]

Rubi steps

\begin {align*} \int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx &=\frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 23, normalized size = 0.72 \begin {gather*} \frac {(a+b x) \left ((a+b x)^2\right )^{5/2}}{6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

((a + b*x)*((a + b*x)^2)^(5/2))/(6*b)

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Maple [A]
time = 0.56, size = 20, normalized size = 0.62

method	result	size
default	\(\frac {\left (b x +a \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{6 b}\)	\(20\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (b x +a \right )^{5}}{6 b}\)	\(22\)
gosper	\(\frac {x \left (b^{5} x^{5}+6 a \,b^{4} x^{4}+15 a^{2} b^{3} x^{3}+20 a^{3} x^{2} b^{2}+15 a^{4} b x +6 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{6 \left (b x +a \right )^{5}}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/6*(b*x+a)*((b*x+a)^2)^(5/2)/b

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Maxima [A]
time = 0.29, size = 46, normalized size = 1.44 \begin {gather*} \frac {1}{6} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} x + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a}{6 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*x + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a/b

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Fricas [A]
time = 4.14, size = 53, normalized size = 1.66 \begin {gather*} \frac {1}{6} \, b^{5} x^{6} + a b^{4} x^{5} + \frac {5}{2} \, a^{2} b^{3} x^{4} + \frac {10}{3} \, a^{3} b^{2} x^{3} + \frac {5}{2} \, a^{4} b x^{2} + a^{5} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/6*b^5*x^6 + a*b^4*x^5 + 5/2*a^2*b^3*x^4 + 10/3*a^3*b^2*x^3 + 5/2*a^4*b*x^2 + a^5*x

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a^{2} + 2 a b x + b^{2} x^{2}\right )^{\frac {5}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Integral((a**2 + 2*a*b*x + b**2*x**2)**(5/2), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (28) = 56\).
time = 1.63, size = 83, normalized size = 2.59 \begin {gather*} \frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )} a^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {a^{6} \mathrm {sgn}\left (b x + a\right )}{6 \, b} + \frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )}^{2} a^{2} b \mathrm {sgn}\left (b x + a\right ) + \frac {1}{6} \, {\left (b x^{2} + 2 \, a x\right )}^{3} b^{2} \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

1/2*(b*x^2 + 2*a*x)*a^4*sgn(b*x + a) + 1/6*a^6*sgn(b*x + a)/b + 1/2*(b*x^2 + 2*a*x)^2*a^2*b*sgn(b*x + a) + 1/6
*(b*x^2 + 2*a*x)^3*b^2*sgn(b*x + a)

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Mupad [B]
time = 0.51, size = 32, normalized size = 1.00 \begin {gather*} \frac {\left (x\,b^2+a\,b\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{6\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)

[Out]

((a*b + b^2*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2))/(6*b^2)

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