3.2.15 \(\int x^4 \sqrt {a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=71 \[ \frac {b x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)}+\frac {a x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \]

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Rubi [A]  time = 0.03, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {646, 43} \begin {gather*} \frac {b x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)}+\frac {a x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int x^4 \sqrt {a^2+2 a b x+b^2 x^2} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^4 \left (a b+b^2 x\right ) \, dx}{a b+b^2 x}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (a b x^4+b^2 x^5\right ) \, dx}{a b+b^2 x}\\ &=\frac {a x^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 (a+b x)}+\frac {b x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.46 \begin {gather*} \frac {x^5 \sqrt {(a+b x)^2} (6 a+5 b x)}{30 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(x^5*Sqrt[(a + b*x)^2]*(6*a + 5*b*x))/(30*(a + b*x))

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IntegrateAlgebraic [F]  time = 0.30, size = 0, normalized size = 0.00 \begin {gather*} \int x^4 \sqrt {a^2+2 a b x+b^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

Defer[IntegrateAlgebraic][x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2], x]

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fricas [A]  time = 0.39, size = 13, normalized size = 0.18 \begin {gather*} \frac {1}{6} \, b x^{6} + \frac {1}{5} \, a x^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/6*b*x^6 + 1/5*a*x^5

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giac [A]  time = 0.20, size = 39, normalized size = 0.55 \begin {gather*} \frac {1}{6} \, b x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, a x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {a^{6} \mathrm {sgn}\left (b x + a\right )}{30 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/6*b*x^6*sgn(b*x + a) + 1/5*a*x^5*sgn(b*x + a) + 1/30*a^6*sgn(b*x + a)/b^5

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maple [A]  time = 0.05, size = 30, normalized size = 0.42 \begin {gather*} \frac {\left (5 b x +6 a \right ) \sqrt {\left (b x +a \right )^{2}}\, x^{5}}{30 b x +30 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*((b*x+a)^2)^(1/2),x)

[Out]

1/30*x^5*(5*b*x+6*a)*((b*x+a)^2)^(1/2)/(b*x+a)

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maxima [B]  time = 1.46, size = 160, normalized size = 2.25 \begin {gather*} \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} x^{3}}{6 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4} x}{2 \, b^{4}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a x^{2}}{10 \, b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{5}}{2 \, b^{5}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} x}{5 \, b^{4}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3}}{15 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

1/6*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*x^3/b^2 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^4*x/b^4 - 3/10*(b^2*x^2 + 2*
a*b*x + a^2)^(3/2)*a*x^2/b^3 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^5/b^5 + 2/5*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)
*a^2*x/b^4 - 7/15*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^3/b^5

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mupad [B]  time = 0.59, size = 117, normalized size = 1.65 \begin {gather*} \frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^5+5\,b^3\,x^3\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-14\,a^3\,b^2\,x^2-13\,a^4\,b\,x-9\,a\,b^2\,x^2\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )+12\,a^2\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )\right )}{30\,b^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*((a + b*x)^2)^(1/2),x)

[Out]

((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(a^5 + 5*b^3*x^3*(a^2 + b^2*x^2 + 2*a*b*x) - 14*a^3*b^2*x^2 - 13*a^4*b*x - 9*
a*b^2*x^2*(a^2 + b^2*x^2 + 2*a*b*x) + 12*a^2*b*x*(a^2 + b^2*x^2 + 2*a*b*x)))/(30*b^5)

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sympy [A]  time = 0.09, size = 12, normalized size = 0.17 \begin {gather*} \frac {a x^{5}}{5} + \frac {b x^{6}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*((b*x+a)**2)**(1/2),x)

[Out]

a*x**5/5 + b*x**6/6

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