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

Optimal. Leaf size=94 \[ \frac {a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}}-\frac {a^2 x \sqrt {a+b x^2}}{16 b^2}+\frac {1}{6} x^5 \sqrt {a+b x^2}+\frac {a x^3 \sqrt {a+b x^2}}{24 b} \]

[Out]

1/16*a^3*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(5/2)-1/16*a^2*x*(b*x^2+a)^(1/2)/b^2+1/24*a*x^3*(b*x^2+a)^(1/2)/
b+1/6*x^5*(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {279, 321, 217, 206} \[ -\frac {a^2 x \sqrt {a+b x^2}}{16 b^2}+\frac {a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}}+\frac {1}{6} x^5 \sqrt {a+b x^2}+\frac {a x^3 \sqrt {a+b x^2}}{24 b} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[a + b*x^2],x]

[Out]

-(a^2*x*Sqrt[a + b*x^2])/(16*b^2) + (a*x^3*Sqrt[a + b*x^2])/(24*b) + (x^5*Sqrt[a + b*x^2])/6 + (a^3*ArcTanh[(S
qrt[b]*x)/Sqrt[a + b*x^2]])/(16*b^(5/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps

\begin {align*} \int x^4 \sqrt {a+b x^2} \, dx &=\frac {1}{6} x^5 \sqrt {a+b x^2}+\frac {1}{6} a \int \frac {x^4}{\sqrt {a+b x^2}} \, dx\\ &=\frac {a x^3 \sqrt {a+b x^2}}{24 b}+\frac {1}{6} x^5 \sqrt {a+b x^2}-\frac {a^2 \int \frac {x^2}{\sqrt {a+b x^2}} \, dx}{8 b}\\ &=-\frac {a^2 x \sqrt {a+b x^2}}{16 b^2}+\frac {a x^3 \sqrt {a+b x^2}}{24 b}+\frac {1}{6} x^5 \sqrt {a+b x^2}+\frac {a^3 \int \frac {1}{\sqrt {a+b x^2}} \, dx}{16 b^2}\\ &=-\frac {a^2 x \sqrt {a+b x^2}}{16 b^2}+\frac {a x^3 \sqrt {a+b x^2}}{24 b}+\frac {1}{6} x^5 \sqrt {a+b x^2}+\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{16 b^2}\\ &=-\frac {a^2 x \sqrt {a+b x^2}}{16 b^2}+\frac {a x^3 \sqrt {a+b x^2}}{24 b}+\frac {1}{6} x^5 \sqrt {a+b x^2}+\frac {a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 77, normalized size = 0.82 \[ \frac {a^3 \log \left (\sqrt {b} \sqrt {a+b x^2}+b x\right )}{16 b^{5/2}}+\sqrt {a+b x^2} \left (-\frac {a^2 x}{16 b^2}+\frac {a x^3}{24 b}+\frac {x^5}{6}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Sqrt[a + b*x^2],x]

[Out]

Sqrt[a + b*x^2]*(-1/16*(a^2*x)/b^2 + (a*x^3)/(24*b) + x^5/6) + (a^3*Log[b*x + Sqrt[b]*Sqrt[a + b*x^2]])/(16*b^
(5/2))

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fricas [A]  time = 1.11, size = 146, normalized size = 1.55 \[ \left [\frac {3 \, a^{3} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8 \, b^{3} x^{5} + 2 \, a b^{2} x^{3} - 3 \, a^{2} b x\right )} \sqrt {b x^{2} + a}}{96 \, b^{3}}, -\frac {3 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, b^{3} x^{5} + 2 \, a b^{2} x^{3} - 3 \, a^{2} b x\right )} \sqrt {b x^{2} + a}}{48 \, b^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(3*a^3*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(8*b^3*x^5 + 2*a*b^2*x^3 - 3*a^2*b*x)
*sqrt(b*x^2 + a))/b^3, -1/48*(3*a^3*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*b^3*x^5 + 2*a*b^2*x^3 - 3
*a^2*b*x)*sqrt(b*x^2 + a))/b^3]

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giac [A]  time = 0.69, size = 64, normalized size = 0.68 \[ \frac {1}{48} \, {\left (2 \, {\left (4 \, x^{2} + \frac {a}{b}\right )} x^{2} - \frac {3 \, a^{2}}{b^{2}}\right )} \sqrt {b x^{2} + a} x - \frac {a^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*(4*x^2 + a/b)*x^2 - 3*a^2/b^2)*sqrt(b*x^2 + a)*x - 1/16*a^3*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(
5/2)

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maple [A]  time = 0.01, size = 77, normalized size = 0.82 \[ \frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} x^{3}}{6 b}+\frac {a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {5}{2}}}+\frac {\sqrt {b \,x^{2}+a}\, a^{2} x}{16 b^{2}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} a x}{8 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3*(b*x^2+a)^(3/2)/b-1/8*a/b^2*x*(b*x^2+a)^(3/2)+1/16*a^2*x*(b*x^2+a)^(1/2)/b^2+1/16*a^3/b^(5/2)*ln(x*b^(
1/2)+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.31, size = 69, normalized size = 0.73 \[ \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} x^{3}}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} a^{2} x}{16 \, b^{2}} + \frac {a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(b*x^2 + a)^(3/2)*x^3/b - 1/8*(b*x^2 + a)^(3/2)*a*x/b^2 + 1/16*sqrt(b*x^2 + a)*a^2*x/b^2 + 1/16*a^3*arcsin
h(b*x/sqrt(a*b))/b^(5/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,\sqrt {b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a + b*x^2)^(1/2),x)

[Out]

int(x^4*(a + b*x^2)^(1/2), x)

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sympy [A]  time = 5.77, size = 117, normalized size = 1.24 \[ - \frac {a^{\frac {5}{2}} x}{16 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {a^{\frac {3}{2}} x^{3}}{48 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 \sqrt {a} x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 b^{\frac {5}{2}}} + \frac {b x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(5/2)*x/(16*b**2*sqrt(1 + b*x**2/a)) - a**(3/2)*x**3/(48*b*sqrt(1 + b*x**2/a)) + 5*sqrt(a)*x**5/(24*sqrt(1
+ b*x**2/a)) + a**3*asinh(sqrt(b)*x/sqrt(a))/(16*b**(5/2)) + b*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a))

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