3.4.89 \(\int x^2 (a+b x^2)^{5/2} \, dx\)

Optimal. Leaf size=112 \[ -\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{3/2}}+\frac {5 a^3 x \sqrt {a+b x^2}}{128 b}+\frac {5}{64} a^2 x^3 \sqrt {a+b x^2}+\frac {5}{48} a x^3 \left (a+b x^2\right )^{3/2}+\frac {1}{8} x^3 \left (a+b x^2\right )^{5/2} \]

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Rubi [A]  time = 0.04, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {279, 321, 217, 206} \begin {gather*} -\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{3/2}}+\frac {5 a^3 x \sqrt {a+b x^2}}{128 b}+\frac {5}{64} a^2 x^3 \sqrt {a+b x^2}+\frac {5}{48} a x^3 \left (a+b x^2\right )^{3/2}+\frac {1}{8} x^3 \left (a+b x^2\right )^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x^2)^(5/2),x]

[Out]

(5*a^3*x*Sqrt[a + b*x^2])/(128*b) + (5*a^2*x^3*Sqrt[a + b*x^2])/64 + (5*a*x^3*(a + b*x^2)^(3/2))/48 + (x^3*(a
+ b*x^2)^(5/2))/8 - (5*a^4*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(128*b^(3/2))

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^2 \left (a+b x^2\right )^{5/2} \, dx &=\frac {1}{8} x^3 \left (a+b x^2\right )^{5/2}+\frac {1}{8} (5 a) \int x^2 \left (a+b x^2\right )^{3/2} \, dx\\ &=\frac {5}{48} a x^3 \left (a+b x^2\right )^{3/2}+\frac {1}{8} x^3 \left (a+b x^2\right )^{5/2}+\frac {1}{16} \left (5 a^2\right ) \int x^2 \sqrt {a+b x^2} \, dx\\ &=\frac {5}{64} a^2 x^3 \sqrt {a+b x^2}+\frac {5}{48} a x^3 \left (a+b x^2\right )^{3/2}+\frac {1}{8} x^3 \left (a+b x^2\right )^{5/2}+\frac {1}{64} \left (5 a^3\right ) \int \frac {x^2}{\sqrt {a+b x^2}} \, dx\\ &=\frac {5 a^3 x \sqrt {a+b x^2}}{128 b}+\frac {5}{64} a^2 x^3 \sqrt {a+b x^2}+\frac {5}{48} a x^3 \left (a+b x^2\right )^{3/2}+\frac {1}{8} x^3 \left (a+b x^2\right )^{5/2}-\frac {\left (5 a^4\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{128 b}\\ &=\frac {5 a^3 x \sqrt {a+b x^2}}{128 b}+\frac {5}{64} a^2 x^3 \sqrt {a+b x^2}+\frac {5}{48} a x^3 \left (a+b x^2\right )^{3/2}+\frac {1}{8} x^3 \left (a+b x^2\right )^{5/2}-\frac {\left (5 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{128 b}\\ &=\frac {5 a^3 x \sqrt {a+b x^2}}{128 b}+\frac {5}{64} a^2 x^3 \sqrt {a+b x^2}+\frac {5}{48} a x^3 \left (a+b x^2\right )^{3/2}+\frac {1}{8} x^3 \left (a+b x^2\right )^{5/2}-\frac {5 a^4 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 94, normalized size = 0.84 \begin {gather*} \frac {\sqrt {a+b x^2} \left (\sqrt {b} x \left (15 a^3+118 a^2 b x^2+136 a b^2 x^4+48 b^3 x^6\right )-\frac {15 a^{7/2} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}\right )}{384 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x^2)^(5/2),x]

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*x*(15*a^3 + 118*a^2*b*x^2 + 136*a*b^2*x^4 + 48*b^3*x^6) - (15*a^(7/2)*ArcSinh[(Sqrt[
b]*x)/Sqrt[a]])/Sqrt[1 + (b*x^2)/a]))/(384*b^(3/2))

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IntegrateAlgebraic [A]  time = 0.10, size = 85, normalized size = 0.76 \begin {gather*} \frac {5 a^4 \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{128 b^{3/2}}+\frac {\sqrt {a+b x^2} \left (15 a^3 x+118 a^2 b x^3+136 a b^2 x^5+48 b^3 x^7\right )}{384 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(15*a^3*x + 118*a^2*b*x^3 + 136*a*b^2*x^5 + 48*b^3*x^7))/(384*b) + (5*a^4*Log[-(Sqrt[b]*x) +
Sqrt[a + b*x^2]])/(128*b^(3/2))

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fricas [A]  time = 0.89, size = 167, normalized size = 1.49 \begin {gather*} \left [\frac {15 \, a^{4} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (48 \, b^{4} x^{7} + 136 \, a b^{3} x^{5} + 118 \, a^{2} b^{2} x^{3} + 15 \, a^{3} b x\right )} \sqrt {b x^{2} + a}}{768 \, b^{2}}, \frac {15 \, a^{4} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (48 \, b^{4} x^{7} + 136 \, a b^{3} x^{5} + 118 \, a^{2} b^{2} x^{3} + 15 \, a^{3} b x\right )} \sqrt {b x^{2} + a}}{384 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/768*(15*a^4*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(48*b^4*x^7 + 136*a*b^3*x^5 + 118*a
^2*b^2*x^3 + 15*a^3*b*x)*sqrt(b*x^2 + a))/b^2, 1/384*(15*a^4*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (48
*b^4*x^7 + 136*a*b^3*x^5 + 118*a^2*b^2*x^3 + 15*a^3*b*x)*sqrt(b*x^2 + a))/b^2]

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giac [A]  time = 1.08, size = 77, normalized size = 0.69 \begin {gather*} \frac {5 \, a^{4} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {3}{2}}} + \frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, b^{2} x^{2} + 17 \, a b\right )} x^{2} + 59 \, a^{2}\right )} x^{2} + \frac {15 \, a^{3}}{b}\right )} \sqrt {b x^{2} + a} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/128*a^4*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2) + 1/384*(2*(4*(6*b^2*x^2 + 17*a*b)*x^2 + 59*a^2)*x^2
+ 15*a^3/b)*sqrt(b*x^2 + a)*x

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maple [A]  time = 0.01, size = 93, normalized size = 0.83 \begin {gather*} -\frac {5 a^{4} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {3}{2}}}-\frac {5 \sqrt {b \,x^{2}+a}\, a^{3} x}{128 b}-\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2} x}{192 b}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} a x}{48 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} x}{8 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x*(b*x^2+a)^(7/2)/b-1/48*a/b*x*(b*x^2+a)^(5/2)-5/192*a^2/b*x*(b*x^2+a)^(3/2)-5/128*a^3*x*(b*x^2+a)^(1/2)/b
-5/128*a^4/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.34, size = 85, normalized size = 0.76 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} x}{8 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} a x}{48 \, b} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} x}{192 \, b} - \frac {5 \, \sqrt {b x^{2} + a} a^{3} x}{128 \, b} - \frac {5 \, a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x^2 + a)^(7/2)*x/b - 1/48*(b*x^2 + a)^(5/2)*a*x/b - 5/192*(b*x^2 + a)^(3/2)*a^2*x/b - 5/128*sqrt(b*x^2
+ a)*a^3*x/b - 5/128*a^4*arcsinh(b*x/sqrt(a*b))/b^(3/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,{\left (b\,x^2+a\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*x^2)^(5/2),x)

[Out]

int(x^2*(a + b*x^2)^(5/2), x)

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sympy [A]  time = 7.22, size = 150, normalized size = 1.34 \begin {gather*} \frac {5 a^{\frac {7}{2}} x}{128 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {133 a^{\frac {5}{2}} x^{3}}{384 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {127 a^{\frac {3}{2}} b x^{5}}{192 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {23 \sqrt {a} b^{2} x^{7}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {5 a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{128 b^{\frac {3}{2}}} + \frac {b^{3} x^{9}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*a**(7/2)*x/(128*b*sqrt(1 + b*x**2/a)) + 133*a**(5/2)*x**3/(384*sqrt(1 + b*x**2/a)) + 127*a**(3/2)*b*x**5/(19
2*sqrt(1 + b*x**2/a)) + 23*sqrt(a)*b**2*x**7/(48*sqrt(1 + b*x**2/a)) - 5*a**4*asinh(sqrt(b)*x/sqrt(a))/(128*b*
*(3/2)) + b**3*x**9/(8*sqrt(a)*sqrt(1 + b*x**2/a))

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