3.4.88 \(\int x^4 (a+b x^2)^{5/2} \, dx\)

Optimal. Leaf size=136 \[ \frac {3 a^5 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{5/2}}-\frac {3 a^4 x \sqrt {a+b x^2}}{256 b^2}+\frac {a^3 x^3 \sqrt {a+b x^2}}{128 b}+\frac {1}{32} a^2 x^5 \sqrt {a+b x^2}+\frac {1}{16} a x^5 \left (a+b x^2\right )^{3/2}+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2} \]

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Rubi [A]  time = 0.05, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, 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 {3 a^4 x \sqrt {a+b x^2}}{256 b^2}+\frac {3 a^5 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{5/2}}+\frac {a^3 x^3 \sqrt {a+b x^2}}{128 b}+\frac {1}{32} a^2 x^5 \sqrt {a+b x^2}+\frac {1}{16} a x^5 \left (a+b x^2\right )^{3/2}+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*x^2)^(5/2),x]

[Out]

(-3*a^4*x*Sqrt[a + b*x^2])/(256*b^2) + (a^3*x^3*Sqrt[a + b*x^2])/(128*b) + (a^2*x^5*Sqrt[a + b*x^2])/32 + (a*x
^5*(a + b*x^2)^(3/2))/16 + (x^5*(a + b*x^2)^(5/2))/10 + (3*a^5*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(256*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 \left (a+b x^2\right )^{5/2} \, dx &=\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}+\frac {1}{2} a \int x^4 \left (a+b x^2\right )^{3/2} \, dx\\ &=\frac {1}{16} a x^5 \left (a+b x^2\right )^{3/2}+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}+\frac {1}{16} \left (3 a^2\right ) \int x^4 \sqrt {a+b x^2} \, dx\\ &=\frac {1}{32} a^2 x^5 \sqrt {a+b x^2}+\frac {1}{16} a x^5 \left (a+b x^2\right )^{3/2}+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}+\frac {1}{32} a^3 \int \frac {x^4}{\sqrt {a+b x^2}} \, dx\\ &=\frac {a^3 x^3 \sqrt {a+b x^2}}{128 b}+\frac {1}{32} a^2 x^5 \sqrt {a+b x^2}+\frac {1}{16} a x^5 \left (a+b x^2\right )^{3/2}+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}-\frac {\left (3 a^4\right ) \int \frac {x^2}{\sqrt {a+b x^2}} \, dx}{128 b}\\ &=-\frac {3 a^4 x \sqrt {a+b x^2}}{256 b^2}+\frac {a^3 x^3 \sqrt {a+b x^2}}{128 b}+\frac {1}{32} a^2 x^5 \sqrt {a+b x^2}+\frac {1}{16} a x^5 \left (a+b x^2\right )^{3/2}+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}+\frac {\left (3 a^5\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{256 b^2}\\ &=-\frac {3 a^4 x \sqrt {a+b x^2}}{256 b^2}+\frac {a^3 x^3 \sqrt {a+b x^2}}{128 b}+\frac {1}{32} a^2 x^5 \sqrt {a+b x^2}+\frac {1}{16} a x^5 \left (a+b x^2\right )^{3/2}+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}+\frac {\left (3 a^5\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{256 b^2}\\ &=-\frac {3 a^4 x \sqrt {a+b x^2}}{256 b^2}+\frac {a^3 x^3 \sqrt {a+b x^2}}{128 b}+\frac {1}{32} a^2 x^5 \sqrt {a+b x^2}+\frac {1}{16} a x^5 \left (a+b x^2\right )^{3/2}+\frac {1}{10} x^5 \left (a+b x^2\right )^{5/2}+\frac {3 a^5 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 105, normalized size = 0.77 \begin {gather*} \frac {\sqrt {a+b x^2} \left (\frac {15 a^{9/2} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}+\sqrt {b} x \left (-15 a^4+10 a^3 b x^2+248 a^2 b^2 x^4+336 a b^3 x^6+128 b^4 x^8\right )\right )}{1280 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*x^2)^(5/2),x]

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*x*(-15*a^4 + 10*a^3*b*x^2 + 248*a^2*b^2*x^4 + 336*a*b^3*x^6 + 128*b^4*x^8) + (15*a^(
9/2)*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[1 + (b*x^2)/a]))/(1280*b^(5/2))

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IntegrateAlgebraic [A]  time = 0.12, size = 96, normalized size = 0.71 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-15 a^4 x+10 a^3 b x^3+248 a^2 b^2 x^5+336 a b^3 x^7+128 b^4 x^9\right )}{1280 b^2}-\frac {3 a^5 \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{256 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^4*(a + b*x^2)^(5/2),x]

[Out]

(Sqrt[a + b*x^2]*(-15*a^4*x + 10*a^3*b*x^3 + 248*a^2*b^2*x^5 + 336*a*b^3*x^7 + 128*b^4*x^9))/(1280*b^2) - (3*a
^5*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(256*b^(5/2))

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fricas [A]  time = 0.97, size = 190, normalized size = 1.40 \begin {gather*} \left [\frac {15 \, a^{5} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (128 \, b^{5} x^{9} + 336 \, a b^{4} x^{7} + 248 \, a^{2} b^{3} x^{5} + 10 \, a^{3} b^{2} x^{3} - 15 \, a^{4} b x\right )} \sqrt {b x^{2} + a}}{2560 \, b^{3}}, -\frac {15 \, a^{5} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (128 \, b^{5} x^{9} + 336 \, a b^{4} x^{7} + 248 \, a^{2} b^{3} x^{5} + 10 \, a^{3} b^{2} x^{3} - 15 \, a^{4} b x\right )} \sqrt {b x^{2} + a}}{1280 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[1/2560*(15*a^5*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(128*b^5*x^9 + 336*a*b^4*x^7 + 248
*a^2*b^3*x^5 + 10*a^3*b^2*x^3 - 15*a^4*b*x)*sqrt(b*x^2 + a))/b^3, -1/1280*(15*a^5*sqrt(-b)*arctan(sqrt(-b)*x/s
qrt(b*x^2 + a)) - (128*b^5*x^9 + 336*a*b^4*x^7 + 248*a^2*b^3*x^5 + 10*a^3*b^2*x^3 - 15*a^4*b*x)*sqrt(b*x^2 + a
))/b^3]

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giac [A]  time = 1.09, size = 91, normalized size = 0.67 \begin {gather*} -\frac {3 \, a^{5} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{256 \, b^{\frac {5}{2}}} + \frac {1}{1280} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, b^{2} x^{2} + 21 \, a b\right )} x^{2} + 31 \, a^{2}\right )} x^{2} + \frac {5 \, a^{3}}{b}\right )} x^{2} - \frac {15 \, a^{4}}{b^{2}}\right )} \sqrt {b x^{2} + a} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

-3/256*a^5*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2) + 1/1280*(2*(4*(2*(8*b^2*x^2 + 21*a*b)*x^2 + 31*a^2)
*x^2 + 5*a^3/b)*x^2 - 15*a^4/b^2)*sqrt(b*x^2 + a)*x

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maple [A]  time = 0.01, size = 113, normalized size = 0.83 \begin {gather*} \frac {3 a^{5} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{256 b^{\frac {5}{2}}}+\frac {3 \sqrt {b \,x^{2}+a}\, a^{4} x}{256 b^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{3} x}{128 b^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} x^{3}}{10 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{2} x}{160 b^{2}}-\frac {3 \left (b \,x^{2}+a \right )^{\frac {7}{2}} a x}{80 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x^2+a)^(5/2),x)

[Out]

1/10*x^3*(b*x^2+a)^(7/2)/b-3/80*a/b^2*x*(b*x^2+a)^(7/2)+1/160*a^2/b^2*x*(b*x^2+a)^(5/2)+1/128*a^3/b^2*x*(b*x^2
+a)^(3/2)+3/256*a^4*x*(b*x^2+a)^(1/2)/b^2+3/256*a^5/b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.26, size = 105, normalized size = 0.77 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} x^{3}}{10 \, b} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} a x}{80 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2} x}{160 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3} x}{128 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} a^{4} x}{256 \, b^{2}} + \frac {3 \, a^{5} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{256 \, b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

1/10*(b*x^2 + a)^(7/2)*x^3/b - 3/80*(b*x^2 + a)^(7/2)*a*x/b^2 + 1/160*(b*x^2 + a)^(5/2)*a^2*x/b^2 + 1/128*(b*x
^2 + a)^(3/2)*a^3*x/b^2 + 3/256*sqrt(b*x^2 + a)*a^4*x/b^2 + 3/256*a^5*arcsinh(b*x/sqrt(a*b))/b^(5/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^4\,{\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^4*(a + b*x^2)^(5/2),x)

[Out]

int(x^4*(a + b*x^2)^(5/2), x)

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sympy [A]  time = 11.11, size = 175, normalized size = 1.29 \begin {gather*} - \frac {3 a^{\frac {9}{2}} x}{256 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {a^{\frac {7}{2}} x^{3}}{256 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {129 a^{\frac {5}{2}} x^{5}}{640 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {73 a^{\frac {3}{2}} b x^{7}}{160 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {29 \sqrt {a} b^{2} x^{9}}{80 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 a^{5} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{256 b^{\frac {5}{2}}} + \frac {b^{3} x^{11}}{10 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(b*x**2+a)**(5/2),x)

[Out]

-3*a**(9/2)*x/(256*b**2*sqrt(1 + b*x**2/a)) - a**(7/2)*x**3/(256*b*sqrt(1 + b*x**2/a)) + 129*a**(5/2)*x**5/(64
0*sqrt(1 + b*x**2/a)) + 73*a**(3/2)*b*x**7/(160*sqrt(1 + b*x**2/a)) + 29*sqrt(a)*b**2*x**9/(80*sqrt(1 + b*x**2
/a)) + 3*a**5*asinh(sqrt(b)*x/sqrt(a))/(256*b**(5/2)) + b**3*x**11/(10*sqrt(a)*sqrt(1 + b*x**2/a))

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