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

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

[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))

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Rubi [A]  time = 0.0536311, antiderivative size = 136, normalized size of antiderivative = 1., 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} \[ -\frac{3 a^4 x \sqrt{a+b x^2}}{256 b^2}+\frac{3 a^5 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{256 b^{5/2}}+\frac{a^3 x^3 \sqrt{a+b x^2}}{128 b}+\frac{1}{32} a^2 x^5 \sqrt{a+b x^2}+\frac{1}{16} a x^5 \left (a+b x^2\right )^{3/2}+\frac{1}{10} x^5 \left (a+b x^2\right )^{5/2} \]

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 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]

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 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])

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*}

Mathematica [A]  time = 0.149906, size = 105, normalized size = 0.77 \[ \frac{\sqrt{a+b x^2} \left (\sqrt{b} x \left (248 a^2 b^2 x^4+10 a^3 b x^2-15 a^4+336 a b^3 x^6+128 b^4 x^8\right )+\frac{15 a^{9/2} \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{\frac{b x^2}{a}+1}}\right )}{1280 b^{5/2}} \]

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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Maple [A]  time = 0.006, size = 113, normalized size = 0.8 \begin{align*}{\frac{{x}^{3}}{10\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}-{\frac{3\,ax}{80\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{{a}^{2}x}{160\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{{a}^{3}x}{128\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{a}^{4}x}{256\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{3\,{a}^{5}}{256}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.71064, size = 460, normalized size = 3.38 \begin{align*} \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{align*}

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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Sympy [A]  time = 10.927, size = 175, normalized size = 1.29 \begin{align*} - \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{align*}

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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Giac [A]  time = 2.34175, size = 123, normalized size = 0.9 \begin{align*} -\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{align*}

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