3.556 \(\int \frac{x^4 (A+B x^2)}{\sqrt{a+b x^2}} \, dx\)

Optimal. Leaf size=122 \[ \frac{a^2 (6 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{7/2}}+\frac{x^3 \sqrt{a+b x^2} (6 A b-5 a B)}{24 b^2}-\frac{a x \sqrt{a+b x^2} (6 A b-5 a B)}{16 b^3}+\frac{B x^5 \sqrt{a+b x^2}}{6 b} \]

[Out]

-(a*(6*A*b - 5*a*B)*x*Sqrt[a + b*x^2])/(16*b^3) + ((6*A*b - 5*a*B)*x^3*Sqrt[a + b*x^2])/(24*b^2) + (B*x^5*Sqrt
[a + b*x^2])/(6*b) + (a^2*(6*A*b - 5*a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*b^(7/2))

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Rubi [A]  time = 0.0529656, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {459, 321, 217, 206} \[ \frac{a^2 (6 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{7/2}}+\frac{x^3 \sqrt{a+b x^2} (6 A b-5 a B)}{24 b^2}-\frac{a x \sqrt{a+b x^2} (6 A b-5 a B)}{16 b^3}+\frac{B x^5 \sqrt{a+b x^2}}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x^2))/Sqrt[a + b*x^2],x]

[Out]

-(a*(6*A*b - 5*a*B)*x*Sqrt[a + b*x^2])/(16*b^3) + ((6*A*b - 5*a*B)*x^3*Sqrt[a + b*x^2])/(24*b^2) + (B*x^5*Sqrt
[a + b*x^2])/(6*b) + (a^2*(6*A*b - 5*a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*b^(7/2))

Rule 459

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

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 \frac{x^4 \left (A+B x^2\right )}{\sqrt{a+b x^2}} \, dx &=\frac{B x^5 \sqrt{a+b x^2}}{6 b}-\frac{(-6 A b+5 a B) \int \frac{x^4}{\sqrt{a+b x^2}} \, dx}{6 b}\\ &=\frac{(6 A b-5 a B) x^3 \sqrt{a+b x^2}}{24 b^2}+\frac{B x^5 \sqrt{a+b x^2}}{6 b}-\frac{(a (6 A b-5 a B)) \int \frac{x^2}{\sqrt{a+b x^2}} \, dx}{8 b^2}\\ &=-\frac{a (6 A b-5 a B) x \sqrt{a+b x^2}}{16 b^3}+\frac{(6 A b-5 a B) x^3 \sqrt{a+b x^2}}{24 b^2}+\frac{B x^5 \sqrt{a+b x^2}}{6 b}+\frac{\left (a^2 (6 A b-5 a B)\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{16 b^3}\\ &=-\frac{a (6 A b-5 a B) x \sqrt{a+b x^2}}{16 b^3}+\frac{(6 A b-5 a B) x^3 \sqrt{a+b x^2}}{24 b^2}+\frac{B x^5 \sqrt{a+b x^2}}{6 b}+\frac{\left (a^2 (6 A b-5 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{16 b^3}\\ &=-\frac{a (6 A b-5 a B) x \sqrt{a+b x^2}}{16 b^3}+\frac{(6 A b-5 a B) x^3 \sqrt{a+b x^2}}{24 b^2}+\frac{B x^5 \sqrt{a+b x^2}}{6 b}+\frac{a^2 (6 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0952979, size = 100, normalized size = 0.82 \[ \frac{\sqrt{b} x \sqrt{a+b x^2} \left (15 a^2 B-2 a b \left (9 A+5 B x^2\right )+4 b^2 x^2 \left (3 A+2 B x^2\right )\right )-3 a^2 (5 a B-6 A b) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{48 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x^2))/Sqrt[a + b*x^2],x]

[Out]

(Sqrt[b]*x*Sqrt[a + b*x^2]*(15*a^2*B + 4*b^2*x^2*(3*A + 2*B*x^2) - 2*a*b*(9*A + 5*B*x^2)) - 3*a^2*(-6*A*b + 5*
a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(48*b^(7/2))

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Maple [A]  time = 0.009, size = 143, normalized size = 1.2 \begin{align*}{\frac{{x}^{5}B}{6\,b}\sqrt{b{x}^{2}+a}}-{\frac{5\,Ba{x}^{3}}{24\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{5\,{a}^{2}Bx}{16\,{b}^{3}}\sqrt{b{x}^{2}+a}}-{\frac{5\,B{a}^{3}}{16}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{7}{2}}}}+{\frac{A{x}^{3}}{4\,b}\sqrt{b{x}^{2}+a}}-{\frac{3\,aAx}{8\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{3\,A{a}^{2}}{8}\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)/(b*x^2+a)^(1/2),x)

[Out]

1/6*B*x^5*(b*x^2+a)^(1/2)/b-5/24*B/b^2*a*x^3*(b*x^2+a)^(1/2)+5/16*B/b^3*a^2*x*(b*x^2+a)^(1/2)-5/16*B/b^(7/2)*a
^3*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+1/4*A*x^3/b*(b*x^2+a)^(1/2)-3/8*A/b^2*a*x*(b*x^2+a)^(1/2)+3/8*A/b^(5/2)*a^2*l
n(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)/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.71548, size = 490, normalized size = 4.02 \begin{align*} \left [-\frac{3 \,{\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \sqrt{b} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) - 2 \,{\left (8 \, B b^{3} x^{5} - 2 \,{\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x^{3} + 3 \,{\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{96 \, b^{4}}, \frac{3 \,{\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) +{\left (8 \, B b^{3} x^{5} - 2 \,{\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x^{3} + 3 \,{\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{48 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(3*(5*B*a^3 - 6*A*a^2*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(8*B*b^3*x^5 - 2*(
5*B*a*b^2 - 6*A*b^3)*x^3 + 3*(5*B*a^2*b - 6*A*a*b^2)*x)*sqrt(b*x^2 + a))/b^4, 1/48*(3*(5*B*a^3 - 6*A*a^2*b)*sq
rt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (8*B*b^3*x^5 - 2*(5*B*a*b^2 - 6*A*b^3)*x^3 + 3*(5*B*a^2*b - 6*A*a*
b^2)*x)*sqrt(b*x^2 + a))/b^4]

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Sympy [B]  time = 9.834, size = 235, normalized size = 1.93 \begin{align*} - \frac{3 A a^{\frac{3}{2}} x}{8 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{A \sqrt{a} x^{3}}{8 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 A a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{8 b^{\frac{5}{2}}} + \frac{A x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{5 B a^{\frac{5}{2}} x}{16 b^{3} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{5 B a^{\frac{3}{2}} x^{3}}{48 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{B \sqrt{a} x^{5}}{24 b \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{5 B a^{3} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{16 b^{\frac{7}{2}}} + \frac{B x^{7}}{6 \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)/(b*x**2+a)**(1/2),x)

[Out]

-3*A*a**(3/2)*x/(8*b**2*sqrt(1 + b*x**2/a)) - A*sqrt(a)*x**3/(8*b*sqrt(1 + b*x**2/a)) + 3*A*a**2*asinh(sqrt(b)
*x/sqrt(a))/(8*b**(5/2)) + A*x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a)) + 5*B*a**(5/2)*x/(16*b**3*sqrt(1 + b*x**2/a))
+ 5*B*a**(3/2)*x**3/(48*b**2*sqrt(1 + b*x**2/a)) - B*sqrt(a)*x**5/(24*b*sqrt(1 + b*x**2/a)) - 5*B*a**3*asinh(
sqrt(b)*x/sqrt(a))/(16*b**(7/2)) + B*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]  time = 1.13658, size = 144, normalized size = 1.18 \begin{align*} \frac{1}{48} \,{\left (2 \,{\left (\frac{4 \, B x^{2}}{b} - \frac{5 \, B a b^{3} - 6 \, A b^{4}}{b^{5}}\right )} x^{2} + \frac{3 \,{\left (5 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )}}{b^{5}}\right )} \sqrt{b x^{2} + a} x + \frac{{\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{16 \, b^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*(4*B*x^2/b - (5*B*a*b^3 - 6*A*b^4)/b^5)*x^2 + 3*(5*B*a^2*b^2 - 6*A*a*b^3)/b^5)*sqrt(b*x^2 + a)*x + 1/1
6*(5*B*a^3 - 6*A*a^2*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)