3.571 \(\int \frac{x^4 (A+B x^2)}{(a+b x^2)^{3/2}} \, dx\)

Optimal. Leaf size=119 \[ -\frac{x^3 (4 A b-5 a B)}{4 b^2 \sqrt{a+b x^2}}+\frac{3 x \sqrt{a+b x^2} (4 A b-5 a B)}{8 b^3}-\frac{3 a (4 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 b^{7/2}}+\frac{B x^5}{4 b \sqrt{a+b x^2}} \]

[Out]

-((4*A*b - 5*a*B)*x^3)/(4*b^2*Sqrt[a + b*x^2]) + (B*x^5)/(4*b*Sqrt[a + b*x^2]) + (3*(4*A*b - 5*a*B)*x*Sqrt[a +
b*x^2])/(8*b^3) - (3*a*(4*A*b - 5*a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(8*b^(7/2))

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Rubi [A]  time = 0.0505837, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {459, 288, 321, 217, 206} \[ -\frac{x^3 (4 A b-5 a B)}{4 b^2 \sqrt{a+b x^2}}+\frac{3 x \sqrt{a+b x^2} (4 A b-5 a B)}{8 b^3}-\frac{3 a (4 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 b^{7/2}}+\frac{B x^5}{4 b \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x^2))/(a + b*x^2)^(3/2),x]

[Out]

-((4*A*b - 5*a*B)*x^3)/(4*b^2*Sqrt[a + b*x^2]) + (B*x^5)/(4*b*Sqrt[a + b*x^2]) + (3*(4*A*b - 5*a*B)*x*Sqrt[a +
b*x^2])/(8*b^3) - (3*a*(4*A*b - 5*a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(8*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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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 \frac{x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx &=\frac{B x^5}{4 b \sqrt{a+b x^2}}-\frac{(-4 A b+5 a B) \int \frac{x^4}{\left (a+b x^2\right )^{3/2}} \, dx}{4 b}\\ &=-\frac{(4 A b-5 a B) x^3}{4 b^2 \sqrt{a+b x^2}}+\frac{B x^5}{4 b \sqrt{a+b x^2}}+\frac{(3 (4 A b-5 a B)) \int \frac{x^2}{\sqrt{a+b x^2}} \, dx}{4 b^2}\\ &=-\frac{(4 A b-5 a B) x^3}{4 b^2 \sqrt{a+b x^2}}+\frac{B x^5}{4 b \sqrt{a+b x^2}}+\frac{3 (4 A b-5 a B) x \sqrt{a+b x^2}}{8 b^3}-\frac{(3 a (4 A b-5 a B)) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{8 b^3}\\ &=-\frac{(4 A b-5 a B) x^3}{4 b^2 \sqrt{a+b x^2}}+\frac{B x^5}{4 b \sqrt{a+b x^2}}+\frac{3 (4 A b-5 a B) x \sqrt{a+b x^2}}{8 b^3}-\frac{(3 a (4 A b-5 a B)) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{8 b^3}\\ &=-\frac{(4 A b-5 a B) x^3}{4 b^2 \sqrt{a+b x^2}}+\frac{B x^5}{4 b \sqrt{a+b x^2}}+\frac{3 (4 A b-5 a B) x \sqrt{a+b x^2}}{8 b^3}-\frac{3 a (4 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.119493, size = 108, normalized size = 0.91 \[ \frac{\sqrt{b} x \left (-15 a^2 B+a b \left (12 A-5 B x^2\right )+2 b^2 x^2 \left (2 A+B x^2\right )\right )+3 a^{3/2} \sqrt{\frac{b x^2}{a}+1} (5 a B-4 A b) \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 b^{7/2} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x^2))/(a + b*x^2)^(3/2),x]

[Out]

(Sqrt[b]*x*(-15*a^2*B + a*b*(12*A - 5*B*x^2) + 2*b^2*x^2*(2*A + B*x^2)) + 3*a^(3/2)*(-4*A*b + 5*a*B)*Sqrt[1 +
(b*x^2)/a]*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/(8*b^(7/2)*Sqrt[a + b*x^2])

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Maple [A]  time = 0.008, size = 141, normalized size = 1.2 \begin{align*}{\frac{{x}^{5}B}{4\,b}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{5\,Ba{x}^{3}}{8\,{b}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{15\,{a}^{2}Bx}{8\,{b}^{3}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{15\,{a}^{2}B}{8}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{7}{2}}}}+{\frac{A{x}^{3}}{2\,b}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{3\,aAx}{2\,{b}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{3\,Aa}{2}\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)^(3/2),x)

[Out]

1/4*B*x^5/b/(b*x^2+a)^(1/2)-5/8*B/b^2*a*x^3/(b*x^2+a)^(1/2)-15/8*B/b^3*a^2*x/(b*x^2+a)^(1/2)+15/8*B/b^(7/2)*a^
2*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+1/2*A*x^3/b/(b*x^2+a)^(1/2)+3/2*A/b^2*a*x/(b*x^2+a)^(1/2)-3/2*A/b^(5/2)*a*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)/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62042, size = 603, normalized size = 5.07 \begin{align*} \left [-\frac{3 \,{\left (5 \, B a^{3} - 4 \, A a^{2} b +{\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} x^{2}\right )} \sqrt{b} \log \left (-2 \, b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) - 2 \,{\left (2 \, B b^{3} x^{5} -{\left (5 \, B a b^{2} - 4 \, A b^{3}\right )} x^{3} - 3 \,{\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{16 \,{\left (b^{5} x^{2} + a b^{4}\right )}}, -\frac{3 \,{\left (5 \, B a^{3} - 4 \, A a^{2} b +{\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} x^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (2 \, B b^{3} x^{5} -{\left (5 \, B a b^{2} - 4 \, A b^{3}\right )} x^{3} - 3 \,{\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{8 \,{\left (b^{5} x^{2} + a b^{4}\right )}}\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)^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(3*(5*B*a^3 - 4*A*a^2*b + (5*B*a^2*b - 4*A*a*b^2)*x^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)
*x - a) - 2*(2*B*b^3*x^5 - (5*B*a*b^2 - 4*A*b^3)*x^3 - 3*(5*B*a^2*b - 4*A*a*b^2)*x)*sqrt(b*x^2 + a))/(b^5*x^2
+ a*b^4), -1/8*(3*(5*B*a^3 - 4*A*a^2*b + (5*B*a^2*b - 4*A*a*b^2)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 +
a)) - (2*B*b^3*x^5 - (5*B*a*b^2 - 4*A*b^3)*x^3 - 3*(5*B*a^2*b - 4*A*a*b^2)*x)*sqrt(b*x^2 + a))/(b^5*x^2 + a*b^
4)]

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Sympy [A]  time = 10.7782, size = 177, normalized size = 1.49 \begin{align*} A \left (\frac{3 \sqrt{a} x}{2 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{3 a \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{2 b^{\frac{5}{2}}} + \frac{x^{3}}{2 \sqrt{a} b \sqrt{1 + \frac{b x^{2}}{a}}}\right ) + B \left (- \frac{15 a^{\frac{3}{2}} x}{8 b^{3} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{5 \sqrt{a} x^{3}}{8 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{15 a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{8 b^{\frac{7}{2}}} + \frac{x^{5}}{4 \sqrt{a} b \sqrt{1 + \frac{b x^{2}}{a}}}\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)**(3/2),x)

[Out]

A*(3*sqrt(a)*x/(2*b**2*sqrt(1 + b*x**2/a)) - 3*a*asinh(sqrt(b)*x/sqrt(a))/(2*b**(5/2)) + x**3/(2*sqrt(a)*b*sqr
t(1 + b*x**2/a))) + B*(-15*a**(3/2)*x/(8*b**3*sqrt(1 + b*x**2/a)) - 5*sqrt(a)*x**3/(8*b**2*sqrt(1 + b*x**2/a))
+ 15*a**2*asinh(sqrt(b)*x/sqrt(a))/(8*b**(7/2)) + x**5/(4*sqrt(a)*b*sqrt(1 + b*x**2/a)))

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Giac [A]  time = 1.11724, size = 140, normalized size = 1.18 \begin{align*} \frac{{\left ({\left (\frac{2 \, B x^{2}}{b} - \frac{5 \, B a b^{3} - 4 \, A b^{4}}{b^{5}}\right )} x^{2} - \frac{3 \,{\left (5 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )}}{b^{5}}\right )} x}{8 \, \sqrt{b x^{2} + a}} - \frac{3 \,{\left (5 \, B a^{2} - 4 \, A a b\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{8 \, 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)^(3/2),x, algorithm="giac")

[Out]

1/8*((2*B*x^2/b - (5*B*a*b^3 - 4*A*b^4)/b^5)*x^2 - 3*(5*B*a^2*b^2 - 4*A*a*b^3)/b^5)*x/sqrt(b*x^2 + a) - 3/8*(5
*B*a^2 - 4*A*a*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)