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

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

[Out]

1/16*a^2*(6*A*b-5*B*a)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(7/2)-1/16*a*(6*A*b-5*B*a)*x*(b*x^2+a)^(1/2)/b^3+1
/24*(6*A*b-5*B*a)*x^3*(b*x^2+a)^(1/2)/b^2+1/6*B*x^5*(b*x^2+a)^(1/2)/b

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Rubi [A]  time = 0.05, antiderivative size = 122, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.10, 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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fricas [A]  time = 1.10, size = 211, normalized size = 1.73 \[ \left [-\frac {3 \, {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, B b^{3} x^{5} - 2 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{4}}, \frac {3 \, {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, B b^{3} x^{5} - 2 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{4}}\right ] \]

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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giac [A]  time = 0.45, size = 107, normalized size = 0.88 \[ \frac {1}{48} \, {\left (2 \, {\left (\frac {4 \, B x^{2}}{b} - \frac {5 \, B a b^{3} - 6 \, A b^{4}}{b^{5}}\right )} x^{2} + \frac {3 \, {\left (5 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )}}{b^{5}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {7}{2}}} \]

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)

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maple [A]  time = 0.01, size = 143, normalized size = 1.17 \[ \frac {\sqrt {b \,x^{2}+a}\, B \,x^{5}}{6 b}+\frac {\sqrt {b \,x^{2}+a}\, A \,x^{3}}{4 b}-\frac {5 \sqrt {b \,x^{2}+a}\, B a \,x^{3}}{24 b^{2}}+\frac {3 A \,a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {5}{2}}}-\frac {5 B \,a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {7}{2}}}-\frac {3 \sqrt {b \,x^{2}+a}\, A a x}{8 b^{2}}+\frac {5 \sqrt {b \,x^{2}+a}\, B \,a^{2} x}{16 b^{3}} \]

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*a/b^2*x^3*(b*x^2+a)^(1/2)+5/16*B*a^2/b^3*x*(b*x^2+a)^(1/2)-5/16*B*a^3/b^(7/
2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))+1/4*A*x^3/b*(b*x^2+a)^(1/2)-3/8*A*a/b^2*x*(b*x^2+a)^(1/2)+3/8*A*a^2/b^(5/2)*l
n(b^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.06, size = 128, normalized size = 1.05 \[ \frac {\sqrt {b x^{2} + a} B x^{5}}{6 \, b} - \frac {5 \, \sqrt {b x^{2} + a} B a x^{3}}{24 \, b^{2}} + \frac {\sqrt {b x^{2} + a} A x^{3}}{4 \, b} + \frac {5 \, \sqrt {b x^{2} + a} B a^{2} x}{16 \, b^{3}} - \frac {3 \, \sqrt {b x^{2} + a} A a x}{8 \, b^{2}} - \frac {5 \, B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {7}{2}}} + \frac {3 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} \]

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]

1/6*sqrt(b*x^2 + a)*B*x^5/b - 5/24*sqrt(b*x^2 + a)*B*a*x^3/b^2 + 1/4*sqrt(b*x^2 + a)*A*x^3/b + 5/16*sqrt(b*x^2
+ a)*B*a^2*x/b^3 - 3/8*sqrt(b*x^2 + a)*A*a*x/b^2 - 5/16*B*a^3*arcsinh(b*x/sqrt(a*b))/b^(7/2) + 3/8*A*a^2*arcs
inh(b*x/sqrt(a*b))/b^(5/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4\,\left (B\,x^2+A\right )}{\sqrt {b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(A + B*x^2))/(a + b*x^2)^(1/2),x)

[Out]

int((x^4*(A + B*x^2))/(a + b*x^2)^(1/2), x)

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sympy [B]  time = 12.43, size = 235, normalized size = 1.93 \[ - \frac {3 A a^{\frac {3}{2}} x}{8 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A \sqrt {a} x^{3}}{8 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 A a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 b^{\frac {5}{2}}} + \frac {A x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 B a^{\frac {5}{2}} x}{16 b^{3} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 B a^{\frac {3}{2}} x^{3}}{48 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B \sqrt {a} x^{5}}{24 b \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {5 B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 b^{\frac {7}{2}}} + \frac {B x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

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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