\(\int \frac {x^4 (A+B x^2)}{\sqrt {a+b x^2}} \, dx\) [556]

Optimal result

Integrand size = 22, antiderivative size = 122 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=-\frac {a (6 A b-5 a B) x \sqrt {a+b x^2}}{16 b^3}+\frac {(6 A b-5 a B) x^3 \sqrt {a+b x^2}}{24 b^2}+\frac {B x^5 \sqrt {a+b x^2}}{6 b}+\frac {a^2 (6 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{7/2}} \]

[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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {470, 327, 223, 212} \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {a^2 (6 A b-5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{7/2}}-\frac {a x \sqrt {a+b x^2} (6 A b-5 a B)}{16 b^3}+\frac {x^3 \sqrt {a+b x^2} (6 A b-5 a B)}{24 b^2}+\frac {B x^5 \sqrt {a+b x^2}}{6 b} \]

[In]

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

[Out]

-1/16*(a*(6*A*b - 5*a*B)*x*Sqrt[a + b*x^2])/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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 327

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 470

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*} \text {integral}& = \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 ) \text {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] (verified)

Time = 0.35 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.89 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {x \sqrt {a+b x^2} \left (-18 a A b+15 a^2 B+12 A b^2 x^2-10 a b B x^2+8 b^2 B x^4\right )}{48 b^3}-\frac {a^2 (-6 A b+5 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{8 b^{7/2}} \]

[In]

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

[Out]

(x*Sqrt[a + b*x^2]*(-18*a*A*b + 15*a^2*B + 12*A*b^2*x^2 - 10*a*b*B*x^2 + 8*b^2*B*x^4))/(48*b^3) - (a^2*(-6*A*b
+ 5*a*B)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/(8*b^(7/2))

Maple [A] (verified)

Time = 2.96 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.72

method	result	size
risch	\(-\frac {x \left (-8 b^{2} B \,x^{4}-12 A \,b^{2} x^{2}+10 B a b \,x^{2}+18 a b A -15 a^{2} B \right ) \sqrt {b \,x^{2}+a}}{48 b^{3}}+\frac {a^{2} \left (6 A b -5 B a \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {7}{2}}}\)	\(88\)
pseudoelliptic	\(\frac {\left (\frac {3}{2} a^{2} b A -\frac {5}{4} a^{3} B \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+x \sqrt {b \,x^{2}+a}\, \left (-\frac {3 a \left (\frac {5 x^{2} B}{9}+A \right ) b^{\frac {3}{2}}}{2}+x^{2} \left (\frac {2 x^{2} B}{3}+A \right ) b^{\frac {5}{2}}+\frac {5 B \,a^{2} \sqrt {b}}{4}\right )}{4 b^{\frac {7}{2}}}\)	\(89\)
default	\(B \left (\frac {x^{5} \sqrt {b \,x^{2}+a}}{6 b}-\frac {5 a \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )+A \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )\)	\(154\)

[In]

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

[Out]

-1/48*x*(-8*B*b^2*x^4-12*A*b^2*x^2+10*B*a*b*x^2+18*A*a*b-15*B*a^2)*(b*x^2+a)^(1/2)/b^3+1/16*a^2*(6*A*b-5*B*a)/
b^(7/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.73 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\left [-\frac {3 \, {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, B b^{3} x^{5} - 2 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{4}}, \frac {3 \, {\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, B b^{3} x^{5} - 2 \, {\left (5 \, B a b^{2} - 6 \, A b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{2} b - 6 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{4}}\right ] \]

[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]

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.11 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\begin {cases} \frac {3 a^{2} \left (A - \frac {5 B a}{6 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b^{2}} + \sqrt {a + b x^{2}} \left (\frac {B x^{5}}{6 b} - \frac {3 a x \left (A - \frac {5 B a}{6 b}\right )}{8 b^{2}} + \frac {x^{3} \left (A - \frac {5 B a}{6 b}\right )}{4 b}\right ) & \text {for}\: b \neq 0 \\\frac {\frac {A x^{5}}{5} + \frac {B x^{7}}{7}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((3*a**2*(A - 5*B*a/(6*b))*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*
log(x)/sqrt(b*x**2), True))/(8*b**2) + sqrt(a + b*x**2)*(B*x**5/(6*b) - 3*a*x*(A - 5*B*a/(6*b))/(8*b**2) + x**
3*(A - 5*B*a/(6*b))/(4*b)), Ne(b, 0)), ((A*x**5/5 + B*x**7/7)/sqrt(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.05 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} B x^{5}}{6 \, b} - \frac {5 \, \sqrt {b x^{2} + a} B a x^{3}}{24 \, b^{2}} + \frac {\sqrt {b x^{2} + a} A x^{3}}{4 \, b} + \frac {5 \, \sqrt {b x^{2} + a} B a^{2} x}{16 \, b^{3}} - \frac {3 \, \sqrt {b x^{2} + a} A a x}{8 \, b^{2}} - \frac {5 \, B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {7}{2}}} + \frac {3 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.88 \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {1}{48} \, {\left (2 \, {\left (\frac {4 \, B x^{2}}{b} - \frac {5 \, B a b^{3} - 6 \, A b^{4}}{b^{5}}\right )} x^{2} + \frac {3 \, {\left (5 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )}}{b^{5}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (5 \, B a^{3} - 6 \, A a^{2} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {7}{2}}} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\int \frac {x^4\,\left (B\,x^2+A\right )}{\sqrt {b\,x^2+a}} \,d x \]

[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)