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

Optimal. Leaf size=155 \[ \frac {a^3 (8 A b-5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{7/2}}-\frac {a^2 x \sqrt {a+b x^2} (8 A b-5 a B)}{128 b^3}+\frac {a x^3 \sqrt {a+b x^2} (8 A b-5 a B)}{192 b^2}+\frac {x^5 \sqrt {a+b x^2} (8 A b-5 a B)}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b} \]

[Out]

1/8*B*x^5*(b*x^2+a)^(3/2)/b+1/128*a^3*(8*A*b-5*B*a)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(7/2)-1/128*a^2*(8*A*
b-5*B*a)*x*(b*x^2+a)^(1/2)/b^3+1/192*a*(8*A*b-5*B*a)*x^3*(b*x^2+a)^(1/2)/b^2+1/48*(8*A*b-5*B*a)*x^5*(b*x^2+a)^
(1/2)/b

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Rubi [A]  time = 0.07, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {459, 279, 321, 217, 206} \[ -\frac {a^2 x \sqrt {a+b x^2} (8 A b-5 a B)}{128 b^3}+\frac {a^3 (8 A b-5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{7/2}}+\frac {a x^3 \sqrt {a+b x^2} (8 A b-5 a B)}{192 b^2}+\frac {x^5 \sqrt {a+b x^2} (8 A b-5 a B)}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(8*A*b - 5*a*B)*x*Sqrt[a + b*x^2])/(128*b^3) + (a*(8*A*b - 5*a*B)*x^3*Sqrt[a + b*x^2])/(192*b^2) + ((8*A
*b - 5*a*B)*x^5*Sqrt[a + b*x^2])/(48*b) + (B*x^5*(a + b*x^2)^(3/2))/(8*b) + (a^3*(8*A*b - 5*a*B)*ArcTanh[(Sqrt
[b]*x)/Sqrt[a + b*x^2]])/(128*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 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 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 x^4 \sqrt {a+b x^2} \left (A+B x^2\right ) \, dx &=\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}-\frac {(-8 A b+5 a B) \int x^4 \sqrt {a+b x^2} \, dx}{8 b}\\ &=\frac {(8 A b-5 a B) x^5 \sqrt {a+b x^2}}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {(a (8 A b-5 a B)) \int \frac {x^4}{\sqrt {a+b x^2}} \, dx}{48 b}\\ &=\frac {a (8 A b-5 a B) x^3 \sqrt {a+b x^2}}{192 b^2}+\frac {(8 A b-5 a B) x^5 \sqrt {a+b x^2}}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}-\frac {\left (a^2 (8 A b-5 a B)\right ) \int \frac {x^2}{\sqrt {a+b x^2}} \, dx}{64 b^2}\\ &=-\frac {a^2 (8 A b-5 a B) x \sqrt {a+b x^2}}{128 b^3}+\frac {a (8 A b-5 a B) x^3 \sqrt {a+b x^2}}{192 b^2}+\frac {(8 A b-5 a B) x^5 \sqrt {a+b x^2}}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {\left (a^3 (8 A b-5 a B)\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{128 b^3}\\ &=-\frac {a^2 (8 A b-5 a B) x \sqrt {a+b x^2}}{128 b^3}+\frac {a (8 A b-5 a B) x^3 \sqrt {a+b x^2}}{192 b^2}+\frac {(8 A b-5 a B) x^5 \sqrt {a+b x^2}}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {\left (a^3 (8 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 )}{128 b^3}\\ &=-\frac {a^2 (8 A b-5 a B) x \sqrt {a+b x^2}}{128 b^3}+\frac {a (8 A b-5 a B) x^3 \sqrt {a+b x^2}}{192 b^2}+\frac {(8 A b-5 a B) x^5 \sqrt {a+b x^2}}{48 b}+\frac {B x^5 \left (a+b x^2\right )^{3/2}}{8 b}+\frac {a^3 (8 A b-5 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 130, normalized size = 0.84 \[ \frac {\sqrt {a+b x^2} \left (\sqrt {b} x \left (15 a^3 B-2 a^2 b \left (12 A+5 B x^2\right )+8 a b^2 x^2 \left (2 A+B x^2\right )+16 b^3 x^4 \left (4 A+3 B x^2\right )\right )-\frac {3 a^{5/2} (5 a B-8 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}\right )}{384 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*x*(15*a^3*B + 8*a*b^2*x^2*(2*A + B*x^2) + 16*b^3*x^4*(4*A + 3*B*x^2) - 2*a^2*b*(12*A
+ 5*B*x^2)) - (3*a^(5/2)*(-8*A*b + 5*a*B)*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[1 + (b*x^2)/a]))/(384*b^(7/2))

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fricas [A]  time = 0.95, size = 257, normalized size = 1.66 \[ \left [-\frac {3 \, {\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (48 \, B b^{4} x^{7} + 8 \, {\left (B a b^{3} + 8 \, A b^{4}\right )} x^{5} - 2 \, {\left (5 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{768 \, b^{4}}, \frac {3 \, {\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (48 \, B b^{4} x^{7} + 8 \, {\left (B a b^{3} + 8 \, A b^{4}\right )} x^{5} - 2 \, {\left (5 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x^{3} + 3 \, {\left (5 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{384 \, b^{4}}\right ] \]

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]

[-1/768*(3*(5*B*a^4 - 8*A*a^3*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(48*B*b^4*x^7 + 8
*(B*a*b^3 + 8*A*b^4)*x^5 - 2*(5*B*a^2*b^2 - 8*A*a*b^3)*x^3 + 3*(5*B*a^3*b - 8*A*a^2*b^2)*x)*sqrt(b*x^2 + a))/b
^4, 1/384*(3*(5*B*a^4 - 8*A*a^3*b)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (48*B*b^4*x^7 + 8*(B*a*b^3 +
8*A*b^4)*x^5 - 2*(5*B*a^2*b^2 - 8*A*a*b^3)*x^3 + 3*(5*B*a^3*b - 8*A*a^2*b^2)*x)*sqrt(b*x^2 + a))/b^4]

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giac [A]  time = 0.39, size = 132, normalized size = 0.85 \[ \frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, B x^{2} + \frac {B a b^{5} + 8 \, A b^{6}}{b^{6}}\right )} x^{2} - \frac {5 \, B a^{2} b^{4} - 8 \, A a b^{5}}{b^{6}}\right )} x^{2} + \frac {3 \, {\left (5 \, B a^{3} b^{3} - 8 \, A a^{2} b^{4}\right )}}{b^{6}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {7}{2}}} \]

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]

1/384*(2*(4*(6*B*x^2 + (B*a*b^5 + 8*A*b^6)/b^6)*x^2 - (5*B*a^2*b^4 - 8*A*a*b^5)/b^6)*x^2 + 3*(5*B*a^3*b^3 - 8*
A*a^2*b^4)/b^6)*sqrt(b*x^2 + a)*x + 1/128*(5*B*a^4 - 8*A*a^3*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)

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maple [A]  time = 0.02, size = 181, normalized size = 1.17 \[ \frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,x^{5}}{8 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A \,x^{3}}{6 b}-\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} B a \,x^{3}}{48 b^{2}}+\frac {A \,a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {5}{2}}}-\frac {5 B \,a^{4} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {7}{2}}}+\frac {\sqrt {b \,x^{2}+a}\, A \,a^{2} x}{16 b^{2}}-\frac {5 \sqrt {b \,x^{2}+a}\, B \,a^{3} x}{128 b^{3}}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A a x}{8 b^{2}}+\frac {5 \left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,a^{2} x}{64 b^{3}} \]

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]

1/8*B*x^5*(b*x^2+a)^(3/2)/b-5/48*B*a/b^2*x^3*(b*x^2+a)^(3/2)+5/64*B*a^2/b^3*x*(b*x^2+a)^(3/2)-5/128*B*a^3/b^3*
x*(b*x^2+a)^(1/2)-5/128*B*a^4/b^(7/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+1/6*A*x^3*(b*x^2+a)^(3/2)/b-1/8*A*a/b^2*x*
(b*x^2+a)^(3/2)+1/16*A*a^2/b^2*x*(b*x^2+a)^(1/2)+1/16*A*a^3/b^(5/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.00, size = 166, normalized size = 1.07 \[ \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{5}}{8 \, b} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x^{3}}{48 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x^{3}}{6 \, b} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} x}{64 \, b^{3}} - \frac {5 \, \sqrt {b x^{2} + a} B a^{3} x}{128 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} A a^{2} x}{16 \, b^{2}} - \frac {5 \, B a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {7}{2}}} + \frac {A a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} \]

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]

1/8*(b*x^2 + a)^(3/2)*B*x^5/b - 5/48*(b*x^2 + a)^(3/2)*B*a*x^3/b^2 + 1/6*(b*x^2 + a)^(3/2)*A*x^3/b + 5/64*(b*x
^2 + a)^(3/2)*B*a^2*x/b^3 - 5/128*sqrt(b*x^2 + a)*B*a^3*x/b^3 - 1/8*(b*x^2 + a)^(3/2)*A*a*x/b^2 + 1/16*sqrt(b*
x^2 + a)*A*a^2*x/b^2 - 5/128*B*a^4*arcsinh(b*x/sqrt(a*b))/b^(7/2) + 1/16*A*a^3*arcsinh(b*x/sqrt(a*b))/b^(5/2)

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

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 [A]  time = 17.65, size = 286, normalized size = 1.85 \[ - \frac {A a^{\frac {5}{2}} x}{16 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A a^{\frac {3}{2}} x^{3}}{48 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 A \sqrt {a} x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {A a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 b^{\frac {5}{2}}} + \frac {A b x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 B a^{\frac {7}{2}} x}{128 b^{3} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 B a^{\frac {5}{2}} x^{3}}{384 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{\frac {3}{2}} x^{5}}{192 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {7 B \sqrt {a} x^{7}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {5 B a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{128 b^{\frac {7}{2}}} + \frac {B b x^{9}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

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]

-A*a**(5/2)*x/(16*b**2*sqrt(1 + b*x**2/a)) - A*a**(3/2)*x**3/(48*b*sqrt(1 + b*x**2/a)) + 5*A*sqrt(a)*x**5/(24*
sqrt(1 + b*x**2/a)) + A*a**3*asinh(sqrt(b)*x/sqrt(a))/(16*b**(5/2)) + A*b*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a))
+ 5*B*a**(7/2)*x/(128*b**3*sqrt(1 + b*x**2/a)) + 5*B*a**(5/2)*x**3/(384*b**2*sqrt(1 + b*x**2/a)) - B*a**(3/2)*
x**5/(192*b*sqrt(1 + b*x**2/a)) + 7*B*sqrt(a)*x**7/(48*sqrt(1 + b*x**2/a)) - 5*B*a**4*asinh(sqrt(b)*x/sqrt(a))
/(128*b**(7/2)) + B*b*x**9/(8*sqrt(a)*sqrt(1 + b*x**2/a))

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