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

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

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Rubi [A]  time = 0.05, antiderivative size = 119, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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 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 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 )}{\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*}

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Mathematica [A]  time = 0.12, size = 108, normalized size = 0.91 \begin {gather*} \frac {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 )+\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 )}{8 b^{7/2} \sqrt {a+b x^2}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.18, size = 101, normalized size = 0.85 \begin {gather*} \frac {-15 a^2 B x+12 a A b x-5 a b B x^3+4 A b^2 x^3+2 b^2 B x^5}{8 b^3 \sqrt {a+b x^2}}-\frac {3 \left (5 a^2 B-4 a A b\right ) \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{8 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.08, size = 274, normalized size = 2.30 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.37, size = 104, normalized size = 0.87 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.01, size = 141, normalized size = 1.18 \begin {gather*} \frac {B \,x^{5}}{4 \sqrt {b \,x^{2}+a}\, b}+\frac {A \,x^{3}}{2 \sqrt {b \,x^{2}+a}\, b}-\frac {5 B a \,x^{3}}{8 \sqrt {b \,x^{2}+a}\, b^{2}}+\frac {3 A a x}{2 \sqrt {b \,x^{2}+a}\, b^{2}}-\frac {15 B \,a^{2} x}{8 \sqrt {b \,x^{2}+a}\, b^{3}}-\frac {3 A a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {5}{2}}}+\frac {15 B \,a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {7}{2}}} \end {gather*}

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*a/b^2*x^3/(b*x^2+a)^(1/2)-15/8*B*a^2/b^3*x/(b*x^2+a)^(1/2)+15/8*B*a^2/b^(7/2
)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))+1/2*A*x^3/b/(b*x^2+a)^(1/2)+3/2*A*a/b^2*x/(b*x^2+a)^(1/2)-3/2*A*a/b^(5/2)*ln(b
^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.15, size = 126, normalized size = 1.06 \begin {gather*} \frac {B x^{5}}{4 \, \sqrt {b x^{2} + a} b} - \frac {5 \, B a x^{3}}{8 \, \sqrt {b x^{2} + a} b^{2}} + \frac {A x^{3}}{2 \, \sqrt {b x^{2} + a} b} - \frac {15 \, B a^{2} x}{8 \, \sqrt {b x^{2} + a} b^{3}} + \frac {3 \, A a x}{2 \, \sqrt {b x^{2} + a} b^{2}} + \frac {15 \, B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {7}{2}}} - \frac {3 \, A a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} \end {gather*}

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4\,\left (B\,x^2+A\right )}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(A + B*x^2))/(a + b*x^2)^(3/2),x)

[Out]

int((x^4*(A + B*x^2))/(a + b*x^2)^(3/2), x)

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sympy [A]  time = 16.63, size = 177, normalized size = 1.49 \begin {gather*} 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 {gather*}

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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