\(\int \frac {x^6 (A+B x^2)}{(a+b x^2)^{3/2}} \, dx\) [569]

Optimal result

Integrand size = 22, antiderivative size = 152 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {(6 A b-7 a B) x^5}{6 b^2 \sqrt {a+b x^2}}+\frac {B x^7}{6 b \sqrt {a+b x^2}}-\frac {5 a (6 A b-7 a B) x \sqrt {a+b x^2}}{16 b^4}+\frac {5 (6 A b-7 a B) x^3 \sqrt {a+b x^2}}{24 b^3}+\frac {5 a^2 (6 A b-7 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{9/2}} \]

[Out]

5/16*a^2*(6*A*b-7*B*a)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(9/2)-1/6*(6*A*b-7*B*a)*x^5/b^2/(b*x^2+a)^(1/2)+1/
6*B*x^7/b/(b*x^2+a)^(1/2)-5/16*a*(6*A*b-7*B*a)*x*(b*x^2+a)^(1/2)/b^4+5/24*(6*A*b-7*B*a)*x^3*(b*x^2+a)^(1/2)/b^
3

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {470, 294, 327, 223, 212} \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {5 a^2 (6 A b-7 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{9/2}}-\frac {5 a x \sqrt {a+b x^2} (6 A b-7 a B)}{16 b^4}+\frac {5 x^3 \sqrt {a+b x^2} (6 A b-7 a B)}{24 b^3}-\frac {x^5 (6 A b-7 a B)}{6 b^2 \sqrt {a+b x^2}}+\frac {B x^7}{6 b \sqrt {a+b x^2}} \]

[In]

Int[(x^6*(A + B*x^2))/(a + b*x^2)^(3/2),x]

[Out]

-1/6*((6*A*b - 7*a*B)*x^5)/(b^2*Sqrt[a + b*x^2]) + (B*x^7)/(6*b*Sqrt[a + b*x^2]) - (5*a*(6*A*b - 7*a*B)*x*Sqrt
[a + b*x^2])/(16*b^4) + (5*(6*A*b - 7*a*B)*x^3*Sqrt[a + b*x^2])/(24*b^3) + (5*a^2*(6*A*b - 7*a*B)*ArcTanh[(Sqr
t[b]*x)/Sqrt[a + b*x^2]])/(16*b^(9/2))

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 294

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 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^7}{6 b \sqrt {a+b x^2}}-\frac {(-6 A b+7 a B) \int \frac {x^6}{\left (a+b x^2\right )^{3/2}} \, dx}{6 b} \\ & = -\frac {(6 A b-7 a B) x^5}{6 b^2 \sqrt {a+b x^2}}+\frac {B x^7}{6 b \sqrt {a+b x^2}}+\frac {(5 (6 A b-7 a B)) \int \frac {x^4}{\sqrt {a+b x^2}} \, dx}{6 b^2} \\ & = -\frac {(6 A b-7 a B) x^5}{6 b^2 \sqrt {a+b x^2}}+\frac {B x^7}{6 b \sqrt {a+b x^2}}+\frac {5 (6 A b-7 a B) x^3 \sqrt {a+b x^2}}{24 b^3}-\frac {(5 a (6 A b-7 a B)) \int \frac {x^2}{\sqrt {a+b x^2}} \, dx}{8 b^3} \\ & = -\frac {(6 A b-7 a B) x^5}{6 b^2 \sqrt {a+b x^2}}+\frac {B x^7}{6 b \sqrt {a+b x^2}}-\frac {5 a (6 A b-7 a B) x \sqrt {a+b x^2}}{16 b^4}+\frac {5 (6 A b-7 a B) x^3 \sqrt {a+b x^2}}{24 b^3}+\frac {\left (5 a^2 (6 A b-7 a B)\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{16 b^4} \\ & = -\frac {(6 A b-7 a B) x^5}{6 b^2 \sqrt {a+b x^2}}+\frac {B x^7}{6 b \sqrt {a+b x^2}}-\frac {5 a (6 A b-7 a B) x \sqrt {a+b x^2}}{16 b^4}+\frac {5 (6 A b-7 a B) x^3 \sqrt {a+b x^2}}{24 b^3}+\frac {\left (5 a^2 (6 A b-7 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^4} \\ & = -\frac {(6 A b-7 a B) x^5}{6 b^2 \sqrt {a+b x^2}}+\frac {B x^7}{6 b \sqrt {a+b x^2}}-\frac {5 a (6 A b-7 a B) x \sqrt {a+b x^2}}{16 b^4}+\frac {5 (6 A b-7 a B) x^3 \sqrt {a+b x^2}}{24 b^3}+\frac {5 a^2 (6 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.87 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {x \left (-90 a^2 A b+105 a^3 B-30 a A b^2 x^2+35 a^2 b B x^2+12 A b^3 x^4-14 a b^2 B x^4+8 b^3 B x^6\right )}{48 b^4 \sqrt {a+b x^2}}-\frac {5 a^2 (-6 A b+7 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{8 b^{9/2}} \]

[In]

Integrate[(x^6*(A + B*x^2))/(a + b*x^2)^(3/2),x]

[Out]

(x*(-90*a^2*A*b + 105*a^3*B - 30*a*A*b^2*x^2 + 35*a^2*b*B*x^2 + 12*A*b^3*x^4 - 14*a*b^2*B*x^4 + 8*b^3*B*x^6))/
(48*b^4*Sqrt[a + b*x^2]) - (5*a^2*(-6*A*b + 7*a*B)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/(8*b^(9/
2))

Maple [A] (verified)

Time = 2.91 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(-\frac {5 \left (-3 \left (A b -\frac {7 B a}{6}\right ) a^{2} \sqrt {b \,x^{2}+a}\, \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+x \left (3 \left (-\frac {7 x^{2} B}{18}+A \right ) a^{2} b^{\frac {3}{2}}+x^{2} a \left (\frac {7 x^{2} B}{15}+A \right ) b^{\frac {5}{2}}-\frac {2 x^{4} \left (\frac {2 x^{2} B}{3}+A \right ) b^{\frac {7}{2}}}{5}-\frac {7 B \,a^{3} \sqrt {b}}{2}\right )\right )}{8 \sqrt {b \,x^{2}+a}\, b^{\frac {9}{2}}}\)	\(115\)
risch	\(-\frac {x \left (-8 b^{2} B \,x^{4}-12 A \,b^{2} x^{2}+22 B a b \,x^{2}+42 a b A -57 a^{2} B \right ) \sqrt {b \,x^{2}+a}}{48 b^{4}}+\frac {a^{2} \left (-\frac {19 a B x}{\sqrt {b \,x^{2}+a}}+\frac {14 b A x}{\sqrt {b \,x^{2}+a}}+\left (30 b^{2} A -35 a b B \right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )\right )}{16 b^{4}}\)	\(141\)
default	\(B \left (\frac {x^{7}}{6 b \sqrt {b \,x^{2}+a}}-\frac {7 a \left (\frac {x^{5}}{4 b \sqrt {b \,x^{2}+a}}-\frac {5 a \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )}{4 b}\right )}{6 b}\right )+A \left (\frac {x^{5}}{4 b \sqrt {b \,x^{2}+a}}-\frac {5 a \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )}{4 b}\right )\)	\(198\)

[In]

int(x^6*(B*x^2+A)/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-5/8/(b*x^2+a)^(1/2)/b^(9/2)*(-3*(A*b-7/6*B*a)*a^2*(b*x^2+a)^(1/2)*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))+x*(3*(-7
/18*x^2*B+A)*a^2*b^(3/2)+x^2*a*(7/15*x^2*B+A)*b^(5/2)-2/5*x^4*(2/3*x^2*B+A)*b^(7/2)-7/2*B*a^3*b^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.14 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\left [-\frac {15 \, {\left (7 \, B a^{4} - 6 \, A a^{3} b + {\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, B b^{4} x^{7} - 2 \, {\left (7 \, B a b^{3} - 6 \, A b^{4}\right )} x^{5} + 5 \, {\left (7 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )} x^{3} + 15 \, {\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, {\left (b^{6} x^{2} + a b^{5}\right )}}, \frac {15 \, {\left (7 \, B a^{4} - 6 \, A a^{3} b + {\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, B b^{4} x^{7} - 2 \, {\left (7 \, B a b^{3} - 6 \, A b^{4}\right )} x^{5} + 5 \, {\left (7 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )} x^{3} + 15 \, {\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, {\left (b^{6} x^{2} + a b^{5}\right )}}\right ] \]

[In]

integrate(x^6*(B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/96*(15*(7*B*a^4 - 6*A*a^3*b + (7*B*a^3*b - 6*A*a^2*b^2)*x^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt
(b)*x - a) - 2*(8*B*b^4*x^7 - 2*(7*B*a*b^3 - 6*A*b^4)*x^5 + 5*(7*B*a^2*b^2 - 6*A*a*b^3)*x^3 + 15*(7*B*a^3*b -
6*A*a^2*b^2)*x)*sqrt(b*x^2 + a))/(b^6*x^2 + a*b^5), 1/48*(15*(7*B*a^4 - 6*A*a^3*b + (7*B*a^3*b - 6*A*a^2*b^2)*
x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (8*B*b^4*x^7 - 2*(7*B*a*b^3 - 6*A*b^4)*x^5 + 5*(7*B*a^2*b^2
- 6*A*a*b^3)*x^3 + 15*(7*B*a^3*b - 6*A*a^2*b^2)*x)*sqrt(b*x^2 + a))/(b^6*x^2 + a*b^5)]

Sympy [A] (verification not implemented)

Time = 18.98 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.53 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=A \left (- \frac {15 a^{\frac {3}{2}} x}{8 b^{3} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {5 \sqrt {a} x^{3}}{8 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {15 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 b^{\frac {7}{2}}} + \frac {x^{5}}{4 \sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + B \left (\frac {35 a^{\frac {5}{2}} x}{16 b^{4} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {35 a^{\frac {3}{2}} x^{3}}{48 b^{3} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {7 \sqrt {a} x^{5}}{24 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {35 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 b^{\frac {9}{2}}} + \frac {x^{7}}{6 \sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \]

[In]

integrate(x**6*(B*x**2+A)/(b*x**2+a)**(3/2),x)

[Out]

A*(-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(sqr
t(b)*x/sqrt(a))/(8*b**(7/2)) + x**5/(4*sqrt(a)*b*sqrt(1 + b*x**2/a))) + B*(35*a**(5/2)*x/(16*b**4*sqrt(1 + b*x
**2/a)) + 35*a**(3/2)*x**3/(48*b**3*sqrt(1 + b*x**2/a)) - 7*sqrt(a)*x**5/(24*b**2*sqrt(1 + b*x**2/a)) - 35*a**
3*asinh(sqrt(b)*x/sqrt(a))/(16*b**(9/2)) + x**7/(6*sqrt(a)*b*sqrt(1 + b*x**2/a)))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.12 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B x^{7}}{6 \, \sqrt {b x^{2} + a} b} - \frac {7 \, B a x^{5}}{24 \, \sqrt {b x^{2} + a} b^{2}} + \frac {A x^{5}}{4 \, \sqrt {b x^{2} + a} b} + \frac {35 \, B a^{2} x^{3}}{48 \, \sqrt {b x^{2} + a} b^{3}} - \frac {5 \, A a x^{3}}{8 \, \sqrt {b x^{2} + a} b^{2}} + \frac {35 \, B a^{3} x}{16 \, \sqrt {b x^{2} + a} b^{4}} - \frac {15 \, A a^{2} x}{8 \, \sqrt {b x^{2} + a} b^{3}} - \frac {35 \, B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {9}{2}}} + \frac {15 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {7}{2}}} \]

[In]

integrate(x^6*(B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

1/6*B*x^7/(sqrt(b*x^2 + a)*b) - 7/24*B*a*x^5/(sqrt(b*x^2 + a)*b^2) + 1/4*A*x^5/(sqrt(b*x^2 + a)*b) + 35/48*B*a
^2*x^3/(sqrt(b*x^2 + a)*b^3) - 5/8*A*a*x^3/(sqrt(b*x^2 + a)*b^2) + 35/16*B*a^3*x/(sqrt(b*x^2 + a)*b^4) - 15/8*
A*a^2*x/(sqrt(b*x^2 + a)*b^3) - 35/16*B*a^3*arcsinh(b*x/sqrt(a*b))/b^(9/2) + 15/8*A*a^2*arcsinh(b*x/sqrt(a*b))
/b^(7/2)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.89 \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (2 \, {\left (\frac {4 \, B x^{2}}{b} - \frac {7 \, B a b^{5} - 6 \, A b^{6}}{b^{7}}\right )} x^{2} + \frac {5 \, {\left (7 \, B a^{2} b^{4} - 6 \, A a b^{5}\right )}}{b^{7}}\right )} x^{2} + \frac {15 \, {\left (7 \, B a^{3} b^{3} - 6 \, A a^{2} b^{4}\right )}}{b^{7}}\right )} x}{48 \, \sqrt {b x^{2} + a}} + \frac {5 \, {\left (7 \, B a^{3} - 6 \, A a^{2} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {9}{2}}} \]

[In]

integrate(x^6*(B*x^2+A)/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

1/48*((2*(4*B*x^2/b - (7*B*a*b^5 - 6*A*b^6)/b^7)*x^2 + 5*(7*B*a^2*b^4 - 6*A*a*b^5)/b^7)*x^2 + 15*(7*B*a^3*b^3
- 6*A*a^2*b^4)/b^7)*x/sqrt(b*x^2 + a) + 5/16*(7*B*a^3 - 6*A*a^2*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(9
/2)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {x^6\,\left (B\,x^2+A\right )}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \]

[In]

int((x^6*(A + B*x^2))/(a + b*x^2)^(3/2),x)

[Out]

int((x^6*(A + B*x^2))/(a + b*x^2)^(3/2), x)