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

Optimal. Leaf size=152 \[ \frac {5 a^2 (6 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{9/2}}-\frac {5 a x \sqrt {a+b x^2} (6 A b-7 a B)}{16 b^4}+\frac {5 x^3 \sqrt {a+b x^2} (6 A b-7 a B)}{24 b^3}-\frac {x^5 (6 A b-7 a B)}{6 b^2 \sqrt {a+b x^2}}+\frac {B x^7}{6 b \sqrt {a+b x^2}} \]

[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

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Rubi [A]  time = 0.07, antiderivative size = 152, 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, 288, 321, 217, 206} \[ \frac {5 a^2 (6 A b-7 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{9/2}}-\frac {x^5 (6 A b-7 a B)}{6 b^2 \sqrt {a+b x^2}}+\frac {5 x^3 \sqrt {a+b x^2} (6 A b-7 a B)}{24 b^3}-\frac {5 a x \sqrt {a+b x^2} (6 A b-7 a B)}{16 b^4}+\frac {B x^7}{6 b \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((6*A*b - 7*a*B)*x^5)/(6*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[(Sqrt[
b]*x)/Sqrt[a + b*x^2]])/(16*b^(9/2))

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^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx &=\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 ) \operatorname {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*}

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Mathematica [A]  time = 0.16, size = 131, normalized size = 0.86 \[ \frac {\sqrt {b} x \left (105 a^3 B+a^2 \left (35 b B x^2-90 A b\right )-2 a b^2 x^2 \left (15 A+7 B x^2\right )+4 b^3 x^4 \left (3 A+2 B x^2\right )\right )-15 a^{5/2} \sqrt {\frac {b x^2}{a}+1} (7 a B-6 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{48 b^{9/2} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*x*(105*a^3*B + 4*b^3*x^4*(3*A + 2*B*x^2) - 2*a*b^2*x^2*(15*A + 7*B*x^2) + a^2*(-90*A*b + 35*b*B*x^2))
- 15*a^(5/2)*(-6*A*b + 7*a*B)*Sqrt[1 + (b*x^2)/a]*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/(48*b^(9/2)*Sqrt[a + b*x^2])

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fricas [A]  time = 0.97, size = 325, normalized size = 2.14 \[ \left [-\frac {15 \, {\left (7 \, B a^{4} - 6 \, A a^{3} b + {\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, B b^{4} x^{7} - 2 \, {\left (7 \, B a b^{3} - 6 \, A b^{4}\right )} x^{5} + 5 \, {\left (7 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )} x^{3} + 15 \, {\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, {\left (b^{6} x^{2} + a b^{5}\right )}}, \frac {15 \, {\left (7 \, B a^{4} - 6 \, A a^{3} b + {\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, B b^{4} x^{7} - 2 \, {\left (7 \, B a b^{3} - 6 \, A b^{4}\right )} x^{5} + 5 \, {\left (7 \, B a^{2} b^{2} - 6 \, A a b^{3}\right )} x^{3} + 15 \, {\left (7 \, B a^{3} b - 6 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, {\left (b^{6} x^{2} + a b^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.37, size = 136, normalized size = 0.89 \[ \frac {{\left ({\left (2 \, {\left (\frac {4 \, B x^{2}}{b} - \frac {7 \, B a b^{5} - 6 \, A b^{6}}{b^{7}}\right )} x^{2} + \frac {5 \, {\left (7 \, B a^{2} b^{4} - 6 \, A a b^{5}\right )}}{b^{7}}\right )} x^{2} + \frac {15 \, {\left (7 \, B a^{3} b^{3} - 6 \, A a^{2} b^{4}\right )}}{b^{7}}\right )} x}{48 \, \sqrt {b x^{2} + a}} + \frac {5 \, {\left (7 \, B a^{3} - 6 \, A a^{2} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.02, size = 185, normalized size = 1.22 \[ \frac {B \,x^{7}}{6 \sqrt {b \,x^{2}+a}\, b}+\frac {A \,x^{5}}{4 \sqrt {b \,x^{2}+a}\, b}-\frac {7 B a \,x^{5}}{24 \sqrt {b \,x^{2}+a}\, b^{2}}-\frac {5 A a \,x^{3}}{8 \sqrt {b \,x^{2}+a}\, b^{2}}+\frac {35 B \,a^{2} x^{3}}{48 \sqrt {b \,x^{2}+a}\, b^{3}}-\frac {15 A \,a^{2} x}{8 \sqrt {b \,x^{2}+a}\, b^{3}}+\frac {35 B \,a^{3} x}{16 \sqrt {b \,x^{2}+a}\, b^{4}}+\frac {15 A \,a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {7}{2}}}-\frac {35 B \,a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.06, size = 170, normalized size = 1.12 \[ \frac {B x^{7}}{6 \, \sqrt {b x^{2} + a} b} - \frac {7 \, B a x^{5}}{24 \, \sqrt {b x^{2} + a} b^{2}} + \frac {A x^{5}}{4 \, \sqrt {b x^{2} + a} b} + \frac {35 \, B a^{2} x^{3}}{48 \, \sqrt {b x^{2} + a} b^{3}} - \frac {5 \, A a x^{3}}{8 \, \sqrt {b x^{2} + a} b^{2}} + \frac {35 \, B a^{3} x}{16 \, \sqrt {b x^{2} + a} b^{4}} - \frac {15 \, A a^{2} x}{8 \, \sqrt {b x^{2} + a} b^{3}} - \frac {35 \, B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {9}{2}}} + \frac {15 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^6\,\left (B\,x^2+A\right )}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 35.61, size = 233, normalized size = 1.53 \[ A \left (- \frac {15 a^{\frac {3}{2}} x}{8 b^{3} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {5 \sqrt {a} x^{3}}{8 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {15 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 b^{\frac {7}{2}}} + \frac {x^{5}}{4 \sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + B \left (\frac {35 a^{\frac {5}{2}} x}{16 b^{4} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {35 a^{\frac {3}{2}} x^{3}}{48 b^{3} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {7 \sqrt {a} x^{5}}{24 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {35 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 b^{\frac {9}{2}}} + \frac {x^{7}}{6 \sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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