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

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

[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))

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Rubi [A]  time = 0.0676372, antiderivative size = 152, normalized size of antiderivative = 1., 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.146117, size = 131, normalized size = 0.86 \[ \frac{\sqrt{b} x \left (a^2 \left (35 b B x^2-90 A b\right )+105 a^3 B-2 a b^2 x^2 \left (15 A+7 B x^2\right )+4 b^3 x^4 \left (3 A+2 B x^2\right )\right )-15 a^{5/2} \sqrt{\frac{b x^2}{a}+1} (7 a B-6 A b) \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{48 b^{9/2} \sqrt{a+b x^2}} \]

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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Maple [A]  time = 0.015, size = 185, normalized size = 1.2 \begin{align*}{\frac{{x}^{7}B}{6\,b}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{7\,Ba{x}^{5}}{24\,{b}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{35\,{a}^{2}B{x}^{3}}{48\,{b}^{3}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{35\,B{a}^{3}x}{16\,{b}^{4}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{35\,B{a}^{3}}{16}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{9}{2}}}}+{\frac{A{x}^{5}}{4\,b}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{5\,aA{x}^{3}}{8\,{b}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{15\,{a}^{2}Ax}{8\,{b}^{3}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{15\,A{a}^{2}}{8}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{7}{2}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.78727, size = 717, normalized size = 4.72 \begin{align*} \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 ] \end{align*}

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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Sympy [A]  time = 17.7981, size = 233, normalized size = 1.53 \begin{align*} 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 ) \end{align*}

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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Giac [A]  time = 1.12643, size = 184, normalized size = 1.21 \begin{align*} \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}}} \end{align*}

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)