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

Optimal. Leaf size=155 \[ -\frac {a^3 (8 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{5/2}}+\frac {a^2 x \sqrt {a+b x^2} (8 A b-3 a B)}{128 b^2}+\frac {a x^3 \sqrt {a+b x^2} (8 A b-3 a B)}{64 b}+\frac {x^3 \left (a+b x^2\right )^{3/2} (8 A b-3 a B)}{48 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b} \]

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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} \begin {gather*} \frac {a^2 x \sqrt {a+b x^2} (8 A b-3 a B)}{128 b^2}-\frac {a^3 (8 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{5/2}}+\frac {a x^3 \sqrt {a+b x^2} (8 A b-3 a B)}{64 b}+\frac {x^3 \left (a+b x^2\right )^{3/2} (8 A b-3 a B)}{48 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(8*A*b - 3*a*B)*x*Sqrt[a + b*x^2])/(128*b^2) + (a*(8*A*b - 3*a*B)*x^3*Sqrt[a + b*x^2])/(64*b) + ((8*A*b -
3*a*B)*x^3*(a + b*x^2)^(3/2))/(48*b) + (B*x^3*(a + b*x^2)^(5/2))/(8*b) - (a^3*(8*A*b - 3*a*B)*ArcTanh[(Sqrt[b
]*x)/Sqrt[a + b*x^2]])/(128*b^(5/2))

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^2 \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx &=\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {(-8 A b+3 a B) \int x^2 \left (a+b x^2\right )^{3/2} \, dx}{8 b}\\ &=\frac {(8 A b-3 a B) x^3 \left (a+b x^2\right )^{3/2}}{48 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}+\frac {(a (8 A b-3 a B)) \int x^2 \sqrt {a+b x^2} \, dx}{16 b}\\ &=\frac {a (8 A b-3 a B) x^3 \sqrt {a+b x^2}}{64 b}+\frac {(8 A b-3 a B) x^3 \left (a+b x^2\right )^{3/2}}{48 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}+\frac {\left (a^2 (8 A b-3 a B)\right ) \int \frac {x^2}{\sqrt {a+b x^2}} \, dx}{64 b}\\ &=\frac {a^2 (8 A b-3 a B) x \sqrt {a+b x^2}}{128 b^2}+\frac {a (8 A b-3 a B) x^3 \sqrt {a+b x^2}}{64 b}+\frac {(8 A b-3 a B) x^3 \left (a+b x^2\right )^{3/2}}{48 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {\left (a^3 (8 A b-3 a B)\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{128 b^2}\\ &=\frac {a^2 (8 A b-3 a B) x \sqrt {a+b x^2}}{128 b^2}+\frac {a (8 A b-3 a B) x^3 \sqrt {a+b x^2}}{64 b}+\frac {(8 A b-3 a B) x^3 \left (a+b x^2\right )^{3/2}}{48 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {\left (a^3 (8 A b-3 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^2}\\ &=\frac {a^2 (8 A b-3 a B) x \sqrt {a+b x^2}}{128 b^2}+\frac {a (8 A b-3 a B) x^3 \sqrt {a+b x^2}}{64 b}+\frac {(8 A b-3 a B) x^3 \left (a+b x^2\right )^{3/2}}{48 b}+\frac {B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac {a^3 (8 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 130, normalized size = 0.84 \begin {gather*} \frac {\sqrt {a+b x^2} \left (\frac {3 a^{5/2} (3 a B-8 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}+\sqrt {b} x \left (-9 a^3 B+6 a^2 b \left (4 A+B x^2\right )+8 a b^2 x^2 \left (14 A+9 B x^2\right )+16 b^3 x^4 \left (4 A+3 B x^2\right )\right )\right )}{384 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*x*(-9*a^3*B + 6*a^2*b*(4*A + B*x^2) + 16*b^3*x^4*(4*A + 3*B*x^2) + 8*a*b^2*x^2*(14*A
+ 9*B*x^2)) + (3*a^(5/2)*(-8*A*b + 3*a*B)*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[1 + (b*x^2)/a]))/(384*b^(5/2))

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IntegrateAlgebraic [A]  time = 0.19, size = 127, normalized size = 0.82 \begin {gather*} \frac {\left (8 a^3 A b-3 a^4 B\right ) \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{128 b^{5/2}}+\frac {\sqrt {a+b x^2} \left (-9 a^3 B x+24 a^2 A b x+6 a^2 b B x^3+112 a A b^2 x^3+72 a b^2 B x^5+64 A b^3 x^5+48 b^3 B x^7\right )}{384 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(24*a^2*A*b*x - 9*a^3*B*x + 112*a*A*b^2*x^3 + 6*a^2*b*B*x^3 + 64*A*b^3*x^5 + 72*a*b^2*B*x^5 +
48*b^3*B*x^7))/(384*b^2) + ((8*a^3*A*b - 3*a^4*B)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(128*b^(5/2))

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fricas [A]  time = 1.25, size = 260, normalized size = 1.68 \begin {gather*} \left [-\frac {3 \, {\left (3 \, 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 (9 \, B a b^{3} + 8 \, A b^{4}\right )} x^{5} + 2 \, {\left (3 \, B a^{2} b^{2} + 56 \, A a b^{3}\right )} x^{3} - 3 \, {\left (3 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{768 \, b^{3}}, -\frac {3 \, {\left (3 \, 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 (9 \, B a b^{3} + 8 \, A b^{4}\right )} x^{5} + 2 \, {\left (3 \, B a^{2} b^{2} + 56 \, A a b^{3}\right )} x^{3} - 3 \, {\left (3 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{384 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/768*(3*(3*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
*(9*B*a*b^3 + 8*A*b^4)*x^5 + 2*(3*B*a^2*b^2 + 56*A*a*b^3)*x^3 - 3*(3*B*a^3*b - 8*A*a^2*b^2)*x)*sqrt(b*x^2 + a)
)/b^3, -1/384*(3*(3*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*(9*B*a*
b^3 + 8*A*b^4)*x^5 + 2*(3*B*a^2*b^2 + 56*A*a*b^3)*x^3 - 3*(3*B*a^3*b - 8*A*a^2*b^2)*x)*sqrt(b*x^2 + a))/b^3]

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giac [A]  time = 0.38, size = 133, normalized size = 0.86 \begin {gather*} \frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, B b x^{2} + \frac {9 \, B a b^{6} + 8 \, A b^{7}}{b^{6}}\right )} x^{2} + \frac {3 \, B a^{2} b^{5} + 56 \, A a b^{6}}{b^{6}}\right )} x^{2} - \frac {3 \, {\left (3 \, B a^{3} b^{4} - 8 \, A a^{2} b^{5}\right )}}{b^{6}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (3 \, 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 {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(2*(4*(6*B*b*x^2 + (9*B*a*b^6 + 8*A*b^7)/b^6)*x^2 + (3*B*a^2*b^5 + 56*A*a*b^6)/b^6)*x^2 - 3*(3*B*a^3*b^4
- 8*A*a^2*b^5)/b^6)*sqrt(b*x^2 + a)*x - 1/128*(3*B*a^4 - 8*A*a^3*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^
(5/2)

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maple [A]  time = 0.01, size = 177, normalized size = 1.14 \begin {gather*} -\frac {A \,a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {3}{2}}}+\frac {3 B \,a^{4} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {5}{2}}}-\frac {\sqrt {b \,x^{2}+a}\, A \,a^{2} x}{16 b}+\frac {3 \sqrt {b \,x^{2}+a}\, B \,a^{3} x}{128 b^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B \,x^{3}}{8 b}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A a x}{24 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,a^{2} x}{64 b^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} A x}{6 b}-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B a x}{16 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*B*x^3*(b*x^2+a)^(5/2)/b-1/16*B*a/b^2*x*(b*x^2+a)^(5/2)+1/64*B*a^2/b^2*x*(b*x^2+a)^(3/2)+3/128*B*a^3/b^2*x*
(b*x^2+a)^(1/2)+3/128*B*a^4/b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))+1/6*A*x*(b*x^2+a)^(5/2)/b-1/24*A*a/b*x*(b*x^
2+a)^(3/2)-1/16*A*a^2/b*x*(b*x^2+a)^(1/2)-1/16*A*a^3/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.05, size = 162, normalized size = 1.05 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x^{3}}{8 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{2} x}{64 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} B a^{3} x}{128 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A a x}{24 \, b} - \frac {\sqrt {b x^{2} + a} A a^{2} x}{16 \, b} + \frac {3 \, B a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {5}{2}}} - \frac {A a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x^2 + a)^(5/2)*B*x^3/b - 1/16*(b*x^2 + a)^(5/2)*B*a*x/b^2 + 1/64*(b*x^2 + a)^(3/2)*B*a^2*x/b^2 + 3/128*
sqrt(b*x^2 + a)*B*a^3*x/b^2 + 1/6*(b*x^2 + a)^(5/2)*A*x/b - 1/24*(b*x^2 + a)^(3/2)*A*a*x/b - 1/16*sqrt(b*x^2 +
a)*A*a^2*x/b + 3/128*B*a^4*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 1/16*A*a^3*arcsinh(b*x/sqrt(a*b))/b^(3/2)

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

[Out]

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

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sympy [B]  time = 26.19, size = 287, normalized size = 1.85 \begin {gather*} \frac {A a^{\frac {5}{2}} x}{16 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {17 A a^{\frac {3}{2}} x^{3}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {11 A \sqrt {a} b 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 {3}{2}}} + \frac {A b^{2} x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 B a^{\frac {7}{2}} x}{128 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{\frac {5}{2}} x^{3}}{128 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {13 B a^{\frac {3}{2}} x^{5}}{64 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {5 B \sqrt {a} b x^{7}}{16 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 B a^{4} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{128 b^{\frac {5}{2}}} + \frac {B b^{2} x^{9}}{8 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**(5/2)*x/(16*b*sqrt(1 + b*x**2/a)) + 17*A*a**(3/2)*x**3/(48*sqrt(1 + b*x**2/a)) + 11*A*sqrt(a)*b*x**5/(24*
sqrt(1 + b*x**2/a)) - A*a**3*asinh(sqrt(b)*x/sqrt(a))/(16*b**(3/2)) + A*b**2*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a
)) - 3*B*a**(7/2)*x/(128*b**2*sqrt(1 + b*x**2/a)) - B*a**(5/2)*x**3/(128*b*sqrt(1 + b*x**2/a)) + 13*B*a**(3/2)
*x**5/(64*sqrt(1 + b*x**2/a)) + 5*B*sqrt(a)*b*x**7/(16*sqrt(1 + b*x**2/a)) + 3*B*a**4*asinh(sqrt(b)*x/sqrt(a))
/(128*b**(5/2)) + B*b**2*x**9/(8*sqrt(a)*sqrt(1 + b*x**2/a))

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