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

Optimal. Leaf size=118 \[ \frac {a^2 (6 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}}+\frac {x \left (a+b x^2\right )^{3/2} (6 A b-a B)}{24 b}+\frac {a x \sqrt {a+b x^2} (6 A b-a B)}{16 b}+\frac {B x \left (a+b x^2\right )^{5/2}}{6 b} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {388, 195, 217, 206} \[ \frac {a^2 (6 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}}+\frac {x \left (a+b x^2\right )^{3/2} (6 A b-a B)}{24 b}+\frac {a x \sqrt {a+b x^2} (6 A b-a B)}{16 b}+\frac {B x \left (a+b x^2\right )^{5/2}}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)*(A + B*x^2),x]

[Out]

(a*(6*A*b - a*B)*x*Sqrt[a + b*x^2])/(16*b) + ((6*A*b - a*B)*x*(a + b*x^2)^(3/2))/(24*b) + (B*x*(a + b*x^2)^(5/
2))/(6*b) + (a^2*(6*A*b - a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*b^(3/2))

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 0.21, size = 109, normalized size = 0.92 \[ \frac {\sqrt {a+b x^2} \left (\sqrt {b} x \left (3 a^2 B+2 a b \left (15 A+7 B x^2\right )+4 b^2 x^2 \left (3 A+2 B x^2\right )\right )-\frac {3 a^{3/2} (a B-6 A b) \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}\right )}{48 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 207, normalized size = 1.75 \[ \left [-\frac {3 \, {\left (B a^{3} - 6 \, A a^{2} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, B b^{3} x^{5} + 2 \, {\left (7 \, B a b^{2} + 6 \, A b^{3}\right )} x^{3} + 3 \, {\left (B a^{2} b + 10 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{2}}, \frac {3 \, {\left (B a^{3} - 6 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, B b^{3} x^{5} + 2 \, {\left (7 \, B a b^{2} + 6 \, A b^{3}\right )} x^{3} + 3 \, {\left (B a^{2} b + 10 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/96*(3*(B*a^3 - 6*A*a^2*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(8*B*b^3*x^5 + 2*(7*
B*a*b^2 + 6*A*b^3)*x^3 + 3*(B*a^2*b + 10*A*a*b^2)*x)*sqrt(b*x^2 + a))/b^2, 1/48*(3*(B*a^3 - 6*A*a^2*b)*sqrt(-b
)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (8*B*b^3*x^5 + 2*(7*B*a*b^2 + 6*A*b^3)*x^3 + 3*(B*a^2*b + 10*A*a*b^2)*x
)*sqrt(b*x^2 + a))/b^2]

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giac [A]  time = 0.42, size = 102, normalized size = 0.86 \[ \frac {1}{48} \, {\left (2 \, {\left (4 \, B b x^{2} + \frac {7 \, B a b^{4} + 6 \, A b^{5}}{b^{4}}\right )} x^{2} + \frac {3 \, {\left (B a^{2} b^{3} + 10 \, A a b^{4}\right )}}{b^{4}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (B a^{3} - 6 \, A a^{2} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(2*(4*B*b*x^2 + (7*B*a*b^4 + 6*A*b^5)/b^4)*x^2 + 3*(B*a^2*b^3 + 10*A*a*b^4)/b^4)*sqrt(b*x^2 + a)*x + 1/16
*(B*a^3 - 6*A*a^2*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2)

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maple [A]  time = 0.01, size = 131, normalized size = 1.11 \[ \frac {3 A \,a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 \sqrt {b}}-\frac {B \,a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {3}{2}}}+\frac {3 \sqrt {b \,x^{2}+a}\, A a x}{8}-\frac {\sqrt {b \,x^{2}+a}\, B \,a^{2} x}{16 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A x}{4}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B a x}{24 b}+\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} B x}{6 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.96, size = 116, normalized size = 0.98 \[ \frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A x + \frac {3}{8} \, \sqrt {b x^{2} + a} A a x + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x}{24 \, b} - \frac {\sqrt {b x^{2} + a} B a^{2} x}{16 \, b} - \frac {B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} + \frac {3 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)*(a + b*x^2)^(3/2),x)

[Out]

int((A + B*x^2)*(a + b*x^2)^(3/2), x)

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sympy [B]  time = 16.54, size = 253, normalized size = 2.14 \[ \frac {A a^{\frac {3}{2}} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {A a^{\frac {3}{2}} x}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 A \sqrt {a} b x^{3}}{8 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 A a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 \sqrt {b}} + \frac {A b^{2} x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {B a^{\frac {5}{2}} x}{16 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {17 B a^{\frac {3}{2}} x^{3}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {11 B \sqrt {a} b x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 b^{\frac {3}{2}}} + \frac {B b^{2} x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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