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]

(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))

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Rubi [A]  time = 0.040886, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, 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 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]

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 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 \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*}

Mathematica [A]  time = 0.210422, 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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Maple [A]  time = 0.003, size = 131, normalized size = 1.1 \begin{align*}{\frac{Bx}{6\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{Bax}{24\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{{a}^{2}Bx}{16\,b}\sqrt{b{x}^{2}+a}}-{\frac{B{a}^{3}}{16}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{3}{2}}}}+{\frac{Ax}{4} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,aAx}{8}\sqrt{b{x}^{2}+a}}+{\frac{3\,A{a}^{2}}{8}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){\frac{1}{\sqrt{b}}}} \end{align*}

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/b*a*x*(b*x^2+a)^(3/2)-1/16*B/b*a^2*x*(b*x^2+a)^(1/2)-1/16*B/b^(3/2)*a^3*ln(x*
b^(1/2)+(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(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((b*x^2+a)^(3/2)*(B*x^2+A),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72901, size = 482, normalized size = 4.08 \begin{align*} \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 ] \end{align*}

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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Sympy [B]  time = 13.3764, size = 253, normalized size = 2.14 \begin{align*} \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}}} \end{align*}

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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Giac [A]  time = 1.11507, size = 138, normalized size = 1.17 \begin{align*} \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}}} \end{align*}

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)