\(\int (a+b x^2)^{3/2} (A+B x^2) \, dx\) [526]

Optimal result

Integrand size = 19, antiderivative size = 118 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {a (6 A b-a B) x \sqrt {a+b x^2}}{16 b}+\frac {(6 A b-a B) x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {B x \left (a+b x^2\right )^{5/2}}{6 b}+\frac {a^2 (6 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}} \]

[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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {396, 201, 223, 212} \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {a^2 (6 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}}+\frac {x \left (a+b x^2\right )^{3/2} (6 A b-a B)}{24 b}+\frac {a x \sqrt {a+b x^2} (6 A b-a B)}{16 b}+\frac {B x \left (a+b x^2\right )^{5/2}}{6 b} \]

[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 201

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 396

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*} \text {integral}& = \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 ) \text {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] (verified)

Time = 0.14 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.84 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {x \sqrt {a+b x^2} \left (30 a A b+3 a^2 B+12 A b^2 x^2+14 a b B x^2+8 b^2 B x^4\right )}{48 b}+\frac {a^2 (-6 A b+a B) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 b^{3/2}} \]

[In]

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

[Out]

(x*Sqrt[a + b*x^2]*(30*a*A*b + 3*a^2*B + 12*A*b^2*x^2 + 14*a*b*B*x^2 + 8*b^2*B*x^4))/(48*b) + (a^2*(-6*A*b + a
*B)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(16*b^(3/2))

Maple [A] (verified)

Time = 2.85 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.75

method	result	size
risch	\(\frac {x \left (8 b^{2} B \,x^{4}+12 A \,b^{2} x^{2}+14 B a b \,x^{2}+30 a b A +3 a^{2} B \right ) \sqrt {b \,x^{2}+a}}{48 b}+\frac {a^{2} \left (6 A b -B a \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {3}{2}}}\)	\(88\)
pseudoelliptic	\(\frac {\left (\frac {3}{2} a^{2} b A -\frac {1}{4} a^{3} B \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+x \sqrt {b \,x^{2}+a}\, \left (\frac {5 \left (\frac {7 x^{2} B}{15}+A \right ) a \,b^{\frac {3}{2}}}{2}+x^{2} \left (\frac {2 x^{2} B}{3}+A \right ) b^{\frac {5}{2}}+\frac {B \,a^{2} \sqrt {b}}{4}\right )}{4 b^{\frac {3}{2}}}\)	\(89\)
default	\(A \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )\)	\(130\)

[In]

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

[Out]

1/48/b*x*(8*B*b^2*x^4+12*A*b^2*x^2+14*B*a*b*x^2+30*A*a*b+3*B*a^2)*(b*x^2+a)^(1/2)+1/16*a^2*(6*A*b-B*a)/b^(3/2)
*ln(x*b^(1/2)+(b*x^2+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.75 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\left [-\frac {3 \, {\left (B a^{3} - 6 \, A a^{2} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, B b^{3} x^{5} + 2 \, {\left (7 \, B a b^{2} + 6 \, A b^{3}\right )} x^{3} + 3 \, {\left (B a^{2} b + 10 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{2}}, \frac {3 \, {\left (B a^{3} - 6 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, B b^{3} x^{5} + 2 \, {\left (7 \, B a b^{2} + 6 \, A b^{3}\right )} x^{3} + 3 \, {\left (B a^{2} b + 10 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{2}}\right ] \]

[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]

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.48 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {B b x^{5}}{6} + \frac {x^{3} \left (A b^{2} + \frac {7 B a b}{6}\right )}{4 b} + \frac {x \left (2 A a b + B a^{2} - \frac {3 a \left (A b^{2} + \frac {7 B a b}{6}\right )}{4 b}\right )}{2 b}\right ) + \left (A a^{2} - \frac {a \left (2 A a b + B a^{2} - \frac {3 a \left (A b^{2} + \frac {7 B a b}{6}\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (A x + \frac {B x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(a + b*x**2)*(B*b*x**5/6 + x**3*(A*b**2 + 7*B*a*b/6)/(4*b) + x*(2*A*a*b + B*a**2 - 3*a*(A*b**2
+ 7*B*a*b/6)/(4*b))/(2*b)) + (A*a**2 - a*(2*A*a*b + B*a**2 - 3*a*(A*b**2 + 7*B*a*b/6)/(4*b))/(2*b))*Piecewise(
(log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True)), Ne(b, 0)), (a**(3
/2)*(A*x + B*x**3/3), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.98 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A x + \frac {3}{8} \, \sqrt {b x^{2} + a} A a x + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x}{24 \, b} - \frac {\sqrt {b x^{2} + a} B a^{2} x}{16 \, b} - \frac {B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} + \frac {3 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.86 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {1}{48} \, {\left (2 \, {\left (4 \, B b x^{2} + \frac {7 \, B a b^{4} + 6 \, A b^{5}}{b^{4}}\right )} x^{2} + \frac {3 \, {\left (B a^{2} b^{3} + 10 \, A a b^{4}\right )}}{b^{4}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (B a^{3} - 6 \, A a^{2} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {3}{2}}} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\int \left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{3/2} \,d x \]

[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)