\(\int x^3 (A+B x) \sqrt {a+b x^2} \, dx\) [1]

Optimal result

Integrand size = 20, antiderivative size = 127 \[ \int x^3 (A+B x) \sqrt {a+b x^2} \, dx=\frac {a^2 B x \sqrt {a+b x^2}}{16 b^2}+\frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {a (16 A+15 B x) \left (a+b x^2\right )^{3/2}}{120 b^2}+\frac {a^3 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}} \]

[Out]

1/5*A*x^2*(b*x^2+a)^(3/2)/b+1/6*B*x^3*(b*x^2+a)^(3/2)/b-1/120*a*(15*B*x+16*A)*(b*x^2+a)^(3/2)/b^2+1/16*a^3*B*a
rctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(5/2)+1/16*a^2*B*x*(b*x^2+a)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {847, 794, 201, 223, 212} \[ \int x^3 (A+B x) \sqrt {a+b x^2} \, dx=\frac {a^3 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}}+\frac {a^2 B x \sqrt {a+b x^2}}{16 b^2}-\frac {a \left (a+b x^2\right )^{3/2} (16 A+15 B x)}{120 b^2}+\frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b} \]

[In]

Int[x^3*(A + B*x)*Sqrt[a + b*x^2],x]

[Out]

(a^2*B*x*Sqrt[a + b*x^2])/(16*b^2) + (A*x^2*(a + b*x^2)^(3/2))/(5*b) + (B*x^3*(a + b*x^2)^(3/2))/(6*b) - (a*(1
6*A + 15*B*x)*(a + b*x^2)^(3/2))/(120*b^2) + (a^3*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*b^(5/2))

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 794

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 847

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^
m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}+\frac {\int x^2 (-3 a B+6 A b x) \sqrt {a+b x^2} \, dx}{6 b} \\ & = \frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}+\frac {\int x (-12 a A b-15 a b B x) \sqrt {a+b x^2} \, dx}{30 b^2} \\ & = \frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {a (16 A+15 B x) \left (a+b x^2\right )^{3/2}}{120 b^2}+\frac {\left (a^2 B\right ) \int \sqrt {a+b x^2} \, dx}{8 b^2} \\ & = \frac {a^2 B x \sqrt {a+b x^2}}{16 b^2}+\frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {a (16 A+15 B x) \left (a+b x^2\right )^{3/2}}{120 b^2}+\frac {\left (a^3 B\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{16 b^2} \\ & = \frac {a^2 B x \sqrt {a+b x^2}}{16 b^2}+\frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {a (16 A+15 B x) \left (a+b x^2\right )^{3/2}}{120 b^2}+\frac {\left (a^3 B\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{16 b^2} \\ & = \frac {a^2 B x \sqrt {a+b x^2}}{16 b^2}+\frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {a (16 A+15 B x) \left (a+b x^2\right )^{3/2}}{120 b^2}+\frac {a^3 B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.80 \[ \int x^3 (A+B x) \sqrt {a+b x^2} \, dx=\frac {\sqrt {a+b x^2} \left (-32 a^2 A-15 a^2 B x+16 a A b x^2+10 a b B x^3+48 A b^2 x^4+40 b^2 B x^5\right )}{240 b^2}-\frac {a^3 B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 b^{5/2}} \]

[In]

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

[Out]

(Sqrt[a + b*x^2]*(-32*a^2*A - 15*a^2*B*x + 16*a*A*b*x^2 + 10*a*b*B*x^3 + 48*A*b^2*x^4 + 40*b^2*B*x^5))/(240*b^
2) - (a^3*B*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(16*b^(5/2))

Maple [A] (verified)

Time = 4.33 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.70

method	result	size
risch	\(-\frac {\left (-40 b^{2} B \,x^{5}-48 A \,b^{2} x^{4}-10 B a b \,x^{3}-16 a A b \,x^{2}+15 a^{2} B x +32 a^{2} A \right ) \sqrt {b \,x^{2}+a}}{240 b^{2}}+\frac {a^{3} B \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {5}{2}}}\)	\(89\)
default	\(B \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )+A \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )\)	\(120\)

[In]

int(x^3*(B*x+A)*(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/240*(-40*B*b^2*x^5-48*A*b^2*x^4-10*B*a*b*x^3-16*A*a*b*x^2+15*B*a^2*x+32*A*a^2)/b^2*(b*x^2+a)^(1/2)+1/16*a^3
*B/b^(5/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.62 \[ \int x^3 (A+B x) \sqrt {a+b x^2} \, dx=\left [\frac {15 \, B a^{3} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 10 \, B a b^{2} x^{3} + 16 \, A a b^{2} x^{2} - 15 \, B a^{2} b x - 32 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{480 \, b^{3}}, -\frac {15 \, B a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 10 \, B a b^{2} x^{3} + 16 \, A a b^{2} x^{2} - 15 \, B a^{2} b x - 32 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{240 \, b^{3}}\right ] \]

[In]

integrate(x^3*(B*x+A)*(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/480*(15*B*a^3*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(40*B*b^3*x^5 + 48*A*b^3*x^4 + 10
*B*a*b^2*x^3 + 16*A*a*b^2*x^2 - 15*B*a^2*b*x - 32*A*a^2*b)*sqrt(b*x^2 + a))/b^3, -1/240*(15*B*a^3*sqrt(-b)*arc
tan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (40*B*b^3*x^5 + 48*A*b^3*x^4 + 10*B*a*b^2*x^3 + 16*A*a*b^2*x^2 - 15*B*a^2*b*
x - 32*A*a^2*b)*sqrt(b*x^2 + a))/b^3]

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.09 \[ \int x^3 (A+B x) \sqrt {a+b x^2} \, dx=\begin {cases} \frac {B a^{3} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{16 b^{2}} + \sqrt {a + b x^{2}} \left (- \frac {2 A a^{2}}{15 b^{2}} + \frac {A a x^{2}}{15 b} + \frac {A x^{4}}{5} - \frac {B a^{2} x}{16 b^{2}} + \frac {B a x^{3}}{24 b} + \frac {B x^{5}}{6}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{4}}{4} + \frac {B x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(x**3*(B*x+A)*(b*x**2+a)**(1/2),x)

[Out]

Piecewise((B*a**3*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))/(16*b**2) + sqrt(a + b*x**2)*(-2*A*a**2/(15*b**2) + A*a*x**2/(15*b) + A*x**4/5 - B*a**2*x/(16*b**2)
+ B*a*x**3/(24*b) + B*x**5/6), Ne(b, 0)), (sqrt(a)*(A*x**4/4 + B*x**5/5), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.84 \[ \int x^3 (A+B x) \sqrt {a+b x^2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{3}}{6 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x^{2}}{5 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} B a^{2} x}{16 \, b^{2}} + \frac {B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a}{15 \, b^{2}} \]

[In]

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

[Out]

1/6*(b*x^2 + a)^(3/2)*B*x^3/b + 1/5*(b*x^2 + a)^(3/2)*A*x^2/b - 1/8*(b*x^2 + a)^(3/2)*B*a*x/b^2 + 1/16*sqrt(b*
x^2 + a)*B*a^2*x/b^2 + 1/16*B*a^3*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 2/15*(b*x^2 + a)^(3/2)*A*a/b^2

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.73 \[ \int x^3 (A+B x) \sqrt {a+b x^2} \, dx=-\frac {B a^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {5}{2}}} + \frac {1}{240} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, B x + 6 \, A\right )} x + \frac {5 \, B a}{b}\right )} x + \frac {8 \, A a}{b}\right )} x - \frac {15 \, B a^{2}}{b^{2}}\right )} x - \frac {32 \, A a^{2}}{b^{2}}\right )} \]

[In]

integrate(x^3*(B*x+A)*(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-1/16*B*a^3*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2) + 1/240*sqrt(b*x^2 + a)*((2*((4*(5*B*x + 6*A)*x + 5
*B*a/b)*x + 8*A*a/b)*x - 15*B*a^2/b^2)*x - 32*A*a^2/b^2)

Mupad [F(-1)]

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

[In]

int(x^3*(a + b*x^2)^(1/2)*(A + B*x),x)

[Out]

int(x^3*(a + b*x^2)^(1/2)*(A + B*x), x)