\(\int x^2 (A+B x) (a+b x^2)^{3/2} \, dx\) [9]

Optimal result

Integrand size = 20, antiderivative size = 127 \[ \int x^2 (A+B x) \left (a+b x^2\right )^{3/2} \, dx=-\frac {a^2 A x \sqrt {a+b x^2}}{16 b}-\frac {a A x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {B x^2 \left (a+b x^2\right )^{5/2}}{7 b}-\frac {(12 a B-35 A b x) \left (a+b x^2\right )^{5/2}}{210 b^2}-\frac {a^3 A \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}} \]

[Out]

-1/24*a*A*x*(b*x^2+a)^(3/2)/b+1/7*B*x^2*(b*x^2+a)^(5/2)/b-1/210*(-35*A*b*x+12*B*a)*(b*x^2+a)^(5/2)/b^2-1/16*a^
3*A*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(3/2)-1/16*a^2*A*x*(b*x^2+a)^(1/2)/b

Rubi [A] (verified)

Time = 0.04 (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} \[ \int x^2 (A+B x) \left (a+b x^2\right )^{3/2} \, dx=-\frac {a^3 A \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}}-\frac {a^2 A x \sqrt {a+b x^2}}{16 b}-\frac {\left (a+b x^2\right )^{5/2} (12 a B-35 A b x)}{210 b^2}-\frac {a A x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {B x^2 \left (a+b x^2\right )^{5/2}}{7 b} \]

[In]

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

[Out]

-1/16*(a^2*A*x*Sqrt[a + b*x^2])/b - (a*A*x*(a + b*x^2)^(3/2))/(24*b) + (B*x^2*(a + b*x^2)^(5/2))/(7*b) - ((12*
a*B - 35*A*b*x)*(a + b*x^2)^(5/2))/(210*b^2) - (a^3*A*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 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^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac {\int x (-2 a B+7 A b x) \left (a+b x^2\right )^{3/2} \, dx}{7 b} \\ & = \frac {B x^2 \left (a+b x^2\right )^{5/2}}{7 b}-\frac {(12 a B-35 A b x) \left (a+b x^2\right )^{5/2}}{210 b^2}-\frac {(a A) \int \left (a+b x^2\right )^{3/2} \, dx}{6 b} \\ & = -\frac {a A x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {B x^2 \left (a+b x^2\right )^{5/2}}{7 b}-\frac {(12 a B-35 A b x) \left (a+b x^2\right )^{5/2}}{210 b^2}-\frac {\left (a^2 A\right ) \int \sqrt {a+b x^2} \, dx}{8 b} \\ & = -\frac {a^2 A x \sqrt {a+b x^2}}{16 b}-\frac {a A x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {B x^2 \left (a+b x^2\right )^{5/2}}{7 b}-\frac {(12 a B-35 A b x) \left (a+b x^2\right )^{5/2}}{210 b^2}-\frac {\left (a^3 A\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{16 b} \\ & = -\frac {a^2 A x \sqrt {a+b x^2}}{16 b}-\frac {a A x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {B x^2 \left (a+b x^2\right )^{5/2}}{7 b}-\frac {(12 a B-35 A b x) \left (a+b x^2\right )^{5/2}}{210 b^2}-\frac {\left (a^3 A\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^2 A x \sqrt {a+b x^2}}{16 b}-\frac {a A x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {B x^2 \left (a+b x^2\right )^{5/2}}{7 b}-\frac {(12 a B-35 A b x) \left (a+b x^2\right )^{5/2}}{210 b^2}-\frac {a^3 A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.84 \[ \int x^2 (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\frac {\sqrt {a+b x^2} \left (-96 a^3 B+40 b^3 x^5 (7 A+6 B x)+3 a^2 b x (35 A+16 B x)+2 a b^2 x^3 (245 A+192 B x)\right )+105 a^3 A \sqrt {b} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{1680 b^2} \]

[In]

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

[Out]

(Sqrt[a + b*x^2]*(-96*a^3*B + 40*b^3*x^5*(7*A + 6*B*x) + 3*a^2*b*x*(35*A + 16*B*x) + 2*a*b^2*x^3*(245*A + 192*
B*x)) + 105*a^3*A*Sqrt[b]*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(1680*b^2)

Maple [A] (verified)

Time = 3.51 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.82

method	result	size
risch	\(\frac {\left (240 b^{3} B \,x^{6}+280 A \,b^{3} x^{5}+384 B a \,b^{2} x^{4}+490 a A \,b^{2} x^{3}+48 B \,a^{2} b \,x^{2}+105 a^{2} A b x -96 a^{3} B \right ) \sqrt {b \,x^{2}+a}}{1680 b^{2}}-\frac {A \,a^{3} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {3}{2}}}\)	\(104\)
default	\(B \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{7 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{35 b^{2}}\right )+A \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )\)	\(112\)

[In]

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

[Out]

1/1680*(240*B*b^3*x^6+280*A*b^3*x^5+384*B*a*b^2*x^4+490*A*a*b^2*x^3+48*B*a^2*b*x^2+105*A*a^2*b*x-96*B*a^3)/b^2
*(b*x^2+a)^(1/2)-1/16*A*a^3/b^(3/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.76 \[ \int x^2 (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\left [\frac {105 \, A a^{3} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (240 \, B b^{3} x^{6} + 280 \, A b^{3} x^{5} + 384 \, B a b^{2} x^{4} + 490 \, A a b^{2} x^{3} + 48 \, B a^{2} b x^{2} + 105 \, A a^{2} b x - 96 \, B a^{3}\right )} \sqrt {b x^{2} + a}}{3360 \, b^{2}}, \frac {105 \, A a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (240 \, B b^{3} x^{6} + 280 \, A b^{3} x^{5} + 384 \, B a b^{2} x^{4} + 490 \, A a b^{2} x^{3} + 48 \, B a^{2} b x^{2} + 105 \, A a^{2} b x - 96 \, B a^{3}\right )} \sqrt {b x^{2} + a}}{1680 \, b^{2}}\right ] \]

[In]

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

[Out]

[1/3360*(105*A*a^3*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(240*B*b^3*x^6 + 280*A*b^3*x^5
+ 384*B*a*b^2*x^4 + 490*A*a*b^2*x^3 + 48*B*a^2*b*x^2 + 105*A*a^2*b*x - 96*B*a^3)*sqrt(b*x^2 + a))/b^2, 1/1680*
(105*A*a^3*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (240*B*b^3*x^6 + 280*A*b^3*x^5 + 384*B*a*b^2*x^4 + 49
0*A*a*b^2*x^3 + 48*B*a^2*b*x^2 + 105*A*a^2*b*x - 96*B*a^3)*sqrt(b*x^2 + a))/b^2]

Sympy [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.18 \[ \int x^2 (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\begin {cases} - \frac {A a^{3} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{16 b} + \sqrt {a + b x^{2}} \left (\frac {A a^{2} x}{16 b} + \frac {7 A a x^{3}}{24} + \frac {A b x^{5}}{6} - \frac {2 B a^{3}}{35 b^{2}} + \frac {B a^{2} x^{2}}{35 b} + \frac {8 B a x^{4}}{35} + \frac {B b x^{6}}{7}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (\frac {A x^{3}}{3} + \frac {B x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-A*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) + sqrt(a + b*x**2)*(A*a**2*x/(16*b) + 7*A*a*x**3/24 + A*b*x**5/6 - 2*B*a**3/(35*b**2) + B*a*
*2*x**2/(35*b) + 8*B*a*x**4/35 + B*b*x**6/7), Ne(b, 0)), (a**(3/2)*(A*x**3/3 + B*x**4/4), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.83 \[ \int x^2 (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x^{2}}{7 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A a x}{24 \, b} - \frac {\sqrt {b x^{2} + a} A a^{2} x}{16 \, b} - \frac {A a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a}{35 \, b^{2}} \]

[In]

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

[Out]

1/7*(b*x^2 + a)^(5/2)*B*x^2/b + 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 - 1/16*A*a^3*arcsinh(b*x/sqrt(a*b))/b^(3/2) - 2/35*(b*x^2 + a)^(5/2)*B*a/b^2

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.81 \[ \int x^2 (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\frac {A a^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {3}{2}}} - \frac {1}{1680} \, \sqrt {b x^{2} + a} {\left (\frac {96 \, B a^{3}}{b^{2}} - {\left (\frac {105 \, A a^{2}}{b} + 2 \, {\left (\frac {24 \, B a^{2}}{b} + {\left (245 \, A a + 4 \, {\left (48 \, B a + 5 \, {\left (6 \, B b x + 7 \, A b\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \]

[In]

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

[Out]

1/16*A*a^3*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2) - 1/1680*sqrt(b*x^2 + a)*(96*B*a^3/b^2 - (105*A*a^2/
b + 2*(24*B*a^2/b + (245*A*a + 4*(48*B*a + 5*(6*B*b*x + 7*A*b)*x)*x)*x)*x)*x)

Mupad [F(-1)]

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

[In]

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

[Out]

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