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

Optimal. Leaf size=127 \[ -\frac {a^3 A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}}-\frac {a^2 A x \sqrt {a+b x^2}}{16 b}-\frac {\left (a+b x^2\right )^{5/2} (12 a B-35 A b x)}{210 b^2}-\frac {a A x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {B x^2 \left (a+b x^2\right )^{5/2}}{7 b} \]

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Rubi [A]  time = 0.06, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {833, 780, 195, 217, 206} \begin {gather*} -\frac {a^3 A \tanh ^{-1}\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*A*x*Sqrt[a + b*x^2])/(16*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 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 780

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 833

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*} \int x^2 (A+B x) \left (a+b x^2\right )^{3/2} \, dx &=\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 ) \operatorname {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*}

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Mathematica [A]  time = 0.24, size = 113, normalized size = 0.89 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-\frac {105 a^{5/2} A \sqrt {b} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {\frac {b x^2}{a}+1}}-96 a^3 B+3 a^2 b x (35 A+16 B x)+2 a b^2 x^3 (245 A+192 B x)+40 b^3 x^5 (7 A+6 B x)\right )}{1680 b^2} \end {gather*}

Antiderivative was successfully verified.

[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^(5/2)*A*Sqrt[b]*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[1 + (b*x^2)/a]))/(1680*b^2)

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IntegrateAlgebraic [A]  time = 0.42, size = 116, normalized size = 0.91 \begin {gather*} \frac {a^3 A \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{16 b^{3/2}}+\frac {\sqrt {a+b x^2} \left (-96 a^3 B+105 a^2 A b x+48 a^2 b B x^2+490 a A b^2 x^3+384 a b^2 B x^4+280 A b^3 x^5+240 b^3 B x^6\right )}{1680 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.91, size = 223, normalized size = 1.76 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [A]  time = 0.51, size = 103, normalized size = 0.81 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.01, size = 113, normalized size = 0.89 \begin {gather*} -\frac {A \,a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {3}{2}}}-\frac {\sqrt {b \,x^{2}+a}\, A \,a^{2} x}{16 b}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A a x}{24 b}+\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 {2 \left (b \,x^{2}+a \right )^{\frac {5}{2}} B a}{35 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*B*x^2*(b*x^2+a)^(5/2)/b-2/35*B*a/b^2*(b*x^2+a)^(5/2)+1/6*A*x*(b*x^2+a)^(5/2)/b-1/24*a*A*x*(b*x^2+a)^(3/2)/
b-1/16*a^2*A*x*(b*x^2+a)^(1/2)/b-1/16*A*a^3/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.37, size = 105, normalized size = 0.83 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 20.41, size = 287, normalized size = 2.26 \begin {gather*} \frac {A a^{\frac {5}{2}} x}{16 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {17 A a^{\frac {3}{2}} x^{3}}{48 \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {11 A \sqrt {a} b x^{5}}{24 \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {A a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 b^{\frac {3}{2}}} + \frac {A b^{2} x^{7}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} + B a \left (\begin {cases} - \frac {2 a^{2} \sqrt {a + b x^{2}}}{15 b^{2}} + \frac {a x^{2} \sqrt {a + b x^{2}}}{15 b} + \frac {x^{4} \sqrt {a + b x^{2}}}{5} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + B b \left (\begin {cases} \frac {8 a^{3} \sqrt {a + b x^{2}}}{105 b^{3}} - \frac {4 a^{2} x^{2} \sqrt {a + b x^{2}}}{105 b^{2}} + \frac {a x^{4} \sqrt {a + b x^{2}}}{35 b} + \frac {x^{6} \sqrt {a + b x^{2}}}{7} & \text {for}\: b \neq 0 \\\frac {\sqrt {a} x^{6}}{6} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**(5/2)*x/(16*b*sqrt(1 + b*x**2/a)) + 17*A*a**(3/2)*x**3/(48*sqrt(1 + b*x**2/a)) + 11*A*sqrt(a)*b*x**5/(24*
sqrt(1 + b*x**2/a)) - A*a**3*asinh(sqrt(b)*x/sqrt(a))/(16*b**(3/2)) + A*b**2*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a
)) + B*a*Piecewise((-2*a**2*sqrt(a + b*x**2)/(15*b**2) + a*x**2*sqrt(a + b*x**2)/(15*b) + x**4*sqrt(a + b*x**2
)/5, Ne(b, 0)), (sqrt(a)*x**4/4, True)) + B*b*Piecewise((8*a**3*sqrt(a + b*x**2)/(105*b**3) - 4*a**2*x**2*sqrt
(a + b*x**2)/(105*b**2) + a*x**4*sqrt(a + b*x**2)/(35*b) + x**6*sqrt(a + b*x**2)/7, Ne(b, 0)), (sqrt(a)*x**6/6
, True))

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