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

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

[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))

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Rubi [A]  time = 0.0620431, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {833, 780, 195, 217, 206} \[ -\frac{a^3 A \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{3/2}}-\frac{a^2 A x \sqrt{a+b x^2}}{16 b}-\frac{\left (a+b x^2\right )^{5/2} (12 a B-35 A b x)}{210 b^2}-\frac{a A x \left (a+b x^2\right )^{3/2}}{24 b}+\frac{B x^2 \left (a+b x^2\right )^{5/2}}{7 b} \]

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 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])

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 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 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*}

Mathematica [A]  time = 0.214517, size = 113, normalized size = 0.89 \[ \frac{\sqrt{a+b x^2} \left (3 a^2 b x (35 A+16 B x)-\frac{105 a^{5/2} A \sqrt{b} \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{\frac{b x^2}{a}+1}}-96 a^3 B+2 a b^2 x^3 (245 A+192 B x)+40 b^3 x^5 (7 A+6 B x)\right )}{1680 b^2} \]

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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Maple [A]  time = 0.007, size = 113, normalized size = 0.9 \begin{align*}{\frac{B{x}^{2}}{7\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{2\,Ba}{35\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{Ax}{6\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{aAx}{24\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{{a}^{2}Ax}{16\,b}\sqrt{b{x}^{2}+a}}-{\frac{A{a}^{3}}{16}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{3}{2}}}} \end{align*}

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.648, size = 559, normalized size = 4.4 \begin{align*} \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{align*}

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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Sympy [A]  time = 10.7177, size = 287, normalized size = 2.26 \begin{align*} \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{align*}

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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Giac [A]  time = 1.25538, size = 139, normalized size = 1.09 \begin{align*} \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{align*}

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)