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

Optimal. Leaf size=150 \[ \frac{3 a^3 B x \sqrt{a+b x^2}}{128 b^2}+\frac{a^2 B x \left (a+b x^2\right )^{3/2}}{64 b^2}+\frac{3 a^4 B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{128 b^{5/2}}-\frac{a \left (a+b x^2\right )^{5/2} (32 A+35 B x)}{560 b^2}+\frac{A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac{B x^3 \left (a+b x^2\right )^{5/2}}{8 b} \]

[Out]

(3*a^3*B*x*Sqrt[a + b*x^2])/(128*b^2) + (a^2*B*x*(a + b*x^2)^(3/2))/(64*b^2) + (A*x^2*(a + b*x^2)^(5/2))/(7*b)
+ (B*x^3*(a + b*x^2)^(5/2))/(8*b) - (a*(32*A + 35*B*x)*(a + b*x^2)^(5/2))/(560*b^2) + (3*a^4*B*ArcTanh[(Sqrt[
b]*x)/Sqrt[a + b*x^2]])/(128*b^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0950358, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{3 a^3 B x \sqrt{a+b x^2}}{128 b^2}+\frac{a^2 B x \left (a+b x^2\right )^{3/2}}{64 b^2}+\frac{3 a^4 B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{128 b^{5/2}}-\frac{a \left (a+b x^2\right )^{5/2} (32 A+35 B x)}{560 b^2}+\frac{A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac{B x^3 \left (a+b x^2\right )^{5/2}}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^3*B*x*Sqrt[a + b*x^2])/(128*b^2) + (a^2*B*x*(a + b*x^2)^(3/2))/(64*b^2) + (A*x^2*(a + b*x^2)^(5/2))/(7*b)
+ (B*x^3*(a + b*x^2)^(5/2))/(8*b) - (a*(32*A + 35*B*x)*(a + b*x^2)^(5/2))/(560*b^2) + (3*a^4*B*ArcTanh[(Sqrt[
b]*x)/Sqrt[a + b*x^2]])/(128*b^(5/2))

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^3 (A+B x) \left (a+b x^2\right )^{3/2} \, dx &=\frac{B x^3 \left (a+b x^2\right )^{5/2}}{8 b}+\frac{\int x^2 (-3 a B+8 A b x) \left (a+b x^2\right )^{3/2} \, dx}{8 b}\\ &=\frac{A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac{B x^3 \left (a+b x^2\right )^{5/2}}{8 b}+\frac{\int x (-16 a A b-21 a b B x) \left (a+b x^2\right )^{3/2} \, dx}{56 b^2}\\ &=\frac{A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac{B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac{a (32 A+35 B x) \left (a+b x^2\right )^{5/2}}{560 b^2}+\frac{\left (a^2 B\right ) \int \left (a+b x^2\right )^{3/2} \, dx}{16 b^2}\\ &=\frac{a^2 B x \left (a+b x^2\right )^{3/2}}{64 b^2}+\frac{A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac{B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac{a (32 A+35 B x) \left (a+b x^2\right )^{5/2}}{560 b^2}+\frac{\left (3 a^3 B\right ) \int \sqrt{a+b x^2} \, dx}{64 b^2}\\ &=\frac{3 a^3 B x \sqrt{a+b x^2}}{128 b^2}+\frac{a^2 B x \left (a+b x^2\right )^{3/2}}{64 b^2}+\frac{A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac{B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac{a (32 A+35 B x) \left (a+b x^2\right )^{5/2}}{560 b^2}+\frac{\left (3 a^4 B\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{128 b^2}\\ &=\frac{3 a^3 B x \sqrt{a+b x^2}}{128 b^2}+\frac{a^2 B x \left (a+b x^2\right )^{3/2}}{64 b^2}+\frac{A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac{B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac{a (32 A+35 B x) \left (a+b x^2\right )^{5/2}}{560 b^2}+\frac{\left (3 a^4 B\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{128 b^2}\\ &=\frac{3 a^3 B x \sqrt{a+b x^2}}{128 b^2}+\frac{a^2 B x \left (a+b x^2\right )^{3/2}}{64 b^2}+\frac{A x^2 \left (a+b x^2\right )^{5/2}}{7 b}+\frac{B x^3 \left (a+b x^2\right )^{5/2}}{8 b}-\frac{a (32 A+35 B x) \left (a+b x^2\right )^{5/2}}{560 b^2}+\frac{3 a^4 B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{128 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.228028, size = 126, normalized size = 0.84 \[ \frac{\sqrt{a+b x^2} \left (\sqrt{b} \left (2 a^2 b x^2 (64 A+35 B x)-a^3 (256 A+105 B x)+8 a b^2 x^4 (128 A+105 B x)+80 b^3 x^6 (8 A+7 B x)\right )+\frac{105 a^{7/2} B \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{\frac{b x^2}{a}+1}}\right )}{4480 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*(80*b^3*x^6*(8*A + 7*B*x) + 2*a^2*b*x^2*(64*A + 35*B*x) + 8*a*b^2*x^4*(128*A + 105*B
*x) - a^3*(256*A + 105*B*x)) + (105*a^(7/2)*B*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[1 + (b*x^2)/a]))/(4480*b^(5/2
))

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 134, normalized size = 0.9 \begin{align*}{\frac{B{x}^{3}}{8\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{Bax}{16\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{{a}^{2}Bx}{64\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{3\,{a}^{3}Bx}{128\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{3\,B{a}^{4}}{128}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}}+{\frac{A{x}^{2}}{7\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{2\,Aa}{35\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*B*x^3*(b*x^2+a)^(5/2)/b-1/16*B/b^2*a*x*(b*x^2+a)^(5/2)+1/64*B/b^2*a^2*x*(b*x^2+a)^(3/2)+3/128*B/b^2*a^3*x*
(b*x^2+a)^(1/2)+3/128*B/b^(5/2)*a^4*ln(x*b^(1/2)+(b*x^2+a)^(1/2))+1/7*A*x^2*(b*x^2+a)^(5/2)/b-2/35*A*a/b^2*(b*
x^2+a)^(5/2)

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.60339, size = 630, normalized size = 4.2 \begin{align*} \left [\frac{105 \, B a^{4} \sqrt{b} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (560 \, B b^{4} x^{7} + 640 \, A b^{4} x^{6} + 840 \, B a b^{3} x^{5} + 1024 \, A a b^{3} x^{4} + 70 \, B a^{2} b^{2} x^{3} + 128 \, A a^{2} b^{2} x^{2} - 105 \, B a^{3} b x - 256 \, A a^{3} b\right )} \sqrt{b x^{2} + a}}{8960 \, b^{3}}, -\frac{105 \, B a^{4} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (560 \, B b^{4} x^{7} + 640 \, A b^{4} x^{6} + 840 \, B a b^{3} x^{5} + 1024 \, A a b^{3} x^{4} + 70 \, B a^{2} b^{2} x^{3} + 128 \, A a^{2} b^{2} x^{2} - 105 \, B a^{3} b x - 256 \, A a^{3} b\right )} \sqrt{b x^{2} + a}}{4480 \, b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8960*(105*B*a^4*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(560*B*b^4*x^7 + 640*A*b^4*x^6
+ 840*B*a*b^3*x^5 + 1024*A*a*b^3*x^4 + 70*B*a^2*b^2*x^3 + 128*A*a^2*b^2*x^2 - 105*B*a^3*b*x - 256*A*a^3*b)*sqr
t(b*x^2 + a))/b^3, -1/4480*(105*B*a^4*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (560*B*b^4*x^7 + 640*A*b^4
*x^6 + 840*B*a*b^3*x^5 + 1024*A*a*b^3*x^4 + 70*B*a^2*b^2*x^3 + 128*A*a^2*b^2*x^2 - 105*B*a^3*b*x - 256*A*a^3*b
)*sqrt(b*x^2 + a))/b^3]

________________________________________________________________________________________

Sympy [A]  time = 15.7861, size = 318, normalized size = 2.12 \begin{align*} A 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 ) + A 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 ) - \frac{3 B a^{\frac{7}{2}} x}{128 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{B a^{\frac{5}{2}} x^{3}}{128 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{13 B a^{\frac{3}{2}} x^{5}}{64 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{5 B \sqrt{a} b x^{7}}{16 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 B a^{4} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{128 b^{\frac{5}{2}}} + \frac{B b^{2} x^{9}}{8 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*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)) + A*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, Tru
e)) - 3*B*a**(7/2)*x/(128*b**2*sqrt(1 + b*x**2/a)) - B*a**(5/2)*x**3/(128*b*sqrt(1 + b*x**2/a)) + 13*B*a**(3/2
)*x**5/(64*sqrt(1 + b*x**2/a)) + 5*B*sqrt(a)*b*x**7/(16*sqrt(1 + b*x**2/a)) + 3*B*a**4*asinh(sqrt(b)*x/sqrt(a)
)/(128*b**(5/2)) + B*b**2*x**9/(8*sqrt(a)*sqrt(1 + b*x**2/a))

________________________________________________________________________________________

Giac [A]  time = 1.22078, size = 155, normalized size = 1.03 \begin{align*} -\frac{3 \, B a^{4} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{128 \, b^{\frac{5}{2}}} - \frac{1}{4480} \, \sqrt{b x^{2} + a}{\left (\frac{256 \, A a^{3}}{b^{2}} +{\left (\frac{105 \, B a^{3}}{b^{2}} - 2 \,{\left (\frac{64 \, A a^{2}}{b} +{\left (\frac{35 \, B a^{2}}{b} + 4 \,{\left (128 \, A a + 5 \,{\left (21 \, B a + 2 \,{\left (7 \, B b x + 8 \, A b\right )} x\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^3*(B*x+A)*(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

-3/128*B*a^4*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2) - 1/4480*sqrt(b*x^2 + a)*(256*A*a^3/b^2 + (105*B*a
^3/b^2 - 2*(64*A*a^2/b + (35*B*a^2/b + 4*(128*A*a + 5*(21*B*a + 2*(7*B*b*x + 8*A*b)*x)*x)*x)*x)*x)*x)