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

Optimal. Leaf size=103 \[ -\frac{a^3 B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{3/2}}-\frac{a^2 B x \sqrt{a+b x^2}}{16 b}+\frac{\left (a+b x^2\right )^{5/2} (6 A+5 B x)}{30 b}-\frac{a B x \left (a+b x^2\right )^{3/2}}{24 b} \]

[Out]

-(a^2*B*x*Sqrt[a + b*x^2])/(16*b) - (a*B*x*(a + b*x^2)^(3/2))/(24*b) + ((6*A + 5
*B*x)*(a + b*x^2)^(5/2))/(30*b) - (a^3*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(
16*b^(3/2))

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Rubi [A]  time = 0.113047, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222 \[ -\frac{a^3 B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{3/2}}-\frac{a^2 B x \sqrt{a+b x^2}}{16 b}+\frac{\left (a+b x^2\right )^{5/2} (6 A+5 B x)}{30 b}-\frac{a B x \left (a+b x^2\right )^{3/2}}{24 b} \]

Antiderivative was successfully verified.

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

[Out]

-(a^2*B*x*Sqrt[a + b*x^2])/(16*b) - (a*B*x*(a + b*x^2)^(3/2))/(24*b) + ((6*A + 5
*B*x)*(a + b*x^2)^(5/2))/(30*b) - (a^3*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(
16*b^(3/2))

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Rubi in Sympy [A]  time = 10.6217, size = 90, normalized size = 0.87 \[ - \frac{B a^{3} \operatorname{atanh}{\left (\frac{\sqrt{b} x}{\sqrt{a + b x^{2}}} \right )}}{16 b^{\frac{3}{2}}} - \frac{B a^{2} x \sqrt{a + b x^{2}}}{16 b} - \frac{B a x \left (a + b x^{2}\right )^{\frac{3}{2}}}{24 b} + \frac{\left (6 A + 5 B x\right ) \left (a + b x^{2}\right )^{\frac{5}{2}}}{30 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x*(B*x+A)*(b*x**2+a)**(3/2),x)

[Out]

-B*a**3*atanh(sqrt(b)*x/sqrt(a + b*x**2))/(16*b**(3/2)) - B*a**2*x*sqrt(a + b*x*
*2)/(16*b) - B*a*x*(a + b*x**2)**(3/2)/(24*b) + (6*A + 5*B*x)*(a + b*x**2)**(5/2
)/(30*b)

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Mathematica [A]  time = 0.113902, size = 100, normalized size = 0.97 \[ \frac{\sqrt{b} \sqrt{a+b x^2} \left (3 a^2 (16 A+5 B x)+2 a b x^2 (48 A+35 B x)+8 b^2 x^4 (6 A+5 B x)\right )-15 a^3 B \log \left (\sqrt{b} \sqrt{a+b x^2}+b x\right )}{240 b^{3/2}} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[b]*Sqrt[a + b*x^2]*(8*b^2*x^4*(6*A + 5*B*x) + 3*a^2*(16*A + 5*B*x) + 2*a*b
*x^2*(48*A + 35*B*x)) - 15*a^3*B*Log[b*x + Sqrt[b]*Sqrt[a + b*x^2]])/(240*b^(3/2
))

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Maple [A]  time = 0.007, size = 94, normalized size = 0.9 \[{\frac{A}{5\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{Bx}{6\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{Bxa}{24\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{Bx{a}^{2}}{16\,b}\sqrt{b{x}^{2}+a}}-{\frac{B{a}^{3}}{16}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{3}{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.277005, size = 1, normalized size = 0.01 \[ \left [\frac{15 \, B a^{3} \log \left (2 \, \sqrt{b x^{2} + a} b x -{\left (2 \, b x^{2} + a\right )} \sqrt{b}\right ) + 2 \,{\left (40 \, B b^{2} x^{5} + 48 \, A b^{2} x^{4} + 70 \, B a b x^{3} + 96 \, A a b x^{2} + 15 \, B a^{2} x + 48 \, A a^{2}\right )} \sqrt{b x^{2} + a} \sqrt{b}}{480 \, b^{\frac{3}{2}}}, -\frac{15 \, B a^{3} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (40 \, B b^{2} x^{5} + 48 \, A b^{2} x^{4} + 70 \, B a b x^{3} + 96 \, A a b x^{2} + 15 \, B a^{2} x + 48 \, A a^{2}\right )} \sqrt{b x^{2} + a} \sqrt{-b}}{240 \, \sqrt{-b} b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

[1/480*(15*B*a^3*log(2*sqrt(b*x^2 + a)*b*x - (2*b*x^2 + a)*sqrt(b)) + 2*(40*B*b^
2*x^5 + 48*A*b^2*x^4 + 70*B*a*b*x^3 + 96*A*a*b*x^2 + 15*B*a^2*x + 48*A*a^2)*sqrt
(b*x^2 + a)*sqrt(b))/b^(3/2), -1/240*(15*B*a^3*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)
) - (40*B*b^2*x^5 + 48*A*b^2*x^4 + 70*B*a*b*x^3 + 96*A*a*b*x^2 + 15*B*a^2*x + 48
*A*a^2)*sqrt(b*x^2 + a)*sqrt(-b))/(sqrt(-b)*b)]

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Sympy [A]  time = 15.4076, size = 223, normalized size = 2.17 \[ A a \left (\begin{cases} \frac{\sqrt{a} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\left (a + b x^{2}\right )^{\frac{3}{2}}}{3 b} & \text{otherwise} \end{cases}\right ) + A b \left (\begin{cases} - \frac{2 a^{2} \sqrt{a + b x^{2}}}{15 b^{2}} + \frac{a x^{2} \sqrt{a + b x^{2}}}{15 b} + \frac{x^{4} \sqrt{a + b x^{2}}}{5} & \text{for}\: b \neq 0 \\\frac{\sqrt{a} x^{4}}{4} & \text{otherwise} \end{cases}\right ) + \frac{B a^{\frac{5}{2}} x}{16 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{17 B a^{\frac{3}{2}} x^{3}}{48 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{11 B \sqrt{a} b x^{5}}{24 \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{B a^{3} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{16 b^{\frac{3}{2}}} + \frac{B b^{2} x^{7}}{6 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

A*a*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)), ((a + b*x**2)**(3/2)/(3*b), True)) + A
*b*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*a**(5/2)*x/(
16*b*sqrt(1 + b*x**2/a)) + 17*B*a**(3/2)*x**3/(48*sqrt(1 + b*x**2/a)) + 11*B*sqr
t(a)*b*x**5/(24*sqrt(1 + b*x**2/a)) - B*a**3*asinh(sqrt(b)*x/sqrt(a))/(16*b**(3/
2)) + B*b**2*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a))

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GIAC/XCAS [A]  time = 0.224078, size = 120, normalized size = 1.17 \[ \frac{B a^{3}{\rm ln}\left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{16 \, b^{\frac{3}{2}}} + \frac{1}{240} \, \sqrt{b x^{2} + a}{\left (\frac{48 \, A a^{2}}{b} +{\left (\frac{15 \, B a^{2}}{b} + 2 \,{\left (48 \, A a +{\left (35 \, B a + 4 \,{\left (5 \, B b x + 6 \, A b\right )} x\right )} x\right )} x\right )} x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/16*B*a^3*ln(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2) + 1/240*sqrt(b*x^2 + a)
*(48*A*a^2/b + (15*B*a^2/b + 2*(48*A*a + (35*B*a + 4*(5*B*b*x + 6*A*b)*x)*x)*x)*
x)