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

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

[Out]

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

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Rubi [A]  time = 0.0924524, 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 \[ -\frac{a^3 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 c^{3/2}}-\frac{a^2 B x \sqrt{a+c x^2}}{16 c}+\frac{\left (a+c x^2\right )^{5/2} (6 A+5 B x)}{30 c}-\frac{a B x \left (a+c x^2\right )^{3/2}}{24 c} \]

Antiderivative was successfully verified.

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

[Out]

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

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Rubi in Sympy [A]  time = 10.2918, size = 90, normalized size = 0.87 \[ - \frac{B a^{3} \operatorname{atanh}{\left (\frac{\sqrt{c} x}{\sqrt{a + c x^{2}}} \right )}}{16 c^{\frac{3}{2}}} - \frac{B a^{2} x \sqrt{a + c x^{2}}}{16 c} - \frac{B a x \left (a + c x^{2}\right )^{\frac{3}{2}}}{24 c} + \frac{\left (6 A + 5 B x\right ) \left (a + c x^{2}\right )^{\frac{5}{2}}}{30 c} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Mathematica [A]  time = 0.110718, size = 100, normalized size = 0.97 \[ \frac{\sqrt{c} \sqrt{a+c x^2} \left (3 a^2 (16 A+5 B x)+2 a c x^2 (48 A+35 B x)+8 c^2 x^4 (6 A+5 B x)\right )-15 a^3 B \log \left (\sqrt{c} \sqrt{a+c x^2}+c x\right )}{240 c^{3/2}} \]

Antiderivative was successfully verified.

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/5*A*(c*x^2+a)^(5/2)/c+1/6*B*x*(c*x^2+a)^(5/2)/c-1/24*a*B*x*(c*x^2+a)^(3/2)/c-1
/16*a^2*B*x*(c*x^2+a)^(1/2)/c-1/16*B*a^3/c^(3/2)*ln(c^(1/2)*x+(c*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((c*x^2 + a)^(3/2)*(B*x + A)*x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.393439, size = 1, normalized size = 0.01 \[ \left [\frac{15 \, B a^{3} \log \left (2 \, \sqrt{c x^{2} + a} c x -{\left (2 \, c x^{2} + a\right )} \sqrt{c}\right ) + 2 \,{\left (40 \, B c^{2} x^{5} + 48 \, A c^{2} x^{4} + 70 \, B a c x^{3} + 96 \, A a c x^{2} + 15 \, B a^{2} x + 48 \, A a^{2}\right )} \sqrt{c x^{2} + a} \sqrt{c}}{480 \, c^{\frac{3}{2}}}, -\frac{15 \, B a^{3} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) -{\left (40 \, B c^{2} x^{5} + 48 \, A c^{2} x^{4} + 70 \, B a c x^{3} + 96 \, A a c x^{2} + 15 \, B a^{2} x + 48 \, A a^{2}\right )} \sqrt{c x^{2} + a} \sqrt{-c}}{240 \, \sqrt{-c} c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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Sympy [A]  time = 30.6655, size = 223, normalized size = 2.17 \[ A a \left (\begin{cases} \frac{\sqrt{a} x^{2}}{2} & \text{for}\: c = 0 \\\frac{\left (a + c x^{2}\right )^{\frac{3}{2}}}{3 c} & \text{otherwise} \end{cases}\right ) + A c \left (\begin{cases} - \frac{2 a^{2} \sqrt{a + c x^{2}}}{15 c^{2}} + \frac{a x^{2} \sqrt{a + c x^{2}}}{15 c} + \frac{x^{4} \sqrt{a + c x^{2}}}{5} & \text{for}\: c \neq 0 \\\frac{\sqrt{a} x^{4}}{4} & \text{otherwise} \end{cases}\right ) + \frac{B a^{\frac{5}{2}} x}{16 c \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{17 B a^{\frac{3}{2}} x^{3}}{48 \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{11 B \sqrt{a} c x^{5}}{24 \sqrt{1 + \frac{c x^{2}}{a}}} - \frac{B a^{3} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )}}{16 c^{\frac{3}{2}}} + \frac{B c^{2} x^{7}}{6 \sqrt{a} \sqrt{1 + \frac{c x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

A*a*Piecewise((sqrt(a)*x**2/2, Eq(c, 0)), ((a + c*x**2)**(3/2)/(3*c), True)) + A
*c*Piecewise((-2*a**2*sqrt(a + c*x**2)/(15*c**2) + a*x**2*sqrt(a + c*x**2)/(15*c
) + x**4*sqrt(a + c*x**2)/5, Ne(c, 0)), (sqrt(a)*x**4/4, True)) + B*a**(5/2)*x/(
16*c*sqrt(1 + c*x**2/a)) + 17*B*a**(3/2)*x**3/(48*sqrt(1 + c*x**2/a)) + 11*B*sqr
t(a)*c*x**5/(24*sqrt(1 + c*x**2/a)) - B*a**3*asinh(sqrt(c)*x/sqrt(a))/(16*c**(3/
2)) + B*c**2*x**7/(6*sqrt(a)*sqrt(1 + c*x**2/a))

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GIAC/XCAS [A]  time = 0.275136, size = 120, normalized size = 1.17 \[ \frac{B a^{3}{\rm ln}\left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{16 \, c^{\frac{3}{2}}} + \frac{1}{240} \, \sqrt{c x^{2} + a}{\left (\frac{48 \, A a^{2}}{c} +{\left (\frac{15 \, B a^{2}}{c} + 2 \,{\left (48 \, A a +{\left (35 \, B a + 4 \,{\left (5 \, B c x + 6 \, A c\right )} x\right )} x\right )} x\right )} x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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