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

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 = 0.031254, 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, Rules used = {780, 195, 217, 206} \[ -\frac{a^3 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 c^{3/2}}-\frac{a^2 B x \sqrt{a+c x^2}}{16 c}+\frac{\left (a+c x^2\right )^{5/2} (6 A+5 B x)}{30 c}-\frac{a B x \left (a+c x^2\right )^{3/2}}{24 c} \]

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))

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 (A+B x) \left (a+c x^2\right )^{3/2} \, dx &=\frac{(6 A+5 B x) \left (a+c x^2\right )^{5/2}}{30 c}-\frac{(a B) \int \left (a+c x^2\right )^{3/2} \, dx}{6 c}\\ &=-\frac{a B x \left (a+c x^2\right )^{3/2}}{24 c}+\frac{(6 A+5 B x) \left (a+c x^2\right )^{5/2}}{30 c}-\frac{\left (a^2 B\right ) \int \sqrt{a+c x^2} \, dx}{8 c}\\ &=-\frac{a^2 B x \sqrt{a+c x^2}}{16 c}-\frac{a B x \left (a+c x^2\right )^{3/2}}{24 c}+\frac{(6 A+5 B x) \left (a+c x^2\right )^{5/2}}{30 c}-\frac{\left (a^3 B\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{16 c}\\ &=-\frac{a^2 B x \sqrt{a+c x^2}}{16 c}-\frac{a B x \left (a+c x^2\right )^{3/2}}{24 c}+\frac{(6 A+5 B x) \left (a+c x^2\right )^{5/2}}{30 c}-\frac{\left (a^3 B\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{a+c x^2}}\right )}{16 c}\\ &=-\frac{a^2 B x \sqrt{a+c x^2}}{16 c}-\frac{a B x \left (a+c x^2\right )^{3/2}}{24 c}+\frac{(6 A+5 B x) \left (a+c x^2\right )^{5/2}}{30 c}-\frac{a^3 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.201848, size = 107, normalized size = 1.04 \[ \frac{\sqrt{a+c x^2} \left (\sqrt{c} \left (3 a^2 (16 A+5 B x)+2 a c x^2 (48 A+35 B x)+8 c^2 x^4 (6 A+5 B x)\right )-\frac{15 a^{5/2} B \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{\frac{c x^2}{a}+1}}\right )}{240 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(Sqrt[c]*(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
^(5/2)*B*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/Sqrt[1 + (c*x^2)/a]))/(240*c^(3/2))

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Maple [A]  time = 0.006, size = 94, normalized size = 0.9 \begin{align*}{\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 ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{A}{5\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*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(x*
c^(1/2)+(c*x^2+a)^(1/2))+1/5*A/c*(c*x^2+a)^(5/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.02663, size = 500, normalized size = 4.85 \begin{align*} \left [\frac{15 \, B a^{3} \sqrt{c} \log \left (-2 \, c x^{2} + 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 2 \,{\left (40 \, B c^{3} x^{5} + 48 \, A c^{3} x^{4} + 70 \, B a c^{2} x^{3} + 96 \, A a c^{2} x^{2} + 15 \, B a^{2} c x + 48 \, A a^{2} c\right )} \sqrt{c x^{2} + a}}{480 \, c^{2}}, \frac{15 \, B a^{3} \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) +{\left (40 \, B c^{3} x^{5} + 48 \, A c^{3} x^{4} + 70 \, B a c^{2} x^{3} + 96 \, A a c^{2} x^{2} + 15 \, B a^{2} c x + 48 \, A a^{2} c\right )} \sqrt{c x^{2} + a}}{240 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 12.5033, size = 223, normalized size = 2.17 \begin{align*} 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}}} \end{align*}

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, Tru
e)) + 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*sqrt(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 [A]  time = 1.15589, size = 120, normalized size = 1.17 \begin{align*} \frac{B a^{3} \log \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*B*a^3*log(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)