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

Optimal. Leaf size=127 \[ \frac{a^2 B x \sqrt{a+c x^2}}{16 c^2}+\frac{a^3 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 c^{5/2}}-\frac{a \left (a+c x^2\right )^{3/2} (16 A+15 B x)}{120 c^2}+\frac{A x^2 \left (a+c x^2\right )^{3/2}}{5 c}+\frac{B x^3 \left (a+c x^2\right )^{3/2}}{6 c} \]

[Out]

(a^2*B*x*Sqrt[a + c*x^2])/(16*c^2) + (A*x^2*(a + c*x^2)^(3/2))/(5*c) + (B*x^3*(a + c*x^2)^(3/2))/(6*c) - (a*(1
6*A + 15*B*x)*(a + c*x^2)^(3/2))/(120*c^2) + (a^3*B*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(16*c^(5/2))

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Rubi [A]  time = 0.0809779, 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} \[ \frac{a^2 B x \sqrt{a+c x^2}}{16 c^2}+\frac{a^3 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 c^{5/2}}-\frac{a \left (a+c x^2\right )^{3/2} (16 A+15 B x)}{120 c^2}+\frac{A x^2 \left (a+c x^2\right )^{3/2}}{5 c}+\frac{B x^3 \left (a+c x^2\right )^{3/2}}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(A + B*x)*Sqrt[a + c*x^2],x]

[Out]

(a^2*B*x*Sqrt[a + c*x^2])/(16*c^2) + (A*x^2*(a + c*x^2)^(3/2))/(5*c) + (B*x^3*(a + c*x^2)^(3/2))/(6*c) - (a*(1
6*A + 15*B*x)*(a + c*x^2)^(3/2))/(120*c^2) + (a^3*B*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(16*c^(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) \sqrt{a+c x^2} \, dx &=\frac{B x^3 \left (a+c x^2\right )^{3/2}}{6 c}+\frac{\int x^2 (-3 a B+6 A c x) \sqrt{a+c x^2} \, dx}{6 c}\\ &=\frac{A x^2 \left (a+c x^2\right )^{3/2}}{5 c}+\frac{B x^3 \left (a+c x^2\right )^{3/2}}{6 c}+\frac{\int x (-12 a A c-15 a B c x) \sqrt{a+c x^2} \, dx}{30 c^2}\\ &=\frac{A x^2 \left (a+c x^2\right )^{3/2}}{5 c}+\frac{B x^3 \left (a+c x^2\right )^{3/2}}{6 c}-\frac{a (16 A+15 B x) \left (a+c x^2\right )^{3/2}}{120 c^2}+\frac{\left (a^2 B\right ) \int \sqrt{a+c x^2} \, dx}{8 c^2}\\ &=\frac{a^2 B x \sqrt{a+c x^2}}{16 c^2}+\frac{A x^2 \left (a+c x^2\right )^{3/2}}{5 c}+\frac{B x^3 \left (a+c x^2\right )^{3/2}}{6 c}-\frac{a (16 A+15 B x) \left (a+c x^2\right )^{3/2}}{120 c^2}+\frac{\left (a^3 B\right ) \int \frac{1}{\sqrt{a+c x^2}} \, dx}{16 c^2}\\ &=\frac{a^2 B x \sqrt{a+c x^2}}{16 c^2}+\frac{A x^2 \left (a+c x^2\right )^{3/2}}{5 c}+\frac{B x^3 \left (a+c x^2\right )^{3/2}}{6 c}-\frac{a (16 A+15 B x) \left (a+c x^2\right )^{3/2}}{120 c^2}+\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^2}\\ &=\frac{a^2 B x \sqrt{a+c x^2}}{16 c^2}+\frac{A x^2 \left (a+c x^2\right )^{3/2}}{5 c}+\frac{B x^3 \left (a+c x^2\right )^{3/2}}{6 c}-\frac{a (16 A+15 B x) \left (a+c x^2\right )^{3/2}}{120 c^2}+\frac{a^3 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{16 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.190143, size = 107, normalized size = 0.84 \[ \frac{\sqrt{a+c x^2} \left (\sqrt{c} \left (-a^2 (32 A+15 B x)+2 a c x^2 (8 A+5 B x)+8 c^2 x^4 (6 A+5 B x)\right )+\frac{15 a^{5/2} B \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{\frac{c x^2}{a}+1}}\right )}{240 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(Sqrt[c]*(8*c^2*x^4*(6*A + 5*B*x) + 2*a*c*x^2*(8*A + 5*B*x) - a^2*(32*A + 15*B*x)) + (15*a^(5
/2)*B*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/Sqrt[1 + (c*x^2)/a]))/(240*c^(5/2))

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Maple [A]  time = 0.009, size = 115, normalized size = 0.9 \begin{align*}{\frac{B{x}^{3}}{6\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{aBx}{8\,{c}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{{a}^{2}Bx}{16\,{c}^{2}}\sqrt{c{x}^{2}+a}}+{\frac{B{a}^{3}}{16}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}}+{\frac{A{x}^{2}}{5\,c} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{2\,aA}{15\,{c}^{2}} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x+A)*(c*x^2+a)^(1/2),x)

[Out]

1/6*B*x^3*(c*x^2+a)^(3/2)/c-1/8*B*a/c^2*x*(c*x^2+a)^(3/2)+1/16*a^2*B*x*(c*x^2+a)^(1/2)/c^2+1/16*B*a^3/c^(5/2)*
ln(x*c^(1/2)+(c*x^2+a)^(1/2))+1/5*A*x^2*(c*x^2+a)^(3/2)/c-2/15*A*a/c^2*(c*x^2+a)^(3/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.6731, size = 501, normalized size = 3.94 \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} + 10 \, B a c^{2} x^{3} + 16 \, A a c^{2} x^{2} - 15 \, B a^{2} c x - 32 \, A a^{2} c\right )} \sqrt{c x^{2} + a}}{480 \, c^{3}}, -\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} + 10 \, B a c^{2} x^{3} + 16 \, A a c^{2} x^{2} - 15 \, B a^{2} c x - 32 \, A a^{2} c\right )} \sqrt{c x^{2} + a}}{240 \, c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)*(c*x^2+a)^(1/2),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 + 10
*B*a*c^2*x^3 + 16*A*a*c^2*x^2 - 15*B*a^2*c*x - 32*A*a^2*c)*sqrt(c*x^2 + a))/c^3, -1/240*(15*B*a^3*sqrt(-c)*arc
tan(sqrt(-c)*x/sqrt(c*x^2 + a)) - (40*B*c^3*x^5 + 48*A*c^3*x^4 + 10*B*a*c^2*x^3 + 16*A*a*c^2*x^2 - 15*B*a^2*c*
x - 32*A*a^2*c)*sqrt(c*x^2 + a))/c^3]

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Sympy [A]  time = 7.70583, size = 192, normalized size = 1.51 \begin{align*} A \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^{2} \sqrt{1 + \frac{c x^{2}}{a}}} - \frac{B a^{\frac{3}{2}} x^{3}}{48 c \sqrt{1 + \frac{c x^{2}}{a}}} + \frac{5 B \sqrt{a} 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{5}{2}}} + \frac{B c 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**3*(B*x+A)*(c*x**2+a)**(1/2),x)

[Out]

A*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**2*sqrt(1 + c*x**2/a)) - B*a**(3/2)*x**3/(48*c*sqrt(1 +
c*x**2/a)) + 5*B*sqrt(a)*x**5/(24*sqrt(1 + c*x**2/a)) + B*a**3*asinh(sqrt(c)*x/sqrt(a))/(16*c**(5/2)) + B*c*x*
*7/(6*sqrt(a)*sqrt(1 + c*x**2/a))

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Giac [A]  time = 1.16024, size = 126, normalized size = 0.99 \begin{align*} -\frac{B a^{3} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{16 \, c^{\frac{5}{2}}} + \frac{1}{240} \, \sqrt{c x^{2} + a}{\left ({\left (2 \,{\left ({\left (4 \,{\left (5 \, B x + 6 \, A\right )} x + \frac{5 \, B a}{c}\right )} x + \frac{8 \, A a}{c}\right )} x - \frac{15 \, B a^{2}}{c^{2}}\right )} x - \frac{32 \, A a^{2}}{c^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x+A)*(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-1/16*B*a^3*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(5/2) + 1/240*sqrt(c*x^2 + a)*((2*((4*(5*B*x + 6*A)*x + 5
*B*a/c)*x + 8*A*a/c)*x - 15*B*a^2/c^2)*x - 32*A*a^2/c^2)