3.4.15 \(\int x^3 (A+B x) \sqrt {a+c x^2} \, dx\) [315]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {847, 794, 201, 223, 212} \begin {gather*} \frac {a^3 B \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{16 c^{5/2}}+\frac {a^2 B x \sqrt {a+c x^2}}{16 c^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} \end {gather*}

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 201

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 794

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 847

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])

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 ) \text {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*}

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Mathematica [A]
time = 0.17, size = 101, normalized size = 0.80 \begin {gather*} \frac {\sqrt {a+c x^2} \left (-32 a^2 A-15 a^2 B x+16 a A c x^2+10 a B c x^3+48 A c^2 x^4+40 B c^2 x^5\right )}{240 c^2}-\frac {a^3 B \log \left (-\sqrt {c} x+\sqrt {a+c x^2}\right )}{16 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + c*x^2]*(-32*a^2*A - 15*a^2*B*x + 16*a*A*c*x^2 + 10*a*B*c*x^3 + 48*A*c^2*x^4 + 40*B*c^2*x^5))/(240*c^
2) - (a^3*B*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(16*c^(5/2))

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Maple [A]
time = 0.54, size = 120, normalized size = 0.94

method	result	size
risch	\(-\frac {\left (-40 B \,c^{2} x^{5}-48 A \,c^{2} x^{4}-10 a B c \,x^{3}-16 a A c \,x^{2}+15 a^{2} B x +32 a^{2} A \right ) \sqrt {c \,x^{2}+a}}{240 c^{2}}+\frac {B \,a^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{16 c^{\frac {5}{2}}}\)	\(89\)
default	\(B \left (\frac {x^{3} \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{6 c}-\frac {a \left (\frac {x \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{4 c}-\frac {a \left (\frac {x \sqrt {c \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}}\right )}{4 c}\right )}{2 c}\right )+A \left (\frac {x^{2} \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{5 c}-\frac {2 a \left (c \,x^{2}+a \right )^{\frac {3}{2}}}{15 c^{2}}\right )\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*(1/6*x^3*(c*x^2+a)^(3/2)/c-1/2*a/c*(1/4*x*(c*x^2+a)^(3/2)/c-1/4*a/c*(1/2*x*(c*x^2+a)^(1/2)+1/2*a/c^(1/2)*ln(
c^(1/2)*x+(c*x^2+a)^(1/2)))))+A*(1/5*x^2*(c*x^2+a)^(3/2)/c-2/15*a/c^2*(c*x^2+a)^(3/2))

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Maxima [A]
time = 0.28, size = 107, normalized size = 0.84 \begin {gather*} \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B x^{3}}{6 \, c} + \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} A x^{2}}{5 \, c} - \frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} B a x}{8 \, c^{2}} + \frac {\sqrt {c x^{2} + a} B a^{2} x}{16 \, c^{2}} + \frac {B a^{3} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{16 \, c^{\frac {5}{2}}} - \frac {2 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}} A a}{15 \, c^{2}} \end {gather*}

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]

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

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Fricas [A]
time = 2.95, size = 206, normalized size = 1.62 \begin {gather*} \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 {gather*}

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 = 8.81, size = 192, normalized size = 1.51 \begin {gather*} 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 {gather*}

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 = 1.21, size = 93, normalized size = 0.73 \begin {gather*} -\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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\sqrt {c\,x^2+a}\,\left (A+B\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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