3.4.54 \(\int \frac {x^4 (A+B x)}{\sqrt {a+c x^2}} \, dx\)

Optimal. Leaf size=129 \[ \frac {3 a^2 A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{5/2}}+\frac {a \sqrt {a+c x^2} (64 a B-45 A c x)}{120 c^3}+\frac {A x^3 \sqrt {a+c x^2}}{4 c}-\frac {4 a B x^2 \sqrt {a+c x^2}}{15 c^2}+\frac {B x^4 \sqrt {a+c x^2}}{5 c} \]

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Rubi [A]  time = 0.11, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {833, 780, 217, 206} \begin {gather*} \frac {3 a^2 A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{5/2}}+\frac {a \sqrt {a+c x^2} (64 a B-45 A c x)}{120 c^3}+\frac {A x^3 \sqrt {a+c x^2}}{4 c}-\frac {4 a B x^2 \sqrt {a+c x^2}}{15 c^2}+\frac {B x^4 \sqrt {a+c x^2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x))/Sqrt[a + c*x^2],x]

[Out]

(-4*a*B*x^2*Sqrt[a + c*x^2])/(15*c^2) + (A*x^3*Sqrt[a + c*x^2])/(4*c) + (B*x^4*Sqrt[a + c*x^2])/(5*c) + (a*(64
*a*B - 45*A*c*x)*Sqrt[a + c*x^2])/(120*c^3) + (3*a^2*A*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(8*c^(5/2))

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])

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 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 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])

Rubi steps

\begin {align*} \int \frac {x^4 (A+B x)}{\sqrt {a+c x^2}} \, dx &=\frac {B x^4 \sqrt {a+c x^2}}{5 c}+\frac {\int \frac {x^3 (-4 a B+5 A c x)}{\sqrt {a+c x^2}} \, dx}{5 c}\\ &=\frac {A x^3 \sqrt {a+c x^2}}{4 c}+\frac {B x^4 \sqrt {a+c x^2}}{5 c}+\frac {\int \frac {x^2 (-15 a A c-16 a B c x)}{\sqrt {a+c x^2}} \, dx}{20 c^2}\\ &=-\frac {4 a B x^2 \sqrt {a+c x^2}}{15 c^2}+\frac {A x^3 \sqrt {a+c x^2}}{4 c}+\frac {B x^4 \sqrt {a+c x^2}}{5 c}+\frac {\int \frac {x \left (32 a^2 B c-45 a A c^2 x\right )}{\sqrt {a+c x^2}} \, dx}{60 c^3}\\ &=-\frac {4 a B x^2 \sqrt {a+c x^2}}{15 c^2}+\frac {A x^3 \sqrt {a+c x^2}}{4 c}+\frac {B x^4 \sqrt {a+c x^2}}{5 c}+\frac {a (64 a B-45 A c x) \sqrt {a+c x^2}}{120 c^3}+\frac {\left (3 a^2 A\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{8 c^2}\\ &=-\frac {4 a B x^2 \sqrt {a+c x^2}}{15 c^2}+\frac {A x^3 \sqrt {a+c x^2}}{4 c}+\frac {B x^4 \sqrt {a+c x^2}}{5 c}+\frac {a (64 a B-45 A c x) \sqrt {a+c x^2}}{120 c^3}+\frac {\left (3 a^2 A\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{8 c^2}\\ &=-\frac {4 a B x^2 \sqrt {a+c x^2}}{15 c^2}+\frac {A x^3 \sqrt {a+c x^2}}{4 c}+\frac {B x^4 \sqrt {a+c x^2}}{5 c}+\frac {a (64 a B-45 A c x) \sqrt {a+c x^2}}{120 c^3}+\frac {3 a^2 A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{8 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 86, normalized size = 0.67 \begin {gather*} \frac {\sqrt {a+c x^2} \left (64 a^2 B-a c x (45 A+32 B x)+6 c^2 x^3 (5 A+4 B x)\right )+45 a^2 A \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{120 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x))/Sqrt[a + c*x^2],x]

[Out]

(Sqrt[a + c*x^2]*(64*a^2*B + 6*c^2*x^3*(5*A + 4*B*x) - a*c*x*(45*A + 32*B*x)) + 45*a^2*A*Sqrt[c]*ArcTanh[(Sqrt
[c]*x)/Sqrt[a + c*x^2]])/(120*c^3)

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IntegrateAlgebraic [A]  time = 0.30, size = 92, normalized size = 0.71 \begin {gather*} \frac {\sqrt {a+c x^2} \left (64 a^2 B-45 a A c x-32 a B c x^2+30 A c^2 x^3+24 B c^2 x^4\right )}{120 c^3}-\frac {3 a^2 A \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{8 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^4*(A + B*x))/Sqrt[a + c*x^2],x]

[Out]

(Sqrt[a + c*x^2]*(64*a^2*B - 45*a*A*c*x - 32*a*B*c*x^2 + 30*A*c^2*x^3 + 24*B*c^2*x^4))/(120*c^3) - (3*a^2*A*Lo
g[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(8*c^(5/2))

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fricas [A]  time = 0.47, size = 176, normalized size = 1.36 \begin {gather*} \left [\frac {45 \, A a^{2} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (24 \, B c^{2} x^{4} + 30 \, A c^{2} x^{3} - 32 \, B a c x^{2} - 45 \, A a c x + 64 \, B a^{2}\right )} \sqrt {c x^{2} + a}}{240 \, c^{3}}, -\frac {45 \, A a^{2} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) - {\left (24 \, B c^{2} x^{4} + 30 \, A c^{2} x^{3} - 32 \, B a c x^{2} - 45 \, A a c x + 64 \, B a^{2}\right )} \sqrt {c x^{2} + a}}{120 \, c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(c*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/240*(45*A*a^2*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(24*B*c^2*x^4 + 30*A*c^2*x^3 - 32
*B*a*c*x^2 - 45*A*a*c*x + 64*B*a^2)*sqrt(c*x^2 + a))/c^3, -1/120*(45*A*a^2*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x
^2 + a)) - (24*B*c^2*x^4 + 30*A*c^2*x^3 - 32*B*a*c*x^2 - 45*A*a*c*x + 64*B*a^2)*sqrt(c*x^2 + a))/c^3]

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giac [A]  time = 0.19, size = 87, normalized size = 0.67 \begin {gather*} \frac {1}{120} \, \sqrt {c x^{2} + a} {\left ({\left (2 \, {\left (3 \, {\left (\frac {4 \, B x}{c} + \frac {5 \, A}{c}\right )} x - \frac {16 \, B a}{c^{2}}\right )} x - \frac {45 \, A a}{c^{2}}\right )} x + \frac {64 \, B a^{2}}{c^{3}}\right )} - \frac {3 \, A a^{2} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{8 \, c^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(c*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/120*sqrt(c*x^2 + a)*((2*(3*(4*B*x/c + 5*A/c)*x - 16*B*a/c^2)*x - 45*A*a/c^2)*x + 64*B*a^2/c^3) - 3/8*A*a^2*l
og(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c^(5/2)

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maple [A]  time = 0.05, size = 117, normalized size = 0.91 \begin {gather*} \frac {\sqrt {c \,x^{2}+a}\, B \,x^{4}}{5 c}+\frac {\sqrt {c \,x^{2}+a}\, A \,x^{3}}{4 c}-\frac {4 \sqrt {c \,x^{2}+a}\, B a \,x^{2}}{15 c^{2}}+\frac {3 A \,a^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{8 c^{\frac {5}{2}}}-\frac {3 \sqrt {c \,x^{2}+a}\, A a x}{8 c^{2}}+\frac {8 \sqrt {c \,x^{2}+a}\, B \,a^{2}}{15 c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x+A)/(c*x^2+a)^(1/2),x)

[Out]

1/5*B*x^4*(c*x^2+a)^(1/2)/c-4/15*a*B*x^2*(c*x^2+a)^(1/2)/c^2+8/15*B*a^2/c^3*(c*x^2+a)^(1/2)+1/4*A*x^3*(c*x^2+a
)^(1/2)/c-3/8*A*a/c^2*x*(c*x^2+a)^(1/2)+3/8*A*a^2/c^(5/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))

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maxima [A]  time = 0.57, size = 109, normalized size = 0.84 \begin {gather*} \frac {\sqrt {c x^{2} + a} B x^{4}}{5 \, c} + \frac {\sqrt {c x^{2} + a} A x^{3}}{4 \, c} - \frac {4 \, \sqrt {c x^{2} + a} B a x^{2}}{15 \, c^{2}} - \frac {3 \, \sqrt {c x^{2} + a} A a x}{8 \, c^{2}} + \frac {3 \, A a^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{8 \, c^{\frac {5}{2}}} + \frac {8 \, \sqrt {c x^{2} + a} B a^{2}}{15 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(c*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

1/5*sqrt(c*x^2 + a)*B*x^4/c + 1/4*sqrt(c*x^2 + a)*A*x^3/c - 4/15*sqrt(c*x^2 + a)*B*a*x^2/c^2 - 3/8*sqrt(c*x^2
+ a)*A*a*x/c^2 + 3/8*A*a^2*arcsinh(c*x/sqrt(a*c))/c^(5/2) + 8/15*sqrt(c*x^2 + a)*B*a^2/c^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^4\,\left (A+B\,x\right )}{\sqrt {c\,x^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(A + B*x))/(a + c*x^2)^(1/2),x)

[Out]

int((x^4*(A + B*x))/(a + c*x^2)^(1/2), x)

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sympy [A]  time = 6.58, size = 173, normalized size = 1.34 \begin {gather*} - \frac {3 A a^{\frac {3}{2}} x}{8 c^{2} \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {A \sqrt {a} x^{3}}{8 c \sqrt {1 + \frac {c x^{2}}{a}}} + \frac {3 A a^{2} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{8 c^{\frac {5}{2}}} + \frac {A x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {c x^{2}}{a}}} + B \left (\begin {cases} \frac {8 a^{2} \sqrt {a + c x^{2}}}{15 c^{3}} - \frac {4 a x^{2} \sqrt {a + c x^{2}}}{15 c^{2}} + \frac {x^{4} \sqrt {a + c x^{2}}}{5 c} & \text {for}\: c \neq 0 \\\frac {x^{6}}{6 \sqrt {a}} & \text {otherwise} \end {cases}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(B*x+A)/(c*x**2+a)**(1/2),x)

[Out]

-3*A*a**(3/2)*x/(8*c**2*sqrt(1 + c*x**2/a)) - A*sqrt(a)*x**3/(8*c*sqrt(1 + c*x**2/a)) + 3*A*a**2*asinh(sqrt(c)
*x/sqrt(a))/(8*c**(5/2)) + A*x**5/(4*sqrt(a)*sqrt(1 + c*x**2/a)) + B*Piecewise((8*a**2*sqrt(a + c*x**2)/(15*c*
*3) - 4*a*x**2*sqrt(a + c*x**2)/(15*c**2) + x**4*sqrt(a + c*x**2)/(5*c), Ne(c, 0)), (x**6/(6*sqrt(a)), True))

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