3.4.65 \(\int \frac {x^4 (A+B x)}{(a+c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=105 \[ -\frac {\sqrt {a+c x^2} (16 a B-9 A c x)}{6 c^3}-\frac {x^3 (A+B x)}{c \sqrt {a+c x^2}}-\frac {3 a A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{5/2}}+\frac {4 B x^2 \sqrt {a+c x^2}}{3 c^2} \]

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Rubi [A]  time = 0.07, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {819, 833, 780, 217, 206} \begin {gather*} -\frac {\sqrt {a+c x^2} (16 a B-9 A c x)}{6 c^3}-\frac {x^3 (A+B x)}{c \sqrt {a+c x^2}}-\frac {3 a A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{5/2}}+\frac {4 B x^2 \sqrt {a+c x^2}}{3 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x))/(a + c*x^2)^(3/2),x]

[Out]

-((x^3*(A + B*x))/(c*Sqrt[a + c*x^2])) + (4*B*x^2*Sqrt[a + c*x^2])/(3*c^2) - ((16*a*B - 9*A*c*x)*Sqrt[a + c*x^
2])/(6*c^3) - (3*a*A*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*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 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

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)}{\left (a+c x^2\right )^{3/2}} \, dx &=-\frac {x^3 (A+B x)}{c \sqrt {a+c x^2}}+\frac {\int \frac {x^2 (3 a A+4 a B x)}{\sqrt {a+c x^2}} \, dx}{a c}\\ &=-\frac {x^3 (A+B x)}{c \sqrt {a+c x^2}}+\frac {4 B x^2 \sqrt {a+c x^2}}{3 c^2}+\frac {\int \frac {x \left (-8 a^2 B+9 a A c x\right )}{\sqrt {a+c x^2}} \, dx}{3 a c^2}\\ &=-\frac {x^3 (A+B x)}{c \sqrt {a+c x^2}}+\frac {4 B x^2 \sqrt {a+c x^2}}{3 c^2}-\frac {(16 a B-9 A c x) \sqrt {a+c x^2}}{6 c^3}-\frac {(3 a A) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{2 c^2}\\ &=-\frac {x^3 (A+B x)}{c \sqrt {a+c x^2}}+\frac {4 B x^2 \sqrt {a+c x^2}}{3 c^2}-\frac {(16 a B-9 A c x) \sqrt {a+c x^2}}{6 c^3}-\frac {(3 a A) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{2 c^2}\\ &=-\frac {x^3 (A+B x)}{c \sqrt {a+c x^2}}+\frac {4 B x^2 \sqrt {a+c x^2}}{3 c^2}-\frac {(16 a B-9 A c x) \sqrt {a+c x^2}}{6 c^3}-\frac {3 a A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 91, normalized size = 0.87 \begin {gather*} \frac {-16 a^2 B+a c x (9 A-8 B x)-9 a A \sqrt {c} \sqrt {a+c x^2} \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )+c^2 x^3 (3 A+2 B x)}{6 c^3 \sqrt {a+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x))/(a + c*x^2)^(3/2),x]

[Out]

(-16*a^2*B + a*c*x*(9*A - 8*B*x) + c^2*x^3*(3*A + 2*B*x) - 9*a*A*Sqrt[c]*Sqrt[a + c*x^2]*ArcTanh[(Sqrt[c]*x)/S
qrt[a + c*x^2]])/(6*c^3*Sqrt[a + c*x^2])

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IntegrateAlgebraic [A]  time = 0.41, size = 90, normalized size = 0.86 \begin {gather*} \frac {-16 a^2 B+9 a A c x-8 a B c x^2+3 A c^2 x^3+2 B c^2 x^4}{6 c^3 \sqrt {a+c x^2}}+\frac {3 a A \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{2 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^4*(A + B*x))/(a + c*x^2)^(3/2),x]

[Out]

(-16*a^2*B + 9*a*A*c*x - 8*a*B*c*x^2 + 3*A*c^2*x^3 + 2*B*c^2*x^4)/(6*c^3*Sqrt[a + c*x^2]) + (3*a*A*Log[-(Sqrt[
c]*x) + Sqrt[a + c*x^2]])/(2*c^(5/2))

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fricas [A]  time = 0.48, size = 217, normalized size = 2.07 \begin {gather*} \left [\frac {9 \, {\left (A a c x^{2} + A a^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (2 \, B c^{2} x^{4} + 3 \, A c^{2} x^{3} - 8 \, B a c x^{2} + 9 \, A a c x - 16 \, B a^{2}\right )} \sqrt {c x^{2} + a}}{12 \, {\left (c^{4} x^{2} + a c^{3}\right )}}, \frac {9 \, {\left (A a c x^{2} + A a^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (2 \, B c^{2} x^{4} + 3 \, A c^{2} x^{3} - 8 \, B a c x^{2} + 9 \, A a c x - 16 \, B a^{2}\right )} \sqrt {c x^{2} + a}}{6 \, {\left (c^{4} x^{2} + a c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x+A)/(c*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[1/12*(9*(A*a*c*x^2 + A*a^2)*sqrt(c)*log(-2*c*x^2 + 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(2*B*c^2*x^4 + 3*A*c^
2*x^3 - 8*B*a*c*x^2 + 9*A*a*c*x - 16*B*a^2)*sqrt(c*x^2 + a))/(c^4*x^2 + a*c^3), 1/6*(9*(A*a*c*x^2 + A*a^2)*sqr
t(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (2*B*c^2*x^4 + 3*A*c^2*x^3 - 8*B*a*c*x^2 + 9*A*a*c*x - 16*B*a^2)*sq
rt(c*x^2 + a))/(c^4*x^2 + a*c^3)]

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giac [A]  time = 0.20, size = 83, normalized size = 0.79 \begin {gather*} \frac {{\left ({\left ({\left (\frac {2 \, B x}{c} + \frac {3 \, A}{c}\right )} x - \frac {8 \, B a}{c^{2}}\right )} x + \frac {9 \, A a}{c^{2}}\right )} x - \frac {16 \, B a^{2}}{c^{3}}}{6 \, \sqrt {c x^{2} + a}} + \frac {3 \, A a \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{2 \, 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)^(3/2),x, algorithm="giac")

[Out]

1/6*((((2*B*x/c + 3*A/c)*x - 8*B*a/c^2)*x + 9*A*a/c^2)*x - 16*B*a^2/c^3)/sqrt(c*x^2 + a) + 3/2*A*a*log(abs(-sq
rt(c)*x + sqrt(c*x^2 + a)))/c^(5/2)

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maple [A]  time = 0.05, size = 115, normalized size = 1.10 \begin {gather*} \frac {B \,x^{4}}{3 \sqrt {c \,x^{2}+a}\, c}+\frac {A \,x^{3}}{2 \sqrt {c \,x^{2}+a}\, c}-\frac {4 B a \,x^{2}}{3 \sqrt {c \,x^{2}+a}\, c^{2}}+\frac {3 A a x}{2 \sqrt {c \,x^{2}+a}\, c^{2}}-\frac {3 A a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 c^{\frac {5}{2}}}-\frac {8 B \,a^{2}}{3 \sqrt {c \,x^{2}+a}\, 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)^(3/2),x)

[Out]

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

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maxima [A]  time = 0.49, size = 107, normalized size = 1.02 \begin {gather*} \frac {B x^{4}}{3 \, \sqrt {c x^{2} + a} c} + \frac {A x^{3}}{2 \, \sqrt {c x^{2} + a} c} - \frac {4 \, B a x^{2}}{3 \, \sqrt {c x^{2} + a} c^{2}} + \frac {3 \, A a x}{2 \, \sqrt {c x^{2} + a} c^{2}} - \frac {3 \, A a \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, c^{\frac {5}{2}}} - \frac {8 \, B a^{2}}{3 \, \sqrt {c x^{2} + a} 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)^(3/2),x, algorithm="maxima")

[Out]

1/3*B*x^4/(sqrt(c*x^2 + a)*c) + 1/2*A*x^3/(sqrt(c*x^2 + a)*c) - 4/3*B*a*x^2/(sqrt(c*x^2 + a)*c^2) + 3/2*A*a*x/
(sqrt(c*x^2 + a)*c^2) - 3/2*A*a*arcsinh(c*x/sqrt(a*c))/c^(5/2) - 8/3*B*a^2/(sqrt(c*x^2 + a)*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 )}{{\left (c\,x^2+a\right )}^{3/2}} \,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)^(3/2),x)

[Out]

int((x^4*(A + B*x))/(a + c*x^2)^(3/2), x)

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sympy [A]  time = 12.31, size = 144, normalized size = 1.37 \begin {gather*} A \left (\frac {3 \sqrt {a} x}{2 c^{2} \sqrt {1 + \frac {c x^{2}}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {a}} \right )}}{2 c^{\frac {5}{2}}} + \frac {x^{3}}{2 \sqrt {a} c \sqrt {1 + \frac {c x^{2}}{a}}}\right ) + B \left (\begin {cases} - \frac {8 a^{2}}{3 c^{3} \sqrt {a + c x^{2}}} - \frac {4 a x^{2}}{3 c^{2} \sqrt {a + c x^{2}}} + \frac {x^{4}}{3 c \sqrt {a + c x^{2}}} & \text {for}\: c \neq 0 \\\frac {x^{6}}{6 a^{\frac {3}{2}}} & \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)**(3/2),x)

[Out]

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

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