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

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

[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))

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Rubi [A]  time = 0.0714329, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {819, 833, 780, 217, 206} \[ -\frac{\sqrt{a+c x^2} (16 a B-9 A c x)}{6 c^3}-\frac{x^3 (A+B x)}{c \sqrt{a+c x^2}}-\frac{3 a A \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{2 c^{5/2}}+\frac{4 B x^2 \sqrt{a+c x^2}}{3 c^2} \]

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 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])

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 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 \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*}

Mathematica [A]  time = 0.0482126, size = 91, normalized size = 0.87 \[ \frac{-16 a^2 B+a c x (9 A-8 B x)-9 a A \sqrt{c} \sqrt{a+c x^2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )+c^2 x^3 (3 A+2 B x)}{6 c^3 \sqrt{a+c x^2}} \]

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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Maple [A]  time = 0.009, size = 115, normalized size = 1.1 \begin{align*}{\frac{{x}^{4}B}{3\,c}{\frac{1}{\sqrt{c{x}^{2}+a}}}}-{\frac{4\,aB{x}^{2}}{3\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+a}}}}-{\frac{8\,B{a}^{2}}{3\,{c}^{3}}{\frac{1}{\sqrt{c{x}^{2}+a}}}}+{\frac{A{x}^{3}}{2\,c}{\frac{1}{\sqrt{c{x}^{2}+a}}}}+{\frac{3\,aAx}{2\,{c}^{2}}{\frac{1}{\sqrt{c{x}^{2}+a}}}}-{\frac{3\,aA}{2}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){c}^{-{\frac{5}{2}}}} \end{align*}

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(x*c^(1/2)+(c*x^2+a)^(1/2))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51209, size = 495, normalized size = 4.71 \begin{align*} \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{align*}

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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Sympy [A]  time = 11.7048, size = 144, normalized size = 1.37 \begin{align*} 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{align*}

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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Giac [A]  time = 1.22249, size = 112, normalized size = 1.07 \begin{align*} \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{align*}

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)