3.24 \(\int (c+d x^2)^2 \cos ^{-1}(a x) \, dx\)

Optimal. Leaf size=135 \[ -\frac{\sqrt{1-a^2 x^2} \left (15 a^4 c^2+10 a^2 c d+3 d^2\right )}{15 a^5}+\frac{2 d \left (1-a^2 x^2\right )^{3/2} \left (5 a^2 c+3 d\right )}{45 a^5}-\frac{d^2 \left (1-a^2 x^2\right )^{5/2}}{25 a^5}+c^2 x \cos ^{-1}(a x)+\frac{2}{3} c d x^3 \cos ^{-1}(a x)+\frac{1}{5} d^2 x^5 \cos ^{-1}(a x) \]

[Out]

-((15*a^4*c^2 + 10*a^2*c*d + 3*d^2)*Sqrt[1 - a^2*x^2])/(15*a^5) + (2*d*(5*a^2*c + 3*d)*(1 - a^2*x^2)^(3/2))/(4
5*a^5) - (d^2*(1 - a^2*x^2)^(5/2))/(25*a^5) + c^2*x*ArcCos[a*x] + (2*c*d*x^3*ArcCos[a*x])/3 + (d^2*x^5*ArcCos[
a*x])/5

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Rubi [A]  time = 0.127258, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {194, 4666, 12, 1247, 698} \[ -\frac{\sqrt{1-a^2 x^2} \left (15 a^4 c^2+10 a^2 c d+3 d^2\right )}{15 a^5}+\frac{2 d \left (1-a^2 x^2\right )^{3/2} \left (5 a^2 c+3 d\right )}{45 a^5}-\frac{d^2 \left (1-a^2 x^2\right )^{5/2}}{25 a^5}+c^2 x \cos ^{-1}(a x)+\frac{2}{3} c d x^3 \cos ^{-1}(a x)+\frac{1}{5} d^2 x^5 \cos ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2)^2*ArcCos[a*x],x]

[Out]

-((15*a^4*c^2 + 10*a^2*c*d + 3*d^2)*Sqrt[1 - a^2*x^2])/(15*a^5) + (2*d*(5*a^2*c + 3*d)*(1 - a^2*x^2)^(3/2))/(4
5*a^5) - (d^2*(1 - a^2*x^2)^(5/2))/(25*a^5) + c^2*x*ArcCos[a*x] + (2*c*d*x^3*ArcCos[a*x])/3 + (d^2*x^5*ArcCos[
a*x])/5

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 4666

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2)
^p, x]}, Dist[a + b*ArcCos[c*x], u, x] + Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]] /; F
reeQ[{a, b, c, d, e}, x] && NeQ[c^2*d + e, 0] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int \left (c+d x^2\right )^2 \cos ^{-1}(a x) \, dx &=c^2 x \cos ^{-1}(a x)+\frac{2}{3} c d x^3 \cos ^{-1}(a x)+\frac{1}{5} d^2 x^5 \cos ^{-1}(a x)+a \int \frac{x \left (15 c^2+10 c d x^2+3 d^2 x^4\right )}{15 \sqrt{1-a^2 x^2}} \, dx\\ &=c^2 x \cos ^{-1}(a x)+\frac{2}{3} c d x^3 \cos ^{-1}(a x)+\frac{1}{5} d^2 x^5 \cos ^{-1}(a x)+\frac{1}{15} a \int \frac{x \left (15 c^2+10 c d x^2+3 d^2 x^4\right )}{\sqrt{1-a^2 x^2}} \, dx\\ &=c^2 x \cos ^{-1}(a x)+\frac{2}{3} c d x^3 \cos ^{-1}(a x)+\frac{1}{5} d^2 x^5 \cos ^{-1}(a x)+\frac{1}{30} a \operatorname{Subst}\left (\int \frac{15 c^2+10 c d x+3 d^2 x^2}{\sqrt{1-a^2 x}} \, dx,x,x^2\right )\\ &=c^2 x \cos ^{-1}(a x)+\frac{2}{3} c d x^3 \cos ^{-1}(a x)+\frac{1}{5} d^2 x^5 \cos ^{-1}(a x)+\frac{1}{30} a \operatorname{Subst}\left (\int \left (\frac{15 a^4 c^2+10 a^2 c d+3 d^2}{a^4 \sqrt{1-a^2 x}}-\frac{2 d \left (5 a^2 c+3 d\right ) \sqrt{1-a^2 x}}{a^4}+\frac{3 d^2 \left (1-a^2 x\right )^{3/2}}{a^4}\right ) \, dx,x,x^2\right )\\ &=-\frac{\left (15 a^4 c^2+10 a^2 c d+3 d^2\right ) \sqrt{1-a^2 x^2}}{15 a^5}+\frac{2 d \left (5 a^2 c+3 d\right ) \left (1-a^2 x^2\right )^{3/2}}{45 a^5}-\frac{d^2 \left (1-a^2 x^2\right )^{5/2}}{25 a^5}+c^2 x \cos ^{-1}(a x)+\frac{2}{3} c d x^3 \cos ^{-1}(a x)+\frac{1}{5} d^2 x^5 \cos ^{-1}(a x)\\ \end{align*}

Mathematica [A]  time = 0.115404, size = 99, normalized size = 0.73 \[ \cos ^{-1}(a x) \left (c^2 x+\frac{2}{3} c d x^3+\frac{d^2 x^5}{5}\right )-\frac{\sqrt{1-a^2 x^2} \left (a^4 \left (225 c^2+50 c d x^2+9 d^2 x^4\right )+4 a^2 d \left (25 c+3 d x^2\right )+24 d^2\right )}{225 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2)^2*ArcCos[a*x],x]

[Out]

-(Sqrt[1 - a^2*x^2]*(24*d^2 + 4*a^2*d*(25*c + 3*d*x^2) + a^4*(225*c^2 + 50*c*d*x^2 + 9*d^2*x^4)))/(225*a^5) +
(c^2*x + (2*c*d*x^3)/3 + (d^2*x^5)/5)*ArcCos[a*x]

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Maple [A]  time = 0.022, size = 169, normalized size = 1.3 \begin{align*}{\frac{1}{a} \left ({\frac{a\arccos \left ( ax \right ){d}^{2}{x}^{5}}{5}}+{\frac{2\,a\arccos \left ( ax \right ) cd{x}^{3}}{3}}+\arccos \left ( ax \right ){c}^{2}ax+{\frac{1}{15\,{a}^{4}} \left ( 3\,{d}^{2} \left ( -1/5\,{a}^{4}{x}^{4}\sqrt{-{a}^{2}{x}^{2}+1}-{\frac{4\,{a}^{2}{x}^{2}\sqrt{-{a}^{2}{x}^{2}+1}}{15}}-{\frac{8\,\sqrt{-{a}^{2}{x}^{2}+1}}{15}} \right ) +10\,{a}^{2}cd \left ( -1/3\,{a}^{2}{x}^{2}\sqrt{-{a}^{2}{x}^{2}+1}-2/3\,\sqrt{-{a}^{2}{x}^{2}+1} \right ) -15\,{a}^{4}{c}^{2}\sqrt{-{a}^{2}{x}^{2}+1} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^2*arccos(a*x),x)

[Out]

1/a*(1/5*a*arccos(a*x)*d^2*x^5+2/3*a*arccos(a*x)*c*d*x^3+arccos(a*x)*c^2*a*x+1/15/a^4*(3*d^2*(-1/5*a^4*x^4*(-a
^2*x^2+1)^(1/2)-4/15*a^2*x^2*(-a^2*x^2+1)^(1/2)-8/15*(-a^2*x^2+1)^(1/2))+10*a^2*c*d*(-1/3*a^2*x^2*(-a^2*x^2+1)
^(1/2)-2/3*(-a^2*x^2+1)^(1/2))-15*a^4*c^2*(-a^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.47587, size = 216, normalized size = 1.6 \begin{align*} -\frac{1}{225} \,{\left (\frac{9 \, \sqrt{-a^{2} x^{2} + 1} d^{2} x^{4}}{a^{2}} + \frac{50 \, \sqrt{-a^{2} x^{2} + 1} c d x^{2}}{a^{2}} + \frac{225 \, \sqrt{-a^{2} x^{2} + 1} c^{2}}{a^{2}} + \frac{12 \, \sqrt{-a^{2} x^{2} + 1} d^{2} x^{2}}{a^{4}} + \frac{100 \, \sqrt{-a^{2} x^{2} + 1} c d}{a^{4}} + \frac{24 \, \sqrt{-a^{2} x^{2} + 1} d^{2}}{a^{6}}\right )} a + \frac{1}{15} \,{\left (3 \, d^{2} x^{5} + 10 \, c d x^{3} + 15 \, c^{2} x\right )} \arccos \left (a x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^2*arccos(a*x),x, algorithm="maxima")

[Out]

-1/225*(9*sqrt(-a^2*x^2 + 1)*d^2*x^4/a^2 + 50*sqrt(-a^2*x^2 + 1)*c*d*x^2/a^2 + 225*sqrt(-a^2*x^2 + 1)*c^2/a^2
+ 12*sqrt(-a^2*x^2 + 1)*d^2*x^2/a^4 + 100*sqrt(-a^2*x^2 + 1)*c*d/a^4 + 24*sqrt(-a^2*x^2 + 1)*d^2/a^6)*a + 1/15
*(3*d^2*x^5 + 10*c*d*x^3 + 15*c^2*x)*arccos(a*x)

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Fricas [A]  time = 2.40604, size = 247, normalized size = 1.83 \begin{align*} \frac{15 \,{\left (3 \, a^{5} d^{2} x^{5} + 10 \, a^{5} c d x^{3} + 15 \, a^{5} c^{2} x\right )} \arccos \left (a x\right ) -{\left (9 \, a^{4} d^{2} x^{4} + 225 \, a^{4} c^{2} + 100 \, a^{2} c d + 2 \,{\left (25 \, a^{4} c d + 6 \, a^{2} d^{2}\right )} x^{2} + 24 \, d^{2}\right )} \sqrt{-a^{2} x^{2} + 1}}{225 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^2*arccos(a*x),x, algorithm="fricas")

[Out]

1/225*(15*(3*a^5*d^2*x^5 + 10*a^5*c*d*x^3 + 15*a^5*c^2*x)*arccos(a*x) - (9*a^4*d^2*x^4 + 225*a^4*c^2 + 100*a^2
*c*d + 2*(25*a^4*c*d + 6*a^2*d^2)*x^2 + 24*d^2)*sqrt(-a^2*x^2 + 1))/a^5

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Sympy [A]  time = 2.61294, size = 197, normalized size = 1.46 \begin{align*} \begin{cases} c^{2} x \operatorname{acos}{\left (a x \right )} + \frac{2 c d x^{3} \operatorname{acos}{\left (a x \right )}}{3} + \frac{d^{2} x^{5} \operatorname{acos}{\left (a x \right )}}{5} - \frac{c^{2} \sqrt{- a^{2} x^{2} + 1}}{a} - \frac{2 c d x^{2} \sqrt{- a^{2} x^{2} + 1}}{9 a} - \frac{d^{2} x^{4} \sqrt{- a^{2} x^{2} + 1}}{25 a} - \frac{4 c d \sqrt{- a^{2} x^{2} + 1}}{9 a^{3}} - \frac{4 d^{2} x^{2} \sqrt{- a^{2} x^{2} + 1}}{75 a^{3}} - \frac{8 d^{2} \sqrt{- a^{2} x^{2} + 1}}{75 a^{5}} & \text{for}\: a \neq 0 \\\frac{\pi \left (c^{2} x + \frac{2 c d x^{3}}{3} + \frac{d^{2} x^{5}}{5}\right )}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**2*acos(a*x),x)

[Out]

Piecewise((c**2*x*acos(a*x) + 2*c*d*x**3*acos(a*x)/3 + d**2*x**5*acos(a*x)/5 - c**2*sqrt(-a**2*x**2 + 1)/a - 2
*c*d*x**2*sqrt(-a**2*x**2 + 1)/(9*a) - d**2*x**4*sqrt(-a**2*x**2 + 1)/(25*a) - 4*c*d*sqrt(-a**2*x**2 + 1)/(9*a
**3) - 4*d**2*x**2*sqrt(-a**2*x**2 + 1)/(75*a**3) - 8*d**2*sqrt(-a**2*x**2 + 1)/(75*a**5), Ne(a, 0)), (pi*(c**
2*x + 2*c*d*x**3/3 + d**2*x**5/5)/2, True))

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Giac [A]  time = 1.14764, size = 216, normalized size = 1.6 \begin{align*} \frac{1}{5} \, d^{2} x^{5} \arccos \left (a x\right ) + \frac{2}{3} \, c d x^{3} \arccos \left (a x\right ) - \frac{\sqrt{-a^{2} x^{2} + 1} d^{2} x^{4}}{25 \, a} + c^{2} x \arccos \left (a x\right ) - \frac{2 \, \sqrt{-a^{2} x^{2} + 1} c d x^{2}}{9 \, a} - \frac{\sqrt{-a^{2} x^{2} + 1} c^{2}}{a} - \frac{4 \, \sqrt{-a^{2} x^{2} + 1} d^{2} x^{2}}{75 \, a^{3}} - \frac{4 \, \sqrt{-a^{2} x^{2} + 1} c d}{9 \, a^{3}} - \frac{8 \, \sqrt{-a^{2} x^{2} + 1} d^{2}}{75 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^2*arccos(a*x),x, algorithm="giac")

[Out]

1/5*d^2*x^5*arccos(a*x) + 2/3*c*d*x^3*arccos(a*x) - 1/25*sqrt(-a^2*x^2 + 1)*d^2*x^4/a + c^2*x*arccos(a*x) - 2/
9*sqrt(-a^2*x^2 + 1)*c*d*x^2/a - sqrt(-a^2*x^2 + 1)*c^2/a - 4/75*sqrt(-a^2*x^2 + 1)*d^2*x^2/a^3 - 4/9*sqrt(-a^
2*x^2 + 1)*c*d/a^3 - 8/75*sqrt(-a^2*x^2 + 1)*d^2/a^5