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

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

[Out]

2/45*d*(5*a^2*c+3*d)*(-a^2*x^2+1)^(3/2)/a^5-1/25*d^2*(-a^2*x^2+1)^(5/2)/a^5+c^2*x*arccos(a*x)+2/3*c*d*x^3*arcc
os(a*x)+1/5*d^2*x^5*arccos(a*x)-1/15*(15*a^4*c^2+10*a^2*c*d+3*d^2)*(-a^2*x^2+1)^(1/2)/a^5

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Rubi [A]  time = 0.13, antiderivative size = 135, normalized size of antiderivative = 1.00, 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 12

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

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 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]))

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 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])

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*}

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Mathematica [A]  time = 0.13, 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]

-1/225*(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)))/a^5 +
(c^2*x + (2*c*d*x^3)/3 + (d^2*x^5)/5)*ArcCos[a*x]

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fricas [A]  time = 0.48, size = 110, normalized size = 0.81 \[ \frac {15 \, {\left (3 \, a^{5} d^{2} x^{5} + 10 \, a^{5} c d x^{3} + 15 \, a^{5} c^{2} x\right )} \arccos \left (a x\right ) - {\left (9 \, a^{4} d^{2} x^{4} + 225 \, a^{4} c^{2} + 100 \, a^{2} c d + 2 \, {\left (25 \, a^{4} c d + 6 \, a^{2} d^{2}\right )} x^{2} + 24 \, d^{2}\right )} \sqrt {-a^{2} x^{2} + 1}}{225 \, a^{5}} \]

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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giac [A]  time = 1.27, size = 160, normalized size = 1.19 \[ \frac {1}{5} \, d^{2} x^{5} \arccos \left (a x\right ) + \frac {2}{3} \, c d x^{3} \arccos \left (a x\right ) - \frac {\sqrt {-a^{2} x^{2} + 1} d^{2} x^{4}}{25 \, a} + c^{2} x \arccos \left (a x\right ) - \frac {2 \, \sqrt {-a^{2} x^{2} + 1} c d x^{2}}{9 \, a} - \frac {\sqrt {-a^{2} x^{2} + 1} c^{2}}{a} - \frac {4 \, \sqrt {-a^{2} x^{2} + 1} d^{2} x^{2}}{75 \, a^{3}} - \frac {4 \, \sqrt {-a^{2} x^{2} + 1} c d}{9 \, a^{3}} - \frac {8 \, \sqrt {-a^{2} x^{2} + 1} d^{2}}{75 \, a^{5}} \]

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

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maple [A]  time = 0.02, size = 169, normalized size = 1.25 \[ \frac {\frac {a \arccos \left (a x \right ) d^{2} x^{5}}{5}+\frac {2 a \arccos \left (a x \right ) c d \,x^{3}}{3}+\arccos \left (a x \right ) c^{2} a x +\frac {3 d^{2} \left (-\frac {a^{4} x^{4} \sqrt {-a^{2} x^{2}+1}}{5}-\frac {4 a^{2} x^{2} \sqrt {-a^{2} x^{2}+1}}{15}-\frac {8 \sqrt {-a^{2} x^{2}+1}}{15}\right )+10 a^{2} c d \left (-\frac {a^{2} x^{2} \sqrt {-a^{2} x^{2}+1}}{3}-\frac {2 \sqrt {-a^{2} x^{2}+1}}{3}\right )-15 a^{4} c^{2} \sqrt {-a^{2} x^{2}+1}}{15 a^{4}}}{a} \]

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 = 0.41, size = 160, normalized size = 1.19 \[ -\frac {1}{225} \, {\left (\frac {9 \, \sqrt {-a^{2} x^{2} + 1} d^{2} x^{4}}{a^{2}} + \frac {50 \, \sqrt {-a^{2} x^{2} + 1} c d x^{2}}{a^{2}} + \frac {225 \, \sqrt {-a^{2} x^{2} + 1} c^{2}}{a^{2}} + \frac {12 \, \sqrt {-a^{2} x^{2} + 1} d^{2} x^{2}}{a^{4}} + \frac {100 \, \sqrt {-a^{2} x^{2} + 1} c d}{a^{4}} + \frac {24 \, \sqrt {-a^{2} x^{2} + 1} d^{2}}{a^{6}}\right )} a + \frac {1}{15} \, {\left (3 \, d^{2} x^{5} + 10 \, c d x^{3} + 15 \, c^{2} x\right )} \arccos \left (a x\right ) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {acos}\left (a\,x\right )\,{\left (d\,x^2+c\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(a*x)*(c + d*x^2)^2,x)

[Out]

int(acos(a*x)*(c + d*x^2)^2, x)

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sympy [A]  time = 1.99, size = 197, normalized size = 1.46 \[ \begin {cases} c^{2} x \operatorname {acos}{\left (a x \right )} + \frac {2 c d x^{3} \operatorname {acos}{\left (a x \right )}}{3} + \frac {d^{2} x^{5} \operatorname {acos}{\left (a x \right )}}{5} - \frac {c^{2} \sqrt {- a^{2} x^{2} + 1}}{a} - \frac {2 c d x^{2} \sqrt {- a^{2} x^{2} + 1}}{9 a} - \frac {d^{2} x^{4} \sqrt {- a^{2} x^{2} + 1}}{25 a} - \frac {4 c d \sqrt {- a^{2} x^{2} + 1}}{9 a^{3}} - \frac {4 d^{2} x^{2} \sqrt {- a^{2} x^{2} + 1}}{75 a^{3}} - \frac {8 d^{2} \sqrt {- a^{2} x^{2} + 1}}{75 a^{5}} & \text {for}\: a \neq 0 \\\frac {\pi \left (c^{2} x + \frac {2 c d x^{3}}{3} + \frac {d^{2} x^{5}}{5}\right )}{2} & \text {otherwise} \end {cases} \]

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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