3.1.32 \(\int \frac {\text {ArcCos}(a x)}{(c+d x^2)^{5/2}} \, dx\) [32]

Optimal. Leaf size=136 \[ -\frac {a \sqrt {1-a^2 x^2}}{3 c \left (a^2 c+d\right ) \sqrt {c+d x^2}}+\frac {x \text {ArcCos}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \text {ArcCos}(a x)}{3 c^2 \sqrt {c+d x^2}}-\frac {2 \text {ArcTan}\left (\frac {\sqrt {d} \sqrt {1-a^2 x^2}}{a \sqrt {c+d x^2}}\right )}{3 c^2 \sqrt {d}} \]

[Out]

1/3*x*arccos(a*x)/c/(d*x^2+c)^(3/2)-2/3*arctan(d^(1/2)*(-a^2*x^2+1)^(1/2)/a/(d*x^2+c)^(1/2))/c^2/d^(1/2)+2/3*x
*arccos(a*x)/c^2/(d*x^2+c)^(1/2)-1/3*a*(-a^2*x^2+1)^(1/2)/c/(a^2*c+d)/(d*x^2+c)^(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {198, 197, 4756, 12, 585, 79, 65, 223, 209} \begin {gather*} -\frac {2 \text {ArcTan}\left (\frac {\sqrt {d} \sqrt {1-a^2 x^2}}{a \sqrt {c+d x^2}}\right )}{3 c^2 \sqrt {d}}-\frac {a \sqrt {1-a^2 x^2}}{3 c \left (a^2 c+d\right ) \sqrt {c+d x^2}}+\frac {2 x \text {ArcCos}(a x)}{3 c^2 \sqrt {c+d x^2}}+\frac {x \text {ArcCos}(a x)}{3 c \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCos[a*x]/(c + d*x^2)^(5/2),x]

[Out]

-1/3*(a*Sqrt[1 - a^2*x^2])/(c*(a^2*c + d)*Sqrt[c + d*x^2]) + (x*ArcCos[a*x])/(3*c*(c + d*x^2)^(3/2)) + (2*x*Ar
cCos[a*x])/(3*c^2*Sqrt[c + d*x^2]) - (2*ArcTan[(Sqrt[d]*Sqrt[1 - a^2*x^2])/(a*Sqrt[c + d*x^2])])/(3*c^2*Sqrt[d
])

Rule 12

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

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 585

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> Dist[1/n, Subst[Int[(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^n], x] /; FreeQ[{a, b, c, d, e,
f, m, n, p, q, r}, x] && EqQ[m - n + 1, 0]

Rule 4756

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 \frac {\cos ^{-1}(a x)}{\left (c+d x^2\right )^{5/2}} \, dx &=\frac {x \cos ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cos ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}+a \int \frac {x \left (3 c+2 d x^2\right )}{3 c^2 \sqrt {1-a^2 x^2} \left (c+d x^2\right )^{3/2}} \, dx\\ &=\frac {x \cos ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cos ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}+\frac {a \int \frac {x \left (3 c+2 d x^2\right )}{\sqrt {1-a^2 x^2} \left (c+d x^2\right )^{3/2}} \, dx}{3 c^2}\\ &=\frac {x \cos ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cos ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}+\frac {a \text {Subst}\left (\int \frac {3 c+2 d x}{\sqrt {1-a^2 x} (c+d x)^{3/2}} \, dx,x,x^2\right )}{6 c^2}\\ &=-\frac {a \sqrt {1-a^2 x^2}}{3 c \left (a^2 c+d\right ) \sqrt {c+d x^2}}+\frac {x \cos ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cos ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}+\frac {a \text {Subst}\left (\int \frac {1}{\sqrt {1-a^2 x} \sqrt {c+d x}} \, dx,x,x^2\right )}{3 c^2}\\ &=-\frac {a \sqrt {1-a^2 x^2}}{3 c \left (a^2 c+d\right ) \sqrt {c+d x^2}}+\frac {x \cos ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cos ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}-\frac {2 \text {Subst}\left (\int \frac {1}{\sqrt {c+\frac {d}{a^2}-\frac {d x^2}{a^2}}} \, dx,x,\sqrt {1-a^2 x^2}\right )}{3 a c^2}\\ &=-\frac {a \sqrt {1-a^2 x^2}}{3 c \left (a^2 c+d\right ) \sqrt {c+d x^2}}+\frac {x \cos ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cos ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}-\frac {2 \text {Subst}\left (\int \frac {1}{1+\frac {d x^2}{a^2}} \, dx,x,\frac {\sqrt {1-a^2 x^2}}{\sqrt {c+d x^2}}\right )}{3 a c^2}\\ &=-\frac {a \sqrt {1-a^2 x^2}}{3 c \left (a^2 c+d\right ) \sqrt {c+d x^2}}+\frac {x \cos ^{-1}(a x)}{3 c \left (c+d x^2\right )^{3/2}}+\frac {2 x \cos ^{-1}(a x)}{3 c^2 \sqrt {c+d x^2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {1-a^2 x^2}}{a \sqrt {c+d x^2}}\right )}{3 c^2 \sqrt {d}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 0.14, size = 120, normalized size = 0.88 \begin {gather*} \frac {-\frac {a c \sqrt {1-a^2 x^2} \left (c+d x^2\right )}{a^2 c+d}+a x^2 \left (c+d x^2\right ) \sqrt {1+\frac {d x^2}{c}} F_1\left (1;\frac {1}{2},\frac {1}{2};2;a^2 x^2,-\frac {d x^2}{c}\right )+\left (3 c x+2 d x^3\right ) \text {ArcCos}(a x)}{3 c^2 \left (c+d x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCos[a*x]/(c + d*x^2)^(5/2),x]

[Out]

(-((a*c*Sqrt[1 - a^2*x^2]*(c + d*x^2))/(a^2*c + d)) + a*x^2*(c + d*x^2)*Sqrt[1 + (d*x^2)/c]*AppellF1[1, 1/2, 1
/2, 2, a^2*x^2, -((d*x^2)/c)] + (3*c*x + 2*d*x^3)*ArcCos[a*x])/(3*c^2*(c + d*x^2)^(3/2))

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Maple [F]
time = 0.42, size = 0, normalized size = 0.00 \[\int \frac {\arccos \left (a x \right )}{\left (d \,x^{2}+c \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(a*x)/(d*x^2+c)^(5/2),x)

[Out]

int(arccos(a*x)/(d*x^2+c)^(5/2),x)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)/(d*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(d-a^2*c>0)', see `assume?` for
more detail

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (112) = 224\).
time = 3.85, size = 580, normalized size = 4.26 \begin {gather*} \left [-\frac {{\left (a^{2} c^{3} + {\left (a^{2} c d^{2} + d^{3}\right )} x^{4} + c^{2} d + 2 \, {\left (a^{2} c^{2} d + c d^{2}\right )} x^{2}\right )} \sqrt {-d} \log \left (8 \, a^{4} d^{2} x^{4} + a^{4} c^{2} - 6 \, a^{2} c d + 8 \, {\left (a^{4} c d - a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a^{3} d x^{2} + a^{3} c - a d\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {d x^{2} + c} \sqrt {-d} + d^{2}\right ) - 2 \, \sqrt {d x^{2} + c} {\left ({\left (2 \, {\left (a^{2} c d^{2} + d^{3}\right )} x^{3} + 3 \, {\left (a^{2} c^{2} d + c d^{2}\right )} x\right )} \arccos \left (a x\right ) - {\left (a c d^{2} x^{2} + a c^{2} d\right )} \sqrt {-a^{2} x^{2} + 1}\right )}}{6 \, {\left (a^{2} c^{5} d + c^{4} d^{2} + {\left (a^{2} c^{3} d^{3} + c^{2} d^{4}\right )} x^{4} + 2 \, {\left (a^{2} c^{4} d^{2} + c^{3} d^{3}\right )} x^{2}\right )}}, -\frac {{\left (a^{2} c^{3} + {\left (a^{2} c d^{2} + d^{3}\right )} x^{4} + c^{2} d + 2 \, {\left (a^{2} c^{2} d + c d^{2}\right )} x^{2}\right )} \sqrt {d} \arctan \left (\frac {{\left (2 \, a^{2} d x^{2} + a^{2} c - d\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {d x^{2} + c} \sqrt {d}}{2 \, {\left (a^{3} d^{2} x^{4} - a c d + {\left (a^{3} c d - a d^{2}\right )} x^{2}\right )}}\right ) - \sqrt {d x^{2} + c} {\left ({\left (2 \, {\left (a^{2} c d^{2} + d^{3}\right )} x^{3} + 3 \, {\left (a^{2} c^{2} d + c d^{2}\right )} x\right )} \arccos \left (a x\right ) - {\left (a c d^{2} x^{2} + a c^{2} d\right )} \sqrt {-a^{2} x^{2} + 1}\right )}}{3 \, {\left (a^{2} c^{5} d + c^{4} d^{2} + {\left (a^{2} c^{3} d^{3} + c^{2} d^{4}\right )} x^{4} + 2 \, {\left (a^{2} c^{4} d^{2} + c^{3} d^{3}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)/(d*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

[-1/6*((a^2*c^3 + (a^2*c*d^2 + d^3)*x^4 + c^2*d + 2*(a^2*c^2*d + c*d^2)*x^2)*sqrt(-d)*log(8*a^4*d^2*x^4 + a^4*
c^2 - 6*a^2*c*d + 8*(a^4*c*d - a^2*d^2)*x^2 - 4*(2*a^3*d*x^2 + a^3*c - a*d)*sqrt(-a^2*x^2 + 1)*sqrt(d*x^2 + c)
*sqrt(-d) + d^2) - 2*sqrt(d*x^2 + c)*((2*(a^2*c*d^2 + d^3)*x^3 + 3*(a^2*c^2*d + c*d^2)*x)*arccos(a*x) - (a*c*d
^2*x^2 + a*c^2*d)*sqrt(-a^2*x^2 + 1)))/(a^2*c^5*d + c^4*d^2 + (a^2*c^3*d^3 + c^2*d^4)*x^4 + 2*(a^2*c^4*d^2 + c
^3*d^3)*x^2), -1/3*((a^2*c^3 + (a^2*c*d^2 + d^3)*x^4 + c^2*d + 2*(a^2*c^2*d + c*d^2)*x^2)*sqrt(d)*arctan(1/2*(
2*a^2*d*x^2 + a^2*c - d)*sqrt(-a^2*x^2 + 1)*sqrt(d*x^2 + c)*sqrt(d)/(a^3*d^2*x^4 - a*c*d + (a^3*c*d - a*d^2)*x
^2)) - sqrt(d*x^2 + c)*((2*(a^2*c*d^2 + d^3)*x^3 + 3*(a^2*c^2*d + c*d^2)*x)*arccos(a*x) - (a*c*d^2*x^2 + a*c^2
*d)*sqrt(-a^2*x^2 + 1)))/(a^2*c^5*d + c^4*d^2 + (a^2*c^3*d^3 + c^2*d^4)*x^4 + 2*(a^2*c^4*d^2 + c^3*d^3)*x^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {acos}{\left (a x \right )}}{\left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(a*x)/(d*x**2+c)**(5/2),x)

[Out]

Integral(acos(a*x)/(c + d*x**2)**(5/2), x)

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Giac [A]
time = 0.52, size = 185, normalized size = 1.36 \begin {gather*} \frac {1}{3} \, a {\left (\frac {2 \, a^{2} {\left | d \right |}}{{\left (a^{2} c d - {\left (\sqrt {-a^{2} d} \sqrt {d x^{2} + c} - \sqrt {-{\left (d x^{2} + c\right )} a^{2} d + a^{2} c d + d^{2}}\right )}^{2} + d^{2}\right )} c \sqrt {-d} {\left | a \right |}} - \frac {{\left | d \right |} \log \left ({\left (\sqrt {-a^{2} d} \sqrt {d x^{2} + c} - \sqrt {-{\left (d x^{2} + c\right )} a^{2} d + a^{2} c d + d^{2}}\right )}^{2}\right )}{c^{2} \sqrt {-d} d {\left | a \right |}}\right )} + \frac {x {\left (\frac {2 \, d x^{2}}{c^{2}} + \frac {3}{c}\right )} \arccos \left (a x\right )}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)/(d*x^2+c)^(5/2),x, algorithm="giac")

[Out]

1/3*a*(2*a^2*abs(d)/((a^2*c*d - (sqrt(-a^2*d)*sqrt(d*x^2 + c) - sqrt(-(d*x^2 + c)*a^2*d + a^2*c*d + d^2))^2 +
d^2)*c*sqrt(-d)*abs(a)) - abs(d)*log((sqrt(-a^2*d)*sqrt(d*x^2 + c) - sqrt(-(d*x^2 + c)*a^2*d + a^2*c*d + d^2))
^2)/(c^2*sqrt(-d)*d*abs(a))) + 1/3*x*(2*d*x^2/c^2 + 3/c)*arccos(a*x)/(d*x^2 + c)^(3/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {acos}\left (a\,x\right )}{{\left (d\,x^2+c\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(a*x)/(c + d*x^2)^(5/2),x)

[Out]

int(acos(a*x)/(c + d*x^2)^(5/2), x)

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