3.31 \(\int \frac {\cos ^{-1}(a x)}{(c+d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=66 \[ \frac {x \cos ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {d} \sqrt {1-a^2 x^2}}{a \sqrt {c+d x^2}}\right )}{c \sqrt {d}} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {191, 4666, 12, 444, 63, 217, 203} \[ \frac {x \cos ^{-1}(a x)}{c \sqrt {c+d x^2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {d} \sqrt {1-a^2 x^2}}{a \sqrt {c+d x^2}}\right )}{c \sqrt {d}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a*x]/(c + d*x^2)^(3/2),x]

[Out]

(x*ArcCos[a*x])/(c*Sqrt[c + d*x^2]) - ArcTan[(Sqrt[d]*Sqrt[1 - a^2*x^2])/(a*Sqrt[c + d*x^2])]/(c*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 63

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 191

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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 444

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

Rubi steps

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

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Mathematica [C]  time = 0.08, size = 68, normalized size = 1.03 \[ \frac {x \left (a x \sqrt {\frac {d x^2}{c}+1} F_1\left (1;\frac {1}{2},\frac {1}{2};2;a^2 x^2,-\frac {d x^2}{c}\right )+2 \cos ^{-1}(a x)\right )}{2 c \sqrt {c+d x^2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCos[a*x]/(c + d*x^2)^(3/2),x]

[Out]

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

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fricas [B]  time = 0.52, size = 276, normalized size = 4.18 \[ \left [\frac {4 \, \sqrt {d x^{2} + c} d x \arccos \left (a x\right ) - {\left (d x^{2} + c\right )} \sqrt {-d} \log \left (8 \, a^{4} d^{2} x^{4} + a^{4} c^{2} - 6 \, a^{2} c d + 8 \, {\left (a^{4} c d - a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a^{3} d x^{2} + a^{3} c - a d\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {d x^{2} + c} \sqrt {-d} + d^{2}\right )}{4 \, {\left (c d^{2} x^{2} + c^{2} d\right )}}, \frac {2 \, \sqrt {d x^{2} + c} d x \arccos \left (a x\right ) - {\left (d x^{2} + c\right )} \sqrt {d} \arctan \left (\frac {{\left (2 \, a^{2} d x^{2} + a^{2} c - d\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {d x^{2} + c} \sqrt {d}}{2 \, {\left (a^{3} d^{2} x^{4} - a c d + {\left (a^{3} c d - a d^{2}\right )} x^{2}\right )}}\right )}{2 \, {\left (c d^{2} x^{2} + c^{2} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)/(d*x^2+c)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(4*sqrt(d*x^2 + c)*d*x*arccos(a*x) - (d*x^2 + c)*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))/(c*d^
2*x^2 + c^2*d), 1/2*(2*sqrt(d*x^2 + c)*d*x*arccos(a*x) - (d*x^2 + c)*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)))/(c*d^2*x^2 + c^
2*d)]

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giac [A]  time = 1.36, size = 75, normalized size = 1.14 \[ \frac {x \arccos \left (a x\right )}{\sqrt {d x^{2} + c} c} + \frac {a \log \left ({\left | -\sqrt {-a^{2} x^{2} + 1} \sqrt {-d} + \sqrt {a^{2} c + {\left (a^{2} x^{2} - 1\right )} d + d} \right |}\right )}{c \sqrt {-d} {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)/(d*x^2+c)^(3/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.43, size = 0, normalized size = 0.00 \[ \int \frac {\arccos \left (a x \right )}{\left (d \,x^{2}+c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(a*x)/(d*x^2+c)^(3/2),x)

[Out]

int(arccos(a*x)/(d*x^2+c)^(3/2),x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)/(d*x^2+c)^(3/2),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 details)Is d-a^2*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {acos}\left (a\,x\right )}{{\left (d\,x^2+c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acos(a*x)/(c + d*x^2)^(3/2),x)

[Out]

int(acos(a*x)/(c + d*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acos}{\left (a x \right )}}{\left (c + d x^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(a*x)/(d*x**2+c)**(3/2),x)

[Out]

Integral(acos(a*x)/(c + d*x**2)**(3/2), x)

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