3.33 \(\int \frac{\cos ^{-1}(a x)}{(c+d x^2)^{7/2}} \, dx\)

Optimal. Leaf size=211 \[ -\frac{2 a \sqrt{1-a^2 x^2} \left (3 a^2 c+2 d\right )}{15 c^2 \left (a^2 c+d\right )^2 \sqrt{c+d x^2}}-\frac{8 \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{1-a^2 x^2}}{a \sqrt{c+d x^2}}\right )}{15 c^3 \sqrt{d}}-\frac{a \sqrt{1-a^2 x^2}}{15 c \left (a^2 c+d\right ) \left (c+d x^2\right )^{3/2}}+\frac{8 x \cos ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}+\frac{4 x \cos ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{x \cos ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}} \]

[Out]

-(a*Sqrt[1 - a^2*x^2])/(15*c*(a^2*c + d)*(c + d*x^2)^(3/2)) - (2*a*(3*a^2*c + 2*d)*Sqrt[1 - a^2*x^2])/(15*c^2*
(a^2*c + d)^2*Sqrt[c + d*x^2]) + (x*ArcCos[a*x])/(5*c*(c + d*x^2)^(5/2)) + (4*x*ArcCos[a*x])/(15*c^2*(c + d*x^
2)^(3/2)) + (8*x*ArcCos[a*x])/(15*c^3*Sqrt[c + d*x^2]) - (8*ArcTan[(Sqrt[d]*Sqrt[1 - a^2*x^2])/(a*Sqrt[c + d*x
^2])])/(15*c^3*Sqrt[d])

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Rubi [A]  time = 0.848433, antiderivative size = 211, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 10, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {192, 191, 4666, 12, 6715, 949, 78, 63, 217, 203} \[ -\frac{2 a \sqrt{1-a^2 x^2} \left (3 a^2 c+2 d\right )}{15 c^2 \left (a^2 c+d\right )^2 \sqrt{c+d x^2}}-\frac{8 \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{1-a^2 x^2}}{a \sqrt{c+d x^2}}\right )}{15 c^3 \sqrt{d}}-\frac{a \sqrt{1-a^2 x^2}}{15 c \left (a^2 c+d\right ) \left (c+d x^2\right )^{3/2}}+\frac{8 x \cos ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}+\frac{4 x \cos ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{x \cos ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcCos[a*x]/(c + d*x^2)^(7/2),x]

[Out]

-(a*Sqrt[1 - a^2*x^2])/(15*c*(a^2*c + d)*(c + d*x^2)^(3/2)) - (2*a*(3*a^2*c + 2*d)*Sqrt[1 - a^2*x^2])/(15*c^2*
(a^2*c + d)^2*Sqrt[c + d*x^2]) + (x*ArcCos[a*x])/(5*c*(c + d*x^2)^(5/2)) + (4*x*ArcCos[a*x])/(15*c^2*(c + d*x^
2)^(3/2)) + (8*x*ArcCos[a*x])/(15*c^3*Sqrt[c + d*x^2]) - (8*ArcTan[(Sqrt[d]*Sqrt[1 - a^2*x^2])/(a*Sqrt[c + d*x
^2])])/(15*c^3*Sqrt[d])

Rule 192

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 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 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 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 949

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{Qx = PolynomialQuotient[(a + b*x + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p,
d + e*x, x]}, Simp[(R*(d + e*x)^(m + 1)*(f + g*x)^(n + 1))/((m + 1)*(e*f - d*g)), x] + Dist[1/((m + 1)*(e*f -
d*g)), Int[(d + e*x)^(m + 1)*(f + g*x)^n*ExpandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), x], x], x]] /;
FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&& IGtQ[p, 0] && LtQ[m, -1]

Rule 78

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] || LtQ
[p, n]))))

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 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 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])

Rubi steps

\begin{align*} \int \frac{\cos ^{-1}(a x)}{\left (c+d x^2\right )^{7/2}} \, dx &=\frac{x \cos ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \cos ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \cos ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}+a \int \frac{x \left (15 c^2+20 c d x^2+8 d^2 x^4\right )}{15 c^3 \sqrt{1-a^2 x^2} \left (c+d x^2\right )^{5/2}} \, dx\\ &=\frac{x \cos ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \cos ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \cos ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}+\frac{a \int \frac{x \left (15 c^2+20 c d x^2+8 d^2 x^4\right )}{\sqrt{1-a^2 x^2} \left (c+d x^2\right )^{5/2}} \, dx}{15 c^3}\\ &=\frac{x \cos ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \cos ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \cos ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}+\frac{a \operatorname{Subst}\left (\int \frac{15 c^2+20 c d x+8 d^2 x^2}{\sqrt{1-a^2 x} (c+d x)^{5/2}} \, dx,x,x^2\right )}{30 c^3}\\ &=-\frac{a \sqrt{1-a^2 x^2}}{15 c \left (a^2 c+d\right ) \left (c+d x^2\right )^{3/2}}+\frac{x \cos ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \cos ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \cos ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}-\frac{a \operatorname{Subst}\left (\int \frac{-3 c \left (7 a^2 c+6 d\right )-12 d \left (a^2 c+d\right ) x}{\sqrt{1-a^2 x} (c+d x)^{3/2}} \, dx,x,x^2\right )}{45 c^3 \left (a^2 c+d\right )}\\ &=-\frac{a \sqrt{1-a^2 x^2}}{15 c \left (a^2 c+d\right ) \left (c+d x^2\right )^{3/2}}-\frac{2 a \left (3 a^2 c+2 d\right ) \sqrt{1-a^2 x^2}}{15 c^2 \left (a^2 c+d\right )^2 \sqrt{c+d x^2}}+\frac{x \cos ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \cos ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \cos ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}+\frac{(4 a) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-a^2 x} \sqrt{c+d x}} \, dx,x,x^2\right )}{15 c^3}\\ &=-\frac{a \sqrt{1-a^2 x^2}}{15 c \left (a^2 c+d\right ) \left (c+d x^2\right )^{3/2}}-\frac{2 a \left (3 a^2 c+2 d\right ) \sqrt{1-a^2 x^2}}{15 c^2 \left (a^2 c+d\right )^2 \sqrt{c+d x^2}}+\frac{x \cos ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \cos ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \cos ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}-\frac{8 \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 )}{15 a c^3}\\ &=-\frac{a \sqrt{1-a^2 x^2}}{15 c \left (a^2 c+d\right ) \left (c+d x^2\right )^{3/2}}-\frac{2 a \left (3 a^2 c+2 d\right ) \sqrt{1-a^2 x^2}}{15 c^2 \left (a^2 c+d\right )^2 \sqrt{c+d x^2}}+\frac{x \cos ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \cos ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \cos ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}-\frac{8 \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 )}{15 a c^3}\\ &=-\frac{a \sqrt{1-a^2 x^2}}{15 c \left (a^2 c+d\right ) \left (c+d x^2\right )^{3/2}}-\frac{2 a \left (3 a^2 c+2 d\right ) \sqrt{1-a^2 x^2}}{15 c^2 \left (a^2 c+d\right )^2 \sqrt{c+d x^2}}+\frac{x \cos ^{-1}(a x)}{5 c \left (c+d x^2\right )^{5/2}}+\frac{4 x \cos ^{-1}(a x)}{15 c^2 \left (c+d x^2\right )^{3/2}}+\frac{8 x \cos ^{-1}(a x)}{15 c^3 \sqrt{c+d x^2}}-\frac{8 \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{1-a^2 x^2}}{a \sqrt{c+d x^2}}\right )}{15 c^3 \sqrt{d}}\\ \end{align*}

Mathematica [C]  time = 0.308923, size = 162, normalized size = 0.77 \[ \frac{4 a x^2 \sqrt{\frac{d x^2}{c}+1} \left (c+d x^2\right )^2 F_1\left (1;\frac{1}{2},\frac{1}{2};2;a^2 x^2,-\frac{d x^2}{c}\right )-\frac{a c \sqrt{1-a^2 x^2} \left (c+d x^2\right ) \left (a^2 c \left (7 c+6 d x^2\right )+d \left (5 c+4 d x^2\right )\right )}{\left (a^2 c+d\right )^2}+x \cos ^{-1}(a x) \left (15 c^2+20 c d x^2+8 d^2 x^4\right )}{15 c^3 \left (c+d x^2\right )^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCos[a*x]/(c + d*x^2)^(7/2),x]

[Out]

(-((a*c*Sqrt[1 - a^2*x^2]*(c + d*x^2)*(d*(5*c + 4*d*x^2) + a^2*c*(7*c + 6*d*x^2)))/(a^2*c + d)^2) + 4*a*x^2*(c
+ d*x^2)^2*Sqrt[1 + (d*x^2)/c]*AppellF1[1, 1/2, 1/2, 2, a^2*x^2, -((d*x^2)/c)] + x*(15*c^2 + 20*c*d*x^2 + 8*d
^2*x^4)*ArcCos[a*x])/(15*c^3*(c + d*x^2)^(5/2))

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Maple [F]  time = 0.181, size = 0, normalized size = 0. \begin{align*} \int{\arccos \left ( ax \right ) \left ( d{x}^{2}+c \right ) ^{-{\frac{7}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccos(a*x)/(d*x^2+c)^(7/2),x)

[Out]

int(arccos(a*x)/(d*x^2+c)^(7/2),x)

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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(arccos(a*x)/(d*x^2+c)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.08224, size = 2145, normalized size = 10.17 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)/(d*x^2+c)^(7/2),x, algorithm="fricas")

[Out]

[-1/15*(2*(a^4*c^5 + 2*a^2*c^4*d + (a^4*c^2*d^3 + 2*a^2*c*d^4 + d^5)*x^6 + c^3*d^2 + 3*(a^4*c^3*d^2 + 2*a^2*c^
2*d^3 + c*d^4)*x^4 + 3*(a^4*c^4*d + 2*a^2*c^3*d^2 + c^2*d^3)*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) - sqrt(d*x^2 + c)*((8*(a^4*c^2*d^3 + 2*a^2*c*d^4 + d^5)*x^5 + 20*(a^4*c^3*d^2 + 2*a^2*c^2*d^3 + c*d^4)*x
^3 + 15*(a^4*c^4*d + 2*a^2*c^3*d^2 + c^2*d^3)*x)*arccos(a*x) - (7*a^3*c^4*d + 5*a*c^3*d^2 + 2*(3*a^3*c^2*d^3 +
2*a*c*d^4)*x^4 + (13*a^3*c^3*d^2 + 9*a*c^2*d^3)*x^2)*sqrt(-a^2*x^2 + 1)))/(a^4*c^8*d + 2*a^2*c^7*d^2 + c^6*d^
3 + (a^4*c^5*d^4 + 2*a^2*c^4*d^5 + c^3*d^6)*x^6 + 3*(a^4*c^6*d^3 + 2*a^2*c^5*d^4 + c^4*d^5)*x^4 + 3*(a^4*c^7*d
^2 + 2*a^2*c^6*d^3 + c^5*d^4)*x^2), -1/15*(4*(a^4*c^5 + 2*a^2*c^4*d + (a^4*c^2*d^3 + 2*a^2*c*d^4 + d^5)*x^6 +
c^3*d^2 + 3*(a^4*c^3*d^2 + 2*a^2*c^2*d^3 + c*d^4)*x^4 + 3*(a^4*c^4*d + 2*a^2*c^3*d^2 + c^2*d^3)*x^2)*sqrt(d)*a
rctan(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)*((8*(a^4*c^2*d^3 + 2*a^2*c*d^4 + d^5)*x^5 + 20*(a^4*c^3*d^2 + 2*a^2*c^2*d^3
+ c*d^4)*x^3 + 15*(a^4*c^4*d + 2*a^2*c^3*d^2 + c^2*d^3)*x)*arccos(a*x) - (7*a^3*c^4*d + 5*a*c^3*d^2 + 2*(3*a^3
*c^2*d^3 + 2*a*c*d^4)*x^4 + (13*a^3*c^3*d^2 + 9*a*c^2*d^3)*x^2)*sqrt(-a^2*x^2 + 1)))/(a^4*c^8*d + 2*a^2*c^7*d^
2 + c^6*d^3 + (a^4*c^5*d^4 + 2*a^2*c^4*d^5 + c^3*d^6)*x^6 + 3*(a^4*c^6*d^3 + 2*a^2*c^5*d^4 + c^4*d^5)*x^4 + 3*
(a^4*c^7*d^2 + 2*a^2*c^6*d^3 + c^5*d^4)*x^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acos(a*x)/(d*x**2+c)**(7/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccos(a*x)/(d*x^2+c)^(7/2),x, algorithm="giac")

[Out]

Timed out