3.208 \(\int \frac {a+b \cos ^{-1}(c x)}{(d x)^{5/2}} \, dx\)

Optimal. Leaf size=125 \[ -\frac {2 \left (a+b \cos ^{-1}(c x)\right )}{3 d (d x)^{3/2}}-\frac {4 b c^{3/2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{3 d^{5/2}}+\frac {4 b c^{3/2} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{3 d^{5/2}}+\frac {4 b c \sqrt {1-c^2 x^2}}{3 d^2 \sqrt {d x}} \]

[Out]

-2/3*(a+b*arccos(c*x))/d/(d*x)^(3/2)+4/3*b*c^(3/2)*EllipticE(c^(1/2)*(d*x)^(1/2)/d^(1/2),I)/d^(5/2)-4/3*b*c^(3
/2)*EllipticF(c^(1/2)*(d*x)^(1/2)/d^(1/2),I)/d^(5/2)+4/3*b*c*(-c^2*x^2+1)^(1/2)/d^2/(d*x)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4628, 325, 329, 307, 221, 1199, 424} \[ -\frac {2 \left (a+b \cos ^{-1}(c x)\right )}{3 d (d x)^{3/2}}+\frac {4 b c \sqrt {1-c^2 x^2}}{3 d^2 \sqrt {d x}}-\frac {4 b c^{3/2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{3 d^{5/2}}+\frac {4 b c^{3/2} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{3 d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*c*Sqrt[1 - c^2*x^2])/(3*d^2*Sqrt[d*x]) - (2*(a + b*ArcCos[c*x]))/(3*d*(d*x)^(3/2)) + (4*b*c^(3/2)*Ellipti
cE[ArcSin[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]], -1])/(3*d^(5/2)) - (4*b*c^(3/2)*EllipticF[ArcSin[(Sqrt[c]*Sqrt[d*x])/S
qrt[d]], -1])/(3*d^(5/2))

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 307

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(b/a), 2]}, -Dist[q^(-1), Int[1/Sqrt[a + b*x^
4], x], x] + Dist[1/q, Int[(1 + q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 4628

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCo
s[c*x])^n)/(d*(m + 1)), x] + Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCos[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b \cos ^{-1}(c x)}{(d x)^{5/2}} \, dx &=-\frac {2 \left (a+b \cos ^{-1}(c x)\right )}{3 d (d x)^{3/2}}-\frac {(2 b c) \int \frac {1}{(d x)^{3/2} \sqrt {1-c^2 x^2}} \, dx}{3 d}\\ &=\frac {4 b c \sqrt {1-c^2 x^2}}{3 d^2 \sqrt {d x}}-\frac {2 \left (a+b \cos ^{-1}(c x)\right )}{3 d (d x)^{3/2}}+\frac {\left (2 b c^3\right ) \int \frac {\sqrt {d x}}{\sqrt {1-c^2 x^2}} \, dx}{3 d^3}\\ &=\frac {4 b c \sqrt {1-c^2 x^2}}{3 d^2 \sqrt {d x}}-\frac {2 \left (a+b \cos ^{-1}(c x)\right )}{3 d (d x)^{3/2}}+\frac {\left (4 b c^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {c^2 x^4}{d^2}}} \, dx,x,\sqrt {d x}\right )}{3 d^4}\\ &=\frac {4 b c \sqrt {1-c^2 x^2}}{3 d^2 \sqrt {d x}}-\frac {2 \left (a+b \cos ^{-1}(c x)\right )}{3 d (d x)^{3/2}}-\frac {\left (4 b c^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {c^2 x^4}{d^2}}} \, dx,x,\sqrt {d x}\right )}{3 d^3}+\frac {\left (4 b c^2\right ) \operatorname {Subst}\left (\int \frac {1+\frac {c x^2}{d}}{\sqrt {1-\frac {c^2 x^4}{d^2}}} \, dx,x,\sqrt {d x}\right )}{3 d^3}\\ &=\frac {4 b c \sqrt {1-c^2 x^2}}{3 d^2 \sqrt {d x}}-\frac {2 \left (a+b \cos ^{-1}(c x)\right )}{3 d (d x)^{3/2}}-\frac {4 b c^{3/2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{3 d^{5/2}}+\frac {\left (4 b c^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {c x^2}{d}}}{\sqrt {1-\frac {c x^2}{d}}} \, dx,x,\sqrt {d x}\right )}{3 d^3}\\ &=\frac {4 b c \sqrt {1-c^2 x^2}}{3 d^2 \sqrt {d x}}-\frac {2 \left (a+b \cos ^{-1}(c x)\right )}{3 d (d x)^{3/2}}+\frac {4 b c^{3/2} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{3 d^{5/2}}-\frac {4 b c^{3/2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{3 d^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 68, normalized size = 0.54 \[ \frac {2 x \left (2 b c^3 x^3 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};c^2 x^2\right )-3 \left (a-2 b c x \sqrt {1-c^2 x^2}+b \cos ^{-1}(c x)\right )\right )}{9 (d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(-3*(a - 2*b*c*x*Sqrt[1 - c^2*x^2] + b*ArcCos[c*x]) + 2*b*c^3*x^3*Hypergeometric2F1[1/2, 3/4, 7/4, c^2*x^
2]))/(9*(d*x)^(5/2))

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d x} {\left (b \arccos \left (c x\right ) + a\right )}}{d^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*x)*(b*arccos(c*x) + a)/(d^3*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 129, normalized size = 1.03 \[ \frac {-\frac {2 a}{3 \left (d x \right )^{\frac {3}{2}}}+2 b \left (-\frac {\arccos \left (c x \right )}{3 \left (d x \right )^{\frac {3}{2}}}-\frac {2 c \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{\sqrt {d x}}+\frac {c \sqrt {-c x +1}\, \sqrt {c x +1}\, \left (\EllipticF \left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )-\EllipticE \left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )\right )}{d \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{3 d}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(-1/3*a/(d*x)^(3/2)+b*(-1/3/(d*x)^(3/2)*arccos(c*x)-2/3*c/d*(-(-c^2*x^2+1)^(1/2)/(d*x)^(1/2)+c/d/(c/d)^(1/
2)*(-c*x+1)^(1/2)*(c*x+1)^(1/2)/(-c^2*x^2+1)^(1/2)*(EllipticF((d*x)^(1/2)*(c/d)^(1/2),I)-EllipticE((d*x)^(1/2)
*(c/d)^(1/2),I)))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, b \arctan \left (\sqrt {c x + 1} \sqrt {-c x + 1}, c x\right ) - {\left (2 \, b c d^{3} x \int \frac {\sqrt {c x + 1} \sqrt {-c x + 1} \sqrt {x}}{c^{2} d^{3} x^{4} - d^{3} x^{2}}\,{d x} + {\left (2 \, b c x \arctan \left (\frac {1}{\sqrt {c} \sqrt {x}}\right ) + b c x \log \left (-\frac {c x - 1}{c x + 2 \, \sqrt {c} \sqrt {x} + 1}\right )\right )} \sqrt {c}\right )} \sqrt {x}}{3 \, d^{\frac {5}{2}} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*b*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) - (6*b*c*d^3*x*integrate(1/3*sqrt(c*x + 1)*sqrt(-c*x + 1)
*sqrt(x)/(c^2*d^3*x^4 - d^3*x^2), x) + (2*b*c*x*arctan(1/(sqrt(c)*sqrt(x))) + b*c*x*log(-(c*x - 1)/(c*x + 2*sq
rt(c)*sqrt(x) + 1)))*sqrt(c))*sqrt(x))/(d^(5/2)*x^(3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(c*x))/(d*x)^(5/2),x)

[Out]

int((a + b*acos(c*x))/(d*x)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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