3.3.5 \(\int \sqrt {d x} (a+b \text {ArcCos}(c x)) \, dx\) [205]

Optimal. Leaf size=88 \[ -\frac {4 b \sqrt {d x} \sqrt {1-c^2 x^2}}{9 c}+\frac {2 (d x)^{3/2} (a+b \text {ArcCos}(c x))}{3 d}+\frac {4 b \sqrt {d} F\left (\left .\text {ArcSin}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{9 c^{3/2}} \]

[Out]

2/3*(d*x)^(3/2)*(a+b*arccos(c*x))/d+4/9*b*EllipticF(c^(1/2)*(d*x)^(1/2)/d^(1/2),I)*d^(1/2)/c^(3/2)-4/9*b*(d*x)
^(1/2)*(-c^2*x^2+1)^(1/2)/c

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Rubi [A]
time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4724, 327, 335, 227} \begin {gather*} \frac {2 (d x)^{3/2} (a+b \text {ArcCos}(c x))}{3 d}+\frac {4 b \sqrt {d} F\left (\left .\text {ArcSin}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{9 c^{3/2}}-\frac {4 b \sqrt {1-c^2 x^2} \sqrt {d x}}{9 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*x]*(a + b*ArcCos[c*x]),x]

[Out]

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

Rule 227

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 327

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

Rule 335

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 4724

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 \sqrt {d x} \left (a+b \cos ^{-1}(c x)\right ) \, dx &=\frac {2 (d x)^{3/2} \left (a+b \cos ^{-1}(c x)\right )}{3 d}+\frac {(2 b c) \int \frac {(d x)^{3/2}}{\sqrt {1-c^2 x^2}} \, dx}{3 d}\\ &=-\frac {4 b \sqrt {d x} \sqrt {1-c^2 x^2}}{9 c}+\frac {2 (d x)^{3/2} \left (a+b \cos ^{-1}(c x)\right )}{3 d}+\frac {(2 b d) \int \frac {1}{\sqrt {d x} \sqrt {1-c^2 x^2}} \, dx}{9 c}\\ &=-\frac {4 b \sqrt {d x} \sqrt {1-c^2 x^2}}{9 c}+\frac {2 (d x)^{3/2} \left (a+b \cos ^{-1}(c x)\right )}{3 d}+\frac {(4 b) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {c^2 x^4}{d^2}}} \, dx,x,\sqrt {d x}\right )}{9 c}\\ &=-\frac {4 b \sqrt {d x} \sqrt {1-c^2 x^2}}{9 c}+\frac {2 (d x)^{3/2} \left (a+b \cos ^{-1}(c x)\right )}{3 d}+\frac {4 b \sqrt {d} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{9 c^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.13, size = 113, normalized size = 1.28 \begin {gather*} \frac {2}{9} \sqrt {d x} \left (3 a x-\frac {2 b \sqrt {1-c^2 x^2}}{c}+3 b x \text {ArcCos}(c x)-\frac {2 i b \sqrt {-\frac {1}{c}} \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {x} F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {-\frac {1}{c}}}{\sqrt {x}}\right )\right |-1\right )}{\sqrt {1-c^2 x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*x]*(a + b*ArcCos[c*x]),x]

[Out]

(2*Sqrt[d*x]*(3*a*x - (2*b*Sqrt[1 - c^2*x^2])/c + 3*b*x*ArcCos[c*x] - ((2*I)*b*Sqrt[-c^(-1)]*Sqrt[1 - 1/(c^2*x
^2)]*Sqrt[x]*EllipticF[I*ArcSinh[Sqrt[-c^(-1)]/Sqrt[x]], -1])/Sqrt[1 - c^2*x^2]))/9

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Maple [A]
time = 0.01, size = 119, normalized size = 1.35

method	result	size
derivativedivides	\(\frac {\frac {2 \left (d x \right )^{\frac {3}{2}} a}{3}+2 b \left (\frac {\left (d x \right )^{\frac {3}{2}} \arccos \left (c x \right )}{3}+\frac {2 c \left (-\frac {d^{2} \sqrt {d x}\, \sqrt {-c^{2} x^{2}+1}}{3 c^{2}}+\frac {d^{2} \sqrt {-c x +1}\, \sqrt {c x +1}\, \EllipticF \left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )}{3 c^{2} \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{3 d}\right )}{d}\)	\(119\)
default	\(\frac {\frac {2 \left (d x \right )^{\frac {3}{2}} a}{3}+2 b \left (\frac {\left (d x \right )^{\frac {3}{2}} \arccos \left (c x \right )}{3}+\frac {2 c \left (-\frac {d^{2} \sqrt {d x}\, \sqrt {-c^{2} x^{2}+1}}{3 c^{2}}+\frac {d^{2} \sqrt {-c x +1}\, \sqrt {c x +1}\, \EllipticF \left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )}{3 c^{2} \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{3 d}\right )}{d}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccos(c*x))*(d*x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccos(c*x))*(d*x)^(1/2),x, algorithm="maxima")

[Out]

1/9*(6*b*c^2*sqrt(d)*x^(3/2)*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) - (18*b*c^3*integrate(1/3*sqrt(c*x + 1
)*sqrt(-c*x + 1)*x^(3/2)/(c^2*x^2 - 1), x) + 4*b*c^2*x^(3/2) + 3*(2*b*arctan(sqrt(c)*sqrt(x)) + b*log((c*x - 1
)/(c*x + 2*sqrt(c)*sqrt(x) + 1)))*sqrt(c))*sqrt(d))/c^2

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.73, size = 69, normalized size = 0.78 \begin {gather*} -\frac {2 \, {\left (2 \, \sqrt {-c^{2} d} b {\rm weierstrassPInverse}\left (\frac {4}{c^{2}}, 0, x\right ) - {\left (3 \, b c^{3} x \arccos \left (c x\right ) + 3 \, a c^{3} x - 2 \, \sqrt {-c^{2} x^{2} + 1} b c^{2}\right )} \sqrt {d x}\right )}}{9 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccos(c*x))*(d*x)^(1/2),x, algorithm="fricas")

[Out]

-2/9*(2*sqrt(-c^2*d)*b*weierstrassPInverse(4/c^2, 0, x) - (3*b*c^3*x*arccos(c*x) + 3*a*c^3*x - 2*sqrt(-c^2*x^2
+ 1)*b*c^2)*sqrt(d*x))/c^3

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Sympy [A]
time = 1.65, size = 76, normalized size = 0.86 \begin {gather*} \frac {2 a \left (d x\right )^{\frac {3}{2}}}{3 d} + \frac {b c \left (d x\right )^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {c^{2} x^{2} e^{2 i \pi }} \right )}}{3 d^{2} \Gamma \left (\frac {9}{4}\right )} + \frac {2 b \left (d x\right )^{\frac {3}{2}} \operatorname {acos}{\left (c x \right )}}{3 d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acos(c*x))*(d*x)**(1/2),x)

[Out]

2*a*(d*x)**(3/2)/(3*d) + b*c*(d*x)**(5/2)*gamma(5/4)*hyper((1/2, 5/4), (9/4,), c**2*x**2*exp_polar(2*I*pi))/(3
*d**2*gamma(9/4)) + 2*b*(d*x)**(3/2)*acos(c*x)/(3*d)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccos(c*x))*(d*x)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(d*x)*(b*arccos(c*x) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(c*x))*(d*x)^(1/2),x)

[Out]

int((a + b*acos(c*x))*(d*x)^(1/2), x)

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