3.3.4 \(\int (d x)^{3/2} (a+b \text {ArcCos}(c x)) \, dx\) [204]

Optimal. Leaf size=124 \[ -\frac {4 b (d x)^{3/2} \sqrt {1-c^2 x^2}}{25 c}+\frac {2 (d x)^{5/2} (a+b \text {ArcCos}(c x))}{5 d}+\frac {12 b d^{3/2} E\left (\left .\text {ArcSin}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{25 c^{5/2}}-\frac {12 b d^{3/2} F\left (\left .\text {ArcSin}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{25 c^{5/2}} \]

[Out]

2/5*(d*x)^(5/2)*(a+b*arccos(c*x))/d+12/25*b*d^(3/2)*EllipticE(c^(1/2)*(d*x)^(1/2)/d^(1/2),I)/c^(5/2)-12/25*b*d
^(3/2)*EllipticF(c^(1/2)*(d*x)^(1/2)/d^(1/2),I)/c^(5/2)-4/25*b*(d*x)^(3/2)*(-c^2*x^2+1)^(1/2)/c

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 124, 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 = {4724, 327, 335, 313, 227, 1213, 435} \begin {gather*} \frac {2 (d x)^{5/2} (a+b \text {ArcCos}(c x))}{5 d}-\frac {12 b d^{3/2} F\left (\left .\text {ArcSin}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{25 c^{5/2}}+\frac {12 b d^{3/2} E\left (\left .\text {ArcSin}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{25 c^{5/2}}-\frac {4 b \sqrt {1-c^2 x^2} (d x)^{3/2}}{25 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*(d*x)^(3/2)*Sqrt[1 - c^2*x^2])/(25*c) + (2*(d*x)^(5/2)*(a + b*ArcCos[c*x]))/(5*d) + (12*b*d^(3/2)*Ellipt
icE[ArcSin[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]], -1])/(25*c^(5/2)) - (12*b*d^(3/2)*EllipticF[ArcSin[(Sqrt[c]*Sqrt[d*x]
)/Sqrt[d]], -1])/(25*c^(5/2))

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 313

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 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 435

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

Rule 1213

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 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 (d x)^{3/2} \left (a+b \cos ^{-1}(c x)\right ) \, dx &=\frac {2 (d x)^{5/2} \left (a+b \cos ^{-1}(c x)\right )}{5 d}+\frac {(2 b c) \int \frac {(d x)^{5/2}}{\sqrt {1-c^2 x^2}} \, dx}{5 d}\\ &=-\frac {4 b (d x)^{3/2} \sqrt {1-c^2 x^2}}{25 c}+\frac {2 (d x)^{5/2} \left (a+b \cos ^{-1}(c x)\right )}{5 d}+\frac {(6 b d) \int \frac {\sqrt {d x}}{\sqrt {1-c^2 x^2}} \, dx}{25 c}\\ &=-\frac {4 b (d x)^{3/2} \sqrt {1-c^2 x^2}}{25 c}+\frac {2 (d x)^{5/2} \left (a+b \cos ^{-1}(c x)\right )}{5 d}+\frac {(12 b) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-\frac {c^2 x^4}{d^2}}} \, dx,x,\sqrt {d x}\right )}{25 c}\\ &=-\frac {4 b (d x)^{3/2} \sqrt {1-c^2 x^2}}{25 c}+\frac {2 (d x)^{5/2} \left (a+b \cos ^{-1}(c x)\right )}{5 d}-\frac {(12 b d) \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {c^2 x^4}{d^2}}} \, dx,x,\sqrt {d x}\right )}{25 c^2}+\frac {(12 b d) \text {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 )}{25 c^2}\\ &=-\frac {4 b (d x)^{3/2} \sqrt {1-c^2 x^2}}{25 c}+\frac {2 (d x)^{5/2} \left (a+b \cos ^{-1}(c x)\right )}{5 d}-\frac {12 b d^{3/2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{25 c^{5/2}}+\frac {(12 b d) \text {Subst}\left (\int \frac {\sqrt {1+\frac {c x^2}{d}}}{\sqrt {1-\frac {c x^2}{d}}} \, dx,x,\sqrt {d x}\right )}{25 c^2}\\ &=-\frac {4 b (d x)^{3/2} \sqrt {1-c^2 x^2}}{25 c}+\frac {2 (d x)^{5/2} \left (a+b \cos ^{-1}(c x)\right )}{5 d}+\frac {12 b d^{3/2} E\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{25 c^{5/2}}-\frac {12 b d^{3/2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{25 c^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.05, size = 66, normalized size = 0.53 \begin {gather*} \frac {2 (d x)^{3/2} \left (5 a c x-2 b \sqrt {1-c^2 x^2}+5 b c x \text {ArcCos}(c x)+2 b \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};c^2 x^2\right )\right )}{25 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*x)^(3/2)*(5*a*c*x - 2*b*Sqrt[1 - c^2*x^2] + 5*b*c*x*ArcCos[c*x] + 2*b*Hypergeometric2F1[1/2, 3/4, 7/4, c
^2*x^2]))/(25*c)

________________________________________________________________________________________

Maple [A]
time = 0.01, size = 138, normalized size = 1.11

method	result	size
derivativedivides	\(\frac {\frac {2 \left (d x \right )^{\frac {5}{2}} a}{5}+2 b \left (\frac {\left (d x \right )^{\frac {5}{2}} \arccos \left (c x \right )}{5}+\frac {2 c \left (-\frac {d^{2} \left (d x \right )^{\frac {3}{2}} \sqrt {-c^{2} x^{2}+1}}{5 c^{2}}-\frac {3 d^{3} \sqrt {-c x +1}\, \sqrt {c x +1}\, \left (\EllipticF \left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )-\EllipticE \left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )\right )}{5 c^{3} \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{5 d}\right )}{d}\)	\(138\)
default	\(\frac {\frac {2 \left (d x \right )^{\frac {5}{2}} a}{5}+2 b \left (\frac {\left (d x \right )^{\frac {5}{2}} \arccos \left (c x \right )}{5}+\frac {2 c \left (-\frac {d^{2} \left (d x \right )^{\frac {3}{2}} \sqrt {-c^{2} x^{2}+1}}{5 c^{2}}-\frac {3 d^{3} \sqrt {-c x +1}\, \sqrt {c x +1}\, \left (\EllipticF \left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )-\EllipticE \left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )\right )}{5 c^{3} \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{5 d}\right )}{d}\)	\(138\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(3/2)*(a+b*arccos(c*x)),x,method=_RETURNVERBOSE)

[Out]

2/d*(1/5*(d*x)^(5/2)*a+b*(1/5*(d*x)^(5/2)*arccos(c*x)+2/5*c/d*(-1/5/c^2*d^2*(d*x)^(3/2)*(-c^2*x^2+1)^(1/2)-3/5
/c^3*d^3/(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)-Ell
ipticE((d*x)^(1/2)*(c/d)^(1/2),I)))))

________________________________________________________________________________________

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

[Out]

1/25*(10*b*c^3*d^(3/2)*x^(5/2)*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) - (4*b*c^3*d*x^(5/2) + 50*b*c^4*d*in
tegrate(1/5*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^(5/2)/(c^2*x^2 - 1), x) + 20*b*c*d*sqrt(x) - 5*(2*b*d*arctan(sqrt(c
)*sqrt(x)) - b*d*log((c*x - 1)/(c*x + 2*sqrt(c)*sqrt(x) + 1)))*sqrt(c))*sqrt(d))/c^3

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.99, size = 84, normalized size = 0.68 \begin {gather*} \frac {2 \, {\left (6 \, \sqrt {-c^{2} d} b d {\rm weierstrassZeta}\left (\frac {4}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {4}{c^{2}}, 0, x\right )\right ) + {\left (5 \, b c^{3} d x^{2} \arccos \left (c x\right ) + 5 \, a c^{3} d x^{2} - 2 \, \sqrt {-c^{2} x^{2} + 1} b c^{2} d x\right )} \sqrt {d x}\right )}}{25 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/25*(6*sqrt(-c^2*d)*b*d*weierstrassZeta(4/c^2, 0, weierstrassPInverse(4/c^2, 0, x)) + (5*b*c^3*d*x^2*arccos(c
*x) + 5*a*c^3*d*x^2 - 2*sqrt(-c^2*x^2 + 1)*b*c^2*d*x)*sqrt(d*x))/c^3

________________________________________________________________________________________

Sympy [A]
time = 12.94, size = 82, normalized size = 0.66 \begin {gather*} a \left (\begin {cases} 0 & \text {for}\: d = 0 \\\frac {2 \left (d x\right )^{\frac {5}{2}}}{5 d} & \text {otherwise} \end {cases}\right ) + b c \left (\begin {cases} 0 & \text {for}\: d = 0 \\\frac {d^{\frac {3}{2}} x^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {c^{2} x^{2} e^{2 i \pi }} \right )}}{5 \Gamma \left (\frac {11}{4}\right )} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} 0 & \text {for}\: d = 0 \\\frac {2 \left (d x\right )^{\frac {5}{2}}}{5 d} & \text {otherwise} \end {cases}\right ) \operatorname {acos}{\left (c x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((0, Eq(d, 0)), (2*(d*x)**(5/2)/(5*d), True)) + b*c*Piecewise((0, Eq(d, 0)), (d**(3/2)*x**(7/2)*gam
ma(7/4)*hyper((1/2, 7/4), (11/4,), c**2*x**2*exp_polar(2*I*pi))/(5*gamma(11/4)), True)) + b*Piecewise((0, Eq(d
, 0)), (2*(d*x)**(5/2)/(5*d), True))*acos(c*x)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (a+b\,\mathrm {acos}\left (c\,x\right )\right )\,{\left (d\,x\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(c*x))*(d*x)^(3/2),x)

[Out]

int((a + b*acos(c*x))*(d*x)^(3/2), x)

________________________________________________________________________________________