3.3.11 \(\int \sqrt {d x} (a+b \text {ArcCos}(c x))^2 \, dx\) [211]

Optimal. Leaf size=109 \[ \frac {2 (d x)^{3/2} (a+b \text {ArcCos}(c x))^2}{3 d}+\frac {8 b c (d x)^{5/2} (a+b \text {ArcCos}(c x)) \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};c^2 x^2\right )}{15 d^2}+\frac {16 b^2 c^2 (d x)^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )}{105 d^3} \]

[Out]

2/3*(d*x)^(3/2)*(a+b*arccos(c*x))^2/d+8/15*b*c*(d*x)^(5/2)*(a+b*arccos(c*x))*hypergeom([1/2, 5/4],[9/4],c^2*x^
2)/d^2+16/105*b^2*c^2*(d*x)^(7/2)*hypergeom([1, 7/4, 7/4],[9/4, 11/4],c^2*x^2)/d^3

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Rubi [A]
time = 0.10, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4724, 4806} \begin {gather*} \frac {16 b^2 c^2 (d x)^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )}{105 d^3}+\frac {8 b c (d x)^{5/2} \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};c^2 x^2\right ) (a+b \text {ArcCos}(c x))}{15 d^2}+\frac {2 (d x)^{3/2} (a+b \text {ArcCos}(c x))^2}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(d*x)^(3/2)*(a + b*ArcCos[c*x])^2)/(3*d) + (8*b*c*(d*x)^(5/2)*(a + b*ArcCos[c*x])*Hypergeometric2F1[1/2, 5/
4, 9/4, c^2*x^2])/(15*d^2) + (16*b^2*c^2*(d*x)^(7/2)*HypergeometricPFQ[{1, 7/4, 7/4}, {9/4, 11/4}, c^2*x^2])/(
105*d^3)

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]

Rule 4806

Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x)
^(m + 1)/(f*(m + 1)))*(a + b*ArcCos[c*x])*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*Hypergeometric2F1[1/2, (1 +
m)/2, (3 + m)/2, c^2*x^2], x] + Simp[b*c*((f*x)^(m + 2)/(f^2*(m + 1)*(m + 2)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d +
e*x^2]]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2], x] /; FreeQ[{a, b, c, d, e,
f, m}, x] && EqQ[c^2*d + e, 0] &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int \sqrt {d x} \left (a+b \cos ^{-1}(c x)\right )^2 \, dx &=\frac {2 (d x)^{3/2} \left (a+b \cos ^{-1}(c x)\right )^2}{3 d}+\frac {(4 b c) \int \frac {(d x)^{3/2} \left (a+b \cos ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{3 d}\\ &=\frac {2 (d x)^{3/2} \left (a+b \cos ^{-1}(c x)\right )^2}{3 d}+\frac {8 b c (d x)^{5/2} \left (a+b \cos ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {5}{4};\frac {9}{4};c^2 x^2\right )}{15 d^2}+\frac {16 b^2 c^2 (d x)^{7/2} \, _3F_2\left (1,\frac {7}{4},\frac {7}{4};\frac {9}{4},\frac {11}{4};c^2 x^2\right )}{105 d^3}\\ \end {align*}

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Mathematica [A]
time = 10.39, size = 202, normalized size = 1.85 \begin {gather*} \frac {1}{27} \sqrt {d x} \left (\frac {2 \left (9 a^2 c x-8 b^2 c x-12 a b \sqrt {1-c^2 x^2}+18 a b c x \text {ArcCos}(c x)-12 b^2 \sqrt {1-c^2 x^2} \text {ArcCos}(c x)+9 b^2 c x \text {ArcCos}(c x)^2+12 a b \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};c^2 x^2\right )+12 b^2 \sqrt {1-c^2 x^2} \text {ArcCos}(c x) \, _2F_1\left (\frac {3}{4},1;\frac {5}{4};c^2 x^2\right )\right )}{c}+\frac {3 \sqrt {2} b^2 \pi x \, _3F_2\left (\frac {3}{4},\frac {3}{4},1;\frac {5}{4},\frac {7}{4};c^2 x^2\right )}{\text {Gamma}\left (\frac {5}{4}\right ) \text {Gamma}\left (\frac {7}{4}\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d*x]*((2*(9*a^2*c*x - 8*b^2*c*x - 12*a*b*Sqrt[1 - c^2*x^2] + 18*a*b*c*x*ArcCos[c*x] - 12*b^2*Sqrt[1 - c^
2*x^2]*ArcCos[c*x] + 9*b^2*c*x*ArcCos[c*x]^2 + 12*a*b*Hypergeometric2F1[1/4, 1/2, 5/4, c^2*x^2] + 12*b^2*Sqrt[
1 - c^2*x^2]*ArcCos[c*x]*Hypergeometric2F1[3/4, 1, 5/4, c^2*x^2]))/c + (3*Sqrt[2]*b^2*Pi*x*HypergeometricPFQ[{
3/4, 3/4, 1}, {5/4, 7/4}, c^2*x^2])/(Gamma[5/4]*Gamma[7/4])))/27

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Maple [F]
time = 0.18, size = 0, normalized size = 0.00 \[\int \left (a +b \arccos \left (c x \right )\right )^{2} \sqrt {d x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))^2*(d*x)^(1/2),x, algorithm="maxima")

[Out]

2/3*b^2*sqrt(d)*x^(3/2)*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)^2 + 1/6*a^2*c^2*sqrt(d)*(4*x^(3/2)/c^2 + 6*
arctan(sqrt(c)*sqrt(x))/c^(7/2) + 3*log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqrt(c)))/c^(7/2)) + 6*a*b*c^2*sqrt
(d)*integrate(1/3*x^(5/2)*arctan(sqrt(c*x + 1)*sqrt(-c*x + 1)/(c*x))/(c^2*x^2 - 1), x) - 4*b^2*c*sqrt(d)*integ
rate(1/3*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^(3/2)*arctan(sqrt(c*x + 1)*sqrt(-c*x + 1)/(c*x))/(c^2*x^2 - 1), x) - 1
/2*a^2*sqrt(d)*(2*arctan(sqrt(c)*sqrt(x))/c^(3/2) + log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqrt(c)))/c^(3/2))
- 6*a*b*sqrt(d)*integrate(1/3*sqrt(x)*arctan(sqrt(c*x + 1)*sqrt(-c*x + 1)/(c*x))/(c^2*x^2 - 1), x)

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Fricas [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))^2*(d*x)^(1/2),x, algorithm="fricas")

[Out]

integral((b^2*arccos(c*x)^2 + 2*a*b*arccos(c*x) + a^2)*sqrt(d*x), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {d x} \left (a + b \operatorname {acos}{\left (c x \right )}\right )^{2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(d*x)*(a + b*acos(c*x))**2, x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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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 )}^2\,\sqrt {d\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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