3.3.12 \(\int \frac {(a+b \text {ArcCos}(c x))^2}{\sqrt {d x}} \, dx\) [212]

Optimal. Leaf size=107 \[ \frac {2 \sqrt {d x} (a+b \text {ArcCos}(c x))^2}{d}+\frac {8 b c (d x)^{3/2} (a+b \text {ArcCos}(c x)) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};c^2 x^2\right )}{3 d^2}+\frac {16 b^2 c^2 (d x)^{5/2} \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};c^2 x^2\right )}{15 d^3} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 107, 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)^{5/2} \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};c^2 x^2\right )}{15 d^3}+\frac {8 b c (d x)^{3/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};c^2 x^2\right ) (a+b \text {ArcCos}(c x))}{3 d^2}+\frac {2 \sqrt {d x} (a+b \text {ArcCos}(c x))^2}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d*x]*(a + b*ArcCos[c*x])^2)/d + (8*b*c*(d*x)^(3/2)*(a + b*ArcCos[c*x])*Hypergeometric2F1[1/2, 3/4, 7/4
, c^2*x^2])/(3*d^2) + (16*b^2*c^2*(d*x)^(5/2)*HypergeometricPFQ[{1, 5/4, 5/4}, {7/4, 9/4}, c^2*x^2])/(15*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 \frac {\left (a+b \cos ^{-1}(c x)\right )^2}{\sqrt {d x}} \, dx &=\frac {2 \sqrt {d x} \left (a+b \cos ^{-1}(c x)\right )^2}{d}+\frac {(4 b c) \int \frac {\sqrt {d x} \left (a+b \cos ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{d}\\ &=\frac {2 \sqrt {d x} \left (a+b \cos ^{-1}(c x)\right )^2}{d}+\frac {8 b c (d x)^{3/2} \left (a+b \cos ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};c^2 x^2\right )}{3 d^2}+\frac {16 b^2 c^2 (d x)^{5/2} \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};c^2 x^2\right )}{15 d^3}\\ \end {align*}

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Mathematica [A]
time = 10.93, size = 142, normalized size = 1.33 \begin {gather*} \frac {3 \sqrt {2} b^2 c^2 \pi x^3 \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};c^2 x^2\right )+8 x \text {Gamma}\left (\frac {7}{4}\right ) \text {Gamma}\left (\frac {9}{4}\right ) \left (3 (a+b \text {ArcCos}(c x))^2+4 a b c x \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};c^2 x^2\right )+2 b^2 \text {ArcCos}(c x) \, _2F_1\left (1,\frac {5}{4};\frac {7}{4};c^2 x^2\right ) \sin (2 \text {ArcCos}(c x))\right )}{12 \sqrt {d x} \text {Gamma}\left (\frac {7}{4}\right ) \text {Gamma}\left (\frac {9}{4}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[2]*b^2*c^2*Pi*x^3*HypergeometricPFQ[{1, 5/4, 5/4}, {7/4, 9/4}, c^2*x^2] + 8*x*Gamma[7/4]*Gamma[9/4]*(3
*(a + b*ArcCos[c*x])^2 + 4*a*b*c*x*Hypergeometric2F1[1/2, 3/4, 7/4, c^2*x^2] + 2*b^2*ArcCos[c*x]*Hypergeometri
c2F1[1, 5/4, 7/4, c^2*x^2]*Sin[2*ArcCos[c*x]]))/(12*Sqrt[d*x]*Gamma[7/4]*Gamma[9/4])

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arccos \left (c x \right )\right )^{2}}{\sqrt {d x}}\, dx\]

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]

1/2*(4*b^2*sqrt(x)*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)^2 + (a^2*c^2*sqrt(d)*(4*sqrt(x)/(c^2*d) - 2*arct
an(sqrt(c)*sqrt(x))/(c^(5/2)*d) + log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqrt(c)))/(c^(5/2)*d)) + 4*a*b*c^2*sq
rt(d)*integrate(x^(5/2)*arctan(sqrt(c*x + 1)*sqrt(-c*x + 1)/(c*x))/(c^2*d*x^3 - d*x), x) - 8*b^2*c*sqrt(d)*int
egrate(sqrt(c*x + 1)*sqrt(-c*x + 1)*x^(3/2)*arctan(sqrt(c*x + 1)*sqrt(-c*x + 1)/(c*x))/(c^2*d*x^3 - d*x), x) +
a^2*sqrt(d)*(2*arctan(sqrt(c)*sqrt(x))/(sqrt(c)*d) - log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqrt(c)))/(sqrt(c
)*d)) - 4*a*b*sqrt(d)*integrate(sqrt(x)*arctan(sqrt(c*x + 1)*sqrt(-c*x + 1)/(c*x))/(c^2*d*x^3 - d*x), x))*sqrt
(d))/sqrt(d)

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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)/(d*x), x)

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

Exception raised: TypeError >> Invalid comparison of non-real zoo

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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))^2/(d*x)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\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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