3.213 \(\int \frac{(a+b \cos ^{-1}(c x))^2}{(d x)^{3/2}} \, dx\)

Optimal. Leaf size=105 \[ -\frac{16 b^2 c^2 (d x)^{3/2} \text{HypergeometricPFQ}\left (\left \{\frac{3}{4},\frac{3}{4},1\right \},\left \{\frac{5}{4},\frac{7}{4}\right \},c^2 x^2\right )}{3 d^3}-\frac{8 b c \sqrt{d x} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},c^2 x^2\right ) \left (a+b \cos ^{-1}(c x)\right )}{d^2}-\frac{2 \left (a+b \cos ^{-1}(c x)\right )^2}{d \sqrt{d x}} \]

[Out]

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

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Rubi [A]  time = 0.130147, antiderivative size = 105, normalized size of antiderivative = 1., 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 = {4628, 4712} \[ -\frac{16 b^2 c^2 (d x)^{3/2} \, _3F_2\left (\frac{3}{4},\frac{3}{4},1;\frac{5}{4},\frac{7}{4};c^2 x^2\right )}{3 d^3}-\frac{8 b c \sqrt{d x} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};c^2 x^2\right ) \left (a+b \cos ^{-1}(c x)\right )}{d^2}-\frac{2 \left (a+b \cos ^{-1}(c x)\right )^2}{d \sqrt{d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 4712

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

Rubi steps

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

Mathematica [A]  time = 0.432541, size = 129, normalized size = 1.23 \[ \frac{x \left (-\frac{\sqrt{2} \pi b^2 c^2 x^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{4},\frac{3}{4},1\right \},\left \{\frac{5}{4},\frac{7}{4}\right \},c^2 x^2\right )}{\text{Gamma}\left (\frac{5}{4}\right ) \text{Gamma}\left (\frac{7}{4}\right )}-2 \left (4 a b c x \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},c^2 x^2\right )+2 b^2 \cos ^{-1}(c x) \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{5}{4},c^2 x^2\right ) \sin \left (2 \cos ^{-1}(c x)\right )+\left (a+b \cos ^{-1}(c x)\right )^2\right )\right )}{(d x)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(-((Sqrt[2]*b^2*c^2*Pi*x^2*HypergeometricPFQ[{3/4, 3/4, 1}, {5/4, 7/4}, c^2*x^2])/(Gamma[5/4]*Gamma[7/4]))
- 2*((a + b*ArcCos[c*x])^2 + 4*a*b*c*x*Hypergeometric2F1[1/4, 1/2, 5/4, c^2*x^2] + 2*b^2*ArcCos[c*x]*Hypergeom
etric2F1[3/4, 1, 5/4, c^2*x^2]*Sin[2*ArcCos[c*x]])))/(d*x)^(3/2)

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Maple [F]  time = 0.339, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\arccos \left ( cx \right ) \right ) ^{2} \left ( dx \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b^{2} \arccos \left (c x\right )^{2} + 2 \, a b \arccos \left (c x\right ) + a^{2}\right )} \sqrt{d x}}{d^{2} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arccos \left (c x\right ) + a\right )}^{2}}{\left (d x\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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