3.121 \(\int (b x)^m \cos ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=150 \[ \frac{2 a^2 (b x)^{m+3} \text{HypergeometricPFQ}\left (\left \{1,\frac{m}{2}+\frac{3}{2},\frac{m}{2}+\frac{3}{2}\right \},\left \{\frac{m}{2}+2,\frac{m}{2}+\frac{5}{2}\right \},a^2 x^2\right )}{b^3 (m+1) (m+2) (m+3)}+\frac{2 a \cos ^{-1}(a x) (b x)^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},a^2 x^2\right )}{b^2 (m+1) (m+2)}+\frac{\cos ^{-1}(a x)^2 (b x)^{m+1}}{b (m+1)} \]

[Out]

((b*x)^(1 + m)*ArcCos[a*x]^2)/(b*(1 + m)) + (2*a*(b*x)^(2 + m)*ArcCos[a*x]*Hypergeometric2F1[1/2, (2 + m)/2, (
4 + m)/2, a^2*x^2])/(b^2*(1 + m)*(2 + m)) + (2*a^2*(b*x)^(3 + m)*HypergeometricPFQ[{1, 3/2 + m/2, 3/2 + m/2},
{2 + m/2, 5/2 + m/2}, a^2*x^2])/(b^3*(1 + m)*(2 + m)*(3 + m))

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Rubi [A]  time = 0.107421, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4628, 4712} \[ \frac{2 a^2 (b x)^{m+3} \, _3F_2\left (1,\frac{m}{2}+\frac{3}{2},\frac{m}{2}+\frac{3}{2};\frac{m}{2}+2,\frac{m}{2}+\frac{5}{2};a^2 x^2\right )}{b^3 (m+1) (m+2) (m+3)}+\frac{2 a \cos ^{-1}(a x) (b x)^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{b^2 (m+1) (m+2)}+\frac{\cos ^{-1}(a x)^2 (b x)^{m+1}}{b (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(b*x)^m*ArcCos[a*x]^2,x]

[Out]

((b*x)^(1 + m)*ArcCos[a*x]^2)/(b*(1 + m)) + (2*a*(b*x)^(2 + m)*ArcCos[a*x]*Hypergeometric2F1[1/2, (2 + m)/2, (
4 + m)/2, a^2*x^2])/(b^2*(1 + m)*(2 + m)) + (2*a^2*(b*x)^(3 + m)*HypergeometricPFQ[{1, 3/2 + m/2, 3/2 + m/2},
{2 + m/2, 5/2 + m/2}, a^2*x^2])/(b^3*(1 + m)*(2 + m)*(3 + m))

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

Mathematica [C]  time = 2.17981, size = 132, normalized size = 0.88 \[ \frac{x (b x)^m \left (a x \left (\sqrt{\pi } a 2^{-m} x \text{Gamma}(m+2) \, _3\tilde{F}_2\left (1,\frac{m+3}{2},\frac{m+3}{2};\frac{m+4}{2},\frac{m+5}{2};a^2 x^2\right )+\frac{8 \sqrt{1-a^2 x^2} \cos ^{-1}(a x) \text{Hypergeometric2F1}\left (1,\frac{m+3}{2},\frac{m+4}{2},a^2 x^2\right )}{m+2}\right )+4 \cos ^{-1}(a x)^2\right )}{4 (m+1)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(b*x)^m*ArcCos[a*x]^2,x]

[Out]

(x*(b*x)^m*(4*ArcCos[a*x]^2 + a*x*((8*Sqrt[1 - a^2*x^2]*ArcCos[a*x]*Hypergeometric2F1[1, (3 + m)/2, (4 + m)/2,
a^2*x^2])/(2 + m) + (a*Sqrt[Pi]*x*Gamma[2 + m]*HypergeometricPFQRegularized[{1, (3 + m)/2, (3 + m)/2}, {(4 +
m)/2, (5 + m)/2}, a^2*x^2])/2^m)))/(4*(1 + m))

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Maple [F]  time = 0.808, size = 0, normalized size = 0. \begin{align*} \int \left ( bx \right ) ^{m} \left ( \arccos \left ( ax \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x)^m*arccos(a*x)^2,x)

[Out]

int((b*x)^m*arccos(a*x)^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((b*x)^m*arccos(a*x)^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 (\left (b x\right )^{m} \arccos \left (a x\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x)^m*arccos(a*x)^2,x, algorithm="fricas")

[Out]

integral((b*x)^m*arccos(a*x)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b x\right )^{m} \operatorname{acos}^{2}{\left (a x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x)**m*acos(a*x)**2,x)

[Out]

Integral((b*x)**m*acos(a*x)**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x)^m*arccos(a*x)^2,x, algorithm="giac")

[Out]

integrate((b*x)^m*arccos(a*x)^2, x)