3.209 \(\int (d x)^{5/2} (a+b \cos ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=109 \[ \frac{16 b^2 c^2 (d x)^{11/2} \text{HypergeometricPFQ}\left (\left \{1,\frac{11}{4},\frac{11}{4}\right \},\left \{\frac{13}{4},\frac{15}{4}\right \},c^2 x^2\right )}{693 d^3}+\frac{8 b c (d x)^{9/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{9}{4},\frac{13}{4},c^2 x^2\right ) \left (a+b \cos ^{-1}(c x)\right )}{63 d^2}+\frac{2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )^2}{7 d} \]

[Out]

(2*(d*x)^(7/2)*(a + b*ArcCos[c*x])^2)/(7*d) + (8*b*c*(d*x)^(9/2)*(a + b*ArcCos[c*x])*Hypergeometric2F1[1/2, 9/
4, 13/4, c^2*x^2])/(63*d^2) + (16*b^2*c^2*(d*x)^(11/2)*HypergeometricPFQ[{1, 11/4, 11/4}, {13/4, 15/4}, c^2*x^
2])/(693*d^3)

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Rubi [A]  time = 0.13904, antiderivative size = 109, 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} \[ \frac{16 b^2 c^2 (d x)^{11/2} \, _3F_2\left (1,\frac{11}{4},\frac{11}{4};\frac{13}{4},\frac{15}{4};c^2 x^2\right )}{693 d^3}+\frac{8 b c (d x)^{9/2} \, _2F_1\left (\frac{1}{2},\frac{9}{4};\frac{13}{4};c^2 x^2\right ) \left (a+b \cos ^{-1}(c x)\right )}{63 d^2}+\frac{2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )^2}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(5/2)*(a + b*ArcCos[c*x])^2,x]

[Out]

(2*(d*x)^(7/2)*(a + b*ArcCos[c*x])^2)/(7*d) + (8*b*c*(d*x)^(9/2)*(a + b*ArcCos[c*x])*Hypergeometric2F1[1/2, 9/
4, 13/4, c^2*x^2])/(63*d^2) + (16*b^2*c^2*(d*x)^(11/2)*HypergeometricPFQ[{1, 11/4, 11/4}, {13/4, 15/4}, c^2*x^
2])/(693*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 (d x)^{5/2} \left (a+b \cos ^{-1}(c x)\right )^2 \, dx &=\frac{2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )^2}{7 d}+\frac{(4 b c) \int \frac{(d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )}{\sqrt{1-c^2 x^2}} \, dx}{7 d}\\ &=\frac{2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )^2}{7 d}+\frac{8 b c (d x)^{9/2} \left (a+b \cos ^{-1}(c x)\right ) \, _2F_1\left (\frac{1}{2},\frac{9}{4};\frac{13}{4};c^2 x^2\right )}{63 d^2}+\frac{16 b^2 c^2 (d x)^{11/2} \, _3F_2\left (1,\frac{11}{4},\frac{11}{4};\frac{13}{4},\frac{15}{4};c^2 x^2\right )}{693 d^3}\\ \end{align*}

Mathematica [B]  time = 1.26082, size = 234, normalized size = 2.15 \[ \frac{(d x)^{5/2} \left (\frac{b^2 \left (\frac{105 \sqrt{2} \pi c x \text{HypergeometricPFQ}\left (\left \{\frac{3}{4},\frac{3}{4},1\right \},\left \{\frac{5}{4},\frac{7}{4}\right \},c^2 x^2\right )}{\text{Gamma}\left (\frac{5}{4}\right ) \text{Gamma}\left (\frac{7}{4}\right )}+840 \sqrt{1-c^2 x^2} \cos ^{-1}(c x) \text{Hypergeometric2F1}\left (\frac{3}{4},1,\frac{5}{4},c^2 x^2\right )-16 c x \left (9 c^2 x^2+35\right )+882 c^3 x^3 \cos ^{-1}(c x)^2-168 \sqrt{1-c^2 x^2} \left (3 c^2 x^2+5\right ) \cos ^{-1}(c x)\right )}{c^3}+\frac{84 a b \left (10 \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},c^2 x^2\right )-2 \sqrt{1-c^2 x^2} \left (3 c^2 x^2+5\right )+21 c^3 x^3 \cos ^{-1}(c x)\right )}{c^3}+882 a^2 x^3\right )}{3087 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*x)^(5/2)*(a + b*ArcCos[c*x])^2,x]

[Out]

((d*x)^(5/2)*(882*a^2*x^3 + (84*a*b*(-2*Sqrt[1 - c^2*x^2]*(5 + 3*c^2*x^2) + 21*c^3*x^3*ArcCos[c*x] + 10*Hyperg
eometric2F1[1/4, 1/2, 5/4, c^2*x^2]))/c^3 + (b^2*(-16*c*x*(35 + 9*c^2*x^2) - 168*Sqrt[1 - c^2*x^2]*(5 + 3*c^2*
x^2)*ArcCos[c*x] + 882*c^3*x^3*ArcCos[c*x]^2 + 840*Sqrt[1 - c^2*x^2]*ArcCos[c*x]*Hypergeometric2F1[3/4, 1, 5/4
, c^2*x^2] + (105*Sqrt[2]*c*Pi*x*HypergeometricPFQ[{3/4, 3/4, 1}, {5/4, 7/4}, c^2*x^2])/(Gamma[5/4]*Gamma[7/4]
)))/c^3))/(3087*x^2)

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Maple [F]  time = 0.358, size = 0, normalized size = 0. \begin{align*} \int \left ( dx \right ) ^{{\frac{5}{2}}} \left ( a+b\arccos \left ( cx \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(5/2)*(a+b*arccos(c*x))^2,x)

[Out]

int((d*x)^(5/2)*(a+b*arccos(c*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((d*x)^(5/2)*(a+b*arccos(c*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^{2} d^{2} x^{2} \arccos \left (c x\right )^{2} + 2 \, a b d^{2} x^{2} \arccos \left (c x\right ) + a^{2} d^{2} x^{2}\right )} \sqrt{d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(5/2)*(a+b*arccos(c*x))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(5/2)*(a+b*acos(c*x))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d x\right )^{\frac{5}{2}}{\left (b \arccos \left (c x\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(5/2)*(a+b*arccos(c*x))^2,x, algorithm="giac")

[Out]

integrate((d*x)^(5/2)*(b*arccos(c*x) + a)^2, x)