3.3.9 \(\int (d x)^{5/2} (a+b \text {ArcCos}(c x))^2 \, dx\) [209]

Optimal. Leaf size=109 \[ \frac {2 (d x)^{7/2} (a+b \text {ArcCos}(c x))^2}{7 d}+\frac {8 b c (d x)^{9/2} (a+b \text {ArcCos}(c x)) \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};c^2 x^2\right )}{63 d^2}+\frac {16 b^2 c^2 (d x)^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};c^2 x^2\right )}{693 d^3} \]

[Out]

2/7*(d*x)^(7/2)*(a+b*arccos(c*x))^2/d+8/63*b*c*(d*x)^(9/2)*(a+b*arccos(c*x))*hypergeom([1/2, 9/4],[13/4],c^2*x
^2)/d^2+16/693*b^2*c^2*(d*x)^(11/2)*hypergeom([1, 11/4, 11/4],[13/4, 15/4],c^2*x^2)/d^3

________________________________________________________________________________________

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)^{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 ) (a+b \text {ArcCos}(c x))}{63 d^2}+\frac {2 (d x)^{7/2} (a+b \text {ArcCos}(c x))^2}{7 d} \end {gather*}

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 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 (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] Leaf count is larger than twice the leaf count of optimal. \(234\) vs. \(2(109)=218\).
time = 10.85, size = 234, normalized size = 2.15 \begin {gather*} \frac {(d x)^{5/2} \left (882 a^2 x^3+\frac {84 a b \left (-2 \sqrt {1-c^2 x^2} \left (5+3 c^2 x^2\right )+21 c^3 x^3 \text {ArcCos}(c x)+10 \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};c^2 x^2\right )\right )}{c^3}+\frac {b^2 \left (-16 c x \left (35+9 c^2 x^2\right )-168 \sqrt {1-c^2 x^2} \left (5+3 c^2 x^2\right ) \text {ArcCos}(c x)+882 c^3 x^3 \text {ArcCos}(c x)^2+840 \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 )+\frac {105 \sqrt {2} c \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 )}{c^3}\right )}{3087 x^2} \end {gather*}

Antiderivative was successfully verified.

[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)

________________________________________________________________________________________

Maple [F]
time = 0.19, size = 0, normalized size = 0.00 \[\int \left (d x \right )^{\frac {5}{2}} \left (a +b \arccos \left (c x \right )\right )^{2}\, dx\]

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)

________________________________________________________________________________________

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

[Out]

2/7*b^2*d^(5/2)*x^(7/2)*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)^2 + 1/42*a^2*c^2*d^(5/2)*(4*(3*c^2*x^(7/2)
+ 7*x^(3/2))/c^4 + 42*arctan(sqrt(c)*sqrt(x))/c^(11/2) + 21*log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqrt(c)))/c
^(11/2)) + 14*a*b*c^2*d^(5/2)*integrate(1/7*x^(9/2)*arctan(sqrt(c*x + 1)*sqrt(-c*x + 1)/(c*x))/(c^2*x^2 - 1),
x) - 4*b^2*c*d^(5/2)*integrate(1/7*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^(7/2)*arctan(sqrt(c*x + 1)*sqrt(-c*x + 1)/(c
*x))/(c^2*x^2 - 1), x) - 1/6*a^2*d^(5/2)*(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)) - 14*a*b*d^(5/2)*integrate(1/7*x^(5/2)*arctan(sqrt(c*x + 1)*sqrt(-
c*x + 1)/(c*x))/(c^2*x^2 - 1), x)

________________________________________________________________________________________

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((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)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

________________________________________________________________________________________

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

________________________________________________________________________________________

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\,{\left (d\,x\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________