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

Optimal. Leaf size=120 \[ \frac {2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )}{7 d}+\frac {20 b d^{5/2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{147 c^{7/2}}-\frac {4 b \sqrt {1-c^2 x^2} (d x)^{5/2}}{49 c}-\frac {20 b d^2 \sqrt {1-c^2 x^2} \sqrt {d x}}{147 c^3} \]

[Out]

2/7*(d*x)^(7/2)*(a+b*arccos(c*x))/d+20/147*b*d^(5/2)*EllipticF(c^(1/2)*(d*x)^(1/2)/d^(1/2),I)/c^(7/2)-4/49*b*(
d*x)^(5/2)*(-c^2*x^2+1)^(1/2)/c-20/147*b*d^2*(d*x)^(1/2)*(-c^2*x^2+1)^(1/2)/c^3

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Rubi [A]  time = 0.07, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4628, 321, 329, 221} \[ \frac {2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )}{7 d}-\frac {20 b d^2 \sqrt {1-c^2 x^2} \sqrt {d x}}{147 c^3}+\frac {20 b d^{5/2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{147 c^{7/2}}-\frac {4 b \sqrt {1-c^2 x^2} (d x)^{5/2}}{49 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-20*b*d^2*Sqrt[d*x]*Sqrt[1 - c^2*x^2])/(147*c^3) - (4*b*(d*x)^(5/2)*Sqrt[1 - c^2*x^2])/(49*c) + (2*(d*x)^(7/2
)*(a + b*ArcCos[c*x]))/(7*d) + (20*b*d^(5/2)*EllipticF[ArcSin[(Sqrt[c]*Sqrt[d*x])/Sqrt[d]], -1])/(147*c^(7/2))

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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]

Rubi steps

\begin {align*} \int (d x)^{5/2} \left (a+b \cos ^{-1}(c x)\right ) \, dx &=\frac {2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )}{7 d}+\frac {(2 b c) \int \frac {(d x)^{7/2}}{\sqrt {1-c^2 x^2}} \, dx}{7 d}\\ &=-\frac {4 b (d x)^{5/2} \sqrt {1-c^2 x^2}}{49 c}+\frac {2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )}{7 d}+\frac {(10 b d) \int \frac {(d x)^{3/2}}{\sqrt {1-c^2 x^2}} \, dx}{49 c}\\ &=-\frac {20 b d^2 \sqrt {d x} \sqrt {1-c^2 x^2}}{147 c^3}-\frac {4 b (d x)^{5/2} \sqrt {1-c^2 x^2}}{49 c}+\frac {2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )}{7 d}+\frac {\left (10 b d^3\right ) \int \frac {1}{\sqrt {d x} \sqrt {1-c^2 x^2}} \, dx}{147 c^3}\\ &=-\frac {20 b d^2 \sqrt {d x} \sqrt {1-c^2 x^2}}{147 c^3}-\frac {4 b (d x)^{5/2} \sqrt {1-c^2 x^2}}{49 c}+\frac {2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )}{7 d}+\frac {\left (20 b d^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {c^2 x^4}{d^2}}} \, dx,x,\sqrt {d x}\right )}{147 c^3}\\ &=-\frac {20 b d^2 \sqrt {d x} \sqrt {1-c^2 x^2}}{147 c^3}-\frac {4 b (d x)^{5/2} \sqrt {1-c^2 x^2}}{49 c}+\frac {2 (d x)^{7/2} \left (a+b \cos ^{-1}(c x)\right )}{7 d}+\frac {20 b d^{5/2} F\left (\left .\sin ^{-1}\left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{147 c^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.29, size = 158, normalized size = 1.32 \[ \frac {2 d^2 \sqrt {d x} \left (21 a c^3 x^3 \sqrt {1-c^2 x^2}+6 b c^4 x^4+4 b c^2 x^2+\frac {10 i b \sqrt {x} \sqrt {1-\frac {1}{c^2 x^2}} F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {-\frac {1}{c}}}{\sqrt {x}}\right )\right |-1\right )}{\sqrt {-\frac {1}{c}}}+21 b c^3 x^3 \sqrt {1-c^2 x^2} \cos ^{-1}(c x)-10 b\right )}{147 c^3 \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*d^2*Sqrt[d*x]*(-10*b + 4*b*c^2*x^2 + 6*b*c^4*x^4 + 21*a*c^3*x^3*Sqrt[1 - c^2*x^2] + 21*b*c^3*x^3*Sqrt[1 - c
^2*x^2]*ArcCos[c*x] + ((10*I)*b*Sqrt[1 - 1/(c^2*x^2)]*Sqrt[x]*EllipticF[I*ArcSinh[Sqrt[-c^(-1)]/Sqrt[x]], -1])
/Sqrt[-c^(-1)]))/(147*c^3*Sqrt[1 - c^2*x^2])

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b d^{2} x^{2} \arccos \left (c x\right ) + a d^{2} x^{2}\right )} \sqrt {d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 144, normalized size = 1.20 \[ \frac {\frac {2 \left (d x \right )^{\frac {7}{2}} a}{7}+2 b \left (\frac {\left (d x \right )^{\frac {7}{2}} \arccos \left (c x \right )}{7}+\frac {2 c \left (-\frac {d^{2} \left (d x \right )^{\frac {5}{2}} \sqrt {-c^{2} x^{2}+1}}{7 c^{2}}-\frac {5 d^{4} \sqrt {d x}\, \sqrt {-c^{2} x^{2}+1}}{21 c^{4}}+\frac {5 d^{4} \sqrt {-c x +1}\, \sqrt {c x +1}\, \EllipticF \left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )}{21 c^{4} \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{7 d}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*(1/7*(d*x)^(7/2)*a+b*(1/7*(d*x)^(7/2)*arccos(c*x)+2/7*c/d*(-1/7/c^2*d^2*(d*x)^(5/2)*(-c^2*x^2+1)^(1/2)-5/2
1/c^4*d^4*(d*x)^(1/2)*(-c^2*x^2+1)^(1/2)+5/21/c^4*d^4/(c/d)^(1/2)*(-c*x+1)^(1/2)*(c*x+1)^(1/2)/(-c^2*x^2+1)^(1
/2)*EllipticF((d*x)^(1/2)*(c/d)^(1/2),I))))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {42 \, b c^{4} d^{\frac {5}{2}} x^{\frac {7}{2}} \arctan \left (\sqrt {c x + 1} \sqrt {-c x + 1}, c x\right ) - {\left (12 \, b c^{4} d^{2} x^{\frac {7}{2}} + 42 \, b c^{5} d^{2} \int \frac {\sqrt {c x + 1} \sqrt {-c x + 1} x^{\frac {7}{2}}}{c^{2} x^{2} - 1}\,{d x} + 28 \, b c^{2} d^{2} x^{\frac {3}{2}} + 21 \, {\left (2 \, b d^{2} \arctan \left (\sqrt {c} \sqrt {x}\right ) + b d^{2} \log \left (\frac {c x - 1}{c x + 2 \, \sqrt {c} \sqrt {x} + 1}\right )\right )} \sqrt {c}\right )} \sqrt {d}}{147 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/147*(42*b*c^4*d^(5/2)*x^(7/2)*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) - (12*b*c^4*d^2*x^(7/2) + 294*b*c^5
*d^2*integrate(1/7*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^(7/2)/(c^2*x^2 - 1), x) + 28*b*c^2*d^2*x^(3/2) + 21*(2*b*d^2
*arctan(sqrt(c)*sqrt(x)) + b*d^2*log((c*x - 1)/(c*x + 2*sqrt(c)*sqrt(x) + 1)))*sqrt(c))*sqrt(d))/c^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (a+b\,\mathrm {acos}\left (c\,x\right )\right )\,{\left (d\,x\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acos(c*x))*(d*x)^(5/2),x)

[Out]

int((a + b*acos(c*x))*(d*x)^(5/2), x)

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sympy [A]  time = 98.83, size = 82, normalized size = 0.68 \[ a \left (\begin {cases} 0 & \text {for}\: d = 0 \\\frac {2 \left (d x\right )^{\frac {7}{2}}}{7 d} & \text {otherwise} \end {cases}\right ) + b c \left (\begin {cases} 0 & \text {for}\: d = 0 \\\frac {d^{\frac {5}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {c^{2} x^{2} e^{2 i \pi }} \right )}}{7 \Gamma \left (\frac {13}{4}\right )} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} 0 & \text {for}\: d = 0 \\\frac {2 \left (d x\right )^{\frac {7}{2}}}{7 d} & \text {otherwise} \end {cases}\right ) \operatorname {acos}{\left (c x \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((0, Eq(d, 0)), (2*(d*x)**(7/2)/(7*d), True)) + b*c*Piecewise((0, Eq(d, 0)), (d**(5/2)*x**(9/2)*gam
ma(9/4)*hyper((1/2, 9/4), (13/4,), c**2*x**2*exp_polar(2*I*pi))/(7*gamma(13/4)), True)) + b*Piecewise((0, Eq(d
, 0)), (2*(d*x)**(7/2)/(7*d), True))*acos(c*x)

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