\(\int (d x)^{5/2} (a+b \arccos (c x)) \, dx\) [203]

Optimal result

Integrand size = 16, antiderivative size = 120 \[ \int (d x)^{5/2} (a+b \arccos (c x)) \, dx=-\frac {20 b d^2 \sqrt {d x} \sqrt {1-c^2 x^2}}{147 c^3}-\frac {4 b (d x)^{5/2} \sqrt {1-c^2 x^2}}{49 c}+\frac {2 (d x)^{7/2} (a+b \arccos (c x))}{7 d}+\frac {20 b d^{5/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),-1\right )}{147 c^{7/2}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 11.21 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.32 \[ \int (d x)^{5/2} (a+b \arccos (c x)) \, dx=\frac {2 d^2 \sqrt {d x} \left (-10 b+4 b c^2 x^2+6 b c^4 x^4+21 a c^3 x^3 \sqrt {1-c^2 x^2}+21 b c^3 x^3 \sqrt {1-c^2 x^2} \arccos (c x)+\frac {10 i b \sqrt {1-\frac {1}{c^2 x^2}} \sqrt {x} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-\frac {1}{c}}}{\sqrt {x}}\right ),-1\right )}{\sqrt {-\frac {1}{c}}}\right )}{147 c^3 \sqrt {1-c^2 x^2}} \] Input:

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

Output:

(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])/S 
qrt[-c^(-1)]))/(147*c^3*Sqrt[1 - c^2*x^2])
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {5139, 262, 262, 266, 762}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (d x)^{5/2} (a+b \arccos (c x)) \, dx\)

\(\Big \downarrow \) 5139

\(\displaystyle \frac {2 b c \int \frac {(d x)^{7/2}}{\sqrt {1-c^2 x^2}}dx}{7 d}+\frac {2 (d x)^{7/2} (a+b \arccos (c x))}{7 d}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {2 b c \left (\frac {5 d^2 \int \frac {(d x)^{3/2}}{\sqrt {1-c^2 x^2}}dx}{7 c^2}-\frac {2 d \sqrt {1-c^2 x^2} (d x)^{5/2}}{7 c^2}\right )}{7 d}+\frac {2 (d x)^{7/2} (a+b \arccos (c x))}{7 d}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {2 b c \left (\frac {5 d^2 \left (\frac {d^2 \int \frac {1}{\sqrt {d x} \sqrt {1-c^2 x^2}}dx}{3 c^2}-\frac {2 d \sqrt {1-c^2 x^2} \sqrt {d x}}{3 c^2}\right )}{7 c^2}-\frac {2 d \sqrt {1-c^2 x^2} (d x)^{5/2}}{7 c^2}\right )}{7 d}+\frac {2 (d x)^{7/2} (a+b \arccos (c x))}{7 d}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {2 b c \left (\frac {5 d^2 \left (\frac {2 d \int \frac {1}{\sqrt {1-c^2 x^2}}d\sqrt {d x}}{3 c^2}-\frac {2 d \sqrt {1-c^2 x^2} \sqrt {d x}}{3 c^2}\right )}{7 c^2}-\frac {2 d \sqrt {1-c^2 x^2} (d x)^{5/2}}{7 c^2}\right )}{7 d}+\frac {2 (d x)^{7/2} (a+b \arccos (c x))}{7 d}\)

\(\Big \downarrow \) 762

\(\displaystyle \frac {2 (d x)^{7/2} (a+b \arccos (c x))}{7 d}+\frac {2 b c \left (\frac {5 d^2 \left (\frac {2 d^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),-1\right )}{3 c^{5/2}}-\frac {2 d \sqrt {1-c^2 x^2} \sqrt {d x}}{3 c^2}\right )}{7 c^2}-\frac {2 d \sqrt {1-c^2 x^2} (d x)^{5/2}}{7 c^2}\right )}{7 d}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 762

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

rule 5139

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] 
:> Simp[(d*x)^(m + 1)*((a + b*ArcCos[c*x])^n/(d*(m + 1))), x] + Simp[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]
 

Maple [A] (verified)

Time = 2.32 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.20

method	result	size
derivativedivides	\(\frac {\frac {2 a \left (d x \right )^{\frac {7}{2}}}{7}+2 b \left (\frac {\left (d x \right )^{\frac {7}{2}} \arccos \left (c x \right )}{7}+\frac {2 c \left (-\frac {d^{2} \left (d x \right )^{\frac {5}{2}} \sqrt {-c^{2} x^{2}+1}}{7 c^{2}}-\frac {5 d^{4} \sqrt {d x}\, \sqrt {-c^{2} x^{2}+1}}{21 c^{4}}+\frac {5 d^{4} \sqrt {-c x +1}\, \sqrt {c x +1}\, \operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )}{21 c^{4} \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{7 d}\right )}{d}\)	\(144\)
default	\(\frac {\frac {2 a \left (d x \right )^{\frac {7}{2}}}{7}+2 b \left (\frac {\left (d x \right )^{\frac {7}{2}} \arccos \left (c x \right )}{7}+\frac {2 c \left (-\frac {d^{2} \left (d x \right )^{\frac {5}{2}} \sqrt {-c^{2} x^{2}+1}}{7 c^{2}}-\frac {5 d^{4} \sqrt {d x}\, \sqrt {-c^{2} x^{2}+1}}{21 c^{4}}+\frac {5 d^{4} \sqrt {-c x +1}\, \sqrt {c x +1}\, \operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )}{21 c^{4} \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{7 d}\right )}{d}\)	\(144\)
parts	\(\frac {2 a \left (d x \right )^{\frac {7}{2}}}{7 d}+\frac {2 b \left (\frac {\left (d x \right )^{\frac {7}{2}} \arccos \left (c x \right )}{7}+\frac {2 c \left (-\frac {d^{2} \left (d x \right )^{\frac {5}{2}} \sqrt {-c^{2} x^{2}+1}}{7 c^{2}}-\frac {5 d^{4} \sqrt {d x}\, \sqrt {-c^{2} x^{2}+1}}{21 c^{4}}+\frac {5 d^{4} \sqrt {-c x +1}\, \sqrt {c x +1}\, \operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )}{21 c^{4} \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{7 d}\right )}{d}\)	\(146\)

Input:

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

Output:

2/d*(1/7*a*(d*x)^(7/2)+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)-5/21/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))))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.83 \[ \int (d x)^{5/2} (a+b \arccos (c x)) \, dx=-\frac {2 \, {\left (10 \, \sqrt {-c^{2} d} b d^{2} {\rm weierstrassPInverse}\left (\frac {4}{c^{2}}, 0, x\right ) - {\left (21 \, b c^{5} d^{2} x^{3} \arccos \left (c x\right ) + 21 \, a c^{5} d^{2} x^{3} - 2 \, {\left (3 \, b c^{4} d^{2} x^{2} + 5 \, b c^{2} d^{2}\right )} \sqrt {-c^{2} x^{2} + 1}\right )} \sqrt {d x}\right )}}{147 \, c^{5}} \] Input:

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

Output:

-2/147*(10*sqrt(-c^2*d)*b*d^2*weierstrassPInverse(4/c^2, 0, x) - (21*b*c^5 
*d^2*x^3*arccos(c*x) + 21*a*c^5*d^2*x^3 - 2*(3*b*c^4*d^2*x^2 + 5*b*c^2*d^2 
)*sqrt(-c^2*x^2 + 1))*sqrt(d*x))/c^5
 

Sympy [A] (verification not implemented)

Time = 62.68 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.71 \[ \int (d x)^{5/2} (a+b \arccos (c x)) \, dx=a \left (\begin {cases} \frac {2 \left (d x\right )^{\frac {7}{2}}}{7 d} & \text {for}\: d \neq 0 \\0 & \text {otherwise} \end {cases}\right ) + b c \left (\begin {cases} \frac {d^{\frac {5}{2}} x^{\frac {9}{2}} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {c^{2} x^{2} e^{2 i \pi }} \right )}}{7 \Gamma \left (\frac {13}{4}\right )} & \text {for}\: d > -\infty \wedge d < \infty \wedge d \neq 0 \\0 & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {2 \left (d x\right )^{\frac {7}{2}}}{7 d} & \text {for}\: d \neq 0 \\0 & \text {otherwise} \end {cases}\right ) \operatorname {acos}{\left (c x \right )} \] Input:

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

Output:

a*Piecewise((2*(d*x)**(7/2)/(7*d), Ne(d, 0)), (0, True)) + b*c*Piecewise(( 
d**(5/2)*x**(9/2)*gamma(9/4)*hyper((1/2, 9/4), (13/4,), c**2*x**2*exp_pola 
r(2*I*pi))/(7*gamma(13/4)), (d > -oo) & (d < oo) & Ne(d, 0)), (0, True)) + 
 b*Piecewise((2*(d*x)**(7/2)/(7*d), Ne(d, 0)), (0, True))*acos(c*x)
 

Maxima [F]

\[ \int (d x)^{5/2} (a+b \arccos (c x)) \, dx=\int { \left (d x\right )^{\frac {5}{2}} {\left (b \arccos \left (c x\right ) + a\right )} \,d x } \] Input:

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

Output:

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*a 
rctan(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
 

Giac [F]

\[ \int (d x)^{5/2} (a+b \arccos (c x)) \, dx=\int { \left (d x\right )^{\frac {5}{2}} {\left (b \arccos \left (c x\right ) + a\right )} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (d x)^{5/2} (a+b \arccos (c x)) \, dx=\int \left (a+b\,\mathrm {acos}\left (c\,x\right )\right )\,{\left (d\,x\right )}^{5/2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (d x)^{5/2} (a+b \arccos (c x)) \, dx=\frac {\sqrt {d}\, d^{2} \left (2 \sqrt {x}\, a \,x^{3}+7 \left (\int \sqrt {x}\, \mathit {acos} \left (c x \right ) x^{2}d x \right ) b \right )}{7} \] Input:

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

Output:

(sqrt(d)*d**2*(2*sqrt(x)*a*x**3 + 7*int(sqrt(x)*acos(c*x)*x**2,x)*b))/7