\(\int \frac {a+b \arccos (c x)}{(d x)^{5/2}} \, dx\) [208]

Optimal result

Integrand size = 16, antiderivative size = 125 \[ \int \frac {a+b \arccos (c x)}{(d x)^{5/2}} \, dx=\frac {4 b c \sqrt {1-c^2 x^2}}{3 d^2 \sqrt {d x}}-\frac {2 (a+b \arccos (c x))}{3 d (d x)^{3/2}}+\frac {4 b c^{3/2} E\left (\left .\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right )\right |-1\right )}{3 d^{5/2}}-\frac {4 b c^{3/2} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {c} \sqrt {d x}}{\sqrt {d}}\right ),-1\right )}{3 d^{5/2}} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.07 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.54 \[ \int \frac {a+b \arccos (c x)}{(d x)^{5/2}} \, dx=\frac {2 x \left (-3 \left (a-2 b c x \sqrt {1-c^2 x^2}+b \arccos (c x)\right )+2 b c^3 x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},c^2 x^2\right )\right )}{9 (d x)^{5/2}} \] Input:

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

Output:

(2*x*(-3*(a - 2*b*c*x*Sqrt[1 - c^2*x^2] + b*ArcCos[c*x]) + 2*b*c^3*x^3*Hyp 
ergeometric2F1[1/2, 3/4, 7/4, c^2*x^2]))/(9*(d*x)^(5/2))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5139, 264, 266, 836, 27, 762, 1389, 327}

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.

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

\(\Big \downarrow \) 5139

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 836

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 762

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

\(\Big \downarrow \) 1389

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

\(\Big \downarrow \) 327

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && 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 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

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 836

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, 
Simp[-q^(-1)   Int[1/Sqrt[a + b*x^4], x], x] + Simp[1/q   Int[(1 + q*x^2)/S 
qrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]
 

rule 1389

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Simp[d/Sq 
rt[a]   Int[Sqrt[1 + e*(x^2/d)]/Sqrt[1 - e*(x^2/d)], x], x] /; FreeQ[{a, c, 
 d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && NegQ[c/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 = 0.58 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(\frac {-\frac {2 a}{3 \left (d x \right )^{\frac {3}{2}}}+2 b \left (-\frac {\arccos \left (c x \right )}{3 \left (d x \right )^{\frac {3}{2}}}-\frac {2 c \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{\sqrt {d x}}+\frac {c \sqrt {-c x +1}\, \sqrt {c x +1}\, \left (\operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )\right )}{d \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{3 d}\right )}{d}\)	\(129\)
default	\(\frac {-\frac {2 a}{3 \left (d x \right )^{\frac {3}{2}}}+2 b \left (-\frac {\arccos \left (c x \right )}{3 \left (d x \right )^{\frac {3}{2}}}-\frac {2 c \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{\sqrt {d x}}+\frac {c \sqrt {-c x +1}\, \sqrt {c x +1}\, \left (\operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )\right )}{d \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{3 d}\right )}{d}\)	\(129\)
parts	\(-\frac {2 a}{3 \left (d x \right )^{\frac {3}{2}} d}+\frac {2 b \left (-\frac {\arccos \left (c x \right )}{3 \left (d x \right )^{\frac {3}{2}}}-\frac {2 c \left (-\frac {\sqrt {-c^{2} x^{2}+1}}{\sqrt {d x}}+\frac {c \sqrt {-c x +1}\, \sqrt {c x +1}\, \left (\operatorname {EllipticF}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )-\operatorname {EllipticE}\left (\sqrt {d x}\, \sqrt {\frac {c}{d}}, i\right )\right )}{d \sqrt {\frac {c}{d}}\, \sqrt {-c^{2} x^{2}+1}}\right )}{3 d}\right )}{d}\)	\(131\)

Input:

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

Output:

2/d*(-1/3*a/(d*x)^(3/2)+b*(-1/3/(d*x)^(3/2)*arccos(c*x)-2/3*c/d*(-(-c^2*x^ 
2+1)^(1/2)/(d*x)^(1/2)+c/d/(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)-EllipticE((d*x)^(1/2)*( 
c/d)^(1/2),I)))))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.58 \[ \int \frac {a+b \arccos (c x)}{(d x)^{5/2}} \, dx=\frac {2 \, {\left (2 \, \sqrt {-c^{2} d} b c x^{2} {\rm weierstrassZeta}\left (\frac {4}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {4}{c^{2}}, 0, x\right )\right ) + {\left (2 \, \sqrt {-c^{2} x^{2} + 1} b c x - b \arccos \left (c x\right ) - a\right )} \sqrt {d x}\right )}}{3 \, d^{3} x^{2}} \] Input:

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

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {a+b \arccos (c x)}{(d x)^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> Invalid comparison of non-real zoo
 

Maxima [F]

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

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

Output:

-1/3*(2*b*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) - (6*b*c*d^3*x*integr 
ate(1/3*sqrt(c*x + 1)*sqrt(-c*x + 1)*sqrt(x)/(c^2*d^3*x^4 - d^3*x^2), x) + 
 (2*b*c*x*arctan(1/(sqrt(c)*sqrt(x))) + b*c*x*log(-(c*x - 1)/(c*x + 2*sqrt 
(c)*sqrt(x) + 1)))*sqrt(c))*sqrt(x))/(d^(5/2)*x^(3/2))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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