\(\int \frac {(a+b \arccos (c x))^2}{(d x)^{5/2}} \, dx\) [214]

Optimal result

Integrand size = 18, antiderivative size = 109 \[ \int \frac {(a+b \arccos (c x))^2}{(d x)^{5/2}} \, dx=-\frac {2 (a+b \arccos (c x))^2}{3 d (d x)^{3/2}}+\frac {8 b c (a+b \arccos (c x)) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},c^2 x^2\right )}{3 d^2 \sqrt {d x}}+\frac {16 b^2 c^2 \sqrt {d x} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};c^2 x^2\right )}{3 d^3} \] Output:

-2/3*(a+b*arccos(c*x))^2/d/(d*x)^(3/2)+8/3*b*c*(a+b*arccos(c*x))*hypergeom 
([-1/4, 1/2],[3/4],c^2*x^2)/d^2/(d*x)^(1/2)+16/3*b^2*c^2*(d*x)^(1/2)*hyper 
geom([1/4, 1/4, 1],[3/4, 5/4],c^2*x^2)/d^3
 

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.82 \[ \int \frac {(a+b \arccos (c x))^2}{(d x)^{5/2}} \, dx=\frac {x \left (-8 \operatorname {Gamma}\left (\frac {7}{4}\right ) \operatorname {Gamma}\left (\frac {9}{4}\right ) \left (3 \left (a^2-8 b^2 c^2 x^2+2 b \left (a-2 b c x \sqrt {1-c^2 x^2}\right ) \arccos (c x)+b^2 \arccos (c x)^2\right )-12 a b c x \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},c^2 x^2\right )-4 b^2 c^3 x^3 \sqrt {1-c^2 x^2} \arccos (c x) \operatorname {Hypergeometric2F1}\left (1,\frac {5}{4},\frac {7}{4},c^2 x^2\right )\right )+3 \sqrt {2} b^2 c^4 \pi x^4 \, _3F_2\left (1,\frac {5}{4},\frac {5}{4};\frac {7}{4},\frac {9}{4};c^2 x^2\right )\right )}{36 (d x)^{5/2} \operatorname {Gamma}\left (\frac {7}{4}\right ) \operatorname {Gamma}\left (\frac {9}{4}\right )} \] Input:

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

Output:

(x*(-8*Gamma[7/4]*Gamma[9/4]*(3*(a^2 - 8*b^2*c^2*x^2 + 2*b*(a - 2*b*c*x*Sq 
rt[1 - c^2*x^2])*ArcCos[c*x] + b^2*ArcCos[c*x]^2) - 12*a*b*c*x*Hypergeomet 
ric2F1[-1/4, 1/2, 3/4, c^2*x^2] - 4*b^2*c^3*x^3*Sqrt[1 - c^2*x^2]*ArcCos[c 
*x]*Hypergeometric2F1[1, 5/4, 7/4, c^2*x^2]) + 3*Sqrt[2]*b^2*c^4*Pi*x^4*Hy 
pergeometricPFQ[{1, 5/4, 5/4}, {7/4, 9/4}, c^2*x^2]))/(36*(d*x)^(5/2)*Gamm 
a[7/4]*Gamma[9/4])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5139, 5221}

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.

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

\(\Big \downarrow \) 5139

\(\displaystyle -\frac {4 b c \int \frac {a+b \arccos (c x)}{(d x)^{3/2} \sqrt {1-c^2 x^2}}dx}{3 d}-\frac {2 (a+b \arccos (c x))^2}{3 d (d x)^{3/2}}\)

\(\Big \downarrow \) 5221

\(\displaystyle -\frac {4 b c \left (-\frac {4 b c \sqrt {d x} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};c^2 x^2\right )}{d^2}-\frac {2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},c^2 x^2\right ) (a+b \arccos (c x))}{d \sqrt {d x}}\right )}{3 d}-\frac {2 (a+b \arccos (c x))^2}{3 d (d x)^{3/2}}\)

Input:

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

Output:

(-2*(a + b*ArcCos[c*x])^2)/(3*d*(d*x)^(3/2)) - (4*b*c*((-2*(a + b*ArcCos[c 
*x])*Hypergeometric2F1[-1/4, 1/2, 3/4, c^2*x^2])/(d*Sqrt[d*x]) - (4*b*c*Sq 
rt[d*x]*HypergeometricPFQ[{1/4, 1/4, 1}, {3/4, 5/4}, c^2*x^2])/d^2))/(3*d)
 

Defintions of rubi rules used

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]
 

rule 5221

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)))*S 
imp[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]
 

Maple [F]

\[\int \frac {\left (a +b \arccos \left (c x \right )\right )^{2}}{\left (d x \right )^{\frac {5}{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((b^2*arccos(c*x)^2 + 2*a*b*arccos(c*x) + a^2)*sqrt(d*x)/(d^3*x^3) 
, x)
 

Sympy [F(-2)]

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

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

Output:

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

Maxima [F]

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

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

Output:

-1/6*((3*a^2*c^2*sqrt(d)*(2*arctan(sqrt(c)*sqrt(x))/(sqrt(c)*d^3) - log((c 
*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqrt(c)))/(sqrt(c)*d^3)) - 36*a*b*c^2*sqr 
t(d)*integrate(1/3*x^(5/2)*arctan(sqrt(c*x + 1)*sqrt(-c*x + 1)/(c*x))/(c^2 
*d^3*x^5 - d^3*x^3), x) - 24*b^2*c*sqrt(d)*integrate(1/3*sqrt(c*x + 1)*sqr 
t(-c*x + 1)*x^(3/2)*arctan(sqrt(c*x + 1)*sqrt(-c*x + 1)/(c*x))/(c^2*d^3*x^ 
5 - d^3*x^3), x) - a^2*sqrt(d)*(6*c^(3/2)*arctan(sqrt(c)*sqrt(x))/d^3 - 3* 
c^(3/2)*log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqrt(c)))/d^3 - 4/(d^3*x^(3 
/2))) + 36*a*b*sqrt(d)*integrate(1/3*sqrt(x)*arctan(sqrt(c*x + 1)*sqrt(-c* 
x + 1)/(c*x))/(c^2*d^3*x^5 - d^3*x^3), x))*d^(5/2)*x^(3/2) + 4*b^2*arctan2 
(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)^2)/(d^(5/2)*x^(3/2))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+b \arccos (c x))^2}{(d x)^{5/2}} \, dx=\frac {6 \sqrt {x}\, \left (\int \frac {\mathit {acos} \left (c x \right )}{\sqrt {x}\, x^{2}}d x \right ) a b x +3 \sqrt {x}\, \left (\int \frac {\mathit {acos} \left (c x \right )^{2}}{\sqrt {x}\, x^{2}}d x \right ) b^{2} x -2 a^{2}}{3 \sqrt {x}\, \sqrt {d}\, d^{2} x} \] Input:

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

Output:

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