Integrand size = 18, antiderivative size = 105 \[ \int \frac {(a+b \arccos (c x))^2}{(d x)^{3/2}} \, dx=-\frac {2 (a+b \arccos (c x))^2}{d \sqrt {d x}}-\frac {8 b c \sqrt {d x} (a+b \arccos (c x)) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},c^2 x^2\right )}{d^2}-\frac {16 b^2 c^2 (d x)^{3/2} \, _3F_2\left (\frac {3}{4},\frac {3}{4},1;\frac {5}{4},\frac {7}{4};c^2 x^2\right )}{3 d^3} \]
-16/3*b^2*c^2*(d*x)^(3/2)*hypergeom([3/4, 3/4, 1],[5/4, 7/4],c^2*x^2)/d^3- 2*(a+b*arccos(c*x))^2/d/(d*x)^(1/2)-8*b*c*(a+b*arccos(c*x))*hypergeom([1/4 , 1/2],[5/4],c^2*x^2)*(d*x)^(1/2)/d^2
Time = 0.40 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.23 \[ \int \frac {(a+b \arccos (c x))^2}{(d x)^{3/2}} \, dx=\frac {x \left (-\frac {\sqrt {2} b^2 c^2 \pi x^2 \, _3F_2\left (\frac {3}{4},\frac {3}{4},1;\frac {5}{4},\frac {7}{4};c^2 x^2\right )}{\operatorname {Gamma}\left (\frac {5}{4}\right ) \operatorname {Gamma}\left (\frac {7}{4}\right )}-2 \left ((a+b \arccos (c x))^2+4 a b c x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},c^2 x^2\right )+2 b^2 \arccos (c x) \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {5}{4},c^2 x^2\right ) \sin (2 \arccos (c x))\right )\right )}{(d x)^{3/2}} \]
(x*(-((Sqrt[2]*b^2*c^2*Pi*x^2*HypergeometricPFQ[{3/4, 3/4, 1}, {5/4, 7/4}, c^2*x^2])/(Gamma[5/4]*Gamma[7/4])) - 2*((a + b*ArcCos[c*x])^2 + 4*a*b*c*x *Hypergeometric2F1[1/4, 1/2, 5/4, c^2*x^2] + 2*b^2*ArcCos[c*x]*Hypergeomet ric2F1[3/4, 1, 5/4, c^2*x^2]*Sin[2*ArcCos[c*x]])))/(d*x)^(3/2)
Time = 0.33 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.02, 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.
\(\displaystyle \int \frac {(a+b \arccos (c x))^2}{(d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 5139 |
\(\displaystyle -\frac {4 b c \int \frac {a+b \arccos (c x)}{\sqrt {d x} \sqrt {1-c^2 x^2}}dx}{d}-\frac {2 (a+b \arccos (c x))^2}{d \sqrt {d x}}\) |
\(\Big \downarrow \) 5221 |
\(\displaystyle -\frac {4 b c \left (\frac {4 b c (d x)^{3/2} \, _3F_2\left (\frac {3}{4},\frac {3}{4},1;\frac {5}{4},\frac {7}{4};c^2 x^2\right )}{3 d^2}+\frac {2 \sqrt {d x} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},c^2 x^2\right ) (a+b \arccos (c x))}{d}\right )}{d}-\frac {2 (a+b \arccos (c x))^2}{d \sqrt {d x}}\) |
(-2*(a + b*ArcCos[c*x])^2)/(d*Sqrt[d*x]) - (4*b*c*((2*Sqrt[d*x]*(a + b*Arc Cos[c*x])*Hypergeometric2F1[1/4, 1/2, 5/4, c^2*x^2])/d + (4*b*c*(d*x)^(3/2 )*HypergeometricPFQ[{3/4, 3/4, 1}, {5/4, 7/4}, c^2*x^2])/(3*d^2)))/d
3.3.13.3.1 Defintions of rubi rules used
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]
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]
\[\int \frac {\left (a +b \arccos \left (c x \right )\right )^{2}}{\left (d x \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(a+b \arccos (c x))^2}{(d x)^{3/2}} \, dx=\int { \frac {{\left (b \arccos \left (c x\right ) + a\right )}^{2}}{\left (d x\right )^{\frac {3}{2}}} \,d x } \]
Exception generated. \[ \int \frac {(a+b \arccos (c x))^2}{(d x)^{3/2}} \, dx=\text {Exception raised: TypeError} \]
\[ \int \frac {(a+b \arccos (c x))^2}{(d x)^{3/2}} \, dx=\int { \frac {{\left (b \arccos \left (c x\right ) + a\right )}^{2}}{\left (d x\right )^{\frac {3}{2}}} \,d x } \]
-1/2*(4*b^2*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x)^2 - (a^2*c^2*sqrt(d )*(2*arctan(sqrt(c)*sqrt(x))/(c^(3/2)*d^2) + log((c*sqrt(x) - sqrt(c))/(c* sqrt(x) + sqrt(c)))/(c^(3/2)*d^2)) + 4*a*b*c^2*sqrt(d)*integrate(x^(5/2)*a rctan(sqrt(c*x + 1)*sqrt(-c*x + 1)/(c*x))/(c^2*d^2*x^4 - d^2*x^2), x) + 8* b^2*c*sqrt(d)*integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)*x^(3/2)*arctan(sqrt(c *x + 1)*sqrt(-c*x + 1)/(c*x))/(c^2*d^2*x^4 - d^2*x^2), x) - a^2*sqrt(d)*(2 *sqrt(c)*arctan(sqrt(c)*sqrt(x))/d^2 + sqrt(c)*log((c*sqrt(x) - sqrt(c))/( c*sqrt(x) + sqrt(c)))/d^2 + 4/(d^2*sqrt(x))) - 4*a*b*sqrt(d)*integrate(sqr t(x)*arctan(sqrt(c*x + 1)*sqrt(-c*x + 1)/(c*x))/(c^2*d^2*x^4 - d^2*x^2), x ))*d^(3/2)*sqrt(x))/(d^(3/2)*sqrt(x))
\[ \int \frac {(a+b \arccos (c x))^2}{(d x)^{3/2}} \, dx=\int { \frac {{\left (b \arccos \left (c x\right ) + a\right )}^{2}}{\left (d x\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \arccos (c x))^2}{(d x)^{3/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}^2}{{\left (d\,x\right )}^{3/2}} \,d x \]