Integrand size = 30, antiderivative size = 265 \[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{5/2}} \, dx=-\frac {b d^2 \left (1-c^2 x^2\right )^{5/2}}{3 c (1-c x) (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {2 d^2 (1+c x) \left (1-c^2 x^2\right ) (a+b \arcsin (c x))}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {d^2 x \left (1-c^2 x^2\right )^2 (a+b \arcsin (c x))}{3 (d+c d x)^{5/2} (f-c f x)^{5/2}}-\frac {b d^2 \left (1-c^2 x^2\right )^{5/2} \text {arctanh}(c x)}{3 c (d+c d x)^{5/2} (f-c f x)^{5/2}}+\frac {b d^2 \left (1-c^2 x^2\right )^{5/2} \log \left (1-c^2 x^2\right )}{6 c (d+c d x)^{5/2} (f-c f x)^{5/2}} \]
-1/3*b*d^2*(-c^2*x^2+1)^(5/2)/c/(-c*x+1)/(c*d*x+d)^(5/2)/(-c*f*x+f)^(5/2)+ 2/3*d^2*(c*x+1)*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(5/2)/(-c*f*x+f )^(5/2)+1/3*d^2*x*(-c^2*x^2+1)^2*(a+b*arcsin(c*x))/(c*d*x+d)^(5/2)/(-c*f*x +f)^(5/2)-1/3*b*d^2*(-c^2*x^2+1)^(5/2)*arctanh(c*x)/c/(c*d*x+d)^(5/2)/(-c* f*x+f)^(5/2)+1/6*b*d^2*(-c^2*x^2+1)^(5/2)*ln(-c^2*x^2+1)/c/(c*d*x+d)^(5/2) /(-c*f*x+f)^(5/2)
Time = 1.04 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.49 \[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{5/2}} \, dx=\frac {\sqrt {d+c d x} \sqrt {f-c f x} \left (-\left ((-2+c x) \left (-b+b c x+a \sqrt {1-c^2 x^2}\right )\right )-b (-2+c x) \sqrt {1-c^2 x^2} \arcsin (c x)+b (-1+c x)^2 \log (f-c f x)\right )}{3 c d f^3 (-1+c x)^2 \sqrt {1-c^2 x^2}} \]
(Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(-((-2 + c*x)*(-b + b*c*x + a*Sqrt[1 - c^ 2*x^2])) - b*(-2 + c*x)*Sqrt[1 - c^2*x^2]*ArcSin[c*x] + b*(-1 + c*x)^2*Log [f - c*f*x]))/(3*c*d*f^3*(-1 + c*x)^2*Sqrt[1 - c^2*x^2])
Time = 0.51 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.60, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {5178, 27, 5260, 2009}
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.
| \(\displaystyle \int \frac {a+b \arcsin (c x)}{\sqrt {c d x+d} (f-c f x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 5178 |
| \(\displaystyle \frac {\left (1-c^2 x^2\right )^{5/2} \int \frac {d^2 (c x+1)^2 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^2 \left (1-c^2 x^2\right )^{5/2} \int \frac {(c x+1)^2 (a+b \arcsin (c x))}{\left (1-c^2 x^2\right )^{5/2}}dx}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 5260 |
| \(\displaystyle \frac {d^2 \left (1-c^2 x^2\right )^{5/2} \left (-b c \int \left (\frac {x}{3 \left (1-c^2 x^2\right )}+\frac {2 (c x+1)}{3 c \left (1-c^2 x^2\right )^2}\right )dx+\frac {x (a+b \arcsin (c x))}{3 \sqrt {1-c^2 x^2}}+\frac {2 (c x+1) (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 \left (1-c^2 x^2\right )^{5/2} \left (\frac {x (a+b \arcsin (c x))}{3 \sqrt {1-c^2 x^2}}+\frac {2 (c x+1) (a+b \arcsin (c x))}{3 c \left (1-c^2 x^2\right )^{3/2}}-b c \left (\frac {\text {arctanh}(c x)}{3 c^2}+\frac {c x+1}{3 c^2 \left (1-c^2 x^2\right )}-\frac {\log \left (1-c^2 x^2\right )}{6 c^2}\right )\right )}{(c d x+d)^{5/2} (f-c f x)^{5/2}}\) |
(d^2*(1 - c^2*x^2)^(5/2)*((2*(1 + c*x)*(a + b*ArcSin[c*x]))/(3*c*(1 - c^2* x^2)^(3/2)) + (x*(a + b*ArcSin[c*x]))/(3*Sqrt[1 - c^2*x^2]) - b*c*((1 + c* x)/(3*c^2*(1 - c^2*x^2)) + ArcTanh[c*x]/(3*c^2) - Log[1 - c^2*x^2]/(6*c^2) )))/((d + c*d*x)^(5/2)*(f - c*f*x)^(5/2))
3.6.37.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> Simp[(d + e*x)^q*((f + g*x)^q/(1 - c^2*x^ 2)^q) Int[(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e _.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p, x]}, Simp[(a + b*ArcSin[c*x]) u, x] - Simp[b*c Int[1/Sqrt[1 - c^2*x^2] u, x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IG tQ[m, 0] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3] )
\[\int \frac {a +b \arcsin \left (c x \right )}{\sqrt {c d x +d}\, \left (-c f x +f \right )^{\frac {5}{2}}}d x\]
Time = 0.34 (sec) , antiderivative size = 527, normalized size of antiderivative = 1.99 \[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{5/2}} \, dx=\left [\frac {{\left (b c^{3} x^{3} - b c^{2} x^{2} - b c x + b\right )} \sqrt {d f} \log \left (\frac {c^{6} d f x^{6} - 4 \, c^{5} d f x^{5} + 5 \, c^{4} d f x^{4} - 4 \, c^{2} d f x^{2} + 4 \, c d f x - {\left (c^{4} x^{4} - 4 \, c^{3} x^{3} + 6 \, c^{2} x^{2} - 4 \, c x\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {d f} - 2 \, d f}{c^{4} x^{4} - 2 \, c^{3} x^{3} + 2 \, c x - 1}\right ) - 2 \, {\left (a c^{2} x^{2} + \sqrt {-c^{2} x^{2} + 1} b c x - a c x + {\left (b c^{2} x^{2} - b c x - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{6 \, {\left (c^{4} d f^{3} x^{3} - c^{3} d f^{3} x^{2} - c^{2} d f^{3} x + c d f^{3}\right )}}, \frac {{\left (b c^{3} x^{3} - b c^{2} x^{2} - b c x + b\right )} \sqrt {-d f} \arctan \left (\frac {{\left (c^{2} x^{2} - 2 \, c x + 2\right )} \sqrt {-c^{2} x^{2} + 1} \sqrt {c d x + d} \sqrt {-c f x + f} \sqrt {-d f}}{c^{4} d f x^{4} - 2 \, c^{3} d f x^{3} - c^{2} d f x^{2} + 2 \, c d f x}\right ) - {\left (a c^{2} x^{2} + \sqrt {-c^{2} x^{2} + 1} b c x - a c x + {\left (b c^{2} x^{2} - b c x - 2 \, b\right )} \arcsin \left (c x\right ) - 2 \, a\right )} \sqrt {c d x + d} \sqrt {-c f x + f}}{3 \, {\left (c^{4} d f^{3} x^{3} - c^{3} d f^{3} x^{2} - c^{2} d f^{3} x + c d f^{3}\right )}}\right ] \]
[1/6*((b*c^3*x^3 - b*c^2*x^2 - b*c*x + b)*sqrt(d*f)*log((c^6*d*f*x^6 - 4*c ^5*d*f*x^5 + 5*c^4*d*f*x^4 - 4*c^2*d*f*x^2 + 4*c*d*f*x - (c^4*x^4 - 4*c^3* x^3 + 6*c^2*x^2 - 4*c*x)*sqrt(-c^2*x^2 + 1)*sqrt(c*d*x + d)*sqrt(-c*f*x + f)*sqrt(d*f) - 2*d*f)/(c^4*x^4 - 2*c^3*x^3 + 2*c*x - 1)) - 2*(a*c^2*x^2 + sqrt(-c^2*x^2 + 1)*b*c*x - a*c*x + (b*c^2*x^2 - b*c*x - 2*b)*arcsin(c*x) - 2*a)*sqrt(c*d*x + d)*sqrt(-c*f*x + f))/(c^4*d*f^3*x^3 - c^3*d*f^3*x^2 - c ^2*d*f^3*x + c*d*f^3), 1/3*((b*c^3*x^3 - b*c^2*x^2 - b*c*x + b)*sqrt(-d*f) *arctan((c^2*x^2 - 2*c*x + 2)*sqrt(-c^2*x^2 + 1)*sqrt(c*d*x + d)*sqrt(-c*f *x + f)*sqrt(-d*f)/(c^4*d*f*x^4 - 2*c^3*d*f*x^3 - c^2*d*f*x^2 + 2*c*d*f*x) ) - (a*c^2*x^2 + sqrt(-c^2*x^2 + 1)*b*c*x - a*c*x + (b*c^2*x^2 - b*c*x - 2 *b)*arcsin(c*x) - 2*a)*sqrt(c*d*x + d)*sqrt(-c*f*x + f))/(c^4*d*f^3*x^3 - c^3*d*f^3*x^2 - c^2*d*f^3*x + c*d*f^3)]
\[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{5/2}} \, dx=\int \frac {a + b \operatorname {asin}{\left (c x \right )}}{\sqrt {d \left (c x + 1\right )} \left (- f \left (c x - 1\right )\right )^{\frac {5}{2}}}\, dx \]
Time = 0.30 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.86 \[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{5/2}} \, dx=\frac {1}{3} \, b c {\left (\frac {1}{c^{3} \sqrt {d} f^{\frac {5}{2}} x - c^{2} \sqrt {d} f^{\frac {5}{2}}} + \frac {\log \left (c x - 1\right )}{c^{2} \sqrt {d} f^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{3} d f^{3} x^{2} - 2 \, c^{2} d f^{3} x + c d f^{3}} - \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} d f^{3} x - c d f^{3}}\right )} \arcsin \left (c x\right ) + \frac {1}{3} \, a {\left (\frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{3} d f^{3} x^{2} - 2 \, c^{2} d f^{3} x + c d f^{3}} - \frac {\sqrt {-c^{2} d f x^{2} + d f}}{c^{2} d f^{3} x - c d f^{3}}\right )} \]
1/3*b*c*(1/(c^3*sqrt(d)*f^(5/2)*x - c^2*sqrt(d)*f^(5/2)) + log(c*x - 1)/(c ^2*sqrt(d)*f^(5/2))) + 1/3*b*(sqrt(-c^2*d*f*x^2 + d*f)/(c^3*d*f^3*x^2 - 2* c^2*d*f^3*x + c*d*f^3) - sqrt(-c^2*d*f*x^2 + d*f)/(c^2*d*f^3*x - c*d*f^3)) *arcsin(c*x) + 1/3*a*(sqrt(-c^2*d*f*x^2 + d*f)/(c^3*d*f^3*x^2 - 2*c^2*d*f^ 3*x + c*d*f^3) - sqrt(-c^2*d*f*x^2 + d*f)/(c^2*d*f^3*x - c*d*f^3))
\[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{5/2}} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{\sqrt {c d x + d} {\left (-c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {a+b \arcsin (c x)}{\sqrt {d+c d x} (f-c f x)^{5/2}} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{\sqrt {d+c\,d\,x}\,{\left (f-c\,f\,x\right )}^{5/2}} \,d x \]