Integrand size = 25, antiderivative size = 164 \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^2} \, dx=-\frac {11}{5} b c d^3 \sqrt {1-c^2 x^2}-\frac {1}{5} b c d^3 \left (1-c^2 x^2\right )^{3/2}-\frac {1}{25} b c d^3 \left (1-c^2 x^2\right )^{5/2}-\frac {d^3 (a+b \arcsin (c x))}{x}-3 c^2 d^3 x (a+b \arcsin (c x))+c^4 d^3 x^3 (a+b \arcsin (c x))-\frac {1}{5} c^6 d^3 x^5 (a+b \arcsin (c x))-b c d^3 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right ) \]
-1/5*b*c*d^3*(-c^2*x^2+1)^(3/2)-1/25*b*c*d^3*(-c^2*x^2+1)^(5/2)-d^3*(a+b*a rcsin(c*x))/x-3*c^2*d^3*x*(a+b*arcsin(c*x))+c^4*d^3*x^3*(a+b*arcsin(c*x))- 1/5*c^6*d^3*x^5*(a+b*arcsin(c*x))-b*c*d^3*arctanh((-c^2*x^2+1)^(1/2))-11/5 *b*c*d^3*(-c^2*x^2+1)^(1/2)
Time = 0.16 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.75 \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^2} \, dx=\frac {1}{10} d^3 \left (-\frac {10 (a+b \arcsin (c x))}{x}-30 c^2 x (a+b \arcsin (c x))+10 c^4 x^3 (a+b \arcsin (c x))-2 c^6 x^5 (a+b \arcsin (c x))-\frac {2}{5} b c \left (\sqrt {1-c^2 x^2} \left (61-7 c^2 x^2+c^4 x^4\right )+25 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right )\right )\right ) \]
(d^3*((-10*(a + b*ArcSin[c*x]))/x - 30*c^2*x*(a + b*ArcSin[c*x]) + 10*c^4* x^3*(a + b*ArcSin[c*x]) - 2*c^6*x^5*(a + b*ArcSin[c*x]) - (2*b*c*(Sqrt[1 - c^2*x^2]*(61 - 7*c^2*x^2 + c^4*x^4) + 25*ArcTanh[Sqrt[1 - c^2*x^2]]))/5)) /10
Time = 0.50 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5192, 27, 2331, 2123, 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.
| \(\displaystyle \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^2} \, dx\) |
| \(\Big \downarrow \) 5192 |
| \(\displaystyle -b c \int -\frac {d^3 \left (c^6 x^6-5 c^4 x^4+15 c^2 x^2+5\right )}{5 x \sqrt {1-c^2 x^2}}dx-\frac {1}{5} c^6 d^3 x^5 (a+b \arcsin (c x))+c^4 d^3 x^3 (a+b \arcsin (c x))-3 c^2 d^3 x (a+b \arcsin (c x))-\frac {d^3 (a+b \arcsin (c x))}{x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} b c d^3 \int \frac {c^6 x^6-5 c^4 x^4+15 c^2 x^2+5}{x \sqrt {1-c^2 x^2}}dx-\frac {1}{5} c^6 d^3 x^5 (a+b \arcsin (c x))+c^4 d^3 x^3 (a+b \arcsin (c x))-3 c^2 d^3 x (a+b \arcsin (c x))-\frac {d^3 (a+b \arcsin (c x))}{x}\) |
| \(\Big \downarrow \) 2331 |
| \(\displaystyle \frac {1}{10} b c d^3 \int \frac {c^6 x^6-5 c^4 x^4+15 c^2 x^2+5}{x^2 \sqrt {1-c^2 x^2}}dx^2-\frac {1}{5} c^6 d^3 x^5 (a+b \arcsin (c x))+c^4 d^3 x^3 (a+b \arcsin (c x))-3 c^2 d^3 x (a+b \arcsin (c x))-\frac {d^3 (a+b \arcsin (c x))}{x}\) |
| \(\Big \downarrow \) 2123 |
| \(\displaystyle \frac {1}{10} b c d^3 \int \left (\left (1-c^2 x^2\right )^{3/2} c^2+3 \sqrt {1-c^2 x^2} c^2+\frac {11 c^2}{\sqrt {1-c^2 x^2}}+\frac {5}{x^2 \sqrt {1-c^2 x^2}}\right )dx^2-\frac {1}{5} c^6 d^3 x^5 (a+b \arcsin (c x))+c^4 d^3 x^3 (a+b \arcsin (c x))-3 c^2 d^3 x (a+b \arcsin (c x))-\frac {d^3 (a+b \arcsin (c x))}{x}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {1}{5} c^6 d^3 x^5 (a+b \arcsin (c x))+c^4 d^3 x^3 (a+b \arcsin (c x))-3 c^2 d^3 x (a+b \arcsin (c x))-\frac {d^3 (a+b \arcsin (c x))}{x}+\frac {1}{10} b c d^3 \left (-10 \text {arctanh}\left (\sqrt {1-c^2 x^2}\right )-\frac {2}{5} \left (1-c^2 x^2\right )^{5/2}-2 \left (1-c^2 x^2\right )^{3/2}-22 \sqrt {1-c^2 x^2}\right )\) |
-((d^3*(a + b*ArcSin[c*x]))/x) - 3*c^2*d^3*x*(a + b*ArcSin[c*x]) + c^4*d^3 *x^3*(a + b*ArcSin[c*x]) - (c^6*d^3*x^5*(a + b*ArcSin[c*x]))/5 + (b*c*d^3* (-22*Sqrt[1 - c^2*x^2] - 2*(1 - c^2*x^2)^(3/2) - (2*(1 - c^2*x^2)^(5/2))/5 - 10*ArcTanh[Sqrt[1 - c^2*x^2]]))/10
3.1.25.3.1 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[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/2 S ubst[Int[x^((m - 1)/2)*SubstFor[x^2, Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_) ^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Simp[ (a + b*ArcSin[c*x]) u, x] - Simp[b*c Int[SimplifyIntegrand[u/Sqrt[1 - c ^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0 ] && IGtQ[p, 0]
Time = 0.05 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.93
| method | result | size |
| parts | \(-d^{3} a \left (\frac {c^{6} x^{5}}{5}-c^{4} x^{3}+3 c^{2} x +\frac {1}{x}\right )-d^{3} b c \left (\frac {\arcsin \left (c x \right ) c^{5} x^{5}}{5}-c^{3} x^{3} \arcsin \left (c x \right )+3 c x \arcsin \left (c x \right )+\frac {\arcsin \left (c x \right )}{c x}+\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{25}-\frac {7 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{25}+\frac {61 \sqrt {-c^{2} x^{2}+1}}{25}+\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )\) | \(152\) |
| derivativedivides | \(c \left (-d^{3} a \left (\frac {c^{5} x^{5}}{5}-c^{3} x^{3}+3 c x +\frac {1}{c x}\right )-d^{3} b \left (\frac {\arcsin \left (c x \right ) c^{5} x^{5}}{5}-c^{3} x^{3} \arcsin \left (c x \right )+3 c x \arcsin \left (c x \right )+\frac {\arcsin \left (c x \right )}{c x}+\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{25}-\frac {7 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{25}+\frac {61 \sqrt {-c^{2} x^{2}+1}}{25}+\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )\right )\) | \(155\) |
| default | \(c \left (-d^{3} a \left (\frac {c^{5} x^{5}}{5}-c^{3} x^{3}+3 c x +\frac {1}{c x}\right )-d^{3} b \left (\frac {\arcsin \left (c x \right ) c^{5} x^{5}}{5}-c^{3} x^{3} \arcsin \left (c x \right )+3 c x \arcsin \left (c x \right )+\frac {\arcsin \left (c x \right )}{c x}+\frac {c^{4} x^{4} \sqrt {-c^{2} x^{2}+1}}{25}-\frac {7 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}}{25}+\frac {61 \sqrt {-c^{2} x^{2}+1}}{25}+\operatorname {arctanh}\left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )\right )\right )\) | \(155\) |
-d^3*a*(1/5*c^6*x^5-c^4*x^3+3*c^2*x+1/x)-d^3*b*c*(1/5*arcsin(c*x)*c^5*x^5- c^3*x^3*arcsin(c*x)+3*c*x*arcsin(c*x)+1/c/x*arcsin(c*x)+1/25*c^4*x^4*(-c^2 *x^2+1)^(1/2)-7/25*c^2*x^2*(-c^2*x^2+1)^(1/2)+61/25*(-c^2*x^2+1)^(1/2)+arc tanh(1/(-c^2*x^2+1)^(1/2)))
Time = 0.30 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.15 \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^2} \, dx=-\frac {10 \, a c^{6} d^{3} x^{6} - 50 \, a c^{4} d^{3} x^{4} + 150 \, a c^{2} d^{3} x^{2} + 25 \, b c d^{3} x \log \left (\sqrt {-c^{2} x^{2} + 1} + 1\right ) - 25 \, b c d^{3} x \log \left (\sqrt {-c^{2} x^{2} + 1} - 1\right ) + 50 \, a d^{3} + 10 \, {\left (b c^{6} d^{3} x^{6} - 5 \, b c^{4} d^{3} x^{4} + 15 \, b c^{2} d^{3} x^{2} + 5 \, b d^{3}\right )} \arcsin \left (c x\right ) + 2 \, {\left (b c^{5} d^{3} x^{5} - 7 \, b c^{3} d^{3} x^{3} + 61 \, b c d^{3} x\right )} \sqrt {-c^{2} x^{2} + 1}}{50 \, x} \]
-1/50*(10*a*c^6*d^3*x^6 - 50*a*c^4*d^3*x^4 + 150*a*c^2*d^3*x^2 + 25*b*c*d^ 3*x*log(sqrt(-c^2*x^2 + 1) + 1) - 25*b*c*d^3*x*log(sqrt(-c^2*x^2 + 1) - 1) + 50*a*d^3 + 10*(b*c^6*d^3*x^6 - 5*b*c^4*d^3*x^4 + 15*b*c^2*d^3*x^2 + 5*b *d^3)*arcsin(c*x) + 2*(b*c^5*d^3*x^5 - 7*b*c^3*d^3*x^3 + 61*b*c*d^3*x)*sqr t(-c^2*x^2 + 1))/x
Time = 3.07 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.77 \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^2} \, dx=- \frac {a c^{6} d^{3} x^{5}}{5} + a c^{4} d^{3} x^{3} - 3 a c^{2} d^{3} x - \frac {a d^{3}}{x} + \frac {b c^{7} d^{3} \left (\begin {cases} - \frac {x^{4} \sqrt {- c^{2} x^{2} + 1}}{5 c^{2}} - \frac {4 x^{2} \sqrt {- c^{2} x^{2} + 1}}{15 c^{4}} - \frac {8 \sqrt {- c^{2} x^{2} + 1}}{15 c^{6}} & \text {for}\: c^{2} \neq 0 \\\frac {x^{6}}{6} & \text {otherwise} \end {cases}\right )}{5} - \frac {b c^{6} d^{3} x^{5} \operatorname {asin}{\left (c x \right )}}{5} - b c^{5} d^{3} \left (\begin {cases} - \frac {x^{2} \sqrt {- c^{2} x^{2} + 1}}{3 c^{2}} - \frac {2 \sqrt {- c^{2} x^{2} + 1}}{3 c^{4}} & \text {for}\: c^{2} \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases}\right ) + b c^{4} d^{3} x^{3} \operatorname {asin}{\left (c x \right )} - 3 b c^{2} d^{3} \left (\begin {cases} 0 & \text {for}\: c = 0 \\x \operatorname {asin}{\left (c x \right )} + \frac {\sqrt {- c^{2} x^{2} + 1}}{c} & \text {otherwise} \end {cases}\right ) + b c d^{3} \left (\begin {cases} - \operatorname {acosh}{\left (\frac {1}{c x} \right )} & \text {for}\: \frac {1}{\left |{c^{2} x^{2}}\right |} > 1 \\i \operatorname {asin}{\left (\frac {1}{c x} \right )} & \text {otherwise} \end {cases}\right ) - \frac {b d^{3} \operatorname {asin}{\left (c x \right )}}{x} \]
-a*c**6*d**3*x**5/5 + a*c**4*d**3*x**3 - 3*a*c**2*d**3*x - a*d**3/x + b*c* *7*d**3*Piecewise((-x**4*sqrt(-c**2*x**2 + 1)/(5*c**2) - 4*x**2*sqrt(-c**2 *x**2 + 1)/(15*c**4) - 8*sqrt(-c**2*x**2 + 1)/(15*c**6), Ne(c**2, 0)), (x* *6/6, True))/5 - b*c**6*d**3*x**5*asin(c*x)/5 - b*c**5*d**3*Piecewise((-x* *2*sqrt(-c**2*x**2 + 1)/(3*c**2) - 2*sqrt(-c**2*x**2 + 1)/(3*c**4), Ne(c** 2, 0)), (x**4/4, True)) + b*c**4*d**3*x**3*asin(c*x) - 3*b*c**2*d**3*Piece wise((0, Eq(c, 0)), (x*asin(c*x) + sqrt(-c**2*x**2 + 1)/c, True)) + b*c*d* *3*Piecewise((-acosh(1/(c*x)), 1/Abs(c**2*x**2) > 1), (I*asin(1/(c*x)), Tr ue)) - b*d**3*asin(c*x)/x
Time = 0.28 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.52 \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^2} \, dx=-\frac {1}{5} \, a c^{6} d^{3} x^{5} - \frac {1}{75} \, {\left (15 \, x^{5} \arcsin \left (c x\right ) + {\left (\frac {3 \, \sqrt {-c^{2} x^{2} + 1} x^{4}}{c^{2}} + \frac {4 \, \sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{4}} + \frac {8 \, \sqrt {-c^{2} x^{2} + 1}}{c^{6}}\right )} c\right )} b c^{6} d^{3} + a c^{4} d^{3} x^{3} + \frac {1}{3} \, {\left (3 \, x^{3} \arcsin \left (c x\right ) + c {\left (\frac {\sqrt {-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b c^{4} d^{3} - 3 \, a c^{2} d^{3} x - 3 \, {\left (c x \arcsin \left (c x\right ) + \sqrt {-c^{2} x^{2} + 1}\right )} b c d^{3} - {\left (c \log \left (\frac {2 \, \sqrt {-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) + \frac {\arcsin \left (c x\right )}{x}\right )} b d^{3} - \frac {a d^{3}}{x} \]
-1/5*a*c^6*d^3*x^5 - 1/75*(15*x^5*arcsin(c*x) + (3*sqrt(-c^2*x^2 + 1)*x^4/ c^2 + 4*sqrt(-c^2*x^2 + 1)*x^2/c^4 + 8*sqrt(-c^2*x^2 + 1)/c^6)*c)*b*c^6*d^ 3 + a*c^4*d^3*x^3 + 1/3*(3*x^3*arcsin(c*x) + c*(sqrt(-c^2*x^2 + 1)*x^2/c^2 + 2*sqrt(-c^2*x^2 + 1)/c^4))*b*c^4*d^3 - 3*a*c^2*d^3*x - 3*(c*x*arcsin(c* x) + sqrt(-c^2*x^2 + 1))*b*c*d^3 - (c*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/ abs(x)) + arcsin(c*x)/x)*b*d^3 - a*d^3/x
Leaf count of result is larger than twice the leaf count of optimal. 5513 vs. \(2 (148) = 296\).
Time = 29.09 (sec) , antiderivative size = 5513, normalized size of antiderivative = 33.62 \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^2} \, dx=\text {Too large to display} \]
-1/2*b*c^13*d^3*x^12*arcsin(c*x)/((c^11*x^11/(sqrt(-c^2*x^2 + 1) + 1)^11 + 5*c^9*x^9/(sqrt(-c^2*x^2 + 1) + 1)^9 + 10*c^7*x^7/(sqrt(-c^2*x^2 + 1) + 1 )^7 + 10*c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 5*c^3*x^3/(sqrt(-c^2*x^2 + 1 ) + 1)^3 + c*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^12) - 1/ 2*a*c^13*d^3*x^12/((c^11*x^11/(sqrt(-c^2*x^2 + 1) + 1)^11 + 5*c^9*x^9/(sqr t(-c^2*x^2 + 1) + 1)^9 + 10*c^7*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 10*c^5*x^ 5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 5*c^3*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c*x/ (sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^12) + b*c^12*d^3*x^11*l og(abs(c)*abs(x))/((c^11*x^11/(sqrt(-c^2*x^2 + 1) + 1)^11 + 5*c^9*x^9/(sqr t(-c^2*x^2 + 1) + 1)^9 + 10*c^7*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 10*c^5*x^ 5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 5*c^3*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c*x/ (sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^11) - b*c^12*d^3*x^11*l og(sqrt(-c^2*x^2 + 1) + 1)/((c^11*x^11/(sqrt(-c^2*x^2 + 1) + 1)^11 + 5*c^9 *x^9/(sqrt(-c^2*x^2 + 1) + 1)^9 + 10*c^7*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 10*c^5*x^5/(sqrt(-c^2*x^2 + 1) + 1)^5 + 5*c^3*x^3/(sqrt(-c^2*x^2 + 1) + 1) ^3 + c*x/(sqrt(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^11) + 61/25*b* c^12*d^3*x^11/((c^11*x^11/(sqrt(-c^2*x^2 + 1) + 1)^11 + 5*c^9*x^9/(sqrt(-c ^2*x^2 + 1) + 1)^9 + 10*c^7*x^7/(sqrt(-c^2*x^2 + 1) + 1)^7 + 10*c^5*x^5/(s qrt(-c^2*x^2 + 1) + 1)^5 + 5*c^3*x^3/(sqrt(-c^2*x^2 + 1) + 1)^3 + c*x/(sqr t(-c^2*x^2 + 1) + 1))*(sqrt(-c^2*x^2 + 1) + 1)^11) - 9*b*c^11*d^3*x^10*...
Timed out. \[ \int \frac {\left (d-c^2 d x^2\right )^3 (a+b \arcsin (c x))}{x^2} \, dx=\int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (d-c^2\,d\,x^2\right )}^3}{x^2} \,d x \]