\(\int \frac {a+b \arcsin (c x^2)}{x^{13}} \, dx\) [351]

Optimal result

Integrand size = 14, antiderivative size = 91 \[ \int \frac {a+b \arcsin \left (c x^2\right )}{x^{13}} \, dx=-\frac {b c \sqrt {1-c^2 x^4}}{60 x^{10}}-\frac {b c^3 \sqrt {1-c^2 x^4}}{45 x^6}-\frac {2 b c^5 \sqrt {1-c^2 x^4}}{45 x^2}-\frac {a+b \arcsin \left (c x^2\right )}{12 x^{12}} \]

[Out]

1/12*(-a-b*arcsin(c*x^2))/x^12-1/60*b*c*(-c^2*x^4+1)^(1/2)/x^10-1/45*b*c^3*(-c^2*x^4+1)^(1/2)/x^6-2/45*b*c^5*(
-c^2*x^4+1)^(1/2)/x^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4926, 12, 277, 270} \[ \int \frac {a+b \arcsin \left (c x^2\right )}{x^{13}} \, dx=-\frac {a+b \arcsin \left (c x^2\right )}{12 x^{12}}-\frac {b c \sqrt {1-c^2 x^4}}{60 x^{10}}-\frac {2 b c^5 \sqrt {1-c^2 x^4}}{45 x^2}-\frac {b c^3 \sqrt {1-c^2 x^4}}{45 x^6} \]

[In]

Int[(a + b*ArcSin[c*x^2])/x^13,x]

[Out]

-1/60*(b*c*Sqrt[1 - c^2*x^4])/x^10 - (b*c^3*Sqrt[1 - c^2*x^4])/(45*x^6) - (2*b*c^5*Sqrt[1 - c^2*x^4])/(45*x^2)
- (a + b*ArcSin[c*x^2])/(12*x^12)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 4926

Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m + 1)*((a + b*ArcSin[
u])/(d*(m + 1))), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[(c + d*x)^(m + 1)*(D[u, x]/Sqrt[1 - u^2]), x]
, x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(
m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {a+b \arcsin \left (c x^2\right )}{12 x^{12}}+\frac {1}{12} b \int \frac {2 c}{x^{11} \sqrt {1-c^2 x^4}} \, dx \\ & = -\frac {a+b \arcsin \left (c x^2\right )}{12 x^{12}}+\frac {1}{6} (b c) \int \frac {1}{x^{11} \sqrt {1-c^2 x^4}} \, dx \\ & = -\frac {b c \sqrt {1-c^2 x^4}}{60 x^{10}}-\frac {a+b \arcsin \left (c x^2\right )}{12 x^{12}}+\frac {1}{15} \left (2 b c^3\right ) \int \frac {1}{x^7 \sqrt {1-c^2 x^4}} \, dx \\ & = -\frac {b c \sqrt {1-c^2 x^4}}{60 x^{10}}-\frac {b c^3 \sqrt {1-c^2 x^4}}{45 x^6}-\frac {a+b \arcsin \left (c x^2\right )}{12 x^{12}}+\frac {1}{45} \left (4 b c^5\right ) \int \frac {1}{x^3 \sqrt {1-c^2 x^4}} \, dx \\ & = -\frac {b c \sqrt {1-c^2 x^4}}{60 x^{10}}-\frac {b c^3 \sqrt {1-c^2 x^4}}{45 x^6}-\frac {2 b c^5 \sqrt {1-c^2 x^4}}{45 x^2}-\frac {a+b \arcsin \left (c x^2\right )}{12 x^{12}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.75 \[ \int \frac {a+b \arcsin \left (c x^2\right )}{x^{13}} \, dx=-\frac {a}{12 x^{12}}+\frac {1}{2} b \left (-\frac {c \sqrt {1-c^2 x^4} \left (3+4 c^2 x^4+8 c^4 x^8\right )}{90 x^{10}}-\frac {\arcsin \left (c x^2\right )}{6 x^{12}}\right ) \]

[In]

Integrate[(a + b*ArcSin[c*x^2])/x^13,x]

[Out]

-1/12*a/x^12 + (b*(-1/90*(c*Sqrt[1 - c^2*x^4]*(3 + 4*c^2*x^4 + 8*c^4*x^8))/x^10 - ArcSin[c*x^2]/(6*x^12)))/2

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.79

method	result	size
default	\(-\frac {a}{12 x^{12}}+b \left (-\frac {\arcsin \left (c \,x^{2}\right )}{12 x^{12}}+\frac {c \left (c \,x^{2}-1\right ) \left (c \,x^{2}+1\right ) \left (8 c^{4} x^{8}+4 c^{2} x^{4}+3\right )}{180 x^{10} \sqrt {-c^{2} x^{4}+1}}\right )\)	\(72\)
parts	\(-\frac {a}{12 x^{12}}+b \left (-\frac {\arcsin \left (c \,x^{2}\right )}{12 x^{12}}+\frac {c \left (c \,x^{2}-1\right ) \left (c \,x^{2}+1\right ) \left (8 c^{4} x^{8}+4 c^{2} x^{4}+3\right )}{180 x^{10} \sqrt {-c^{2} x^{4}+1}}\right )\)	\(72\)

[In]

int((a+b*arcsin(c*x^2))/x^13,x,method=_RETURNVERBOSE)

[Out]

-1/12*a/x^12+b*(-1/12/x^12*arcsin(c*x^2)+1/180*c*(c*x^2-1)*(c*x^2+1)*(8*c^4*x^8+4*c^2*x^4+3)/x^10/(-c^2*x^4+1)
^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.70 \[ \int \frac {a+b \arcsin \left (c x^2\right )}{x^{13}} \, dx=\frac {15 \, a x^{12} - 15 \, b \arcsin \left (c x^{2}\right ) - {\left (8 \, b c^{5} x^{10} + 4 \, b c^{3} x^{6} + 3 \, b c x^{2}\right )} \sqrt {-c^{2} x^{4} + 1} - 15 \, a}{180 \, x^{12}} \]

[In]

integrate((a+b*arcsin(c*x^2))/x^13,x, algorithm="fricas")

[Out]

1/180*(15*a*x^12 - 15*b*arcsin(c*x^2) - (8*b*c^5*x^10 + 4*b*c^3*x^6 + 3*b*c*x^2)*sqrt(-c^2*x^4 + 1) - 15*a)/x^
12

Sympy [A] (verification not implemented)

Time = 5.26 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.87 \[ \int \frac {a+b \arcsin \left (c x^2\right )}{x^{13}} \, dx=- \frac {a}{12 x^{12}} + \frac {b c \left (\begin {cases} - \frac {4 c^{5} \sqrt {-1 + \frac {1}{c^{2} x^{4}}}}{15} - \frac {2 c^{3} \sqrt {-1 + \frac {1}{c^{2} x^{4}}}}{15 x^{4}} - \frac {c \sqrt {-1 + \frac {1}{c^{2} x^{4}}}}{10 x^{8}} & \text {for}\: \frac {1}{\left |{c^{2} x^{4}}\right |} > 1 \\- \frac {4 i c^{5} \sqrt {1 - \frac {1}{c^{2} x^{4}}}}{15} - \frac {2 i c^{3} \sqrt {1 - \frac {1}{c^{2} x^{4}}}}{15 x^{4}} - \frac {i c \sqrt {1 - \frac {1}{c^{2} x^{4}}}}{10 x^{8}} & \text {otherwise} \end {cases}\right )}{6} - \frac {b \operatorname {asin}{\left (c x^{2} \right )}}{12 x^{12}} \]

[In]

integrate((a+b*asin(c*x**2))/x**13,x)

[Out]

-a/(12*x**12) + b*c*Piecewise((-4*c**5*sqrt(-1 + 1/(c**2*x**4))/15 - 2*c**3*sqrt(-1 + 1/(c**2*x**4))/(15*x**4)
- c*sqrt(-1 + 1/(c**2*x**4))/(10*x**8), 1/Abs(c**2*x**4) > 1), (-4*I*c**5*sqrt(1 - 1/(c**2*x**4))/15 - 2*I*c*
*3*sqrt(1 - 1/(c**2*x**4))/(15*x**4) - I*c*sqrt(1 - 1/(c**2*x**4))/(10*x**8), True))/6 - b*asin(c*x**2)/(12*x*
*12)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.90 \[ \int \frac {a+b \arcsin \left (c x^2\right )}{x^{13}} \, dx=-\frac {1}{180} \, {\left ({\left (\frac {15 \, \sqrt {-c^{2} x^{4} + 1} c^{4}}{x^{2}} + \frac {10 \, {\left (-c^{2} x^{4} + 1\right )}^{\frac {3}{2}} c^{2}}{x^{6}} + \frac {3 \, {\left (-c^{2} x^{4} + 1\right )}^{\frac {5}{2}}}{x^{10}}\right )} c + \frac {15 \, \arcsin \left (c x^{2}\right )}{x^{12}}\right )} b - \frac {a}{12 \, x^{12}} \]

[In]

integrate((a+b*arcsin(c*x^2))/x^13,x, algorithm="maxima")

[Out]

-1/180*((15*sqrt(-c^2*x^4 + 1)*c^4/x^2 + 10*(-c^2*x^4 + 1)^(3/2)*c^2/x^6 + 3*(-c^2*x^4 + 1)^(5/2)/x^10)*c + 15
*arcsin(c*x^2)/x^12)*b - 1/12*a/x^12

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 504 vs. \(2 (77) = 154\).

Time = 0.31 (sec) , antiderivative size = 504, normalized size of antiderivative = 5.54 \[ \int \frac {a+b \arcsin \left (c x^2\right )}{x^{13}} \, dx=-\frac {\frac {15 \, b c^{13} x^{12} \arcsin \left (c x^{2}\right )}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{6}} + \frac {15 \, a c^{13} x^{12}}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{6}} - \frac {6 \, b c^{12} x^{10}}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{5}} + \frac {90 \, b c^{11} x^{8} \arcsin \left (c x^{2}\right )}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{4}} + \frac {90 \, a c^{11} x^{8}}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{4}} - \frac {50 \, b c^{10} x^{6}}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{3}} + \frac {225 \, b c^{9} x^{4} \arcsin \left (c x^{2}\right )}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{2}} + \frac {225 \, a c^{9} x^{4}}{{\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{2}} - \frac {300 \, b c^{8} x^{2}}{\sqrt {-c^{2} x^{4} + 1} + 1} + 300 \, b c^{7} \arcsin \left (c x^{2}\right ) + 300 \, a c^{7} + \frac {300 \, b c^{6} {\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}}{x^{2}} + \frac {225 \, b c^{5} {\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{2} \arcsin \left (c x^{2}\right )}{x^{4}} + \frac {225 \, a c^{5} {\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{2}}{x^{4}} + \frac {50 \, b c^{4} {\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{3}}{x^{6}} + \frac {90 \, b c^{3} {\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{4} \arcsin \left (c x^{2}\right )}{x^{8}} + \frac {90 \, a c^{3} {\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{4}}{x^{8}} + \frac {6 \, b c^{2} {\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{5}}{x^{10}} + \frac {15 \, b c {\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{6} \arcsin \left (c x^{2}\right )}{x^{12}} + \frac {15 \, a c {\left (\sqrt {-c^{2} x^{4} + 1} + 1\right )}^{6}}{x^{12}}}{11520 \, c} \]

[In]

integrate((a+b*arcsin(c*x^2))/x^13,x, algorithm="giac")

[Out]

-1/11520*(15*b*c^13*x^12*arcsin(c*x^2)/(sqrt(-c^2*x^4 + 1) + 1)^6 + 15*a*c^13*x^12/(sqrt(-c^2*x^4 + 1) + 1)^6
- 6*b*c^12*x^10/(sqrt(-c^2*x^4 + 1) + 1)^5 + 90*b*c^11*x^8*arcsin(c*x^2)/(sqrt(-c^2*x^4 + 1) + 1)^4 + 90*a*c^1
1*x^8/(sqrt(-c^2*x^4 + 1) + 1)^4 - 50*b*c^10*x^6/(sqrt(-c^2*x^4 + 1) + 1)^3 + 225*b*c^9*x^4*arcsin(c*x^2)/(sqr
t(-c^2*x^4 + 1) + 1)^2 + 225*a*c^9*x^4/(sqrt(-c^2*x^4 + 1) + 1)^2 - 300*b*c^8*x^2/(sqrt(-c^2*x^4 + 1) + 1) + 3
00*b*c^7*arcsin(c*x^2) + 300*a*c^7 + 300*b*c^6*(sqrt(-c^2*x^4 + 1) + 1)/x^2 + 225*b*c^5*(sqrt(-c^2*x^4 + 1) +
1)^2*arcsin(c*x^2)/x^4 + 225*a*c^5*(sqrt(-c^2*x^4 + 1) + 1)^2/x^4 + 50*b*c^4*(sqrt(-c^2*x^4 + 1) + 1)^3/x^6 +
90*b*c^3*(sqrt(-c^2*x^4 + 1) + 1)^4*arcsin(c*x^2)/x^8 + 90*a*c^3*(sqrt(-c^2*x^4 + 1) + 1)^4/x^8 + 6*b*c^2*(sqr
t(-c^2*x^4 + 1) + 1)^5/x^10 + 15*b*c*(sqrt(-c^2*x^4 + 1) + 1)^6*arcsin(c*x^2)/x^12 + 15*a*c*(sqrt(-c^2*x^4 + 1
) + 1)^6/x^12)/c

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arcsin \left (c x^2\right )}{x^{13}} \, dx=\int \frac {a+b\,\mathrm {asin}\left (c\,x^2\right )}{x^{13}} \,d x \]

[In]

int((a + b*asin(c*x^2))/x^13,x)

[Out]

int((a + b*asin(c*x^2))/x^13, x)