Integrand size = 14, antiderivative size = 82 \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {3 b x^2 \sqrt {1-c^2 x^4}}{64 c^3}+\frac {b x^6 \sqrt {1-c^2 x^4}}{32 c}-\frac {3 b \arcsin \left (c x^2\right )}{64 c^4}+\frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right ) \]
-3/64*b*arcsin(c*x^2)/c^4+1/8*x^8*(a+b*arcsin(c*x^2))+3/64*b*x^2*(-c^2*x^4 +1)^(1/2)/c^3+1/32*b*x^6*(-c^2*x^4+1)^(1/2)/c
Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.06 \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {a x^8}{8}+\frac {3 b x^2 \sqrt {1-c^2 x^4}}{64 c^3}+\frac {b x^6 \sqrt {1-c^2 x^4}}{32 c}-\frac {3 b \arcsin \left (c x^2\right )}{64 c^4}+\frac {1}{8} b x^8 \arcsin \left (c x^2\right ) \]
(a*x^8)/8 + (3*b*x^2*Sqrt[1 - c^2*x^4])/(64*c^3) + (b*x^6*Sqrt[1 - c^2*x^4 ])/(32*c) - (3*b*ArcSin[c*x^2])/(64*c^4) + (b*x^8*ArcSin[c*x^2])/8
Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5341, 27, 807, 262, 262, 223}
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.
\(\displaystyle \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx\) |
\(\Big \downarrow \) 5341 |
\(\displaystyle \frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {1}{8} b \int \frac {2 c x^9}{\sqrt {1-c^2 x^4}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {1}{4} b c \int \frac {x^9}{\sqrt {1-c^2 x^4}}dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {1}{8} b c \int \frac {x^8}{\sqrt {1-c^2 x^4}}dx^2\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {1}{8} b c \left (\frac {3 \int \frac {x^4}{\sqrt {1-c^2 x^4}}dx^2}{4 c^2}-\frac {x^6 \sqrt {1-c^2 x^4}}{4 c^2}\right )\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {1}{8} b c \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {1-c^2 x^4}}dx^2}{2 c^2}-\frac {x^2 \sqrt {1-c^2 x^4}}{2 c^2}\right )}{4 c^2}-\frac {x^6 \sqrt {1-c^2 x^4}}{4 c^2}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{8} x^8 \left (a+b \arcsin \left (c x^2\right )\right )-\frac {1}{8} b c \left (\frac {3 \left (\frac {\arcsin \left (c x^2\right )}{2 c^3}-\frac {x^2 \sqrt {1-c^2 x^4}}{2 c^2}\right )}{4 c^2}-\frac {x^6 \sqrt {1-c^2 x^4}}{4 c^2}\right )\) |
(x^8*(a + b*ArcSin[c*x^2]))/8 - (b*c*(-1/4*(x^6*Sqrt[1 - c^2*x^4])/c^2 + ( 3*(-1/2*(x^2*Sqrt[1 - c^2*x^4])/c^2 + ArcSin[c*x^2]/(2*c^3)))/(4*c^2)))/8
3.4.41.3.1 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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
Int[((a_.) + ArcSin[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Sim p[(c + d*x)^(m + 1)*((a + b*ArcSin[u])/(d*(m + 1))), x] - Simp[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]
Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.16
method | result | size |
default | \(\frac {a \,x^{8}}{8}+\frac {b \,x^{8} \arcsin \left (c \,x^{2}\right )}{8}+\frac {b \,x^{6} \sqrt {-c^{2} x^{4}+1}}{32 c}+\frac {3 b \,x^{2} \sqrt {-c^{2} x^{4}+1}}{64 c^{3}}-\frac {3 b \arctan \left (\frac {\sqrt {c^{2}}\, x^{2}}{\sqrt {-c^{2} x^{4}+1}}\right )}{64 c^{3} \sqrt {c^{2}}}\) | \(95\) |
parts | \(\frac {a \,x^{8}}{8}+\frac {b \,x^{8} \arcsin \left (c \,x^{2}\right )}{8}+\frac {b \,x^{6} \sqrt {-c^{2} x^{4}+1}}{32 c}+\frac {3 b \,x^{2} \sqrt {-c^{2} x^{4}+1}}{64 c^{3}}-\frac {3 b \arctan \left (\frac {\sqrt {c^{2}}\, x^{2}}{\sqrt {-c^{2} x^{4}+1}}\right )}{64 c^{3} \sqrt {c^{2}}}\) | \(95\) |
1/8*a*x^8+1/8*b*x^8*arcsin(c*x^2)+1/32*b*x^6*(-c^2*x^4+1)^(1/2)/c+3/64*b*x ^2*(-c^2*x^4+1)^(1/2)/c^3-3/64*b/c^3/(c^2)^(1/2)*arctan((c^2)^(1/2)*x^2/(- c^2*x^4+1)^(1/2))
Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79 \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {8 \, a c^{4} x^{8} + {\left (8 \, b c^{4} x^{8} - 3 \, b\right )} \arcsin \left (c x^{2}\right ) + {\left (2 \, b c^{3} x^{6} + 3 \, b c x^{2}\right )} \sqrt {-c^{2} x^{4} + 1}}{64 \, c^{4}} \]
1/64*(8*a*c^4*x^8 + (8*b*c^4*x^8 - 3*b)*arcsin(c*x^2) + (2*b*c^3*x^6 + 3*b *c*x^2)*sqrt(-c^2*x^4 + 1))/c^4
Time = 0.94 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.04 \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\begin {cases} \frac {a x^{8}}{8} + \frac {b x^{8} \operatorname {asin}{\left (c x^{2} \right )}}{8} + \frac {b x^{6} \sqrt {- c^{2} x^{4} + 1}}{32 c} + \frac {3 b x^{2} \sqrt {- c^{2} x^{4} + 1}}{64 c^{3}} - \frac {3 b \operatorname {asin}{\left (c x^{2} \right )}}{64 c^{4}} & \text {for}\: c \neq 0 \\\frac {a x^{8}}{8} & \text {otherwise} \end {cases} \]
Piecewise((a*x**8/8 + b*x**8*asin(c*x**2)/8 + b*x**6*sqrt(-c**2*x**4 + 1)/ (32*c) + 3*b*x**2*sqrt(-c**2*x**4 + 1)/(64*c**3) - 3*b*asin(c*x**2)/(64*c* *4), Ne(c, 0)), (a*x**8/8, True))
Time = 0.31 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.59 \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {1}{8} \, a x^{8} + \frac {1}{64} \, {\left (8 \, x^{8} \arcsin \left (c x^{2}\right ) + c {\left (\frac {\frac {5 \, \sqrt {-c^{2} x^{4} + 1} c^{2}}{x^{2}} + \frac {3 \, {\left (-c^{2} x^{4} + 1\right )}^{\frac {3}{2}}}{x^{6}}}{c^{8} - \frac {2 \, {\left (c^{2} x^{4} - 1\right )} c^{6}}{x^{4}} + \frac {{\left (c^{2} x^{4} - 1\right )}^{2} c^{4}}{x^{8}}} + \frac {3 \, \arctan \left (\frac {\sqrt {-c^{2} x^{4} + 1}}{c x^{2}}\right )}{c^{5}}\right )}\right )} b \]
1/8*a*x^8 + 1/64*(8*x^8*arcsin(c*x^2) + c*((5*sqrt(-c^2*x^4 + 1)*c^2/x^2 + 3*(-c^2*x^4 + 1)^(3/2)/x^6)/(c^8 - 2*(c^2*x^4 - 1)*c^6/x^4 + (c^2*x^4 - 1 )^2*c^4/x^8) + 3*arctan(sqrt(-c^2*x^4 + 1)/(c*x^2))/c^5))*b
Time = 0.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.34 \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\frac {8 \, a c x^{8} - {\left (\frac {2 \, {\left (-c^{2} x^{4} + 1\right )}^{\frac {3}{2}} x^{2}}{c^{2}} - \frac {5 \, \sqrt {-c^{2} x^{4} + 1} x^{2}}{c^{2}} - \frac {8 \, {\left (c^{2} x^{4} - 1\right )}^{2} \arcsin \left (c x^{2}\right )}{c^{3}} - \frac {16 \, {\left (c^{2} x^{4} - 1\right )} \arcsin \left (c x^{2}\right )}{c^{3}} - \frac {5 \, \arcsin \left (c x^{2}\right )}{c^{3}}\right )} b}{64 \, c} \]
1/64*(8*a*c*x^8 - (2*(-c^2*x^4 + 1)^(3/2)*x^2/c^2 - 5*sqrt(-c^2*x^4 + 1)*x ^2/c^2 - 8*(c^2*x^4 - 1)^2*arcsin(c*x^2)/c^3 - 16*(c^2*x^4 - 1)*arcsin(c*x ^2)/c^3 - 5*arcsin(c*x^2)/c^3)*b)/c
Timed out. \[ \int x^7 \left (a+b \arcsin \left (c x^2\right )\right ) \, dx=\int x^7\,\left (a+b\,\mathrm {asin}\left (c\,x^2\right )\right ) \,d x \]